Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 463857 12 pageshttpdxdoiorg1011552013463857

Research ArticleOn the Laws of Total Local Times for ℎ-Paths and Bridges ofSymmetric Leacutevy Processes

Masafumi Hayashi1 and Kouji Yano2

1 Faculty of Science University of the Ryukyus 1 Senbaru Okinawa Nishihara 903-0213 Japan2Graduate School of Science Kyoto University Sakyo-ku Kyoto 606-8502 Japan

Correspondence should be addressed to Kouji Yano kyanomathkyoto-uacjp

Received 12 May 2012 Accepted 29 November 2012

Academic Editor Dumitru Baleanu

Copyright copy 2013 M Hayashi and K Yano This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

The joint lawof the total local times at two levels forℎ-paths of symmetric Levy processes is shown to admit an explicit representationin terms of the laws of the squared Bessel processes of dimensions two and zero The law of the total local time at a single level forbridges is also discussed

1 Introduction

Markov processes associated to heat semigroups generatedby fractional derivatives are called symmetric stable Levyprocesses (cf eg [1]) or Levy flights (cf eg [2]) Thepurpose of the present paper is to study the laws of thetotal local times for ℎ-paths and bridges of (one-dimensional)symmetric Levy processesWe give an explicit representation(Theorem 16) of the joint law as a weighted sum of the lawof the squared Bessel process of dimension two and thegeneralized excursionmeasure for the squared Bessel processof dimension zero We also give an expression (Theorem 20)of the law of the total local time at a single level for bridges

It is well known as one of the Ray-Knight theorems (seeeg [3 Chapter XI] and [4 Chapter 3]) that the total localtime process with space parameter for a Bessel process ofdimension three is a squared Bessel process of dimensiontwo Since the Bessel process of dimension three is the ℎ-pathprocess of a reflected Brownian motion Theorem 16 may beconsidered to be a slight generalization of this result

Eisenbaum and Kaspi [5] have proved that the total localtime of a Markov process with discontinuous paths is nolonger Markov As an analogue of Ray-Knight theoremsEisenbaum et al [6] have recently characterized the law of thelocal time process with space parameter at inverse local timein terms of some Gaussian process whose covariance is givenby the resolvent density of the potential kernel Moreover

if the Levy process is a symmetric stable process then thecorresponding Gaussian process is a fractional BrownianmotionTheir results are based on a version of Feynman-Kacformulae which characterizes the Laplace transform of thejoint laws of total local times of Markov processes at severallevels

In this paper we first focus on the ℎ-path process of asymmetric Levy process which has been introduced in therecent works [7ndash9] by Yano et al The ℎ-path process maybe obtained as the process conditioned to avoid the originduring the whole time (see [10]) We will also start froma version of Feynman-Kac formulae and obtain an explicitrepresentation of the joint law of the total local times at twolevels (For some discussions of the joint law of the total localtimes see Blumenthal-Getoor [11 pages 221ndash226] and Pitman[12]) Unfortunately we have no better result on the law of thetotal local time process with space parameter The difficultywill be explained in Remark 3

In comparisonwith the results by Pitman [13] and Pitmanand Yor [14] about the Brownian and Bessel bridges we alsoinvestigate the law of the total local time at a single pointfor bridges of symmetric Levy process which we call Levybridges in short and also for bridges of the ℎ-paths which wecall ℎ-bridges in short We will prove a version of Feynman-Kac formulae (Theorem 7) for Levy bridges with the helpof the general theorems by Fitzsimmons et al [15] As anapplication of the Feynman-Kac theorem we will give an

2 Abstract and Applied Analysis

expression of the law of the total local time at a single levelfor the Levy bridges while unfortunately we do not have anynice formula for the ℎ-bridges

The present paper is organized as follows In Section 2 wegive two versions of Feynman-Kac formulae in general set-tings In Section 3 we recall several formulae about squaredBessel processes and generalized excursion measures InSection 4 we recall several facts about symmetric Levyprocesses In Section 5 we deal with the joint law of the totallocal times at two levels for the ℎ-paths of symmetric Levyprocesses In Section 6 we study the laws of the total localtimes for the Levy bridges and for the ℎ-bridges

2 Feynman-Kac Formulae

In order to study the laws of total local times we preparetwo versions of Feynman-Kac formulae which describe theirLaplace transforms One is for transient Markov processesand the other is for Markovian bridges

Let D denote the space of cadlag paths 120596 [0infin) rarr

R cup Δ with lifetime 120577 = 120577(120596)

forall119905 lt 120577 120596 (119905) isin R forall119905 ge 120577 120596 (119905) = Δ (1)

Let (119883119905) denote the canonical process119883

119905(120596) = 120596(119905) Let (F

119905)

denote its natural filtration and Finfin

= 120590(cup119905F

119905) For 119886 isin R

we write 119879119886

for the first hitting time of the point 119886

119879119886

= inf 119905 gt 0 119883119905= 119886 (2)

The set of all nonnegative Borel functions on R will bedenoted byB

+(R)

Let (P119909

119909 isin R) denote the laws on D of a right Markovprocess We assume that the transition kernels have jointlymeasurable densities 119901

119905(119909 119910) with respect to a reference

measure 120583(119889119910)

P119909(119883

119905isin 119889119910) = 119901

119905(119909 119910) 120583 (119889119910) (3)

We define

119906119902(119909 119910) = int

infin

0

119890minus119902119905

119901119905(119909 119910) 119889119905 119902 ge 0 (4)

which are resolvent densities if they are finiteWe also assumethat there exists a local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905120583 (119889119910) 119905 gt 0 119891 isin B

+(R) (5)

holds with P119909-probability one for any 119909 isin R

21 Feynman-Kac Formula for TransientMarkov Processes Inthis section we prove Feynman-Kac formula for transientMarkov processes We assume the following conditions

(i) the process is transient(ii) 119906

0(119909 119910) lt infin for any 119909 119910 isin R with 119909 = 0 or 119910 = 0

Note that 1199060(0 0) may be infinite We note that

P119909(forall119910 isin R 119871

119910

infinlt infin) = 1 for any 119909 isin R (6)

By formula (5) it is easy to see that

P119909[119871

119910

infin] = 119906

0(119909 119910) 119909 isin R 119910 isin R 0 (7)

We will prove a version of Feynman-Kac formulae followingMarcus-Rosenrsquos book [16] where it is assumed that 119906

0(0 0) lt

infinFor 119905 ge 0 and 119909

1 119909

2 119909

119899isin R 0 we set

119869119905(x) = int

infin

119905

1198891198711199091

1199051

int

infin

1199051

1198891198711199092

1199052

sdot sdot sdot int

infin

119905119899minus1

119889119871119909119899

119905119899

(8)

where x = (1199091 119909

119899)

Theorem 1 (Kacrsquos moment formula) Let 1199090

isin R and1199091 119909

2 119909

119899isin R 0 Then we has

P1199090

[1198690(x)] = 119906

0(119909

0 119909

1) 119906

0(119909

1 119909

2) sdot sdot sdot 119906

0(119909

119899minus1 119909

119899) (9)

The proof is essentially the same to that of [16 Theorem253] but we give it for completeness of the paper

Proof Note that

1198690(x) = int

infin

0

119869119905(x1015840) 119889119871

1199091

119905 (10)

where x1015840 = (1199092 119909

119899) Denote 120591

1199091

119897= inf119905 gt 0 119871

1199091

119905gt 119897

Since 119869119905(x1015840) = 119869

0(x1015840) ∘ 120579

119905 the strong Markov property yields

that

P1199090

[1198690(x)] = P

1199090

[int

infin

0

1198690(x1015840) ∘ 120579

1205911199091

119897

11205911199091

119897ltinfin

119889119897]

= P1199090

[int

infin

0

11205911199091

119897ltinfin

119889119897] P1199091

[1198690(x1015840)]

= P1199090

[1198711199091

infin]P

1199091

[1198690(x1015840)]

(11)

This yields (9) from (7)

Theorem 2 (Feynman-Kac formula) Let 1199091 119909

119899isin R0

Set

Σ = (

1199060(119909

1 119909

1) sdot sdot sdot 119906

0(119909

1 119909

119899)

1199060(119909

119899 119909

1) sdot sdot sdot 119906

0(119909

119899 119909

119899)

)

Σ0= (

1199060(0 119909

1) sdot sdot sdot 119906

0(0 119909

119899)

1199060(0 119909

1) sdot sdot sdot 119906

0(0 119909

119899)

)

(12)

Then for any diagonal matrix Λ = (120582119894120575119894119895)119899

119894119895=1with nonnega-

tive entries we have

P0[expminus

119899

sum

119894=1

120582119894119871119909119894

infin] =

det (119868 + (Σ minus Σ0)Λ)

det (119868 + ΣΛ) (13)

The proof is almost parallel to that of [16 Lemma 262]but we give it for completeness of the paper

Abstract and Applied Analysis 3

Proof Let 1205821 120582

119899isin R For 119896 isin N we have

P0[

[

(

119899

sum

119895=1

120582119895119871119909119895

infin)

119896

]

]

=

119899

sum

1198951119895119896=1

1205821198951

sdot sdot sdot 120582119895119896

P0[119871

1199091198951

infin sdot sdot sdot 119871119909119895119896

infin ]

(14)

= 119896

119899

sum

1198951119895119896=1

1205821198951

sdot sdot sdot 120582119895119896

P0[1198690(119909

1198951

119909119895119896

)]

(15)

It follows fromTheorem 1 that

(15) = 119896

119899

sum

1198951119895119896=1

1199060(0 119909

1198951

) 1205821198951

sdot 1199060(119909

1198951

1199091198952

) 1205821198952

sdot sdot sdot 1199060(119909

119895119896minus1

119909119895119896

) 120582119895119896

= 119896(ΣΛ)119896

10

(16)

where 1 =⊤(1 1) v

0= 119907

0for v =

⊤(1199070 1199071 119907

119899)

Σ = (

0 1199060(0 119909

1) sdot sdot sdot 119906

0(0 119909

119899)

0 1199060(119909

1 119909

1) sdot sdot sdot 119906

0(119909

1 119909

119899)

0 1199060(119909

119899 119909

1) sdot sdot sdot 119906

0(119909

119899 119909

119899)

)

Λ = (

0 0 sdot sdot sdot 0

0 1205821

sdot sdot sdot 0

0 0 sdot sdot sdot 120582119899

)

(17)

Hence for all 1205821 120582

119899isin R such that |120582

119894|rsquos are small enough

we have

P0[exp

119899

sum

119894=1

120582119894119871119909119894

infin]

=

infin

sum

119896=0

(ΣΛ)119896

10

= (119868 minus ΣΛ)minus1

10

(18)

By Cramerrsquos formula we obtain

(119868 minus ΣΛ)minus1

10

=

det ((119868 minus ΣΛ)(1)

)

det (119868 minus ΣΛ)=det (119868 minus (Σ minus Σ

0)Λ)

det (119868 minus ΣΛ)

(19)

Here for a matrix 119860 we denote by 119860(1) the matrix which

is obtained by replacing each entry in the first column of 119860by number 1 Since Σ is nonnegative definite we obtain thedesired result (13) by analytic continuation

Remark 3 Eisenbaum et al [6] have proved an analogue ofRay-Knight theorem for the total local time of a symmetricLevy process killed at an independent exponential time Wemay say that the key to the proof is that Σ minus Σ

0 is a constantmatrix which is positive definite The difficulty in the case ofthe ℎ-path process of a symmetric Levy process is that thematrix Σ minus Σ

0 no longer has such a nice property

22 Feynman-Kac Formula for Markovian Bridges In thissection we show Feynman-Kac formula for Markovianbridges For this we recall several theorems for Markovianbridges from Fitzsimmons et al [15] See [15] for details

For 119905 gt 0 119909 119910 isin R let P119905119909119910

denote the bridge law whichserves as a version of the regular conditional distribution for119883

119904 0 le 119904 le 119905 under P

119909given 119883

119905minus= 119910 In this section we

assume the following condition

(i) 0 lt 119901119905(119909 119910) lt infin for any 119905 gt 0 119909 119910 isin R

We also assume that there exists a local time (119871119909

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119910) 119871

119910

119904120583 (119889119910) 0 le 119904 le 119905

119891 isin B+(R)

(20)

holds with P119905119909119910

-probability one for any 119905 gt 0 and 119909 119910 isin R

Theorem 4 (see [15 Lemma 1]) Let 119905 gt 0 119909 119910 119911 isin R Thenone has

P119905119909119910

[int

119905

0

119891 (119904 119883119904) 119889119871

119911

119904] = int

119905

0

119889119904119901119904(119909 119911) 119901

119905minus119904(119911 119910)

119901119905(119909 119910)

119891 (119904 119911)

(21)

for any nonnegative Borel function 119891

We will also use the following conditioning formula

Theorem 5 (see [15 Proposition 3]) Let 119905 gt 0 119909 119910 119911 isin RThen one has

P119905119909119910

[int

119905

0

119891 (119904 119883119904)119867

119904119889119871

119911

119904]

= P119905119909119910

[int

119905

0

119891 (119904 119911)P119904119909119911

[119867119904] 119889119871

119911

119904]

(22)

for any nonnegative Borel function 119891 and any nonnegativepredictable process 119867

119904

For 119904 ge 0 and 1199111 119911

119899isin R we define

119867119904(z(119899)) = int

119904

0

119889119871119911119899

119904119899

int

119904119899

0

119889119871119911119899minus1

119904119899minus1

sdot sdot sdot int

1199042

0

1198891198711199111

1199041

(23)

where z(119899) = (1199111 119911

119899) The following theorem is a version

of Kacrsquos moment formulae

4 Abstract and Applied Analysis

Theorem 6 For any 119902 gt 0 119899 isin N and for any 1199111 119911

119899isin R

one has

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(z(119899))] 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119910)

(24)

Proof Let us prove the claim by induction For 119899 = 1 theassertion follows fromTheorem 4 Suppose that formula (24)holds for a given 119899 ge 2 Note that

119867119905(z(119899+1)) = int

119905

0

119867119904(z(119899)) 119889119871

119911119899+1

119904 (25)

Since119867119904(z(119899)) is a nonnegative predictable processTheorems

5 and 4 show that

P119905119909119910

[119867119905(z(119899+1))]

= int

119905

0

119889119904119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)

119901119905(119909 119910)

P119905119909119911119899+1

[119867119904(119911

(119899))]

(26)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(119911

(119899+1))] 119889119905

= int

infin

0

119890minus119902119905

119889119905

times int

119905

0

119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

= int

infin

0

119890minus119902119904

119901119904(119909 119911

119899+1)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

times int

infin

0

119890minus119902119905

119901119905(119911

119899+1 119910) 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119911

119899+1) sdot 119906

119902(119911

119899+1 119910)

(27)

by the assumption of the induction Nowwe have proved thatformula (24) is valid also for 119899+1 which completes the proof

The following theorem is a version of Feynman-Kacformulae

Theorem 7 Let 1199111= 0 119911

2 119911

119899isin R and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (28)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895) Then for

any diagonal matrix Λ = (120582119894120575119894119895)119899

119894119895=1with nonnegative entries

one has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

= (119868 + Σ(119902)

Λ)minus1

Σ(119902)

11

(29)

Proof We have

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

120582119895119896

sdot sdot sdot 1205821198951

times int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119867

119905(119911

119895119896

1199111198951

)] 119889119905

(30)

UsingTheorem 6 we see that the above quantity is equal to

119896

119899

sum

1198951119895119896=1

119906119902(119911

1 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

1199111)

(31)

which amounts to 119896(Σ(119902)

Λ)119896Σ(119902)

11 Hence for all 120582

1

120582119899gt 0 sufficiently small we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

]

]

119889119905

= 119906119902(0 0) +

infin

sum

119896=1

(Σ(119902)

Λ)119896

Σ(119902)

11

= (119868 minus Σ(119902)

Λ)minus1

Σ(119902)

11

(32)

Since Σ(119902) is nonnegative definite we obtain the desired result

(29) by analytic continuation

The following theorem is valid even if

119906119902(0 119911

119895) = 119906

119902(119911

119895 0) = infin 119902 gt 0 119895 = 1 119899 (33)

Theorem 8 Let 1199111 119911

119899isin R 0 and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (34)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895)

u(119902) = (

119906119902(0 119911

1)

119906119902(0 119911

119899)

) v(119902) = (

119906119902(119911

1 0)

119906119902(119911

119899 0)

) (35)

Abstract and Applied Analysis 5

and let Λ be the matrix with elements Λ119894119895

= 120582119894120575119894119895 Then one

has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[1 minus 119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

=⊤u(119902)Λ(119868 + Σ

(119902)Λ)

minus1

v(119902)(36)

Proof UsingTheorem 6 we see that

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

119906119902(0 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

0)

(37)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

minus 1]

]

119889119905

=

infin

sum

119896=1

⊤u(119902)Λ(Σ

(119902)Λ)

119896minus1

v(119902)

=⊤u(119902)Λ(119868 minus Σ

(119902)Λ)

minus1

v(119902)

(38)

The rest of the proof is now obvious

3 Preliminaries Squared Bessel Processes andGeneralized Excursion Measures

In this section we recall squared Bessel processes andgeneralized excursion measures

First we introduce several notations about squared Besselprocesses for which we follow [3 XI1] For 120575 ge 0 let (Q120575

119911

119911 ge 0) denote the law of the 120575-dimensional squared Besselprocess where the origin is a trap when 120575 = 0 Then theLaplace transform of a one-dimensional marginal is given by

Q120575

119911[exp minus120582119883

119905] =

1

(1 + 2120582119905)1205752

expminus120582119911

1 + 2120582119905 (39)

We may obtain the transition kernels Q120575

119911(119883

119905isin 119889119908) by the

Laplace inversion

(i) For 120575 gt 0 and 119911 gt 0 we have

Q120575

119911(119883

119905isin 119889119908)

=1

2119905(119908

119911)

(12)(1205752minus1)

exp minus119911 + 119908

2119905 119868

1205752minus1(radic119911119908

119905) 119889119908

(40)

where 119868120584stands for the modified Bessel function of

order 120584

(ii) For 120575 gt 0 and 119911 = 0 we have

Q120575

0(119883

119905isin 119889119908) =

1

(2119905)1205752

Γ (1205752)1199081205752minus1 exp minus

119908

2119905 119889119908

(41)

where Γ stands for the gamma function

(iii) For 120575 = 0 and 119911 ge 0 we have

Q0

119911(119883

119905isin 119889119908) = exp minus

119911

2119905 120575

0(119889119908)

+1

2119905(119908

119911)

minus12

exp minus119911 + 119908

2119905

times 1198681(radic119911119908

119905) 119889119908

(42)

The squared Bessel process satisfies the scaling property for120575 ge 0 119911 ge 0 and 119888 gt 0 it holds that

(119888119883119905119888

) under Q120575

119911119888

law= (119883

119905) under Q120575

119911 (43)

Second we recall the notion of the generalized excursionmeasure By formula (39) we have

Q4

0[

1

1198832

119904+119905

119883119904+119905

isin 119861] = Q4

0[

1

1198832

119904

sdot Q0

119883119904

(119883119905isin 119861)] (44)

for 119904 119905 gt 0 and 119861 isin B([0infin)) If we put 120583119905(119889119909) =

(11199092)Q4

0(119883

119905isin 119889119909) we have

120583119904+119905

(119861) = int120583119904(119889119909)Q0

119909(119883

119905isin 119861) (45)

This shows that the family of laws 120583119905

119905 gt 0 is an entrancelaw for Q0

119909 119909 gt 0 In fact there exists a unique 120590-finite

measure n(0) on D such that

n(0) (1198831199051

isin 1198611 119883

119905119899

isin 119861119899)

= int1198611

1205831199051

(1198891199091) int

1198612

Q0

119909(119883

1199052minus1199051

isin 1198891199092)

sdot sdot sdot int119861119899

Q0

119909(119883

119905119899minus119905119899minus1

isin 119889119909119899)

(46)

for 0 lt 1199051

lt sdot sdot sdot lt 119905119899and 119861

1 119861

119899isin B([0infin)) Note

that to construct such a measure n(0) we can not appeal toKolmogorovrsquos extension theorem because the entrance lawshave infinite total mass However we can actually constructn(0) via the agreement formula (see Pitman-Yor [17 Cor 3]with 120575 = 4) or via the time change of a Brownian excursion(see Fitzsimmons-Yano [18 Theorem 25] with change ofscales)Wemay calln(0) the generalized excursionmeasure forthe squared Bessel process of dimension 0 See the referencesabove for several characteristic formulae of n(0)

6 Abstract and Applied Analysis

4 Symmetric Leacutevy Processes

Let us confine ourselves to one-dimensional symmetric Levyprocesses We recall general facts and state several resultsfrom [7]

In what follows we assume that (P119909) is the law of a

one-dimensional conservative Levy process Throughout thepresent paper we assume the following conditions whichwillbe referred to as (A) are satisfied

(i) the process is symmetric(ii) the origin (and consequently any point) is regular for

itself(iii) the process is not a compound Poisson

Under the condition (A) we have the following The charac-teristic exponent is given by

120579 (120582) = minus logP0[1198901198941205821198831] = 119907120582

2+ 2int

infin

0

(1 minus cos 120582119909) 120584 (119889119909)

(47)

for some 119907 ge 0 and some positive Radonmeasure 120584 on (0infin)

such that

int(0infin)

min 1199092 1 120584 (119889119909) lt infin (48)

The reference measure is 120583(119889119909) = 119889119909 and we have

119901119905(119909 119910) = 119901

119905(119910 minus 119909) =

1

120587int

infin

0

(cos 120582 (119910 minus 119909)) 119890minus119905120579(120582)

119889120582

(49)

119906119902(119909 119910) = 119906

119902(119910 minus 119909) =

1

120587int

infin

0

cos 120582 (119910 minus 119909)

119902 + 120579 (120582)119889120582 (50)

There exists a local time (119871119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905119889119910 119891 isin B

+(R) (51)

with P119909-probability one for any 119909 isin R Then it holds that

P119909[int

infin

0

119890minus119902119904

119889119871119910

119904] = 119906

119902(119910 minus 119909) 119909 119910 isin R (52)

Let n denote the excursion measure associated to the localtime 119871

0

119905 We denote by (P0

119909 119909 isin R 0) the law of the

process killed upon hitting the origin that is

P0119909(119860 120577 gt 119905) = P

119909(119860 119879

0gt 119905) 119909 isin R 0

119905 gt 0 119860 isin F119905

(53)

Then the excursion measure n satisfies the Markov propertyin the following sense for any 119905 gt 0 and for any nonnegativeF

119905-measurable functional 119885

119905and for any nonnegative F

infin-

measurable functional 119865 it holds that

n [119885119905119865 (119883

119905+sdot)] = intn [119885

119905 119883

119905isin 119889119909]P0

119909[119865 (119883)] (54)

We need the following additional conditions

(R) the process is recurrent(T) the function 120579(120582) is nondecreasing in 120582 gt 120582

0for some

1205820gt 0

Under the condition (A) the condition (R) is equivalent to

int

infin

0

119889120582

120579 (120582)= infin (55)

All of the conditions (A) (R) and (T) are obviously satisfiedif the process is a symmetric stable Levy process of index 120572 isin

(1 2]

120579 (120582) = |120582|120572 (56)

In what follows we assume as well as the condition (A) thatthe conditions (R) and (T) are also satisfied

The Laplace transform of the law of 1198790

is given by

P119911[119890minus1199021198790] =

119906119902(119911)

119906119902(0)

(57)

see for example [19 pp 64] It is easy to see that the entrancelaw has the space density

120588 (119905 119909) =n (119883

119905isin 119889119909)

119889119909 (58)

In view of [7 Theorem 210] the law of the hitting time 1198790

is absolutely continuous relative to the Lebesgue measure 119889119905

and the time density coincides with the space density of theentrance law

120588119909(119905) =

P119909(119879

0isin 119889119905)

119889119905= 120588 (119905 119909) (59)

41 Absolute Continuity of the Law of the Inverse Local TimeLet 120591(119897) denote the inverse local time at the origin

120591 (119897) = inf 119905 gt 0 1198710

119905gt 119897 (60)

We prove the absolute continuity of the law of inverse localtime Note that 120591(119897) is a subordinator such that

P0[119890minus119902120591(119897)

] = 119890minus119897119906119902(0)

(61)

see for example [19 pp 131]

Lemma 9 For fixed 119897 gt 0 the law of 120591(119897) under P0has a

density 120574119897(119905)

P0(120591 (119897) isin 119889119905) = 120574

119897(119905) 119889119905 (62)

Furthermore 120574119897(119905) may be chosen to be jointly continuous in

(119897 119905) isin (0infin) times (0infin)

Proof Following [7 Sec 33] we define a positive Borelmeasure 120590 on [0infin) as

120590 (119860) =1

120587int

infin

0

1119860(120579 (120582)) 119889120582 119860 isin B ([0infin)) (63)

Abstract and Applied Analysis 7

Then we have 119906119902(0) = int

[0infin)(120590(119889120585)(119902 + 120585)) for 119902 gt 0

and hence there exists a Radon measure 120590lowast on [0infin) with

int[0infin)

(120590lowast(119889120585)(1 + 120585)) lt infin such that

1

119902119906119902(0)

= int[0infin)

1

119902 + 120585120590lowast(119889120585) 119902 gt 0 (64)

Hence the Laplace exponent 1119906119902(0) may be represented as

1

119906119902(0)

= int

infin

0

(1 minus 119890minus119902119906

) 120584 (119906) 119889119906 (65)

where 120584(119906) = int(0infin)

119890minus119906120585

120585120590lowast(119889120585) Since int

infin

0(1 and 119906

2)120584(119906)119889119906 lt

infin we may appeal to analytic continuation of both sides offormula (61) and obtain

P0[119890119894120582120591(119897)

] = exp119897 int

infin

0

(119890119894120582119906

minus 1) 120584 (119906) 119889119906 (66)

Following [20 Theorem 31] we may invert the Fouriertransform of the law of 120591(119897) and obtain the desired result

42 ℎ-Paths of Symmetric Levy Processes We follow [7]for the notations concerning ℎ-paths of symmetric Levyprocesses For the interpretation of the ℎ-paths as some kindof conditioning see [10]

We define

ℎ (119909) = lim119902rarr0+

119906119902(0) minus 119906

119902(119909)

=1

120587int

infin

0

1 minus cos 120582119909120579 (120582)

119889120582 119909 isin R

(67)

The second equality follows from (50) Then the function ℎ

satisfies the following

(i) ℎ(119909) is continuous(ii) ℎ(0) = 0 ℎ(119909) gt 0 for all 119909 isin R 0(iii) ℎ(119909) rarr infin as |119909| rarr infin (since the condition (R) is

satisfied)

See [7 Lemma 42] for the proof Moreover the function ℎ isharmonic with respect to the killed process

P0119909[ℎ (119883

119905)] = ℎ (119909) if 119909 isin R 0 119905 gt 0

n [ℎ (119883119905)] = 1 if 119905 gt 0

(68)

See [7 Theorems 11 and 12] for the proof We define the ℎ-path process (Pℎ

119909 119909 isin R) by the following local equivalence

relations

119889Pℎ119909

10038161003816100381610038161003816F119905

=

ℎ (119883119905)

ℎ (119909)119889P0

119909

100381610038161003816100381610038161003816100381610038161003816F119905

if 119909 isin R 0

ℎ (119883119905) 119889n1003816100381610038161003816F

119905

if 119909 = 0

(69)

Remark that from the strong Markov properties of (119883119905)

under P0119909and n the family Pℎ

119909|F119905

119905 ge 0 is consistent andhence the probability measure Pℎ

119909is well defined

The ℎ-path process is then symmetric more preciselythe transition kernel has a symmetric density 119901

ℎ

119905(119909 119910) with

respect to the measure ℎ(119910)2119889119910 Here the density 119901

ℎ

119905(119909 119910) is

given by

119901ℎ

119905(119909 119910) =

1

ℎ (119909) ℎ (119910)119901

119905(119910 minus 119909) minus

119901119905(119909) 119901

119905(119910)

119901119905(0)

if 119909 119910 isin R 0

119901ℎ

119905(119909 0) = 119901

ℎ

119905(0 119909) =

120588 (119905 119909)

ℎ (119909)

if 119909 isin R 0

119901ℎ

119905(0 0) = int

(0infin)

119890minus119905120585

120585120590lowast(119889120585)

(70)

By (65) we see that 119901ℎ119905(0 0) is characterized by

int

infin

0

(1 minus 119890minus119902119905

) 119901ℎ

119905(0 0) 119889119905 =

1

119906119902(0)

119902 gt 0 (71)

See [7 Section 5] for the details The ℎ-path process alsosatisfies the following conditions

(i) the process is conservative(ii) any point is regular for itself(iii) the process is transient (since the condition (T) is

satisfied)

We can easily prove regularity of any point by the localequivalence (69) See [7 Theorem 14] for the proof oftransience

The resolvent density of the ℎ-path process with respectto ℎ(119910)

2119889119910 is given by

119906ℎ

119902(119909 119910) =

1

ℎ (119909) ℎ (119910)119906

119902(119909 minus 119910) minus

119906119902(119909) 119906

119902(119910)

119906119902(0)

119902 gt 0 119909 119910 isin R 0

119906ℎ

119902(119909 0) = 119906

ℎ

119902(0 119909) =

1

ℎ (119909)sdot119906119902(119909)

119906119902(0)

119902 gt 0 119909 isin R 0

(72)

We remark here that since lim119902rarrinfin

119906119902(0) = 0 we see by (71)

that

119906ℎ

119902(0 0) = infin (73)

The Green function 119906ℎ

0(119909 119910) = lim

119902rarr0+119906ℎ

119902(119909 119910) exists and is

given by

119906ℎ

0(119909 119910) =

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910) 119909 119910 isin R 0

119906ℎ

0(119909 0) = 119906

ℎ

0(0 119909) =

1

ℎ (119909) 119909 isin R 0

(74)

8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

expression of the law of the total local time at a single levelfor the Levy bridges while unfortunately we do not have anynice formula for the ℎ-bridges

The present paper is organized as follows In Section 2 wegive two versions of Feynman-Kac formulae in general set-tings In Section 3 we recall several formulae about squaredBessel processes and generalized excursion measures InSection 4 we recall several facts about symmetric Levyprocesses In Section 5 we deal with the joint law of the totallocal times at two levels for the ℎ-paths of symmetric Levyprocesses In Section 6 we study the laws of the total localtimes for the Levy bridges and for the ℎ-bridges

2 Feynman-Kac Formulae

In order to study the laws of total local times we preparetwo versions of Feynman-Kac formulae which describe theirLaplace transforms One is for transient Markov processesand the other is for Markovian bridges

Let D denote the space of cadlag paths 120596 [0infin) rarr

R cup Δ with lifetime 120577 = 120577(120596)

forall119905 lt 120577 120596 (119905) isin R forall119905 ge 120577 120596 (119905) = Δ (1)

Let (119883119905) denote the canonical process119883

119905(120596) = 120596(119905) Let (F

119905)

denote its natural filtration and Finfin

= 120590(cup119905F

119905) For 119886 isin R

we write 119879119886

for the first hitting time of the point 119886

119879119886

= inf 119905 gt 0 119883119905= 119886 (2)

The set of all nonnegative Borel functions on R will bedenoted byB

+(R)

Let (P119909

119909 isin R) denote the laws on D of a right Markovprocess We assume that the transition kernels have jointlymeasurable densities 119901

119905(119909 119910) with respect to a reference

measure 120583(119889119910)

P119909(119883

119905isin 119889119910) = 119901

119905(119909 119910) 120583 (119889119910) (3)

We define

119906119902(119909 119910) = int

infin

0

119890minus119902119905

119901119905(119909 119910) 119889119905 119902 ge 0 (4)

which are resolvent densities if they are finiteWe also assumethat there exists a local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905120583 (119889119910) 119905 gt 0 119891 isin B

+(R) (5)

holds with P119909-probability one for any 119909 isin R

21 Feynman-Kac Formula for TransientMarkov Processes Inthis section we prove Feynman-Kac formula for transientMarkov processes We assume the following conditions

(i) the process is transient(ii) 119906

0(119909 119910) lt infin for any 119909 119910 isin R with 119909 = 0 or 119910 = 0

Note that 1199060(0 0) may be infinite We note that

P119909(forall119910 isin R 119871

119910

infinlt infin) = 1 for any 119909 isin R (6)

By formula (5) it is easy to see that

P119909[119871

119910

infin] = 119906

0(119909 119910) 119909 isin R 119910 isin R 0 (7)

We will prove a version of Feynman-Kac formulae followingMarcus-Rosenrsquos book [16] where it is assumed that 119906

0(0 0) lt

infinFor 119905 ge 0 and 119909

1 119909

2 119909

119899isin R 0 we set

119869119905(x) = int

infin

119905

1198891198711199091

1199051

int

infin

1199051

1198891198711199092

1199052

sdot sdot sdot int

infin

119905119899minus1

119889119871119909119899

119905119899

(8)

where x = (1199091 119909

119899)

Theorem 1 (Kacrsquos moment formula) Let 1199090

isin R and1199091 119909

2 119909

119899isin R 0 Then we has

P1199090

[1198690(x)] = 119906

0(119909

0 119909

1) 119906

0(119909

1 119909

2) sdot sdot sdot 119906

0(119909

119899minus1 119909

119899) (9)

The proof is essentially the same to that of [16 Theorem253] but we give it for completeness of the paper

Proof Note that

1198690(x) = int

infin

0

119869119905(x1015840) 119889119871

1199091

119905 (10)

where x1015840 = (1199092 119909

119899) Denote 120591

1199091

119897= inf119905 gt 0 119871

1199091

119905gt 119897

Since 119869119905(x1015840) = 119869

0(x1015840) ∘ 120579

119905 the strong Markov property yields

that

P1199090

[1198690(x)] = P

1199090

[int

infin

0

1198690(x1015840) ∘ 120579

1205911199091

119897

11205911199091

119897ltinfin

119889119897]

= P1199090

[int

infin

0

11205911199091

119897ltinfin

119889119897] P1199091

[1198690(x1015840)]

= P1199090

[1198711199091

infin]P

1199091

[1198690(x1015840)]

(11)

This yields (9) from (7)

Theorem 2 (Feynman-Kac formula) Let 1199091 119909

119899isin R0

Set

Σ = (

1199060(119909

1 119909

1) sdot sdot sdot 119906

0(119909

1 119909

119899)

1199060(119909

119899 119909

1) sdot sdot sdot 119906

0(119909

119899 119909

119899)

)

Σ0= (

1199060(0 119909

1) sdot sdot sdot 119906

0(0 119909

119899)

1199060(0 119909

1) sdot sdot sdot 119906

0(0 119909

119899)

)

(12)

Then for any diagonal matrix Λ = (120582119894120575119894119895)119899

119894119895=1with nonnega-

tive entries we have

P0[expminus

119899

sum

119894=1

120582119894119871119909119894

infin] =

det (119868 + (Σ minus Σ0)Λ)

det (119868 + ΣΛ) (13)

The proof is almost parallel to that of [16 Lemma 262]but we give it for completeness of the paper

Abstract and Applied Analysis 3

Proof Let 1205821 120582

119899isin R For 119896 isin N we have

P0[

[

(

119899

sum

119895=1

120582119895119871119909119895

infin)

119896

]

]

=

119899

sum

1198951119895119896=1

1205821198951

sdot sdot sdot 120582119895119896

P0[119871

1199091198951

infin sdot sdot sdot 119871119909119895119896

infin ]

(14)

= 119896

119899

sum

1198951119895119896=1

1205821198951

sdot sdot sdot 120582119895119896

P0[1198690(119909

1198951

119909119895119896

)]

(15)

It follows fromTheorem 1 that

(15) = 119896

119899

sum

1198951119895119896=1

1199060(0 119909

1198951

) 1205821198951

sdot 1199060(119909

1198951

1199091198952

) 1205821198952

sdot sdot sdot 1199060(119909

119895119896minus1

119909119895119896

) 120582119895119896

= 119896(ΣΛ)119896

10

(16)

where 1 =⊤(1 1) v

0= 119907

0for v =

⊤(1199070 1199071 119907

119899)

Σ = (

0 1199060(0 119909

1) sdot sdot sdot 119906

0(0 119909

119899)

0 1199060(119909

1 119909

1) sdot sdot sdot 119906

0(119909

1 119909

119899)

0 1199060(119909

119899 119909

1) sdot sdot sdot 119906

0(119909

119899 119909

119899)

)

Λ = (

0 0 sdot sdot sdot 0

0 1205821

sdot sdot sdot 0

0 0 sdot sdot sdot 120582119899

)

(17)

Hence for all 1205821 120582

119899isin R such that |120582

119894|rsquos are small enough

we have

P0[exp

119899

sum

119894=1

120582119894119871119909119894

infin]

=

infin

sum

119896=0

(ΣΛ)119896

10

= (119868 minus ΣΛ)minus1

10

(18)

By Cramerrsquos formula we obtain

(119868 minus ΣΛ)minus1

10

=

det ((119868 minus ΣΛ)(1)

)

det (119868 minus ΣΛ)=det (119868 minus (Σ minus Σ

0)Λ)

det (119868 minus ΣΛ)

(19)

Here for a matrix 119860 we denote by 119860(1) the matrix which

is obtained by replacing each entry in the first column of 119860by number 1 Since Σ is nonnegative definite we obtain thedesired result (13) by analytic continuation

Remark 3 Eisenbaum et al [6] have proved an analogue ofRay-Knight theorem for the total local time of a symmetricLevy process killed at an independent exponential time Wemay say that the key to the proof is that Σ minus Σ

0 is a constantmatrix which is positive definite The difficulty in the case ofthe ℎ-path process of a symmetric Levy process is that thematrix Σ minus Σ

0 no longer has such a nice property

22 Feynman-Kac Formula for Markovian Bridges In thissection we show Feynman-Kac formula for Markovianbridges For this we recall several theorems for Markovianbridges from Fitzsimmons et al [15] See [15] for details

For 119905 gt 0 119909 119910 isin R let P119905119909119910

denote the bridge law whichserves as a version of the regular conditional distribution for119883

119904 0 le 119904 le 119905 under P

119909given 119883

119905minus= 119910 In this section we

assume the following condition

(i) 0 lt 119901119905(119909 119910) lt infin for any 119905 gt 0 119909 119910 isin R

We also assume that there exists a local time (119871119909

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119910) 119871

119910

119904120583 (119889119910) 0 le 119904 le 119905

119891 isin B+(R)

(20)

holds with P119905119909119910

-probability one for any 119905 gt 0 and 119909 119910 isin R

Theorem 4 (see [15 Lemma 1]) Let 119905 gt 0 119909 119910 119911 isin R Thenone has

P119905119909119910

[int

119905

0

119891 (119904 119883119904) 119889119871

119911

119904] = int

119905

0

119889119904119901119904(119909 119911) 119901

119905minus119904(119911 119910)

119901119905(119909 119910)

119891 (119904 119911)

(21)

for any nonnegative Borel function 119891

We will also use the following conditioning formula

Theorem 5 (see [15 Proposition 3]) Let 119905 gt 0 119909 119910 119911 isin RThen one has

P119905119909119910

[int

119905

0

119891 (119904 119883119904)119867

119904119889119871

119911

119904]

= P119905119909119910

[int

119905

0

119891 (119904 119911)P119904119909119911

[119867119904] 119889119871

119911

119904]

(22)

for any nonnegative Borel function 119891 and any nonnegativepredictable process 119867

119904

For 119904 ge 0 and 1199111 119911

119899isin R we define

119867119904(z(119899)) = int

119904

0

119889119871119911119899

119904119899

int

119904119899

0

119889119871119911119899minus1

119904119899minus1

sdot sdot sdot int

1199042

0

1198891198711199111

1199041

(23)

where z(119899) = (1199111 119911

119899) The following theorem is a version

of Kacrsquos moment formulae

4 Abstract and Applied Analysis

Theorem 6 For any 119902 gt 0 119899 isin N and for any 1199111 119911

119899isin R

one has

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(z(119899))] 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119910)

(24)

Proof Let us prove the claim by induction For 119899 = 1 theassertion follows fromTheorem 4 Suppose that formula (24)holds for a given 119899 ge 2 Note that

119867119905(z(119899+1)) = int

119905

0

119867119904(z(119899)) 119889119871

119911119899+1

119904 (25)

Since119867119904(z(119899)) is a nonnegative predictable processTheorems

5 and 4 show that

P119905119909119910

[119867119905(z(119899+1))]

= int

119905

0

119889119904119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)

119901119905(119909 119910)

P119905119909119911119899+1

[119867119904(119911

(119899))]

(26)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(119911

(119899+1))] 119889119905

= int

infin

0

119890minus119902119905

119889119905

times int

119905

0

119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

= int

infin

0

119890minus119902119904

119901119904(119909 119911

119899+1)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

times int

infin

0

119890minus119902119905

119901119905(119911

119899+1 119910) 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119911

119899+1) sdot 119906

119902(119911

119899+1 119910)

(27)

by the assumption of the induction Nowwe have proved thatformula (24) is valid also for 119899+1 which completes the proof

The following theorem is a version of Feynman-Kacformulae

Theorem 7 Let 1199111= 0 119911

2 119911

119899isin R and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (28)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895) Then for

any diagonal matrix Λ = (120582119894120575119894119895)119899

119894119895=1with nonnegative entries

one has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

= (119868 + Σ(119902)

Λ)minus1

Σ(119902)

11

(29)

Proof We have

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

120582119895119896

sdot sdot sdot 1205821198951

times int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119867

119905(119911

119895119896

1199111198951

)] 119889119905

(30)

UsingTheorem 6 we see that the above quantity is equal to

119896

119899

sum

1198951119895119896=1

119906119902(119911

1 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

1199111)

(31)

which amounts to 119896(Σ(119902)

Λ)119896Σ(119902)

11 Hence for all 120582

1

120582119899gt 0 sufficiently small we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

]

]

119889119905

= 119906119902(0 0) +

infin

sum

119896=1

(Σ(119902)

Λ)119896

Σ(119902)

11

= (119868 minus Σ(119902)

Λ)minus1

Σ(119902)

11

(32)

Since Σ(119902) is nonnegative definite we obtain the desired result

(29) by analytic continuation

The following theorem is valid even if

119906119902(0 119911

119895) = 119906

119902(119911

119895 0) = infin 119902 gt 0 119895 = 1 119899 (33)

Theorem 8 Let 1199111 119911

119899isin R 0 and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (34)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895)

u(119902) = (

119906119902(0 119911

1)

119906119902(0 119911

119899)

) v(119902) = (

119906119902(119911

1 0)

119906119902(119911

119899 0)

) (35)

Abstract and Applied Analysis 5

and let Λ be the matrix with elements Λ119894119895

= 120582119894120575119894119895 Then one

has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[1 minus 119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

=⊤u(119902)Λ(119868 + Σ

(119902)Λ)

minus1

v(119902)(36)

Proof UsingTheorem 6 we see that

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

119906119902(0 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

0)

(37)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

minus 1]

]

119889119905

=

infin

sum

119896=1

⊤u(119902)Λ(Σ

(119902)Λ)

119896minus1

v(119902)

=⊤u(119902)Λ(119868 minus Σ

(119902)Λ)

minus1

v(119902)

(38)

The rest of the proof is now obvious

3 Preliminaries Squared Bessel Processes andGeneralized Excursion Measures

In this section we recall squared Bessel processes andgeneralized excursion measures

First we introduce several notations about squared Besselprocesses for which we follow [3 XI1] For 120575 ge 0 let (Q120575

119911

119911 ge 0) denote the law of the 120575-dimensional squared Besselprocess where the origin is a trap when 120575 = 0 Then theLaplace transform of a one-dimensional marginal is given by

Q120575

119911[exp minus120582119883

119905] =

1

(1 + 2120582119905)1205752

expminus120582119911

1 + 2120582119905 (39)

We may obtain the transition kernels Q120575

119911(119883

119905isin 119889119908) by the

Laplace inversion

(i) For 120575 gt 0 and 119911 gt 0 we have

Q120575

119911(119883

119905isin 119889119908)

=1

2119905(119908

119911)

(12)(1205752minus1)

exp minus119911 + 119908

2119905 119868

1205752minus1(radic119911119908

119905) 119889119908

(40)

where 119868120584stands for the modified Bessel function of

order 120584

(ii) For 120575 gt 0 and 119911 = 0 we have

Q120575

0(119883

119905isin 119889119908) =

1

(2119905)1205752

Γ (1205752)1199081205752minus1 exp minus

119908

2119905 119889119908

(41)

where Γ stands for the gamma function

(iii) For 120575 = 0 and 119911 ge 0 we have

Q0

119911(119883

119905isin 119889119908) = exp minus

119911

2119905 120575

0(119889119908)

+1

2119905(119908

119911)

minus12

exp minus119911 + 119908

2119905

times 1198681(radic119911119908

119905) 119889119908

(42)

The squared Bessel process satisfies the scaling property for120575 ge 0 119911 ge 0 and 119888 gt 0 it holds that

(119888119883119905119888

) under Q120575

119911119888

law= (119883

119905) under Q120575

119911 (43)

Second we recall the notion of the generalized excursionmeasure By formula (39) we have

Q4

0[

1

1198832

119904+119905

119883119904+119905

isin 119861] = Q4

0[

1

1198832

119904

sdot Q0

119883119904

(119883119905isin 119861)] (44)

for 119904 119905 gt 0 and 119861 isin B([0infin)) If we put 120583119905(119889119909) =

(11199092)Q4

0(119883

119905isin 119889119909) we have

120583119904+119905

(119861) = int120583119904(119889119909)Q0

119909(119883

119905isin 119861) (45)

This shows that the family of laws 120583119905

119905 gt 0 is an entrancelaw for Q0

119909 119909 gt 0 In fact there exists a unique 120590-finite

measure n(0) on D such that

n(0) (1198831199051

isin 1198611 119883

119905119899

isin 119861119899)

= int1198611

1205831199051

(1198891199091) int

1198612

Q0

119909(119883

1199052minus1199051

isin 1198891199092)

sdot sdot sdot int119861119899

Q0

119909(119883

119905119899minus119905119899minus1

isin 119889119909119899)

(46)

for 0 lt 1199051

lt sdot sdot sdot lt 119905119899and 119861

1 119861

119899isin B([0infin)) Note

that to construct such a measure n(0) we can not appeal toKolmogorovrsquos extension theorem because the entrance lawshave infinite total mass However we can actually constructn(0) via the agreement formula (see Pitman-Yor [17 Cor 3]with 120575 = 4) or via the time change of a Brownian excursion(see Fitzsimmons-Yano [18 Theorem 25] with change ofscales)Wemay calln(0) the generalized excursionmeasure forthe squared Bessel process of dimension 0 See the referencesabove for several characteristic formulae of n(0)

6 Abstract and Applied Analysis

4 Symmetric Leacutevy Processes

Let us confine ourselves to one-dimensional symmetric Levyprocesses We recall general facts and state several resultsfrom [7]

In what follows we assume that (P119909) is the law of a

one-dimensional conservative Levy process Throughout thepresent paper we assume the following conditions whichwillbe referred to as (A) are satisfied

(i) the process is symmetric(ii) the origin (and consequently any point) is regular for

itself(iii) the process is not a compound Poisson

Under the condition (A) we have the following The charac-teristic exponent is given by

120579 (120582) = minus logP0[1198901198941205821198831] = 119907120582

2+ 2int

infin

0

(1 minus cos 120582119909) 120584 (119889119909)

(47)

for some 119907 ge 0 and some positive Radonmeasure 120584 on (0infin)

such that

int(0infin)

min 1199092 1 120584 (119889119909) lt infin (48)

The reference measure is 120583(119889119909) = 119889119909 and we have

119901119905(119909 119910) = 119901

119905(119910 minus 119909) =

1

120587int

infin

0

(cos 120582 (119910 minus 119909)) 119890minus119905120579(120582)

119889120582

(49)

119906119902(119909 119910) = 119906

119902(119910 minus 119909) =

1

120587int

infin

0

cos 120582 (119910 minus 119909)

119902 + 120579 (120582)119889120582 (50)

There exists a local time (119871119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905119889119910 119891 isin B

+(R) (51)

with P119909-probability one for any 119909 isin R Then it holds that

P119909[int

infin

0

119890minus119902119904

119889119871119910

119904] = 119906

119902(119910 minus 119909) 119909 119910 isin R (52)

Let n denote the excursion measure associated to the localtime 119871

0

119905 We denote by (P0

119909 119909 isin R 0) the law of the

process killed upon hitting the origin that is

P0119909(119860 120577 gt 119905) = P

119909(119860 119879

0gt 119905) 119909 isin R 0

119905 gt 0 119860 isin F119905

(53)

Then the excursion measure n satisfies the Markov propertyin the following sense for any 119905 gt 0 and for any nonnegativeF

119905-measurable functional 119885

119905and for any nonnegative F

infin-

measurable functional 119865 it holds that

n [119885119905119865 (119883

119905+sdot)] = intn [119885

119905 119883

119905isin 119889119909]P0

119909[119865 (119883)] (54)

We need the following additional conditions

(R) the process is recurrent(T) the function 120579(120582) is nondecreasing in 120582 gt 120582

0for some

1205820gt 0

Under the condition (A) the condition (R) is equivalent to

int

infin

0

119889120582

120579 (120582)= infin (55)

All of the conditions (A) (R) and (T) are obviously satisfiedif the process is a symmetric stable Levy process of index 120572 isin

(1 2]

120579 (120582) = |120582|120572 (56)

In what follows we assume as well as the condition (A) thatthe conditions (R) and (T) are also satisfied

The Laplace transform of the law of 1198790

is given by

P119911[119890minus1199021198790] =

119906119902(119911)

119906119902(0)

(57)

see for example [19 pp 64] It is easy to see that the entrancelaw has the space density

120588 (119905 119909) =n (119883

119905isin 119889119909)

119889119909 (58)

In view of [7 Theorem 210] the law of the hitting time 1198790

is absolutely continuous relative to the Lebesgue measure 119889119905

and the time density coincides with the space density of theentrance law

120588119909(119905) =

P119909(119879

0isin 119889119905)

119889119905= 120588 (119905 119909) (59)

41 Absolute Continuity of the Law of the Inverse Local TimeLet 120591(119897) denote the inverse local time at the origin

120591 (119897) = inf 119905 gt 0 1198710

119905gt 119897 (60)

We prove the absolute continuity of the law of inverse localtime Note that 120591(119897) is a subordinator such that

P0[119890minus119902120591(119897)

] = 119890minus119897119906119902(0)

(61)

see for example [19 pp 131]

Lemma 9 For fixed 119897 gt 0 the law of 120591(119897) under P0has a

density 120574119897(119905)

P0(120591 (119897) isin 119889119905) = 120574

119897(119905) 119889119905 (62)

Furthermore 120574119897(119905) may be chosen to be jointly continuous in

(119897 119905) isin (0infin) times (0infin)

Proof Following [7 Sec 33] we define a positive Borelmeasure 120590 on [0infin) as

120590 (119860) =1

120587int

infin

0

1119860(120579 (120582)) 119889120582 119860 isin B ([0infin)) (63)

Abstract and Applied Analysis 7

Then we have 119906119902(0) = int

[0infin)(120590(119889120585)(119902 + 120585)) for 119902 gt 0

and hence there exists a Radon measure 120590lowast on [0infin) with

int[0infin)

(120590lowast(119889120585)(1 + 120585)) lt infin such that

1

119902119906119902(0)

= int[0infin)

1

119902 + 120585120590lowast(119889120585) 119902 gt 0 (64)

Hence the Laplace exponent 1119906119902(0) may be represented as

1

119906119902(0)

= int

infin

0

(1 minus 119890minus119902119906

) 120584 (119906) 119889119906 (65)

where 120584(119906) = int(0infin)

119890minus119906120585

120585120590lowast(119889120585) Since int

infin

0(1 and 119906

2)120584(119906)119889119906 lt

infin we may appeal to analytic continuation of both sides offormula (61) and obtain

P0[119890119894120582120591(119897)

] = exp119897 int

infin

0

(119890119894120582119906

minus 1) 120584 (119906) 119889119906 (66)

Following [20 Theorem 31] we may invert the Fouriertransform of the law of 120591(119897) and obtain the desired result

42 ℎ-Paths of Symmetric Levy Processes We follow [7]for the notations concerning ℎ-paths of symmetric Levyprocesses For the interpretation of the ℎ-paths as some kindof conditioning see [10]

We define

ℎ (119909) = lim119902rarr0+

119906119902(0) minus 119906

119902(119909)

=1

120587int

infin

0

1 minus cos 120582119909120579 (120582)

119889120582 119909 isin R

(67)

The second equality follows from (50) Then the function ℎ

satisfies the following

(i) ℎ(119909) is continuous(ii) ℎ(0) = 0 ℎ(119909) gt 0 for all 119909 isin R 0(iii) ℎ(119909) rarr infin as |119909| rarr infin (since the condition (R) is

satisfied)

See [7 Lemma 42] for the proof Moreover the function ℎ isharmonic with respect to the killed process

P0119909[ℎ (119883

119905)] = ℎ (119909) if 119909 isin R 0 119905 gt 0

n [ℎ (119883119905)] = 1 if 119905 gt 0

(68)

See [7 Theorems 11 and 12] for the proof We define the ℎ-path process (Pℎ

119909 119909 isin R) by the following local equivalence

relations

119889Pℎ119909

10038161003816100381610038161003816F119905

=

ℎ (119883119905)

ℎ (119909)119889P0

119909

100381610038161003816100381610038161003816100381610038161003816F119905

if 119909 isin R 0

ℎ (119883119905) 119889n1003816100381610038161003816F

119905

if 119909 = 0

(69)

Remark that from the strong Markov properties of (119883119905)

under P0119909and n the family Pℎ

119909|F119905

119905 ge 0 is consistent andhence the probability measure Pℎ

119909is well defined

The ℎ-path process is then symmetric more preciselythe transition kernel has a symmetric density 119901

ℎ

119905(119909 119910) with

respect to the measure ℎ(119910)2119889119910 Here the density 119901

ℎ

119905(119909 119910) is

given by

119901ℎ

119905(119909 119910) =

1

ℎ (119909) ℎ (119910)119901

119905(119910 minus 119909) minus

119901119905(119909) 119901

119905(119910)

119901119905(0)

if 119909 119910 isin R 0

119901ℎ

119905(119909 0) = 119901

ℎ

119905(0 119909) =

120588 (119905 119909)

ℎ (119909)

if 119909 isin R 0

119901ℎ

119905(0 0) = int

(0infin)

119890minus119905120585

120585120590lowast(119889120585)

(70)

By (65) we see that 119901ℎ119905(0 0) is characterized by

int

infin

0

(1 minus 119890minus119902119905

) 119901ℎ

119905(0 0) 119889119905 =

1

119906119902(0)

119902 gt 0 (71)

See [7 Section 5] for the details The ℎ-path process alsosatisfies the following conditions

(i) the process is conservative(ii) any point is regular for itself(iii) the process is transient (since the condition (T) is

satisfied)

We can easily prove regularity of any point by the localequivalence (69) See [7 Theorem 14] for the proof oftransience

The resolvent density of the ℎ-path process with respectto ℎ(119910)

2119889119910 is given by

119906ℎ

119902(119909 119910) =

1

ℎ (119909) ℎ (119910)119906

119902(119909 minus 119910) minus

119906119902(119909) 119906

119902(119910)

119906119902(0)

119902 gt 0 119909 119910 isin R 0

119906ℎ

119902(119909 0) = 119906

ℎ

119902(0 119909) =

1

ℎ (119909)sdot119906119902(119909)

119906119902(0)

119902 gt 0 119909 isin R 0

(72)

We remark here that since lim119902rarrinfin

119906119902(0) = 0 we see by (71)

that

119906ℎ

119902(0 0) = infin (73)

The Green function 119906ℎ

0(119909 119910) = lim

119902rarr0+119906ℎ

119902(119909 119910) exists and is

given by

119906ℎ

0(119909 119910) =

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910) 119909 119910 isin R 0

119906ℎ

0(119909 0) = 119906

ℎ

0(0 119909) =

1

ℎ (119909) 119909 isin R 0

(74)

8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

Proof Let 1205821 120582

119899isin R For 119896 isin N we have

P0[

[

(

119899

sum

119895=1

120582119895119871119909119895

infin)

119896

]

]

=

119899

sum

1198951119895119896=1

1205821198951

sdot sdot sdot 120582119895119896

P0[119871

1199091198951

infin sdot sdot sdot 119871119909119895119896

infin ]

(14)

= 119896

119899

sum

1198951119895119896=1

1205821198951

sdot sdot sdot 120582119895119896

P0[1198690(119909

1198951

119909119895119896

)]

(15)

It follows fromTheorem 1 that

(15) = 119896

119899

sum

1198951119895119896=1

1199060(0 119909

1198951

) 1205821198951

sdot 1199060(119909

1198951

1199091198952

) 1205821198952

sdot sdot sdot 1199060(119909

119895119896minus1

119909119895119896

) 120582119895119896

= 119896(ΣΛ)119896

10

(16)

where 1 =⊤(1 1) v

0= 119907

0for v =

⊤(1199070 1199071 119907

119899)

Σ = (

0 1199060(0 119909

1) sdot sdot sdot 119906

0(0 119909

119899)

0 1199060(119909

1 119909

1) sdot sdot sdot 119906

0(119909

1 119909

119899)

0 1199060(119909

119899 119909

1) sdot sdot sdot 119906

0(119909

119899 119909

119899)

)

Λ = (

0 0 sdot sdot sdot 0

0 1205821

sdot sdot sdot 0

0 0 sdot sdot sdot 120582119899

)

(17)

Hence for all 1205821 120582

119899isin R such that |120582

119894|rsquos are small enough

we have

P0[exp

119899

sum

119894=1

120582119894119871119909119894

infin]

=

infin

sum

119896=0

(ΣΛ)119896

10

= (119868 minus ΣΛ)minus1

10

(18)

By Cramerrsquos formula we obtain

(119868 minus ΣΛ)minus1

10

=

det ((119868 minus ΣΛ)(1)

)

det (119868 minus ΣΛ)=det (119868 minus (Σ minus Σ

0)Λ)

det (119868 minus ΣΛ)

(19)

Here for a matrix 119860 we denote by 119860(1) the matrix which

is obtained by replacing each entry in the first column of 119860by number 1 Since Σ is nonnegative definite we obtain thedesired result (13) by analytic continuation

Remark 3 Eisenbaum et al [6] have proved an analogue ofRay-Knight theorem for the total local time of a symmetricLevy process killed at an independent exponential time Wemay say that the key to the proof is that Σ minus Σ

0 is a constantmatrix which is positive definite The difficulty in the case ofthe ℎ-path process of a symmetric Levy process is that thematrix Σ minus Σ

0 no longer has such a nice property

22 Feynman-Kac Formula for Markovian Bridges In thissection we show Feynman-Kac formula for Markovianbridges For this we recall several theorems for Markovianbridges from Fitzsimmons et al [15] See [15] for details

For 119905 gt 0 119909 119910 isin R let P119905119909119910

denote the bridge law whichserves as a version of the regular conditional distribution for119883

119904 0 le 119904 le 119905 under P

119909given 119883

119905minus= 119910 In this section we

assume the following condition

(i) 0 lt 119901119905(119909 119910) lt infin for any 119905 gt 0 119909 119910 isin R

We also assume that there exists a local time (119871119909

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119910) 119871

119910

119904120583 (119889119910) 0 le 119904 le 119905

119891 isin B+(R)

(20)

holds with P119905119909119910

-probability one for any 119905 gt 0 and 119909 119910 isin R

Theorem 4 (see [15 Lemma 1]) Let 119905 gt 0 119909 119910 119911 isin R Thenone has

P119905119909119910

[int

119905

0

119891 (119904 119883119904) 119889119871

119911

119904] = int

119905

0

119889119904119901119904(119909 119911) 119901

119905minus119904(119911 119910)

119901119905(119909 119910)

119891 (119904 119911)

(21)

for any nonnegative Borel function 119891

We will also use the following conditioning formula

Theorem 5 (see [15 Proposition 3]) Let 119905 gt 0 119909 119910 119911 isin RThen one has

P119905119909119910

[int

119905

0

119891 (119904 119883119904)119867

119904119889119871

119911

119904]

= P119905119909119910

[int

119905

0

119891 (119904 119911)P119904119909119911

[119867119904] 119889119871

119911

119904]

(22)

for any nonnegative Borel function 119891 and any nonnegativepredictable process 119867

119904

For 119904 ge 0 and 1199111 119911

119899isin R we define

119867119904(z(119899)) = int

119904

0

119889119871119911119899

119904119899

int

119904119899

0

119889119871119911119899minus1

119904119899minus1

sdot sdot sdot int

1199042

0

1198891198711199111

1199041

(23)

where z(119899) = (1199111 119911

119899) The following theorem is a version

of Kacrsquos moment formulae

4 Abstract and Applied Analysis

Theorem 6 For any 119902 gt 0 119899 isin N and for any 1199111 119911

119899isin R

one has

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(z(119899))] 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119910)

(24)

Proof Let us prove the claim by induction For 119899 = 1 theassertion follows fromTheorem 4 Suppose that formula (24)holds for a given 119899 ge 2 Note that

119867119905(z(119899+1)) = int

119905

0

119867119904(z(119899)) 119889119871

119911119899+1

119904 (25)

Since119867119904(z(119899)) is a nonnegative predictable processTheorems

5 and 4 show that

P119905119909119910

[119867119905(z(119899+1))]

= int

119905

0

119889119904119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)

119901119905(119909 119910)

P119905119909119911119899+1

[119867119904(119911

(119899))]

(26)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(119911

(119899+1))] 119889119905

= int

infin

0

119890minus119902119905

119889119905

times int

119905

0

119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

= int

infin

0

119890minus119902119904

119901119904(119909 119911

119899+1)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

times int

infin

0

119890minus119902119905

119901119905(119911

119899+1 119910) 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119911

119899+1) sdot 119906

119902(119911

119899+1 119910)

(27)

by the assumption of the induction Nowwe have proved thatformula (24) is valid also for 119899+1 which completes the proof

The following theorem is a version of Feynman-Kacformulae

Theorem 7 Let 1199111= 0 119911

2 119911

119899isin R and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (28)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895) Then for

any diagonal matrix Λ = (120582119894120575119894119895)119899

119894119895=1with nonnegative entries

one has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

= (119868 + Σ(119902)

Λ)minus1

Σ(119902)

11

(29)

Proof We have

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

120582119895119896

sdot sdot sdot 1205821198951

times int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119867

119905(119911

119895119896

1199111198951

)] 119889119905

(30)

UsingTheorem 6 we see that the above quantity is equal to

119896

119899

sum

1198951119895119896=1

119906119902(119911

1 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

1199111)

(31)

which amounts to 119896(Σ(119902)

Λ)119896Σ(119902)

11 Hence for all 120582

1

120582119899gt 0 sufficiently small we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

]

]

119889119905

= 119906119902(0 0) +

infin

sum

119896=1

(Σ(119902)

Λ)119896

Σ(119902)

11

= (119868 minus Σ(119902)

Λ)minus1

Σ(119902)

11

(32)

Since Σ(119902) is nonnegative definite we obtain the desired result

(29) by analytic continuation

The following theorem is valid even if

119906119902(0 119911

119895) = 119906

119902(119911

119895 0) = infin 119902 gt 0 119895 = 1 119899 (33)

Theorem 8 Let 1199111 119911

119899isin R 0 and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (34)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895)

u(119902) = (

119906119902(0 119911

1)

119906119902(0 119911

119899)

) v(119902) = (

119906119902(119911

1 0)

119906119902(119911

119899 0)

) (35)

Abstract and Applied Analysis 5

and let Λ be the matrix with elements Λ119894119895

= 120582119894120575119894119895 Then one

has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[1 minus 119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

=⊤u(119902)Λ(119868 + Σ

(119902)Λ)

minus1

v(119902)(36)

Proof UsingTheorem 6 we see that

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

119906119902(0 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

0)

(37)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

minus 1]

]

119889119905

=

infin

sum

119896=1

⊤u(119902)Λ(Σ

(119902)Λ)

119896minus1

v(119902)

=⊤u(119902)Λ(119868 minus Σ

(119902)Λ)

minus1

v(119902)

(38)

The rest of the proof is now obvious

3 Preliminaries Squared Bessel Processes andGeneralized Excursion Measures

In this section we recall squared Bessel processes andgeneralized excursion measures

First we introduce several notations about squared Besselprocesses for which we follow [3 XI1] For 120575 ge 0 let (Q120575

119911

119911 ge 0) denote the law of the 120575-dimensional squared Besselprocess where the origin is a trap when 120575 = 0 Then theLaplace transform of a one-dimensional marginal is given by

Q120575

119911[exp minus120582119883

119905] =

1

(1 + 2120582119905)1205752

expminus120582119911

1 + 2120582119905 (39)

We may obtain the transition kernels Q120575

119911(119883

119905isin 119889119908) by the

Laplace inversion

(i) For 120575 gt 0 and 119911 gt 0 we have

Q120575

119911(119883

119905isin 119889119908)

=1

2119905(119908

119911)

(12)(1205752minus1)

exp minus119911 + 119908

2119905 119868

1205752minus1(radic119911119908

119905) 119889119908

(40)

where 119868120584stands for the modified Bessel function of

order 120584

(ii) For 120575 gt 0 and 119911 = 0 we have

Q120575

0(119883

119905isin 119889119908) =

1

(2119905)1205752

Γ (1205752)1199081205752minus1 exp minus

119908

2119905 119889119908

(41)

where Γ stands for the gamma function

(iii) For 120575 = 0 and 119911 ge 0 we have

Q0

119911(119883

119905isin 119889119908) = exp minus

119911

2119905 120575

0(119889119908)

+1

2119905(119908

119911)

minus12

exp minus119911 + 119908

2119905

times 1198681(radic119911119908

119905) 119889119908

(42)

The squared Bessel process satisfies the scaling property for120575 ge 0 119911 ge 0 and 119888 gt 0 it holds that

(119888119883119905119888

) under Q120575

119911119888

law= (119883

119905) under Q120575

119911 (43)

Second we recall the notion of the generalized excursionmeasure By formula (39) we have

Q4

0[

1

1198832

119904+119905

119883119904+119905

isin 119861] = Q4

0[

1

1198832

119904

sdot Q0

119883119904

(119883119905isin 119861)] (44)

for 119904 119905 gt 0 and 119861 isin B([0infin)) If we put 120583119905(119889119909) =

(11199092)Q4

0(119883

119905isin 119889119909) we have

120583119904+119905

(119861) = int120583119904(119889119909)Q0

119909(119883

119905isin 119861) (45)

This shows that the family of laws 120583119905

119905 gt 0 is an entrancelaw for Q0

119909 119909 gt 0 In fact there exists a unique 120590-finite

measure n(0) on D such that

n(0) (1198831199051

isin 1198611 119883

119905119899

isin 119861119899)

= int1198611

1205831199051

(1198891199091) int

1198612

Q0

119909(119883

1199052minus1199051

isin 1198891199092)

sdot sdot sdot int119861119899

Q0

119909(119883

119905119899minus119905119899minus1

isin 119889119909119899)

(46)

for 0 lt 1199051

lt sdot sdot sdot lt 119905119899and 119861

1 119861

119899isin B([0infin)) Note

that to construct such a measure n(0) we can not appeal toKolmogorovrsquos extension theorem because the entrance lawshave infinite total mass However we can actually constructn(0) via the agreement formula (see Pitman-Yor [17 Cor 3]with 120575 = 4) or via the time change of a Brownian excursion(see Fitzsimmons-Yano [18 Theorem 25] with change ofscales)Wemay calln(0) the generalized excursionmeasure forthe squared Bessel process of dimension 0 See the referencesabove for several characteristic formulae of n(0)

6 Abstract and Applied Analysis

4 Symmetric Leacutevy Processes

Let us confine ourselves to one-dimensional symmetric Levyprocesses We recall general facts and state several resultsfrom [7]

In what follows we assume that (P119909) is the law of a

one-dimensional conservative Levy process Throughout thepresent paper we assume the following conditions whichwillbe referred to as (A) are satisfied

(i) the process is symmetric(ii) the origin (and consequently any point) is regular for

itself(iii) the process is not a compound Poisson

Under the condition (A) we have the following The charac-teristic exponent is given by

120579 (120582) = minus logP0[1198901198941205821198831] = 119907120582

2+ 2int

infin

0

(1 minus cos 120582119909) 120584 (119889119909)

(47)

for some 119907 ge 0 and some positive Radonmeasure 120584 on (0infin)

such that

int(0infin)

min 1199092 1 120584 (119889119909) lt infin (48)

The reference measure is 120583(119889119909) = 119889119909 and we have

119901119905(119909 119910) = 119901

119905(119910 minus 119909) =

1

120587int

infin

0

(cos 120582 (119910 minus 119909)) 119890minus119905120579(120582)

119889120582

(49)

119906119902(119909 119910) = 119906

119902(119910 minus 119909) =

1

120587int

infin

0

cos 120582 (119910 minus 119909)

119902 + 120579 (120582)119889120582 (50)

There exists a local time (119871119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905119889119910 119891 isin B

+(R) (51)

with P119909-probability one for any 119909 isin R Then it holds that

P119909[int

infin

0

119890minus119902119904

119889119871119910

119904] = 119906

119902(119910 minus 119909) 119909 119910 isin R (52)

Let n denote the excursion measure associated to the localtime 119871

0

119905 We denote by (P0

119909 119909 isin R 0) the law of the

process killed upon hitting the origin that is

P0119909(119860 120577 gt 119905) = P

119909(119860 119879

0gt 119905) 119909 isin R 0

119905 gt 0 119860 isin F119905

(53)

Then the excursion measure n satisfies the Markov propertyin the following sense for any 119905 gt 0 and for any nonnegativeF

119905-measurable functional 119885

119905and for any nonnegative F

infin-

measurable functional 119865 it holds that

n [119885119905119865 (119883

119905+sdot)] = intn [119885

119905 119883

119905isin 119889119909]P0

119909[119865 (119883)] (54)

We need the following additional conditions

(R) the process is recurrent(T) the function 120579(120582) is nondecreasing in 120582 gt 120582

0for some

1205820gt 0

Under the condition (A) the condition (R) is equivalent to

int

infin

0

119889120582

120579 (120582)= infin (55)

All of the conditions (A) (R) and (T) are obviously satisfiedif the process is a symmetric stable Levy process of index 120572 isin

(1 2]

120579 (120582) = |120582|120572 (56)

In what follows we assume as well as the condition (A) thatthe conditions (R) and (T) are also satisfied

The Laplace transform of the law of 1198790

is given by

P119911[119890minus1199021198790] =

119906119902(119911)

119906119902(0)

(57)

see for example [19 pp 64] It is easy to see that the entrancelaw has the space density

120588 (119905 119909) =n (119883

119905isin 119889119909)

119889119909 (58)

In view of [7 Theorem 210] the law of the hitting time 1198790

is absolutely continuous relative to the Lebesgue measure 119889119905

and the time density coincides with the space density of theentrance law

120588119909(119905) =

P119909(119879

0isin 119889119905)

119889119905= 120588 (119905 119909) (59)

41 Absolute Continuity of the Law of the Inverse Local TimeLet 120591(119897) denote the inverse local time at the origin

120591 (119897) = inf 119905 gt 0 1198710

119905gt 119897 (60)

We prove the absolute continuity of the law of inverse localtime Note that 120591(119897) is a subordinator such that

P0[119890minus119902120591(119897)

] = 119890minus119897119906119902(0)

(61)

see for example [19 pp 131]

Lemma 9 For fixed 119897 gt 0 the law of 120591(119897) under P0has a

density 120574119897(119905)

P0(120591 (119897) isin 119889119905) = 120574

119897(119905) 119889119905 (62)

Furthermore 120574119897(119905) may be chosen to be jointly continuous in

(119897 119905) isin (0infin) times (0infin)

Proof Following [7 Sec 33] we define a positive Borelmeasure 120590 on [0infin) as

120590 (119860) =1

120587int

infin

0

1119860(120579 (120582)) 119889120582 119860 isin B ([0infin)) (63)

Abstract and Applied Analysis 7

Then we have 119906119902(0) = int

[0infin)(120590(119889120585)(119902 + 120585)) for 119902 gt 0

and hence there exists a Radon measure 120590lowast on [0infin) with

int[0infin)

(120590lowast(119889120585)(1 + 120585)) lt infin such that

1

119902119906119902(0)

= int[0infin)

1

119902 + 120585120590lowast(119889120585) 119902 gt 0 (64)

Hence the Laplace exponent 1119906119902(0) may be represented as

1

119906119902(0)

= int

infin

0

(1 minus 119890minus119902119906

) 120584 (119906) 119889119906 (65)

where 120584(119906) = int(0infin)

119890minus119906120585

120585120590lowast(119889120585) Since int

infin

0(1 and 119906

2)120584(119906)119889119906 lt

infin we may appeal to analytic continuation of both sides offormula (61) and obtain

P0[119890119894120582120591(119897)

] = exp119897 int

infin

0

(119890119894120582119906

minus 1) 120584 (119906) 119889119906 (66)

Following [20 Theorem 31] we may invert the Fouriertransform of the law of 120591(119897) and obtain the desired result

42 ℎ-Paths of Symmetric Levy Processes We follow [7]for the notations concerning ℎ-paths of symmetric Levyprocesses For the interpretation of the ℎ-paths as some kindof conditioning see [10]

We define

ℎ (119909) = lim119902rarr0+

119906119902(0) minus 119906

119902(119909)

=1

120587int

infin

0

1 minus cos 120582119909120579 (120582)

119889120582 119909 isin R

(67)

The second equality follows from (50) Then the function ℎ

satisfies the following

(i) ℎ(119909) is continuous(ii) ℎ(0) = 0 ℎ(119909) gt 0 for all 119909 isin R 0(iii) ℎ(119909) rarr infin as |119909| rarr infin (since the condition (R) is

satisfied)

See [7 Lemma 42] for the proof Moreover the function ℎ isharmonic with respect to the killed process

P0119909[ℎ (119883

119905)] = ℎ (119909) if 119909 isin R 0 119905 gt 0

n [ℎ (119883119905)] = 1 if 119905 gt 0

(68)

See [7 Theorems 11 and 12] for the proof We define the ℎ-path process (Pℎ

119909 119909 isin R) by the following local equivalence

relations

119889Pℎ119909

10038161003816100381610038161003816F119905

=

ℎ (119883119905)

ℎ (119909)119889P0

119909

100381610038161003816100381610038161003816100381610038161003816F119905

if 119909 isin R 0

ℎ (119883119905) 119889n1003816100381610038161003816F

119905

if 119909 = 0

(69)

Remark that from the strong Markov properties of (119883119905)

under P0119909and n the family Pℎ

119909|F119905

119905 ge 0 is consistent andhence the probability measure Pℎ

119909is well defined

The ℎ-path process is then symmetric more preciselythe transition kernel has a symmetric density 119901

ℎ

119905(119909 119910) with

respect to the measure ℎ(119910)2119889119910 Here the density 119901

ℎ

119905(119909 119910) is

given by

119901ℎ

119905(119909 119910) =

1

ℎ (119909) ℎ (119910)119901

119905(119910 minus 119909) minus

119901119905(119909) 119901

119905(119910)

119901119905(0)

if 119909 119910 isin R 0

119901ℎ

119905(119909 0) = 119901

ℎ

119905(0 119909) =

120588 (119905 119909)

ℎ (119909)

if 119909 isin R 0

119901ℎ

119905(0 0) = int

(0infin)

119890minus119905120585

120585120590lowast(119889120585)

(70)

By (65) we see that 119901ℎ119905(0 0) is characterized by

int

infin

0

(1 minus 119890minus119902119905

) 119901ℎ

119905(0 0) 119889119905 =

1

119906119902(0)

119902 gt 0 (71)

See [7 Section 5] for the details The ℎ-path process alsosatisfies the following conditions

(i) the process is conservative(ii) any point is regular for itself(iii) the process is transient (since the condition (T) is

satisfied)

We can easily prove regularity of any point by the localequivalence (69) See [7 Theorem 14] for the proof oftransience

The resolvent density of the ℎ-path process with respectto ℎ(119910)

2119889119910 is given by

119906ℎ

119902(119909 119910) =

1

ℎ (119909) ℎ (119910)119906

119902(119909 minus 119910) minus

119906119902(119909) 119906

119902(119910)

119906119902(0)

119902 gt 0 119909 119910 isin R 0

119906ℎ

119902(119909 0) = 119906

ℎ

119902(0 119909) =

1

ℎ (119909)sdot119906119902(119909)

119906119902(0)

119902 gt 0 119909 isin R 0

(72)

We remark here that since lim119902rarrinfin

119906119902(0) = 0 we see by (71)

that

119906ℎ

119902(0 0) = infin (73)

The Green function 119906ℎ

0(119909 119910) = lim

119902rarr0+119906ℎ

119902(119909 119910) exists and is

given by

119906ℎ

0(119909 119910) =

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910) 119909 119910 isin R 0

119906ℎ

0(119909 0) = 119906

ℎ

0(0 119909) =

1

ℎ (119909) 119909 isin R 0

(74)

8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Theorem 6 For any 119902 gt 0 119899 isin N and for any 1199111 119911

119899isin R

one has

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(z(119899))] 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119910)

(24)

Proof Let us prove the claim by induction For 119899 = 1 theassertion follows fromTheorem 4 Suppose that formula (24)holds for a given 119899 ge 2 Note that

119867119905(z(119899+1)) = int

119905

0

119867119904(z(119899)) 119889119871

119911119899+1

119904 (25)

Since119867119904(z(119899)) is a nonnegative predictable processTheorems

5 and 4 show that

P119905119909119910

[119867119905(z(119899+1))]

= int

119905

0

119889119904119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)

119901119905(119909 119910)

P119905119909119911119899+1

[119867119904(119911

(119899))]

(26)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(119909 119910)P119905

119909119910[119867

119905(119911

(119899+1))] 119889119905

= int

infin

0

119890minus119902119905

119889119905

times int

119905

0

119901119904(119909 119911

119899+1) 119901

119905minus119904(119911

119899+1 119910)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

= int

infin

0

119890minus119902119904

119901119904(119909 119911

119899+1)P119904

119909119911119899+1

[119867119904(z(119899))] 119889119904

times int

infin

0

119890minus119902119905

119901119905(119911

119899+1 119910) 119889119905

= 119906119902(119909 119911

1) sdot

119899minus1

prod

119895=1

119906119902(119911

119895 119911

119895+1)

sdot 119906119902(119911

119899 119911

119899+1) sdot 119906

119902(119911

119899+1 119910)

(27)

by the assumption of the induction Nowwe have proved thatformula (24) is valid also for 119899+1 which completes the proof

The following theorem is a version of Feynman-Kacformulae

Theorem 7 Let 1199111= 0 119911

2 119911

119899isin R and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (28)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895) Then for

any diagonal matrix Λ = (120582119894120575119894119895)119899

119894119895=1with nonnegative entries

one has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

= (119868 + Σ(119902)

Λ)minus1

Σ(119902)

11

(29)

Proof We have

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

120582119895119896

sdot sdot sdot 1205821198951

times int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[119867

119905(119911

119895119896

1199111198951

)] 119889119905

(30)

UsingTheorem 6 we see that the above quantity is equal to

119896

119899

sum

1198951119895119896=1

119906119902(119911

1 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

1199111)

(31)

which amounts to 119896(Σ(119902)

Λ)119896Σ(119902)

11 Hence for all 120582

1

120582119899gt 0 sufficiently small we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

]

]

119889119905

= 119906119902(0 0) +

infin

sum

119896=1

(Σ(119902)

Λ)119896

Σ(119902)

11

= (119868 minus Σ(119902)

Λ)minus1

Σ(119902)

11

(32)

Since Σ(119902) is nonnegative definite we obtain the desired result

(29) by analytic continuation

The following theorem is valid even if

119906119902(0 119911

119895) = 119906

119902(119911

119895 0) = infin 119902 gt 0 119895 = 1 119899 (33)

Theorem 8 Let 1199111 119911

119899isin R 0 and let 120582

1 120582

119899ge 0

Suppose that

119906119902(119911

119894 119911

119895) lt infin 119902 gt 0 119894 119895 = 1 119899 (34)

Let Σ(119902) be the matrix with elements Σ(119902)119894119895

= 119906119902(119911119894 119911

119895)

u(119902) = (

119906119902(0 119911

1)

119906119902(0 119911

119899)

) v(119902) = (

119906119902(119911

1 0)

119906119902(119911

119899 0)

) (35)

Abstract and Applied Analysis 5

and let Λ be the matrix with elements Λ119894119895

= 120582119894120575119894119895 Then one

has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[1 minus 119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

=⊤u(119902)Λ(119868 + Σ

(119902)Λ)

minus1

v(119902)(36)

Proof UsingTheorem 6 we see that

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

119906119902(0 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

0)

(37)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

minus 1]

]

119889119905

=

infin

sum

119896=1

⊤u(119902)Λ(Σ

(119902)Λ)

119896minus1

v(119902)

=⊤u(119902)Λ(119868 minus Σ

(119902)Λ)

minus1

v(119902)

(38)

The rest of the proof is now obvious

3 Preliminaries Squared Bessel Processes andGeneralized Excursion Measures

In this section we recall squared Bessel processes andgeneralized excursion measures

First we introduce several notations about squared Besselprocesses for which we follow [3 XI1] For 120575 ge 0 let (Q120575

119911

119911 ge 0) denote the law of the 120575-dimensional squared Besselprocess where the origin is a trap when 120575 = 0 Then theLaplace transform of a one-dimensional marginal is given by

Q120575

119911[exp minus120582119883

119905] =

1

(1 + 2120582119905)1205752

expminus120582119911

1 + 2120582119905 (39)

We may obtain the transition kernels Q120575

119911(119883

119905isin 119889119908) by the

Laplace inversion

(i) For 120575 gt 0 and 119911 gt 0 we have

Q120575

119911(119883

119905isin 119889119908)

=1

2119905(119908

119911)

(12)(1205752minus1)

exp minus119911 + 119908

2119905 119868

1205752minus1(radic119911119908

119905) 119889119908

(40)

where 119868120584stands for the modified Bessel function of

order 120584

(ii) For 120575 gt 0 and 119911 = 0 we have

Q120575

0(119883

119905isin 119889119908) =

1

(2119905)1205752

Γ (1205752)1199081205752minus1 exp minus

119908

2119905 119889119908

(41)

where Γ stands for the gamma function

(iii) For 120575 = 0 and 119911 ge 0 we have

Q0

119911(119883

119905isin 119889119908) = exp minus

119911

2119905 120575

0(119889119908)

+1

2119905(119908

119911)

minus12

exp minus119911 + 119908

2119905

times 1198681(radic119911119908

119905) 119889119908

(42)

The squared Bessel process satisfies the scaling property for120575 ge 0 119911 ge 0 and 119888 gt 0 it holds that

(119888119883119905119888

) under Q120575

119911119888

law= (119883

119905) under Q120575

119911 (43)

Second we recall the notion of the generalized excursionmeasure By formula (39) we have

Q4

0[

1

1198832

119904+119905

119883119904+119905

isin 119861] = Q4

0[

1

1198832

119904

sdot Q0

119883119904

(119883119905isin 119861)] (44)

for 119904 119905 gt 0 and 119861 isin B([0infin)) If we put 120583119905(119889119909) =

(11199092)Q4

0(119883

119905isin 119889119909) we have

120583119904+119905

(119861) = int120583119904(119889119909)Q0

119909(119883

119905isin 119861) (45)

This shows that the family of laws 120583119905

119905 gt 0 is an entrancelaw for Q0

119909 119909 gt 0 In fact there exists a unique 120590-finite

measure n(0) on D such that

n(0) (1198831199051

isin 1198611 119883

119905119899

isin 119861119899)

= int1198611

1205831199051

(1198891199091) int

1198612

Q0

119909(119883

1199052minus1199051

isin 1198891199092)

sdot sdot sdot int119861119899

Q0

119909(119883

119905119899minus119905119899minus1

isin 119889119909119899)

(46)

for 0 lt 1199051

lt sdot sdot sdot lt 119905119899and 119861

1 119861

119899isin B([0infin)) Note

that to construct such a measure n(0) we can not appeal toKolmogorovrsquos extension theorem because the entrance lawshave infinite total mass However we can actually constructn(0) via the agreement formula (see Pitman-Yor [17 Cor 3]with 120575 = 4) or via the time change of a Brownian excursion(see Fitzsimmons-Yano [18 Theorem 25] with change ofscales)Wemay calln(0) the generalized excursionmeasure forthe squared Bessel process of dimension 0 See the referencesabove for several characteristic formulae of n(0)

6 Abstract and Applied Analysis

4 Symmetric Leacutevy Processes

Let us confine ourselves to one-dimensional symmetric Levyprocesses We recall general facts and state several resultsfrom [7]

In what follows we assume that (P119909) is the law of a

one-dimensional conservative Levy process Throughout thepresent paper we assume the following conditions whichwillbe referred to as (A) are satisfied

(i) the process is symmetric(ii) the origin (and consequently any point) is regular for

itself(iii) the process is not a compound Poisson

Under the condition (A) we have the following The charac-teristic exponent is given by

120579 (120582) = minus logP0[1198901198941205821198831] = 119907120582

2+ 2int

infin

0

(1 minus cos 120582119909) 120584 (119889119909)

(47)

for some 119907 ge 0 and some positive Radonmeasure 120584 on (0infin)

such that

int(0infin)

min 1199092 1 120584 (119889119909) lt infin (48)

The reference measure is 120583(119889119909) = 119889119909 and we have

119901119905(119909 119910) = 119901

119905(119910 minus 119909) =

1

120587int

infin

0

(cos 120582 (119910 minus 119909)) 119890minus119905120579(120582)

119889120582

(49)

119906119902(119909 119910) = 119906

119902(119910 minus 119909) =

1

120587int

infin

0

cos 120582 (119910 minus 119909)

119902 + 120579 (120582)119889120582 (50)

There exists a local time (119871119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905119889119910 119891 isin B

+(R) (51)

with P119909-probability one for any 119909 isin R Then it holds that

P119909[int

infin

0

119890minus119902119904

119889119871119910

119904] = 119906

119902(119910 minus 119909) 119909 119910 isin R (52)

Let n denote the excursion measure associated to the localtime 119871

0

119905 We denote by (P0

119909 119909 isin R 0) the law of the

process killed upon hitting the origin that is

P0119909(119860 120577 gt 119905) = P

119909(119860 119879

0gt 119905) 119909 isin R 0

119905 gt 0 119860 isin F119905

(53)

Then the excursion measure n satisfies the Markov propertyin the following sense for any 119905 gt 0 and for any nonnegativeF

119905-measurable functional 119885

119905and for any nonnegative F

infin-

measurable functional 119865 it holds that

n [119885119905119865 (119883

119905+sdot)] = intn [119885

119905 119883

119905isin 119889119909]P0

119909[119865 (119883)] (54)

We need the following additional conditions

(R) the process is recurrent(T) the function 120579(120582) is nondecreasing in 120582 gt 120582

0for some

1205820gt 0

Under the condition (A) the condition (R) is equivalent to

int

infin

0

119889120582

120579 (120582)= infin (55)

All of the conditions (A) (R) and (T) are obviously satisfiedif the process is a symmetric stable Levy process of index 120572 isin

(1 2]

120579 (120582) = |120582|120572 (56)

In what follows we assume as well as the condition (A) thatthe conditions (R) and (T) are also satisfied

The Laplace transform of the law of 1198790

is given by

P119911[119890minus1199021198790] =

119906119902(119911)

119906119902(0)

(57)

see for example [19 pp 64] It is easy to see that the entrancelaw has the space density

120588 (119905 119909) =n (119883

119905isin 119889119909)

119889119909 (58)

In view of [7 Theorem 210] the law of the hitting time 1198790

is absolutely continuous relative to the Lebesgue measure 119889119905

and the time density coincides with the space density of theentrance law

120588119909(119905) =

P119909(119879

0isin 119889119905)

119889119905= 120588 (119905 119909) (59)

41 Absolute Continuity of the Law of the Inverse Local TimeLet 120591(119897) denote the inverse local time at the origin

120591 (119897) = inf 119905 gt 0 1198710

119905gt 119897 (60)

We prove the absolute continuity of the law of inverse localtime Note that 120591(119897) is a subordinator such that

P0[119890minus119902120591(119897)

] = 119890minus119897119906119902(0)

(61)

see for example [19 pp 131]

Lemma 9 For fixed 119897 gt 0 the law of 120591(119897) under P0has a

density 120574119897(119905)

P0(120591 (119897) isin 119889119905) = 120574

119897(119905) 119889119905 (62)

Furthermore 120574119897(119905) may be chosen to be jointly continuous in

(119897 119905) isin (0infin) times (0infin)

Proof Following [7 Sec 33] we define a positive Borelmeasure 120590 on [0infin) as

120590 (119860) =1

120587int

infin

0

1119860(120579 (120582)) 119889120582 119860 isin B ([0infin)) (63)

Abstract and Applied Analysis 7

Then we have 119906119902(0) = int

[0infin)(120590(119889120585)(119902 + 120585)) for 119902 gt 0

and hence there exists a Radon measure 120590lowast on [0infin) with

int[0infin)

(120590lowast(119889120585)(1 + 120585)) lt infin such that

1

119902119906119902(0)

= int[0infin)

1

119902 + 120585120590lowast(119889120585) 119902 gt 0 (64)

Hence the Laplace exponent 1119906119902(0) may be represented as

1

119906119902(0)

= int

infin

0

(1 minus 119890minus119902119906

) 120584 (119906) 119889119906 (65)

where 120584(119906) = int(0infin)

119890minus119906120585

120585120590lowast(119889120585) Since int

infin

0(1 and 119906

2)120584(119906)119889119906 lt

infin we may appeal to analytic continuation of both sides offormula (61) and obtain

P0[119890119894120582120591(119897)

] = exp119897 int

infin

0

(119890119894120582119906

minus 1) 120584 (119906) 119889119906 (66)

Following [20 Theorem 31] we may invert the Fouriertransform of the law of 120591(119897) and obtain the desired result

42 ℎ-Paths of Symmetric Levy Processes We follow [7]for the notations concerning ℎ-paths of symmetric Levyprocesses For the interpretation of the ℎ-paths as some kindof conditioning see [10]

We define

ℎ (119909) = lim119902rarr0+

119906119902(0) minus 119906

119902(119909)

=1

120587int

infin

0

1 minus cos 120582119909120579 (120582)

119889120582 119909 isin R

(67)

The second equality follows from (50) Then the function ℎ

satisfies the following

(i) ℎ(119909) is continuous(ii) ℎ(0) = 0 ℎ(119909) gt 0 for all 119909 isin R 0(iii) ℎ(119909) rarr infin as |119909| rarr infin (since the condition (R) is

satisfied)

See [7 Lemma 42] for the proof Moreover the function ℎ isharmonic with respect to the killed process

P0119909[ℎ (119883

119905)] = ℎ (119909) if 119909 isin R 0 119905 gt 0

n [ℎ (119883119905)] = 1 if 119905 gt 0

(68)

See [7 Theorems 11 and 12] for the proof We define the ℎ-path process (Pℎ

119909 119909 isin R) by the following local equivalence

relations

119889Pℎ119909

10038161003816100381610038161003816F119905

=

ℎ (119883119905)

ℎ (119909)119889P0

119909

100381610038161003816100381610038161003816100381610038161003816F119905

if 119909 isin R 0

ℎ (119883119905) 119889n1003816100381610038161003816F

119905

if 119909 = 0

(69)

Remark that from the strong Markov properties of (119883119905)

under P0119909and n the family Pℎ

119909|F119905

119905 ge 0 is consistent andhence the probability measure Pℎ

119909is well defined

The ℎ-path process is then symmetric more preciselythe transition kernel has a symmetric density 119901

ℎ

119905(119909 119910) with

respect to the measure ℎ(119910)2119889119910 Here the density 119901

ℎ

119905(119909 119910) is

given by

119901ℎ

119905(119909 119910) =

1

ℎ (119909) ℎ (119910)119901

119905(119910 minus 119909) minus

119901119905(119909) 119901

119905(119910)

119901119905(0)

if 119909 119910 isin R 0

119901ℎ

119905(119909 0) = 119901

ℎ

119905(0 119909) =

120588 (119905 119909)

ℎ (119909)

if 119909 isin R 0

119901ℎ

119905(0 0) = int

(0infin)

119890minus119905120585

120585120590lowast(119889120585)

(70)

By (65) we see that 119901ℎ119905(0 0) is characterized by

int

infin

0

(1 minus 119890minus119902119905

) 119901ℎ

119905(0 0) 119889119905 =

1

119906119902(0)

119902 gt 0 (71)

See [7 Section 5] for the details The ℎ-path process alsosatisfies the following conditions

(i) the process is conservative(ii) any point is regular for itself(iii) the process is transient (since the condition (T) is

satisfied)

We can easily prove regularity of any point by the localequivalence (69) See [7 Theorem 14] for the proof oftransience

The resolvent density of the ℎ-path process with respectto ℎ(119910)

2119889119910 is given by

119906ℎ

119902(119909 119910) =

1

ℎ (119909) ℎ (119910)119906

119902(119909 minus 119910) minus

119906119902(119909) 119906

119902(119910)

119906119902(0)

119902 gt 0 119909 119910 isin R 0

119906ℎ

119902(119909 0) = 119906

ℎ

119902(0 119909) =

1

ℎ (119909)sdot119906119902(119909)

119906119902(0)

119902 gt 0 119909 isin R 0

(72)

We remark here that since lim119902rarrinfin

119906119902(0) = 0 we see by (71)

that

119906ℎ

119902(0 0) = infin (73)

The Green function 119906ℎ

0(119909 119910) = lim

119902rarr0+119906ℎ

119902(119909 119910) exists and is

given by

119906ℎ

0(119909 119910) =

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910) 119909 119910 isin R 0

119906ℎ

0(119909 0) = 119906

ℎ

0(0 119909) =

1

ℎ (119909) 119909 isin R 0

(74)

8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

and let Λ be the matrix with elements Λ119894119895

= 120582119894120575119894119895 Then one

has

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[1 minus 119890

minussum119899

119895=1120582119895119871119911119895

119905 ] 119889119905

=⊤u(119902)Λ(119868 + Σ

(119902)Λ)

minus1

v(119902)(36)

Proof UsingTheorem 6 we see that

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

(

119899

sum

119895=1

120582119895119871119911119895

119905)

119896

]

]

119889119905

= 119896

119899

sum

1198951119895119896=1

119906119902(0 119911

1198951

) 1205821198951

sdot

119896minus1

prod

119894=1

119906119902(119911

119895119894

119911119895119894+1

) 120582119895119894+1

sdot 119906119902(119911

119895119896

0)

(37)

Hence we obtain

int

infin

0

119890minus119902119905

119901119905(0 0)P119905

00[

[

exp

119899

sum

119895=1

120582119895119871119911119895

119905

minus 1]

]

119889119905

=

infin

sum

119896=1

⊤u(119902)Λ(Σ

(119902)Λ)

119896minus1

v(119902)

=⊤u(119902)Λ(119868 minus Σ

(119902)Λ)

minus1

v(119902)

(38)

The rest of the proof is now obvious

3 Preliminaries Squared Bessel Processes andGeneralized Excursion Measures

In this section we recall squared Bessel processes andgeneralized excursion measures

First we introduce several notations about squared Besselprocesses for which we follow [3 XI1] For 120575 ge 0 let (Q120575

119911

119911 ge 0) denote the law of the 120575-dimensional squared Besselprocess where the origin is a trap when 120575 = 0 Then theLaplace transform of a one-dimensional marginal is given by

Q120575

119911[exp minus120582119883

119905] =

1

(1 + 2120582119905)1205752

expminus120582119911

1 + 2120582119905 (39)

We may obtain the transition kernels Q120575

119911(119883

119905isin 119889119908) by the

Laplace inversion

(i) For 120575 gt 0 and 119911 gt 0 we have

Q120575

119911(119883

119905isin 119889119908)

=1

2119905(119908

119911)

(12)(1205752minus1)

exp minus119911 + 119908

2119905 119868

1205752minus1(radic119911119908

119905) 119889119908

(40)

where 119868120584stands for the modified Bessel function of

order 120584

(ii) For 120575 gt 0 and 119911 = 0 we have

Q120575

0(119883

119905isin 119889119908) =

1

(2119905)1205752

Γ (1205752)1199081205752minus1 exp minus

119908

2119905 119889119908

(41)

where Γ stands for the gamma function

(iii) For 120575 = 0 and 119911 ge 0 we have

Q0

119911(119883

119905isin 119889119908) = exp minus

119911

2119905 120575

0(119889119908)

+1

2119905(119908

119911)

minus12

exp minus119911 + 119908

2119905

times 1198681(radic119911119908

119905) 119889119908

(42)

The squared Bessel process satisfies the scaling property for120575 ge 0 119911 ge 0 and 119888 gt 0 it holds that

(119888119883119905119888

) under Q120575

119911119888

law= (119883

119905) under Q120575

119911 (43)

Second we recall the notion of the generalized excursionmeasure By formula (39) we have

Q4

0[

1

1198832

119904+119905

119883119904+119905

isin 119861] = Q4

0[

1

1198832

119904

sdot Q0

119883119904

(119883119905isin 119861)] (44)

for 119904 119905 gt 0 and 119861 isin B([0infin)) If we put 120583119905(119889119909) =

(11199092)Q4

0(119883

119905isin 119889119909) we have

120583119904+119905

(119861) = int120583119904(119889119909)Q0

119909(119883

119905isin 119861) (45)

This shows that the family of laws 120583119905

119905 gt 0 is an entrancelaw for Q0

119909 119909 gt 0 In fact there exists a unique 120590-finite

measure n(0) on D such that

n(0) (1198831199051

isin 1198611 119883

119905119899

isin 119861119899)

= int1198611

1205831199051

(1198891199091) int

1198612

Q0

119909(119883

1199052minus1199051

isin 1198891199092)

sdot sdot sdot int119861119899

Q0

119909(119883

119905119899minus119905119899minus1

isin 119889119909119899)

(46)

for 0 lt 1199051

lt sdot sdot sdot lt 119905119899and 119861

1 119861

119899isin B([0infin)) Note

that to construct such a measure n(0) we can not appeal toKolmogorovrsquos extension theorem because the entrance lawshave infinite total mass However we can actually constructn(0) via the agreement formula (see Pitman-Yor [17 Cor 3]with 120575 = 4) or via the time change of a Brownian excursion(see Fitzsimmons-Yano [18 Theorem 25] with change ofscales)Wemay calln(0) the generalized excursionmeasure forthe squared Bessel process of dimension 0 See the referencesabove for several characteristic formulae of n(0)

6 Abstract and Applied Analysis

4 Symmetric Leacutevy Processes

Let us confine ourselves to one-dimensional symmetric Levyprocesses We recall general facts and state several resultsfrom [7]

In what follows we assume that (P119909) is the law of a

one-dimensional conservative Levy process Throughout thepresent paper we assume the following conditions whichwillbe referred to as (A) are satisfied

(i) the process is symmetric(ii) the origin (and consequently any point) is regular for

itself(iii) the process is not a compound Poisson

Under the condition (A) we have the following The charac-teristic exponent is given by

120579 (120582) = minus logP0[1198901198941205821198831] = 119907120582

2+ 2int

infin

0

(1 minus cos 120582119909) 120584 (119889119909)

(47)

for some 119907 ge 0 and some positive Radonmeasure 120584 on (0infin)

such that

int(0infin)

min 1199092 1 120584 (119889119909) lt infin (48)

The reference measure is 120583(119889119909) = 119889119909 and we have

119901119905(119909 119910) = 119901

119905(119910 minus 119909) =

1

120587int

infin

0

(cos 120582 (119910 minus 119909)) 119890minus119905120579(120582)

119889120582

(49)

119906119902(119909 119910) = 119906

119902(119910 minus 119909) =

1

120587int

infin

0

cos 120582 (119910 minus 119909)

119902 + 120579 (120582)119889120582 (50)

There exists a local time (119871119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905119889119910 119891 isin B

+(R) (51)

with P119909-probability one for any 119909 isin R Then it holds that

P119909[int

infin

0

119890minus119902119904

119889119871119910

119904] = 119906

119902(119910 minus 119909) 119909 119910 isin R (52)

Let n denote the excursion measure associated to the localtime 119871

0

119905 We denote by (P0

119909 119909 isin R 0) the law of the

process killed upon hitting the origin that is

P0119909(119860 120577 gt 119905) = P

119909(119860 119879

0gt 119905) 119909 isin R 0

119905 gt 0 119860 isin F119905

(53)

Then the excursion measure n satisfies the Markov propertyin the following sense for any 119905 gt 0 and for any nonnegativeF

119905-measurable functional 119885

119905and for any nonnegative F

infin-

measurable functional 119865 it holds that

n [119885119905119865 (119883

119905+sdot)] = intn [119885

119905 119883

119905isin 119889119909]P0

119909[119865 (119883)] (54)

We need the following additional conditions

(R) the process is recurrent(T) the function 120579(120582) is nondecreasing in 120582 gt 120582

0for some

1205820gt 0

Under the condition (A) the condition (R) is equivalent to

int

infin

0

119889120582

120579 (120582)= infin (55)

All of the conditions (A) (R) and (T) are obviously satisfiedif the process is a symmetric stable Levy process of index 120572 isin

(1 2]

120579 (120582) = |120582|120572 (56)

In what follows we assume as well as the condition (A) thatthe conditions (R) and (T) are also satisfied

The Laplace transform of the law of 1198790

is given by

P119911[119890minus1199021198790] =

119906119902(119911)

119906119902(0)

(57)

see for example [19 pp 64] It is easy to see that the entrancelaw has the space density

120588 (119905 119909) =n (119883

119905isin 119889119909)

119889119909 (58)

In view of [7 Theorem 210] the law of the hitting time 1198790

is absolutely continuous relative to the Lebesgue measure 119889119905

and the time density coincides with the space density of theentrance law

120588119909(119905) =

P119909(119879

0isin 119889119905)

119889119905= 120588 (119905 119909) (59)

41 Absolute Continuity of the Law of the Inverse Local TimeLet 120591(119897) denote the inverse local time at the origin

120591 (119897) = inf 119905 gt 0 1198710

119905gt 119897 (60)

We prove the absolute continuity of the law of inverse localtime Note that 120591(119897) is a subordinator such that

P0[119890minus119902120591(119897)

] = 119890minus119897119906119902(0)

(61)

see for example [19 pp 131]

Lemma 9 For fixed 119897 gt 0 the law of 120591(119897) under P0has a

density 120574119897(119905)

P0(120591 (119897) isin 119889119905) = 120574

119897(119905) 119889119905 (62)

Furthermore 120574119897(119905) may be chosen to be jointly continuous in

(119897 119905) isin (0infin) times (0infin)

Proof Following [7 Sec 33] we define a positive Borelmeasure 120590 on [0infin) as

120590 (119860) =1

120587int

infin

0

1119860(120579 (120582)) 119889120582 119860 isin B ([0infin)) (63)

Abstract and Applied Analysis 7

Then we have 119906119902(0) = int

[0infin)(120590(119889120585)(119902 + 120585)) for 119902 gt 0

and hence there exists a Radon measure 120590lowast on [0infin) with

int[0infin)

(120590lowast(119889120585)(1 + 120585)) lt infin such that

1

119902119906119902(0)

= int[0infin)

1

119902 + 120585120590lowast(119889120585) 119902 gt 0 (64)

Hence the Laplace exponent 1119906119902(0) may be represented as

1

119906119902(0)

= int

infin

0

(1 minus 119890minus119902119906

) 120584 (119906) 119889119906 (65)

where 120584(119906) = int(0infin)

119890minus119906120585

120585120590lowast(119889120585) Since int

infin

0(1 and 119906

2)120584(119906)119889119906 lt

infin we may appeal to analytic continuation of both sides offormula (61) and obtain

P0[119890119894120582120591(119897)

] = exp119897 int

infin

0

(119890119894120582119906

minus 1) 120584 (119906) 119889119906 (66)

Following [20 Theorem 31] we may invert the Fouriertransform of the law of 120591(119897) and obtain the desired result

42 ℎ-Paths of Symmetric Levy Processes We follow [7]for the notations concerning ℎ-paths of symmetric Levyprocesses For the interpretation of the ℎ-paths as some kindof conditioning see [10]

We define

ℎ (119909) = lim119902rarr0+

119906119902(0) minus 119906

119902(119909)

=1

120587int

infin

0

1 minus cos 120582119909120579 (120582)

119889120582 119909 isin R

(67)

The second equality follows from (50) Then the function ℎ

satisfies the following

(i) ℎ(119909) is continuous(ii) ℎ(0) = 0 ℎ(119909) gt 0 for all 119909 isin R 0(iii) ℎ(119909) rarr infin as |119909| rarr infin (since the condition (R) is

satisfied)

See [7 Lemma 42] for the proof Moreover the function ℎ isharmonic with respect to the killed process

P0119909[ℎ (119883

119905)] = ℎ (119909) if 119909 isin R 0 119905 gt 0

n [ℎ (119883119905)] = 1 if 119905 gt 0

(68)

See [7 Theorems 11 and 12] for the proof We define the ℎ-path process (Pℎ

119909 119909 isin R) by the following local equivalence

relations

119889Pℎ119909

10038161003816100381610038161003816F119905

=

ℎ (119883119905)

ℎ (119909)119889P0

119909

100381610038161003816100381610038161003816100381610038161003816F119905

if 119909 isin R 0

ℎ (119883119905) 119889n1003816100381610038161003816F

119905

if 119909 = 0

(69)

Remark that from the strong Markov properties of (119883119905)

under P0119909and n the family Pℎ

119909|F119905

119905 ge 0 is consistent andhence the probability measure Pℎ

119909is well defined

The ℎ-path process is then symmetric more preciselythe transition kernel has a symmetric density 119901

ℎ

119905(119909 119910) with

respect to the measure ℎ(119910)2119889119910 Here the density 119901

ℎ

119905(119909 119910) is

given by

119901ℎ

119905(119909 119910) =

1

ℎ (119909) ℎ (119910)119901

119905(119910 minus 119909) minus

119901119905(119909) 119901

119905(119910)

119901119905(0)

if 119909 119910 isin R 0

119901ℎ

119905(119909 0) = 119901

ℎ

119905(0 119909) =

120588 (119905 119909)

ℎ (119909)

if 119909 isin R 0

119901ℎ

119905(0 0) = int

(0infin)

119890minus119905120585

120585120590lowast(119889120585)

(70)

By (65) we see that 119901ℎ119905(0 0) is characterized by

int

infin

0

(1 minus 119890minus119902119905

) 119901ℎ

119905(0 0) 119889119905 =

1

119906119902(0)

119902 gt 0 (71)

See [7 Section 5] for the details The ℎ-path process alsosatisfies the following conditions

(i) the process is conservative(ii) any point is regular for itself(iii) the process is transient (since the condition (T) is

satisfied)

We can easily prove regularity of any point by the localequivalence (69) See [7 Theorem 14] for the proof oftransience

The resolvent density of the ℎ-path process with respectto ℎ(119910)

2119889119910 is given by

119906ℎ

119902(119909 119910) =

1

ℎ (119909) ℎ (119910)119906

119902(119909 minus 119910) minus

119906119902(119909) 119906

119902(119910)

119906119902(0)

119902 gt 0 119909 119910 isin R 0

119906ℎ

119902(119909 0) = 119906

ℎ

119902(0 119909) =

1

ℎ (119909)sdot119906119902(119909)

119906119902(0)

119902 gt 0 119909 isin R 0

(72)

We remark here that since lim119902rarrinfin

119906119902(0) = 0 we see by (71)

that

119906ℎ

119902(0 0) = infin (73)

The Green function 119906ℎ

0(119909 119910) = lim

119902rarr0+119906ℎ

119902(119909 119910) exists and is

given by

119906ℎ

0(119909 119910) =

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910) 119909 119910 isin R 0

119906ℎ

0(119909 0) = 119906

ℎ

0(0 119909) =

1

ℎ (119909) 119909 isin R 0

(74)

8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

4 Symmetric Leacutevy Processes

Let us confine ourselves to one-dimensional symmetric Levyprocesses We recall general facts and state several resultsfrom [7]

In what follows we assume that (P119909) is the law of a

one-dimensional conservative Levy process Throughout thepresent paper we assume the following conditions whichwillbe referred to as (A) are satisfied

(i) the process is symmetric(ii) the origin (and consequently any point) is regular for

itself(iii) the process is not a compound Poisson

Under the condition (A) we have the following The charac-teristic exponent is given by

120579 (120582) = minus logP0[1198901198941205821198831] = 119907120582

2+ 2int

infin

0

(1 minus cos 120582119909) 120584 (119889119909)

(47)

for some 119907 ge 0 and some positive Radonmeasure 120584 on (0infin)

such that

int(0infin)

min 1199092 1 120584 (119889119909) lt infin (48)

The reference measure is 120583(119889119909) = 119889119909 and we have

119901119905(119909 119910) = 119901

119905(119910 minus 119909) =

1

120587int

infin

0

(cos 120582 (119910 minus 119909)) 119890minus119905120579(120582)

119889120582

(49)

119906119902(119909 119910) = 119906

119902(119910 minus 119909) =

1

120587int

infin

0

cos 120582 (119910 minus 119909)

119902 + 120579 (120582)119889120582 (50)

There exists a local time (119871119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905119889119910 119891 isin B

+(R) (51)

with P119909-probability one for any 119909 isin R Then it holds that

P119909[int

infin

0

119890minus119902119904

119889119871119910

119904] = 119906

119902(119910 minus 119909) 119909 119910 isin R (52)

Let n denote the excursion measure associated to the localtime 119871

0

119905 We denote by (P0

119909 119909 isin R 0) the law of the

process killed upon hitting the origin that is

P0119909(119860 120577 gt 119905) = P

119909(119860 119879

0gt 119905) 119909 isin R 0

119905 gt 0 119860 isin F119905

(53)

Then the excursion measure n satisfies the Markov propertyin the following sense for any 119905 gt 0 and for any nonnegativeF

119905-measurable functional 119885

119905and for any nonnegative F

infin-

measurable functional 119865 it holds that

n [119885119905119865 (119883

119905+sdot)] = intn [119885

119905 119883

119905isin 119889119909]P0

119909[119865 (119883)] (54)

We need the following additional conditions

(R) the process is recurrent(T) the function 120579(120582) is nondecreasing in 120582 gt 120582

0for some

1205820gt 0

Under the condition (A) the condition (R) is equivalent to

int

infin

0

119889120582

120579 (120582)= infin (55)

All of the conditions (A) (R) and (T) are obviously satisfiedif the process is a symmetric stable Levy process of index 120572 isin

(1 2]

120579 (120582) = |120582|120572 (56)

In what follows we assume as well as the condition (A) thatthe conditions (R) and (T) are also satisfied

The Laplace transform of the law of 1198790

is given by

P119911[119890minus1199021198790] =

119906119902(119911)

119906119902(0)

(57)

see for example [19 pp 64] It is easy to see that the entrancelaw has the space density

120588 (119905 119909) =n (119883

119905isin 119889119909)

119889119909 (58)

In view of [7 Theorem 210] the law of the hitting time 1198790

is absolutely continuous relative to the Lebesgue measure 119889119905

and the time density coincides with the space density of theentrance law

120588119909(119905) =

P119909(119879

0isin 119889119905)

119889119905= 120588 (119905 119909) (59)

41 Absolute Continuity of the Law of the Inverse Local TimeLet 120591(119897) denote the inverse local time at the origin

120591 (119897) = inf 119905 gt 0 1198710

119905gt 119897 (60)

We prove the absolute continuity of the law of inverse localtime Note that 120591(119897) is a subordinator such that

P0[119890minus119902120591(119897)

] = 119890minus119897119906119902(0)

(61)

see for example [19 pp 131]

Lemma 9 For fixed 119897 gt 0 the law of 120591(119897) under P0has a

density 120574119897(119905)

P0(120591 (119897) isin 119889119905) = 120574

119897(119905) 119889119905 (62)

Furthermore 120574119897(119905) may be chosen to be jointly continuous in

(119897 119905) isin (0infin) times (0infin)

Proof Following [7 Sec 33] we define a positive Borelmeasure 120590 on [0infin) as

120590 (119860) =1

120587int

infin

0

1119860(120579 (120582)) 119889120582 119860 isin B ([0infin)) (63)

Abstract and Applied Analysis 7

Then we have 119906119902(0) = int

[0infin)(120590(119889120585)(119902 + 120585)) for 119902 gt 0

and hence there exists a Radon measure 120590lowast on [0infin) with

int[0infin)

(120590lowast(119889120585)(1 + 120585)) lt infin such that

1

119902119906119902(0)

= int[0infin)

1

119902 + 120585120590lowast(119889120585) 119902 gt 0 (64)

Hence the Laplace exponent 1119906119902(0) may be represented as

1

119906119902(0)

= int

infin

0

(1 minus 119890minus119902119906

) 120584 (119906) 119889119906 (65)

where 120584(119906) = int(0infin)

119890minus119906120585

120585120590lowast(119889120585) Since int

infin

0(1 and 119906

2)120584(119906)119889119906 lt

infin we may appeal to analytic continuation of both sides offormula (61) and obtain

P0[119890119894120582120591(119897)

] = exp119897 int

infin

0

(119890119894120582119906

minus 1) 120584 (119906) 119889119906 (66)

Following [20 Theorem 31] we may invert the Fouriertransform of the law of 120591(119897) and obtain the desired result

42 ℎ-Paths of Symmetric Levy Processes We follow [7]for the notations concerning ℎ-paths of symmetric Levyprocesses For the interpretation of the ℎ-paths as some kindof conditioning see [10]

We define

ℎ (119909) = lim119902rarr0+

119906119902(0) minus 119906

119902(119909)

=1

120587int

infin

0

1 minus cos 120582119909120579 (120582)

119889120582 119909 isin R

(67)

The second equality follows from (50) Then the function ℎ

satisfies the following

(i) ℎ(119909) is continuous(ii) ℎ(0) = 0 ℎ(119909) gt 0 for all 119909 isin R 0(iii) ℎ(119909) rarr infin as |119909| rarr infin (since the condition (R) is

satisfied)

See [7 Lemma 42] for the proof Moreover the function ℎ isharmonic with respect to the killed process

P0119909[ℎ (119883

119905)] = ℎ (119909) if 119909 isin R 0 119905 gt 0

n [ℎ (119883119905)] = 1 if 119905 gt 0

(68)

See [7 Theorems 11 and 12] for the proof We define the ℎ-path process (Pℎ

119909 119909 isin R) by the following local equivalence

relations

119889Pℎ119909

10038161003816100381610038161003816F119905

=

ℎ (119883119905)

ℎ (119909)119889P0

119909

100381610038161003816100381610038161003816100381610038161003816F119905

if 119909 isin R 0

ℎ (119883119905) 119889n1003816100381610038161003816F

119905

if 119909 = 0

(69)

Remark that from the strong Markov properties of (119883119905)

under P0119909and n the family Pℎ

119909|F119905

119905 ge 0 is consistent andhence the probability measure Pℎ

119909is well defined

The ℎ-path process is then symmetric more preciselythe transition kernel has a symmetric density 119901

ℎ

119905(119909 119910) with

respect to the measure ℎ(119910)2119889119910 Here the density 119901

ℎ

119905(119909 119910) is

given by

119901ℎ

119905(119909 119910) =

1

ℎ (119909) ℎ (119910)119901

119905(119910 minus 119909) minus

119901119905(119909) 119901

119905(119910)

119901119905(0)

if 119909 119910 isin R 0

119901ℎ

119905(119909 0) = 119901

ℎ

119905(0 119909) =

120588 (119905 119909)

ℎ (119909)

if 119909 isin R 0

119901ℎ

119905(0 0) = int

(0infin)

119890minus119905120585

120585120590lowast(119889120585)

(70)

By (65) we see that 119901ℎ119905(0 0) is characterized by

int

infin

0

(1 minus 119890minus119902119905

) 119901ℎ

119905(0 0) 119889119905 =

1

119906119902(0)

119902 gt 0 (71)

See [7 Section 5] for the details The ℎ-path process alsosatisfies the following conditions

(i) the process is conservative(ii) any point is regular for itself(iii) the process is transient (since the condition (T) is

satisfied)

We can easily prove regularity of any point by the localequivalence (69) See [7 Theorem 14] for the proof oftransience

The resolvent density of the ℎ-path process with respectto ℎ(119910)

2119889119910 is given by

119906ℎ

119902(119909 119910) =

1

ℎ (119909) ℎ (119910)119906

119902(119909 minus 119910) minus

119906119902(119909) 119906

119902(119910)

119906119902(0)

119902 gt 0 119909 119910 isin R 0

119906ℎ

119902(119909 0) = 119906

ℎ

119902(0 119909) =

1

ℎ (119909)sdot119906119902(119909)

119906119902(0)

119902 gt 0 119909 isin R 0

(72)

We remark here that since lim119902rarrinfin

119906119902(0) = 0 we see by (71)

that

119906ℎ

119902(0 0) = infin (73)

The Green function 119906ℎ

0(119909 119910) = lim

119902rarr0+119906ℎ

119902(119909 119910) exists and is

given by

119906ℎ

0(119909 119910) =

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910) 119909 119910 isin R 0

119906ℎ

0(119909 0) = 119906

ℎ

0(0 119909) =

1

ℎ (119909) 119909 isin R 0

(74)

8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

Then we have 119906119902(0) = int

[0infin)(120590(119889120585)(119902 + 120585)) for 119902 gt 0

and hence there exists a Radon measure 120590lowast on [0infin) with

int[0infin)

(120590lowast(119889120585)(1 + 120585)) lt infin such that

1

119902119906119902(0)

= int[0infin)

1

119902 + 120585120590lowast(119889120585) 119902 gt 0 (64)

Hence the Laplace exponent 1119906119902(0) may be represented as

1

119906119902(0)

= int

infin

0

(1 minus 119890minus119902119906

) 120584 (119906) 119889119906 (65)

where 120584(119906) = int(0infin)

119890minus119906120585

120585120590lowast(119889120585) Since int

infin

0(1 and 119906

2)120584(119906)119889119906 lt

infin we may appeal to analytic continuation of both sides offormula (61) and obtain

P0[119890119894120582120591(119897)

] = exp119897 int

infin

0

(119890119894120582119906

minus 1) 120584 (119906) 119889119906 (66)

Following [20 Theorem 31] we may invert the Fouriertransform of the law of 120591(119897) and obtain the desired result

42 ℎ-Paths of Symmetric Levy Processes We follow [7]for the notations concerning ℎ-paths of symmetric Levyprocesses For the interpretation of the ℎ-paths as some kindof conditioning see [10]

We define

ℎ (119909) = lim119902rarr0+

119906119902(0) minus 119906

119902(119909)

=1

120587int

infin

0

1 minus cos 120582119909120579 (120582)

119889120582 119909 isin R

(67)

The second equality follows from (50) Then the function ℎ

satisfies the following

(i) ℎ(119909) is continuous(ii) ℎ(0) = 0 ℎ(119909) gt 0 for all 119909 isin R 0(iii) ℎ(119909) rarr infin as |119909| rarr infin (since the condition (R) is

satisfied)

See [7 Lemma 42] for the proof Moreover the function ℎ isharmonic with respect to the killed process

P0119909[ℎ (119883

119905)] = ℎ (119909) if 119909 isin R 0 119905 gt 0

n [ℎ (119883119905)] = 1 if 119905 gt 0

(68)

See [7 Theorems 11 and 12] for the proof We define the ℎ-path process (Pℎ

119909 119909 isin R) by the following local equivalence

relations

119889Pℎ119909

10038161003816100381610038161003816F119905

=

ℎ (119883119905)

ℎ (119909)119889P0

119909

100381610038161003816100381610038161003816100381610038161003816F119905

if 119909 isin R 0

ℎ (119883119905) 119889n1003816100381610038161003816F

119905

if 119909 = 0

(69)

Remark that from the strong Markov properties of (119883119905)

under P0119909and n the family Pℎ

119909|F119905

119905 ge 0 is consistent andhence the probability measure Pℎ

119909is well defined

The ℎ-path process is then symmetric more preciselythe transition kernel has a symmetric density 119901

ℎ

119905(119909 119910) with

respect to the measure ℎ(119910)2119889119910 Here the density 119901

ℎ

119905(119909 119910) is

given by

119901ℎ

119905(119909 119910) =

1

ℎ (119909) ℎ (119910)119901

119905(119910 minus 119909) minus

119901119905(119909) 119901

119905(119910)

119901119905(0)

if 119909 119910 isin R 0

119901ℎ

119905(119909 0) = 119901

ℎ

119905(0 119909) =

120588 (119905 119909)

ℎ (119909)

if 119909 isin R 0

119901ℎ

119905(0 0) = int

(0infin)

119890minus119905120585

120585120590lowast(119889120585)

(70)

By (65) we see that 119901ℎ119905(0 0) is characterized by

int

infin

0

(1 minus 119890minus119902119905

) 119901ℎ

119905(0 0) 119889119905 =

1

119906119902(0)

119902 gt 0 (71)

See [7 Section 5] for the details The ℎ-path process alsosatisfies the following conditions

(i) the process is conservative(ii) any point is regular for itself(iii) the process is transient (since the condition (T) is

satisfied)

We can easily prove regularity of any point by the localequivalence (69) See [7 Theorem 14] for the proof oftransience

The resolvent density of the ℎ-path process with respectto ℎ(119910)

2119889119910 is given by

119906ℎ

119902(119909 119910) =

1

ℎ (119909) ℎ (119910)119906

119902(119909 minus 119910) minus

119906119902(119909) 119906

119902(119910)

119906119902(0)

119902 gt 0 119909 119910 isin R 0

119906ℎ

119902(119909 0) = 119906

ℎ

119902(0 119909) =

1

ℎ (119909)sdot119906119902(119909)

119906119902(0)

119902 gt 0 119909 isin R 0

(72)

We remark here that since lim119902rarrinfin

119906119902(0) = 0 we see by (71)

that

119906ℎ

119902(0 0) = infin (73)

The Green function 119906ℎ

0(119909 119910) = lim

119902rarr0+119906ℎ

119902(119909 119910) exists and is

given by

119906ℎ

0(119909 119910) =

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910) 119909 119910 isin R 0

119906ℎ

0(119909 0) = 119906

ℎ

0(0 119909) =

1

ℎ (119909) 119909 isin R 0

(74)

8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

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8 Abstract and Applied Analysis

See [7 Section 53] for the proof Since 119906ℎ

0(119909 119910) ge 0 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) ge 0 119909 119910 isin R (75)

It follows from the local equivalence (69) that there existsa local time (119871

119909

119905) such that

int

119905

0

119891 (119883119904) 119889119904 = int119891 (119910) 119871

119910

119905ℎ(119910)

2

119889119910 119905 gt 0 119891 isin B+(R)

(76)

with Pℎ119909-probability one for any 119909 isin R We have

Pℎ119909[119871

119910

infin] = 119906

ℎ

0(119909 119910) 119909 isin R 119910 isin R 0 (77)

Example 10 If the process is the symmetric stable process ofindex 120572 isin (1 2] then the harmonic function ℎ(119909) may becomputed as

ℎ (119909) = 119862 (120572) |119909|120572minus1

(78)

where 119862(120572) is given as follows (see [9 Appendix])

119862 (120572) =1

120587int

infin

0

1 minus cos 120582120582120572

119889120582 =1

2Γ (120572) sin (120587 (120572 minus 1) 2)

(79)

5 The Laws of the Total LocalTimes for ℎ-Paths

In this section we state and prove our main theoremsconcerning the laws of the total local times of ℎ-paths

51 Laplace Transform Formula for ℎ-Paths In this sectionwe prove Laplace transform formula for ℎ-paths at two levels

Lemma 11 For 119909 119910 isin R 0 and 1205821 120582

2ge 0 one has

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(80)

where

119863 = 119863 (119909 119910) = ℎ (119909 minus 119910) sdotℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909) ℎ (119910)ge 0

119864 = 119864 (119909 119910) = 1 minus(ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910))

2

4ℎ (119909) ℎ (119910)ge 0

(81)

Proof Let us apply Theorem 2 with

119868 + (Σ minus Σ0)Λ

= (

1 + 1205821

ℎ (119910) minus ℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909) minus ℎ (119909 minus 119910)

ℎ (119910)1205821

1 + 1205822

)

119868 + ΣΛ

=(

1+21205821

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119909)1205822

ℎ (119909)+ℎ (119910)minusℎ (119909 minus 119910)

ℎ (119910)1205821

1+21205822

)

(82)

Then we obtain (80) by an easy computationBy (75) we have 119863 ge 0 Since

119864 =ℎ (119909) ℎ (119910)

4det(

119906ℎ

0(119909 119909) 119906

ℎ

0(119909 119910)

119906ℎ

0(119910 119909) 119906

ℎ

0(119910 119910)

) (83)

we obtain 119864 ge 0 by nonnegative definiteness of the abovematrix The proof is now complete

52 The Law of 119871119909infin Using formula (80) we can determine

the law of 119871119909infin see [16 Example 3105] for the formula in a

more general case

Theorem 12 For any 119909 isin R 0 one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897) =

1

2120575

0(119889119897) + 119890

minus1198972 119889119897

2 (84)

where 1205750stands for the Dirac measure concentrated at 0

Consequently one has

Pℎ0(119871

119909

infin= 0) =

1

2 (85)

Proof Letting 1205822= 0 in Lemma 11 we have

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infin] =

1 + 1205821

1 + 21205821

=1

2(1 +

1

1 + 21205821

)

(86)

which proves the claim

Remark 13 Since 119871119909

infin= 0 if and only if 119879

119909= infin the identity

(85) is equivalent to

Pℎ0(119879

119909= infin) =

1

2 (87)

This formula may also be obtained from the followingformula (see [9 Proposition 510])

n (119879119909

lt 120577) =1

2ℎ (119909) (88)

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

Suppose that in the definition (69) we may replace the fixedtime 119905 with the stopping time 119879

119909 Then we have

Pℎ0(119879

119909lt infin) = n [ℎ (119883

119879119909

) 119879119909

lt 120577]

= ℎ (119909)n (119879119909

lt 120577) =1

2

(89)

53 The Probability That Two Levels Are Attained Let usdiscuss the probability that the total local times at two givenlevels are positive

Theorem 14 Let 119909 119910 isin R 0 such that 119909 = 119910 Then one has119864 gt 0 and

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (90)

Pℎ0(119871

119909

infingt 0 119871

119910

infin= 0) = Pℎ

0(119871

119909

infin= 0 119871

119910

infingt 0) =

1

2minus

119863

4119864

(91)

Consequently one has 119863 le 2119864

Proof Letting 1205821= 120582

2= 120582 ge 0 in formula (80) we have

Pℎ0[exp minus120582ℎ (119909) 119871

119909

infinminus 120582ℎ (119910) 119871

119910

infin] =

1 + 2120582 + 1198631205822

1 + 4120582 + 41198641205822

(92)

If 119864 were zero then 119863 would be positive and hence theright-hand side of (92) would diverge as 120582 rarr infin whichcontradicts the fact that the left-hand side of (92) is boundedin 120582 gt 0 Hence we obtain 119864 gt 0

Taking the limit as 120582 rarr infin in both sides of formula (92)we have

Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

119863

4119864 (93)

which is nothing else but the second equality of (90) Byformula (85) we obtain

Pℎ0(119871

119909

infin= 0 119871

119910

infingt 0) = Pℎ

0(119871

119909

infin= 0)

minus Pℎ0(119871

119909

infin= 119871

119910

infin= 0) =

1

2minus

119863

4119864

(94)

Thus we obtain (91) Therefore we obtain

Pℎ0(119871

119909

infingt 0 119871

119910

infingt 0) = 1 minus

119863

4119864minus 2

1

2minus

119863

4119864 =

119863

4119864 (95)

which is nothing else but the first equality of (90) The proofis now complete

54 Joint Law of 119871119909infin

and 119871119910

infin Let us discuss the joint law of

119871119909

infinand 119871

119910

infinfor 119909 119910 isin R 0 such that 119909 = 119910

By Lemma 11 we know that 119863 ge 0 First we discuss thecase of 119863 = 0

Theorem 15 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) = 0 Then

Pℎ0(ℎ (119909) 119871

119909

infinisin 119889119897

1 ℎ (119910) 119871

119910

infinisin 119889119897

2)

=1

2119890

minus11989712 1198891198971

2sdot 120575

0(119889119897

2) + 120575

0(119889119897

1) sdot 119890

minus11989722 1198891198972

2

(96)

Proof Since 119863 = 0 and 119864 = 1 formula (80) implies

Pℎ0[exp minus120582

1ℎ (119909) 119871

119909

infinminus 120582

2ℎ (119910) 119871

119910

infin]

=1 + 120582

1+ 120582

2

1 + 21205821+ 2120582

2+ 4120582

11205822

(97)

We may rewrite the right-hand side as

1

2(

1

1 + 21205821

+1

1 + 21205822

) (98)

which proves the claim

Second we discuss the case of 119863 gt 0

Theorem 16 Suppose that ℎ(119909) + ℎ(119910) minus ℎ(119909 minus 119910) gt 0 Set

119898 = 119898(119909 119910) = ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

119872 = 119872(119909 119910) =4ℎ (119909) ℎ (119910)

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910)

(99)

Then one has 119864 gt 0 and 0 lt 119898 lt 119872 For any 119860 119861 isin

B([0infin)) one has

Pℎ0(ℎ (119909) 119871

119909

infinisin 119860 ℎ (119910) 119871

119910

infinisin 119861) =

4

sum

119896=1

119862119896Φ119896(119860 times 119861)

(100)

where 119862119896= 119862

119896(119909 119910) 119896 = 1 2 3 4 are constants given as

1198621=

119863

4119864 119862

3= 119862

4=

1

2119864(1 minus

119863

2119864)

1198622= 1 minus 119862

1minus 119862

3minus 119862

4

(101)

and Φ119896 119896 = 1 2 3 4 are positive measures on [0infin)

2 suchthat

Φ1(119860 times 119861) = 120575

0(119860) 120575

0(119861) (102)

Φ2(119860 times 119861) = Q2

0(119883119898

119898isin 119860

119883119872

119872isin 119861)

= Q2

0(119883119872

119872isin 119860

119883119898

119898isin 119861)

(103)

Φ3(119860 times 119861) = Φ

4(119861 times 119860)

= Q2

0[119883119898

119898isin 119860Q0

119883119898

(119883119872minus119898

119872isin 119861)]

(104)

Remark 17 The expression (104) coincides with

n(0) [2119898119883119898119883119898

119898isin 119860

119883119872

119872isin 119861] (105)

where n(0) is the generalized excursion measure introducedin Section 3

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

The proof ofTheorem 16 will be given in the next section

Remark 18 In the case where 120572 = 2 the process((119883

119905radic2) Pℎ

0) is the symmetrized three-dimensional Bessel

process In other words if we set

Ω+

= 119908 isin D forall119905 gt 0 119908 (119905) gt 0

Ωminus

= 119908 isin D forall119905 gt 0 119908 (119905) lt 0

(106)

then we have

Pℎ0(Ω

+) = Pℎ

0(Ω

minus) =

1

2(107)

and the processes ((119883119905radic2) Pℎ

0(sdot | Ω

+)) and ((minus119883

119905radic2)

Pℎ0(sdot | Ω

minus)) are one-sided three-dimensional Bessel processes

Hence the Ray-Knight theorem implies that the process(119909

2119871119909

infin 119909 ge 0) conditional on Ω

+is the squared Bessel

process of dimension two Let us check thatTheorems 15 and16 are consistent with this fact Since ℎ(119909) = |119909|2 we have

ℎ (119909) + ℎ (119910) minus ℎ (119909 minus 119910) =|119909| +

10038161003816100381610038161199101003816100381610038161003816 minus

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

2

= min |119909|

10038161003816100381610038161199101003816100381610038161003816 if 119909119910 gt 0

0 if 119909119910 lt 0

(108)

If 119909 gt 0 gt 119910 then we should look at Theorem 15 whichimplies that

Pℎ0(119871

119909

infinisin 119860 119871

119910

infinisin 119861 | Ω

+) = Q2

0(119883

1isin 119860) sdot 120575

0(119861) (109)

If 119909 119910 gt 0 then we should look at Theorem 16 Note that

119898(119909 119910) = min 119909 119910 119872 (119909 119910) = max 119909 119910

119863 =21003816100381610038161003816119909 minus 119910

1003816100381610038161003816

max 119909 119910 119864 =

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

max 119909 119910

119863

4119864=

1

2

(110)

and that

1198621= 119862

2=

1

2 119862

3= 119862

4= 0 (111)

Hence Theorem 16 implies that

Pℎ0(119909

2119871119909

infinisin 119860 119910

2119871119910

infinisin 119861 | Ω

+) = Q2

0(119883

119909isin 119860119883

119910isin 119861)

(112)

55 Proof of Theorem 16 We give the proof of Theorem 16We divide the proofs into several steps

Step 1 Since

0 lt 119863 le 2119864 = 2 (1 minus119898

119872) (113)

we have 0 lt 119898 lt 119872

Step 2 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898minus 120582

2

119883119872

119872] (114)

By the Markov property the right-hand side is equal to

Q2

0[exp minus120582

1

119883119898

119898Q2

119883119898

[exp minus1205822

119883119872minus119898

119872]] (115)

By formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

times Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898]

(116)

Again by formula (39) this expectation is equal to

1

1 + 2 (1205822119872) (119872 minus 119898)

sdot1

1 + 2 (1205821119898 + (120582

2119872) (1 + 2 (120582

2119872) (119872 minus 119898)))119898

(117)

Simplifying this quantity with 119864 = 1 minus 119898119872 we see that

∬119890minus12058211198971minus12058221198972Φ

2(119889119897

1times 119889119897

2) =

1

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(118)

Note that this expression is invariant under interchangebetween 120582

1and 120582

2 which proves the second equality of (103)

Step 3 Let us compute the Laplace transform

Q2

0[exp minus120582

1

119883119898

119898Q0

119883119898

[exp minus1205822

119883119872minus119898

119872]] (119)

By formula (39) this expectation is equal to

Q2

0[expminus(

1205821

119898+

1205822119872

1 + 2 (1205822119872) (119872 minus 119898)

)119883119898] (120)

Using the equality between (116) and (118) we see that

∬119890minus12058211198971minus12058221198972Φ

3(119889119897

1times 119889119897

2) =

1 + 21198641205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(121)

Now we also obtain

∬119890minus12058211198971minus12058221198972Φ

4(119889119897

1times 119889119897

2) =

1 + 21198641205821

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(122)

Step 4 Noting that

∬119890minus12058211198971minus12058221198972Φ

1(119889119897

1times 119889119897

2) = 1 (123)

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 11

we sum up formulae (123) (118) (121) and (122) and weobtain

4

sum

119896=1

119862119896∬119890

minus12058211198971minus12058221198972Φ

119896(119889119897

1times 119889119897

2)

=1 + 120582

1+ 120582

2+ 119863120582

11205822

1 + 21205821+ 2120582

2+ 4119864120582

11205822

(124)

By Lemma 11 we see that the right-hand side coincides withthe Laplace transform of the joint law of (119871

119909

infin 119871

119910

infin) under

Pℎ0 By the uniqueness of Laplace transforms we obtain the

desired conclusion

6 The Laws of Total Local Times for Bridges

In this section we study the total local time of Levy bridgesand ℎ-bridges

61 The Laws of the Total Local Times for Levy Bridges Let uswork with the Levy bridge P119905

00and its local time (119871

119911

119904 0 le 119904 le

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(125)

withP11990500-probability one Let us study the lawof the total local

time 119871119911

119905under P119905

00

Theorem 19 For 119905 gt 0 it holds that

P11990500

(1198710

119905isin 119889119897) =

120574119897(119905)

119901119905(0)

119889119897 (126)

Proof UsingTheorem 7 with 119899 = 1 and 1199111= 0 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890

minus1205821198710

119905] 119889119905 =1

1119906119902(0) + 120582

(127)

By formula (61) we have

1

1119906119902(0) + 120582

= int

infin

0

119890minus(1119906

119902(0)minus120582)119897

119889119897

= int

infin

0

P0[119890minus119902120591(119897)

] 119890minus120582119897

119889119897

(128)

Hence using Lemma 9 we obtain (126) by the Laplaceinversion The proof is now complete

Theorem 20 For any 119911 isin R 0 one has

P11990500

(119871119911

119905= 0) =

119901119905(0) minus (120588

119911lowast 120588

119911lowast 119901

sdot(0)) (119905)

119901119905(0)

(129)

P11990500

(119871119911

119905isin 119889119897) =

(120588119911lowast 120588

119911lowast 120574

119897) (119905)

119901119905(0)

119889119897 for 119897 = 0 (130)

where the symbol lowast stands for the convolution operation

Proof Using Theorem 7 with 119899 = 2 1205821= 0 120582

2= 120582 119911

1= 0

and 1199112= 119911 we have

int

infin

0

119890minus119902119905

119901119905(0)P119905

00[119890minus120582119871119911

119905] 119889119905

= 119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2) +

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

(131)

On the one hand it follows from (57) that

119906119902(0) (1 minus

119906119902(119911)

2

119906119902(0)

2)

= (int

infin

0

119890minus119902119905

119901119905(0) 119889119905) (1 minus P

119911[119890minus1199021198790]

2

)

(132)

This implies (129) On the other hand by (61) and (57) wehave

119906119902(119911)

2

119906119902(0)

2

1

1119906119902(0) + 120582

= P119911[119890minus1199021198790]

2

int

infin

0

P0(119890

minus119902120591(119897)) 119890

minus120582119897119889119897

(133)

This implies (130) The proof is now complete

62 The Laws of the Total Local Times for ℎ-Bridges Let uswork with the ℎ-bridge Pℎ119905

00and its local time (119871

119911

119905) such that

int

119904

0

119891 (119883119906) 119889119906 = int119891 (119911) 119871

119911

119904ℎ(119911)

2119889119911 0 le 119904 le 119905 119891 isin B

+(R)

(134)

with Pℎ11990500-probability one We give the Laplace transform

formula for the law of the total local time 119871119911

119905under Pℎ119905

00

Lemma 21 For 119911 isin R 0 and 120582 ge 0 one has

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582ℎ(119911)2119871119911

119905] 119889119905

=(119906

119902(119911)

2119906

119902(0)

2) 120582

1 + 119906119902(0) 1 minus 119906

119902(119911)

2119906

119902(0)

2 120582

(135)

Proof Using Theorem 8 with 119899 = 1 1205821

= 120582 and 1199111

= 119911 wehave

int

infin

0

119890minus119902119905

119901ℎ

119905(0 0)Pℎ119905

00[1 minus 119890

minus120582119871119911

119905] 119889119905 =119906ℎ

119902(0 119911)

2120582

1 + 119906ℎ119902(119911 119911) 120582

(136)

By formulae (72) we obtain the desired formula

7 Concluding Remark

We gave an explicit formula which describes the jointdistribution of the total local times at two levels and wediscussed several formulae related to the law of the totallocal times However we could not obtain any better result

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Abstract and Applied Analysis

on the law of the total local time with space parameter Aswe noted in Remark 3 a difficulty arises in the case of ℎ-paths which comes from the asymmetry of thematrix ΣminusΣ

0We also remark that we have no better result related to thelaw of total local time in the case where the Markov processis asymmetric We left the further study of the law of thetotal local time for asymmetric Markov process with spaceparameter for future work

Acknowledgment

K Yano is partially supported by Inamori Foundation

References

[1] D Applebaum Levy Processes and Stochastic Calculus vol 116of Cambridge Studies in Advanced Mathematics CambridgeUniversity Press Cambridge UK 2nd edition 2009

[2] P Imkeller and I Pavlyukevich ldquoLevy flights transitions andmeta-stabilityrdquo Journal of Physics A vol 39 no 15 pp L237ndashL246 2006

[3] D Revuz and M Yor Continuous Martingales and BrownianMotion Springer Berlin Germany 3rd edition 1999

[4] M Yor Some Aspects of Brownian Motion Part I Lectures inMathematics ETH Zurich Birkhauser Basel Switzerland 1992

[5] N Eisenbaum and H Kaspi ldquoA necessary and sufficient con-dition for the Markov property of the local time processrdquo TheAnnals of Probability vol 21 no 3 pp 1591ndash1598 1993

[6] N Eisenbaum H Kaspi M B Marcus J Rosen and Z ShildquoA Ray-Knight theorem for symmetric Markov processesrdquo TheAnnals of Probability vol 28 no 4 pp 1781ndash1796 2000

[7] K Yano ldquoExcursions away from a regular point for one-dimensional symmetric Levy processes without Gaussian partrdquoPotential Analysis vol 32 no 4 pp 305ndash341 2010

[8] K Yano Y Yano andM Yor ldquoPenalising symmetric stable Levypathsrdquo Journal of the Mathematical Society of Japan vol 61 no3 pp 757ndash798 2009

[9] KYano Y Yano andMYor ldquoOn the laws of first hitting times ofpoints for one-dimensional symmetric stable Levy processesrdquoin Seminaire de Probabilites XLII vol 1979 of Lecture Notes inMathematics pp 187ndash227 Springer Berlin Germany 2009

[10] K Yano ldquoTwo kinds of conditionings for stable Levy processesrdquoin Probabilistic Approach to Geometry vol 57 of AdvancedStudies in Pure Mathematic pp 493ndash503 Mathematical Societyof Japan Tokyo Japan 2010

[11] R M Blumenthal and R K Getoor Markov Processes andPotential Theory vol 29 of Pure and Applied MathematicsAcademic Press New York NY USA 1968

[12] J Pitman ldquoCyclically stationary Brownian local time processesrdquoProbability Theory and Related Fields vol 106 no 3 pp 299ndash329 1996

[13] J Pitman ldquoThedistribution of local times of a Brownian bridgerdquoin Seminaire de Probabilites XXXIII vol 1709 of Lecture Notesin Mathematics pp 388ndash394 Springer Berlin Germany 1999

[14] J Pitman and M Yor ldquoA decomposition of Bessel bridgesrdquoZeitschrift fur Wahrscheinlichkeitstheorie und Verwandte Gebi-ete vol 59 no 4 pp 425ndash457 1982

[15] P Fitzsimmons J Pitman and M Yor ldquoMarkovian bridgesconstruction Palm interpretation and splicingrdquo in Seminar onStochastic Processes Proceedings from 12th Seminar on Stochastic

Processes 1992 University of Washington Seattle Wash USA1992 vol 33 of Progress in Probability Series pp 101ndash134Birkhaauser Boston Mass USA 1993

[16] M B Marcus and J Rosen Markov Processes Gaussian Pro-cesses and Local Times vol 100 of Cambridge Studies inAdvanced Mathematics Cambridge University Press Cam-bridge UK 2006

[17] J Pitman and M Yor ldquoDecomposition at the maximum forexcursions and bridges of one-dimensional diffusionsrdquo inItorsquos Stochastic Calculus and Probability Theory pp 293ndash310Springer Tokyo Japan 1996

[18] P J Fitzsimmons and K Yano ldquoTime change approach togeneralized excursion measures and its application to limittheoremsrdquo Journal of Theoretical Probability vol 21 no 1 pp246ndash265 2008

[19] J Bertoin Levy Processes vol 121 of Cambridge Tracts inMathematics Cambridge University Press Cambridge UK1996

[20] S Watanabe K Yano and Y Yano ldquoA density formula forthe law of time spent on the positive side of one-dimensionaldiffusion processesrdquo Journal of Mathematics of Kyoto Universityvol 45 no 4 pp 781ndash806 2005

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of