processes discrete time geiss

8/11/2019 Processes Discrete Time Geiss

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Stochastic processes in discrete time

Stefan Geiss

June 1, 2009


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Contents

1 Introduction 51.1 The gamblers ruin problem . . . . . . . . . . . . . . . . . . . . 51.2 Branching processes . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 Some notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Sums of independent random variables 112.1 Zero-One laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Convergence of sums . . . . . . . . . . . . . . . . . . . . . . . 20

2.2.1 The Two-Series-Theorem and its applications . . . . . 202.2.2 Proof of the Two-Series-Theorem . . . . . . . . . . . . 27

2.3 The law of iterated logarithm . . . . . . . . . . . . . . . . . . 33

3 Martingales in discrete time 37

3.1 Some prerequisites . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Definition and examples of martingales . . . . . . . . . . . . . 463.3 Some elementary properties of martingales . . . . . . . . . . . 513.4 Stopping times . . . . . . . . . . . . . . . . . . . . . . . . . . 543.5 Doob-decomposition . . . . . . . . . . . . . . . . . . . . . . . 583.6 Optional stopping theorem . . . . . . . . . . . . . . . . . . . . 603.7 Doobs maximal inequalities . . . . . . . . . . . . . . . . . . . 643.8 Uniformly integrable martingales . . . . . . . . . . . . . . . . 673.9 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.9.1 A martingale proof forKolmogorovs zero-one law . 743.9.2 The branching process example . . . . . . . . . . . . . 75

3.9.3 The Theorem of Radon-Nikodym. . . . . . . . . . . . 773.9.4 On a theorem ofKakutani . . . . . . . . . . . . . . . 80

3.10 Backward martingales . . . . . . . . . . . . . . . . . . . . . . 85

4 Exercises 914.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.2 Sums of independent random variables . . . . . . . . . . . . . 914.3 Martingales in discrete time . . . . . . . . . . . . . . . . . . . 93

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4 CONTENTS

Important remark for the reading: These notes are only lecturenotes which combines material from different sources, for example from thebeautiful books [6] and [7]. Consequently, these lecture notes cannot replacethe careful study of the literature as the mentioned books of Williams [7]and Shirjaev [6], and of Neveu [5]. Furthermore, there is a lecture notes ofHitczenko [3] which concerns the Central Limit Theorem for martingales nottouched in these notes.


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Chapter 1

Introduction

In the Introduction we want to motivate by two examples the main two partsof the lecture which deal with

sums of independent random variables and the martingale theory.

The examples are given at this stage in an intuitive way without being rig-orous. In the course we will come back to the examples and treat them in arigorous way.

Throughout this lecture we assume that the students are familiar withthe basic probability theory. In particular, we assume that the material ofthe lecture notes [1] is known and will refer from time to time to these notes.

1.1 The gamblers ruin problem

We assume that there is a gambler starting with x0 Euro as his startingcapital at time n= 0. The game is as follows: we fix some p(0, 1) as ourprobability of lossand at each time-step n = 1, 2,...the gambler

loses with probability pthe amount of 1 Euro,

wins with probability 1 pthe amount of 1 Euro.To realize this we can assume a coin which gives by flipping

tails with probability p, heads with probability 1 p,

so that the gambler loses or wins if tails or heads, respectively, occurs. Ofcourse, ifp= 1/2, then the coin is not fair.

Iffn is the amount of capital of the gambler at time n, then we obtain a(random) sequence (fn)n=1 with f0 = x0 such that

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6 CHAPTER 1. INTRODUCTION

f1 = x0 + 1

f1 = x0 1

f2 = x0 + 2

f2 = x0

f2 = x0 2

f3 = x0 + 3

f3 = x0 + 1

f3 = x0 1

f3 = x0 3

f0 =x0

To make our example more realistic, one would need to have at least alower bound forfnto bound the losses. Here we take two bounds < A 0, q0 (0, 1), andMn := fnn . Let(Fn)n=0 be the filtration generated by(fn)n=0. Then one has the following:

(i) (Mn)n=0 is a martingale with respect to the filtration (Fn)n=0, thatmeansIE(Mn+1|Fn) =Mn a.s.,

(ii) M:= limn Mn exists almost surely.

(iii) If1, thenM = 0 a.s. and limnIP(Mn= 0) = 1.(iv) If >1, thenIEM = 1 and limnIP(Mn= 0)(0, 1).

The interpretation of is the expected number of sons in each generationof a single family. Then, n can be seen as some kind of average number of

families with name Hakkinen in the n-th generation. So fn/

n

should bethe right quantity which stabilizes after some time.


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10 CHAPTER 1. INTRODUCTION

1.3 Some notation

The symbol IN stands for the natural numbers

{1, 2,...

}. For a sequence of

real numbers (n)n=1 we recall that

lim supn

n := limn

supkn

k,

lim infn

n := limn

infkn

k.

Ifa, b IR, then we let a b := min {a, b} and a+ := max {a, 0}. Besidesthis notation we will need

Lemma 1.3.1 [Lemma of Borel-Cantelli] 1 2 Let(, F, IP)be a prob-ability space andA1, A2,... F. Then one has the following:(1) If

n=1IP(An)


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Chapter 2

Sums of independent randomvariables

First let us recall the independence of random variables.

Definition 2.0.1 Let(, F, IP) be a probability space andf: IR with I be a family of random variables. The family is called independentprovided that for all pairwise different1,...,nIand allB1,...,Bn B(IR)one has that

IP(f1

B1,...,fn

Bn) = IP(f1

B1)

IP(fn

Bn).

2.1 Zero-One laws

It is known that

n=11n

= but thatn=1(1)n1n does converge. Whatis going to happen if we take a random sequence of signs?

Definition 2.1.1 For p (0, 1) we denote by (p)1 , (p)2 ,... : IR a se-quence of independent random variables such that

IP (p)n =1 =p and IP (p)n = 1 = 1 p.If p = 1/2 we write n =

(1/2)n and call this sequence a sequence of

Bernoulli 1 random variables.

Now we are interested in

IP

n=1

nn

converges

.

1

Jacob Bernoulli, 27/12/1654 (Basel, Switzerland)- 16/08/1705 (Basel, Switzerland),Swiss mathematician.

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12 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES

Remark 2.1.2 Let 1, 2,... : IR be random variables over (, F, IP).Then

A:= : n=1

n() converges F.This is not difficult to verify because A if and only if

N=1,2,...

n0=1,2,...

m>nn0

:

m

k=n+1

k()

< 1N

.

What is a typical property of the set A={ : n=1 n() converges}?The condition does not depend on the first realizations 1(),...,N() sincethe convergence ofn=1 n() andn=N+1 n() are equivalent so that

A=

:

n=N+1

n() converges

.

We shall formulate this in a more abstract way. For this we need

Definition 2.1.3 Let (, F, IP) be a probability space and : IR bea family of random variables. Then( : I) is the smallest-algebrawhich contains all sets of form

{

:

()

B}

where

I and B B

(IR).

Exercise 2.1.4 Show that (:I) is the smallest -algebra on suchthat all random variables : IR are measurable.Example 2.1.5 For random variables 1, 2,...: IR and

A=

:

n=1

n() converges

one has that A

N=1 (N, N+1, N+2,...).

The above example leads us straight to the famous Zero-One law ofKol-mogorov 2:

Proposition 2.1.6 (Zero-One law ofKolmogorov) Assume inde-pendent random variables 1, 2,... : IR on some probability space(, F, IP) and let

FN :=(N, N+1,...) and F :=

Fn .

ThenIP(A) {0, 1} for allA F.2

Andrey Nikolaevich Kolmogorov 25/04/1903 (Tambov, Russia) - 20/10/1987 (Mos-cow, Russia), one of the founders of modern probability theory, Wolf prize 1980.


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2.1. ZERO-ONE LAWS 13

For the proof the following lemma is needed:

Lemma 2.1.7 Let

A Fbe an algebra and let(

A) =

F. Then, for all

>0 andB F, there is anA Asuch thatIP(AB)< .

Proof of Proposition 2.1.6. The idea of the proof is to show that IP(A) =IP(A)2. Define the algebra

A:=

n=1

(1,...,n).

We have thatF (A). Hence Lemma 2.1.7 implies that for A Fthere are An(1,...,Nn) such that

IP(AAn)n 0.We get also that

IP(An A)n IP(A) and IP(An)nIP(A).The first relation can be seen as follows: since

IP(An A) + IP(AnA) = IP(An A)IP(A)we get that

lim infn

IP(An A)IP(A)lim sup IP(An A).

The second relation can be also checked easily. But now we get, by indepen-dence, that

IP(A) = limn

IP(A An) = limn

IP(A)IP(An) = IP(A)2

so that IP(A) {0, 1}. Corollary 2.1.8 Let 1, 2,... : IR be independent random variablesover(, F, IP). Then

IP

n=1

nconverges

{0, 1} .

We also obtain an early version of the law of the iterated logarithm (LIL).

But for this we need the central limit theorem (CLT) for the Bernoullirandom variables, which we will not prove here.


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Proposition 2.1.9 (CLT) Let 1, 2,... : IR be independent Ber-noulli random variables andcIR. Then one has that

limn

IP1+ + nn

c = c

ex22 dx

2.

As a second application ofKolmogorovs Zero-One law we get

Corollary 2.1.10 Let 1, 2,... : IR be independentBernoulli ran-dom variables. Then one has that

IP

lim sup

n

fnn

=, lim infn

fnn

=

= 1

wherefn:=1+ + n.

Proof. The assertion is equivalent to

IP

lim sup

n

fnn

=

= 1 and IP

lim inf

n

fnn

=

= 1.

By symmetry it is sufficient to prove the first equality only. Letting

Ac:= lim sup

n

fnnc

for c0 we get that

IP

lim sup

n

fnn

=

= IP

m=1

Am

= lim

mIP

Am

because ofAmAm+1. Hence we have to prove that IP

Am

= 1. Since

lim supn

1() + + n()n

= lim s upn,nN

1() + + N1()n

+N() + + n()n

= lim s up

n,nN

N() + + n()n

because of

limn,nN

1() + + N1()n

= 0,

where we assume that N2, we get that

Am

N=1 (N, N+1,...) .


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Consequently, in order to prove IP

Am

= 1 we are in a position to ap-plyKolmogorovs Zero-One law (Proposition 2.1.6) and the only thing toverify is

IP Am >0.To get this, we first observe that

Am=

lim sup

n

fnnm

n=1

k=n

fk

km

since n=1 k=n fkk m implies that for all n = 1, 2,... there is ak n such that fk()/

k m, which gives a subsequence (kl)l=1 with

fkl

()/

kl

m and finally lim supn

fn()/

n

m. Applying Fatou 3slemma we derive from that

IP(Am)IP

n=1

k=n

fk

km

lim sup

nIP

fn

nm

.

Since the central limit theorem (Proposition 2.1.9) gives

limn

IP

fn

nm

=

m

ex2

2dx

2>0

we are done. Now let us come back to question (Q6) of the ruin problem in Section 1.1.

Let p, q (0, 1), p+ q= 1, and fn :=n

i=1 (p)i where n = 1, 2,... and the

random variables(p)i were defined in Definition 2.1.1. Consider the event

A:={ : # {n: fn() = 0}=} .

In words, A is the event, that the path of the process (fn)n=1 reaches 0 in-

finitely many often. We would like to have a Zero-One law for this event.However, we are not able to apply Kolmogorovs Zero-One law (Proposi-

tion 2.1.6). But what is the typical property ofA? We can rearrangefinitelymany elements of the sum

ni=1

(p)i and we will get the same event since

from some Non, this rearrangement does not influence the sum anymore.To give a formal definition of this property we need

Definition 2.1.11 A map : ININ is calledfinite permutation if(i) the map is a bijection,

(ii) there is someN IN such that(n) =n for allnN.3

Pierre Joseph Louis Fatou, 28/02/1878 (Lorient, France) - 10/08/1929 (Pornichet,France), French mathematician, dynamical systems, Mandelbrot set.


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Moreover, we recall that B(IRIN) is the smallest -algebra on IRIN ={(1, 2,...) :nIR} which contains all cylinder sets Zof form

Z :={(1, 2,...) :an < n< bn, n= 1, 2,...}for some< an < bn ) = IP(l> )

for all k, lIN and IR.

Proposition 2.1.14 (Zero-One law ofHewitt and Savage) 4 As-

sume a sequence of independent and identically distributed random variables1, 2,...: IR over some probability space(, F, IP). If the eventA Fis symmetric, thenIP(A) {0, 1}.

Proof. (a) We are going to use the permutations

n(k) :=

n + k : 1knk n : n + 1k2n

k : k >2n.

4

Leonard Jimmie Savage 20/11/1917 (Detroit, USA) - 1/11/1971 (New Haven, USA),American mathematician and statistician.


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Now we approximate the set A. By Lemma 2.1.7 we find Bn B(IRn) suchthat

IP (AnA)n0 for An:={: (1(), . . . , n())Bn} .(b) Our goal is to show that

IP (An)IP(A),IP (n(An))IP(A),

IP (An n(An))IP(A),as n , where

n(An) := : n(k)()n

k=1

Bn= {: (n+1(), . . . , 2n())Bn} .

(c) Assuming for a moment that (b) is proved we derive

IP (An n(An)) = IP(An)IP(n(An))sinceAnis a condition on1,...,n,n(An) is a condition on n+1,...,2n, and1,...,2n are independent. By n this equality turns into

IP(A) = IP(A)IP(A)

so that IP(A)

{0, 1

}.

(d) Now we prove (b). The convergence

IP(An)IP(A)is obvious since IP(AnA)0. This implies

IP(n(An))IP(A)since IP(n(An)) = IP(An) which follows from the fact that

(n+1,...,2n, 1,...,n) and (1,...,n, n+1,...,2n)

have the same law in IR2n as a consequence that we have an identicallydistributed sequence of random variables. Finally

IP(AAn) = IP (n(AAn))

= IP (n(A)n(An))

= IP (An(An))

where the first equality follows because the random variables (1, 2,...) havethe same distribution and the last equality is a consequence of the symmetryofA. Hence

IP (AAn)n0 and IP (An(An))n0


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which implies

IP (A (An

n(An)))

n0 and IP (An

n(An))

IP(A).

As an application we consider

Proposition 2.1.15 Let1, 2,...: IRbe independentBernoulliran-dom variables andfn =

ni=1 i, n= 1, 2,... Then

IP(# {n: fn= 0}=) = 1

Proof. Consider the sets

A+ := { : # {n: fn() = 0} 0} ,A := { : # {n: fn() = 0}=} ,A := { : # {n: fn() = 0}


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(b) The set A:={ : n() = 0 for all n= 1, 2, . . .} does not belongtoF but is symmetric, because

A = { : (n())n=1B}=

: (n)()n=1B

for B={(0, 0, 0, . . . )} B(IRIN).(c) The set A := { : n=1|n()|exists and n=1n()< 1} does

not belong toF but is symmetric, becauseA = { : (n())n=1B}

=

: (n)()n=1BifB is the set of all summable sequences with sum strictly less than

one.

We finish by the non-symmetric random walk. As preparation we needStir-ling 5s formula

n! =

2nn

e

ne 12n

for n= 1, 2,...and some (0, 1) depending on n. This gives2n

n

=

(2n)!

(n!)2 =

2(2n)

2ne

2ne 1122n

(

2n)2

ne

2n

e 212n

2

4nn

.

Proposition 2.1.17 Letp= 1/2 andfn=n

i=1 (p)i , n= 1, 2,..., where the

random variables(p)1 ,

(p)2 ,...: IR are given by Definition 2.1.1. ThenIP (# {n: fn = 0}=) = 0.

Proof. Letting Bn:={f2n = 0} we obtain

IP(Bn) =

2n

n

(pq)n (4pq)

n

n

byStirlings formula. Since p=qgives 4pq


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2.2 Convergence of sums

If 1, 2, . . . are independent Bernoulli random variables, that means

IP (n = 1) = IP (i=1) = 12 , then we know from Section 2.1 that

IP

n=1

nn

converges

{0, 1} .

But, do we get probability zero or one? For this there is a beautiful completeanswer: It consists of the Three-Series-Theorem ofKolmogorovwhich wewill deduce from the Two-Series-Theorem ofKolmogorov. First we formu-late the Two-Series-Theorem, then we give some examples and consequences(including the Three-Series-Theorem), and finally we prove the Two-Series-

Theorem.

Besides the investigation of sums of independent random variables thissection also provides some basic concepts going into the direction of themartingale-theory.

2.2.1 The Two-Series-Theorem and its applications

Proposition 2.2.1 (Two-Series-Theorem ofKolmogorov) Assumea probability space(, F, IP) and random variables1, 2,...: IR whichare assumed to be independent.

(i) If

n=1IEn converges and

n=1IE (n IEn)2 0 is someconstant. Then the following assertions are equivalent:

(a) IP (n=1 n converges) = 1.

(b)n=1IEn converges andn=1IE (n IEn)2


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2.2. CONVERGENCE OF SUMS 21

Proof. Assuming the conditions on (n)n=1 and (n)

n=1 we get, for n :=

n+ nn, that

n=1

IEn=

n=1

[n+ nIEn] =

n=1

n

and n=1

IE[n IEn]2 =

n=1

IE[nn]2 =

n=1

2n


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converges, we get that supn |n|= supn |n||n(0)| cc : f()0 the following three conditions are satisfied:

(a)

n=1IEcn converges,

(b)

n=1IE (cn IEcn)2 c)


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The lemma ofBorel-Cantelli implies that

IP (

{

: #

{n:

Bn

}=

}) = 0.

This implies that IP (n=1 n converges ) = 1 so thatIP

n=1

n converges

= IP

n=1

[n+ cn] converges

= 1.

(ii) =(iii) is trivial.(i) = (ii) The almost sure convergence ofn=1 n() implies that

limn |n()|= 0 a.s. so that

IPlim supn { :|n()| c} = 0.The Lemma ofBorel-Cantelli (note that the random variables n areindependent) gives that

n=1

IP (|n| c)


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and

n=1 IP(|n| 1)2

n=1pn 0 such that

(i) n ,

(ii) IEn= 0,

(iii)

n=1IE2n2n

0, n ,and(xn)

n=1IR with limn xn = xIR. Then one has that

1

n(1x1+ + nxn)nx.

Proof. Let >0 and find some n0 such that|xn x|< for nn0. Then,for n > n0,

1nn

i=1 ixi x = 1nn

i=1 ixi 1nn

i=1 ix 1

n

ni=1

i|xi x|

= 1

n

n0i=1

i|xi x| + 1n

ni=n0+1

i|xi x|

1n

n0i=1

i|xi x| + .

6

Otto Toplitz 01/08/1881 (Breslau, Germany) - 15/02/1940 (Jerusalem), worked oninfinite linear and quadratic forms.


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Since n there is some n1 > n0 such that for all nn1 one has that1

n

n0

i=1

i|xi x| which implies that 1n

ni=1

ixi x 2

for nn1.

Lemma 2.2.7 (Kronecker) 7 Letn>0,n , and(xn)n=1IRsuchthat

n=1 xn exists. Then

1

n

ni=1

ixin 0.

Proof. Let 0 = S0 = 0 and Sn := x1+ + xn. Then1

n

ni=1

ixi = 1

n

ni=1

i(Si Si1)

= 1

n nSn 0S0

n

i=1Si1(i i1)

= Sn 0S0n

1n

ni=1

Si1(i i1).

Observing that limn Sn=

n=1 xnIR,0S0/n0 since n , and

limn

1

n

ni=1

Si1(i i1) = limn

Sn =

n=1

xn

by Lemma 2.2.6 we finally obtain that

limn

1n

ni=1

ixi = 0.

Proofof Proposition 2.2.5. From the Two-Series-Theorem we know that

IP

n=1

nn

converges

= 1.

7Leopold Kronecker, 7/12/1823 (Liegnitz, Prussia ; now Legnica, Poland)- 29/12/1891

(Berlin, Germany), major contributions in elliptic functions and the theory of algebraicnumbers.


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dyadic expansion 0, 123 where we do not allow an infinite series ofdigits 1. Then considering 1, 2,...: IR we get random variables whichsatisfy the assumptions of Example 2.2.9. The assertion of this propositionsays that zeros and ones are equally likely for the dyadic representation of anumberx[0, 1).

Earlier we have shown for independent Bernoulli random variables1, 2,...: IR that

IP

lim sup

n

fnn

=, lim infn

fnn

=

= 1

wherefn:=1+ + n. As an application of Proposition 2.2.5 we show anopposite statement now.

Proposition 2.2.10 Let1, 2,...: IRbe independentBernoulliran-dom variables,fn:=1+ + n, and >0. Then

IP

limn

fnn(log(n + 1))

12+

= 0

= 1.

Proof. Let n :=

n(log(n + 1))12+ nso that

n=1

IE2nn(log(n + 1))1+2 =

n=1

1

n(log(n + 1))1+2


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for >0 so that

IP (|f| ) IE |f|2

2 .

Assuming independent random variables 1, 2,... : IR and fn := 1+ + n this gives that

IP (|fn| ) IE|fn|2

2 .

Now one can enlarge the left-hand side by replacing|fn|by supk=1,...,n |fk|sothat we get a maximal inequality.

Lemma 2.2.11 (Inequality ofKolmogorov) Let 1, 2,... : IR bea sequence of independent random variables such that

(i) IEn= 0 forn= 1, 2,...,(ii) IE2n0, andn= 1, 2,..., then one has that

IP

max

1kn|fk|

IEf

2n

2 .

Proof. Let n= 1, 2, . . . be fixed, A :={max1kn |fk| }, andBk:={|f1|< , . . . , |fk1|< , |fk| }

for k2 and B1 :={|f1| }. We get a disjoint union A= nk=1 Bk. Now2IP(A) = 2 IP

nk=1

Bk

=n

k=1

2 IP(Bk) =n

k=1

Bk

2dIP

n

k=1

Bk

f2kdIPn

k=1

Bk

f2ndIP

IEf2

n,

which proves our assertion, where the second to the last inequality can beverified as follows:

Bk

f2ndIP =

Bk

(fk + k+1 + + n)2 dIP

=

Bk

f2kdIP + 2

Bk

fk(k+1+ + n) dIP

+

Bk

(k+1+ + n)2 dIP

= Bk

f2kdIP + Bk

(k+1+ + n)2 dIP


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since, by independence,

Bk fk(k+1+ + n) dIP = (fkBk) (k+1+ + n) dIP=

fkBkdIP

(k+1+ + n) dIP

= 0.

Remark 2.2.12 There are other inequalities, called Levy 10 -Octavianiinequalities: assuming independent random variables1, 2,...: IR and >0 one has

(i) IP (max1kn |fk| > )3 max1kn IP |fk|> 3,(ii) IP (max1kn |fk| > )2IP (|fn|> ) if the sequence (n)n=1 is addi-

tionally symmetric that means that for all signs1,...,n {1, 1}thedistributions of the vectors (1,...,n) and (11,...,nn) are the same.

Example 2.2.13 For Bernoulli variables (n)n=1 it follows from Lemma

2.2.11 that

IP

max

1kn|1+ + k|

IEf

2n

2 =

n

2.

Letting = n, >0, this givesIP

max

1kn|1+ + k|

n

1

2.

The left-hand side describes the probability that the random walk exceedsn or n upto step n (not onlyat step n).

Now we need the converse of the above inequality:

Lemma 2.2.14 (Converse inequality ofKolmogorov) Let 1, 2,... :

IR be a sequence of independent random variables such that

(i) IEn= 0 forn= 1, 2,...,

(ii) there exists a constantc > 0 such that|n()| c for all andn= 1, 2,....

Iffn:=1+ + n, >0, n {1, 2,...}, andIEf2n >0, then

IP

max

1kn|fk|

1 (c + )

2

IEf2n.

10Paul Pierre Levy, 15/09/1886 (Paris, France) - 15/12/1971 (Paris, France), influenced

greatly probability theory, also worked in functional analysis and partial differential equa-tions.


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The next lemma, we need, describes Cauchy 11 sequences with respect tothe convergence in probability.

Lemma 2.2.15 Letf1, f2,... : IR be a sequence of random variables.Then the following conditions are equivalent:

(i) IP ({ : (fn())n=1 is aCauchy sequence}) = 1,(ii) For all >0 one has that

limn

IP

supk,ln

|fk fl|

= 0.

(iii) For all >0 one has that

limn

IP

supkn

|fk fn|

= 0.

Proof. (ii)(iii) follows fromsupkn

|fk fn| supk,ln

|fk fl| sup

kn|fk fn| + sup

ln|fn fl|

= 2 supkn |fk fn|.(i)(ii) Let

A:={ : (fn())n=1 is a Cauchysequence} .Then we get that

A=

N=1,2,...

n=1,2,...

k>ln

:|fk() fl()| 1

N

.

Consequently, we have that IP(A) = 1 if and only if

IP

n=1,2,...

k>ln

:|fk() fl()| 1

N

= 1

for all N= 1, 2,..., if and only if

limn

IP

k>ln

:|fk() fl()| 1

N

= 1

11

Augustin Louis Cauchy, 21/08/1789 (Paris, France)- 23/05/1857 (Sceaux, France),study of real and complex analysis, theory of permutation groups.


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for allN= 1, 2,..., if and only if

limn IPk>ln

:|fk() fl()|> 1N = 0for allN= 1, 2,.... We can finish the proof by remarking that

k>ln

:|fk() fl()|> 1

N

=

supk>ln

|fk() fl()|> 1N

.

Proofof Proposition 2.2.1. (i) We letn:= n IEnso that IEn = 0 and

n=1

IE2n=

n=1

IE(n IEn)2 0. But this follows from

IP

supkn

|gk gn|

= limN

IP

sup

nkn+N|gk gn|

lim

NIE(gn+N gn)2

2

= limN Nk=1IE2n+k2

l=n+1IE

2l

2

where we have used theKolmogorovinequality Lemma 2.2.11. Obviously,the last term converges to zero as n .

(ii) Because of step (i) we only have to prove that (a)(b). We use againa symmetrization argument and consider a new sequence 1,

2,...:

IRof independent random variables on (, F, IP) having the same distributionas the original sequence 1, 2,..., that means

IP(n) = IP(n)


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for all n = 1, 2,... and all IR. We also may assume that|n()| d for all and n = 1, 2, .... Taking the product space (M, , ) =(,

F, IP)

(,

F, IP) we may considern,

n

:M

IR with the conventionthat n(,

) =n() and n(, ) =n(

). Now we let

n(, ) =n(, ) n(, )

and get

IEn = IEn IEn= 0,|n(, )| |n(, )| + |n(, )| 2d,

and

(, )M :

n=1

n(, ) converges = IP IP

(, ) :

n=1

(n() n()) converges

= 1.

Letting gn:=1+ + n and >0, Lemma 2.2.15 implies that there is ann {1, 2,...} such that

supk

n|gk gn|

0

IP(g > )

2

e 2

22

= 1.

Proof. By the change of variables y= xwe get that

lim>0

IP(g > )

2

e 2

22

= lim>0

e y2

22 dy

2

2

e 2

22

= lim>0

ex2

2 dx

2

2

e 2

22

= lim

>0

ex2

2 dx

2

12e

2

2

= 1

where we apply the rule ofLHospital 13.

Next we need again a maximal inequality similar to one of that alreadymentioned in Remark 2.2.12.

Lemma 2.3.4 Let1,...,n: IRbe independent random variables whichare symmetric, that means

IP(k) = IP(k)for allk= 1,...,nandIR. Then one has that

IP

maxk=1,...,n

fk>

2IP(fn> )

wherefk :=1+ + k and >0.12Norbert Wiener, 26/11/1894 (Columbia, USA) - 18/03/1964 (Stockholm, Sweden),

worked on Brownian motion, from where he progressed to harmonic analysis, and won theBocher prize from his studies on Tauberian theorems.

13Guillaume Francois Antoine Marquis de LHopital, 1661 (Paris, France)- 2/2/1704

(Paris, France), French mathematician who wrote the first textbook on calculus, whichconsisted of the lectures of his teacher Johann Bernoulli.


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2.3. THE LAW OF ITERATED LOGARITHM 35

Proof. Let 1kn. Because of IP(fn fk >0) = IP(fn fk 0) + IP(fn

fk } and

Bk :={f1, ..., fk1, fk > }for k= 2,...,n. Then we get that, where 0/0 := 1,

IP

maxk=1,...,n

fk>

=

nk=1

IP(Bk)

=

n

k=1

IP(Bk)

IP(fn

fk)

IP(fnfk)

=n

k=1

IP(Bk {fnfk})IP(fnfk)

where we used the independence ofBk andfn fk=k+1 + + n ifk < n.Using that

1

IP(fnfk)2we end up with

IP maxk=1,...,n

fk > 2 n

k=1

IP(Bk {fnfk})

2IP(fn> ).

Proofof Proposition 2.3.1 for n = gn. By symmetry we only need toshow that

IP

lim sup

nn3

fn(n)

1

= 1 and IP

lim sup

nn3

fn(n)

1

= 1.

This is equivalent that for all (0, 1) one has that

IP

:n03 nn0 fn()(1 + )(n)

= 1 (2.2)

and

IP

: # {n3 fn()(1 )(n)}

=

= 1. (2.3)

First we turn to (2.2): let := 1 + and nk := k for k k0 such that

nk0

3. Define

Ak :={ :n(nk, nk+1] fn()> (n)}


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for k k0. Then IP(lim supkAk) = 0 would imply (2.2). According to theLemma ofBorel-Cantelliit is sufficient to show that

k=k0

IP(Ak) (nk)) 2IP f[nk+1]> (nk) 2c (f[nk+1])

2((nk))e ((nk))

2

2(f[nk+1])2

where we have used the maximal inequality from Lemma 2.3.4 and the esti-mate from Lemma 2.3.3 and where (fn)

2 is the variance offn. Finally, by(fn)

2 =n we get (after some computation) that

k=k0

IP(Ak)

k=k0

c1k k0 and := 1 with(0, 1). Assume

that we can show thatYk > (nk) + 2(nk1) (2.5)

happens infinitely often with probability one, which means by definition that

f[nk] f[nk1]> (nk) + 2(nk1)

happens infinitely often with probability one. Together with (2.4) this wouldimply that

f[nk]> (nk)

happens infinitely often with probability one. Hence we have to show (2.5).For this one can prove that

IP (Yk > (nk) + 2(nk1)) c2k log k

so that k=k0+1

IP (Yk> (nk) + 2(nk1)) =.

An application of the Lemma ofBorel-Cantelli implies (2.5).

Now we have also answered questions (Q5) and (Q7) from Section 1.1.


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Chapter 3

Martingales in discrete time

Originally martingales were intended to model fair games. Having this inmind, it is not surprising that martingale theory plays an important role inareas connected to stochastic modeling such as Stochastic Finance, Biology,and Physics. On the other hand side, martingale theory enters many branchesof pure mathematics as well: for example it is a powerful tool in harmonicand functional analysis and gives new insight into certain phenomena.

In this lecture notes we want to develop some basic parts of the mar-tingale theory. One can distinguish between two cases: discrete time andcontinuous time. The first case deals with sequences of random variablesand is sometimes easier to treat. In the second case we have to work with

families of random variables having uncountably many elements, which istechnically partially more advanced, is however based on the discrete timecase in many cases.

3.1 Some prerequisites

First we summarize some notation and basic facts used from now on.

The Lebesgue spaces. The Lebesgue spaces are named after HenriLebesgue, 1875-1941.

Definition 3.1.1 (i) Let (, F, IP) be a probability space and 1p


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38 CHAPTER 3. MARTINGALES IN DISCRETE TIME

Example 3.1.2

(a) Let = [0, 1),F =B([0, 1)) and IP = be the Lebesgue-measure,that means

([a, b)) =b afor 0 a < b 1. Assume a continuous f : [0, 1) IR. Thenf Lp([0, 1)) if and only if

fp= 1

0

|f(x)|pdx 1

p

0,

f(x) := 1

x Lp([0, 1))

if and only ifp


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3.1. SOME PREREQUISITES 39

(iv) Equivalence-classes: forf Lp(, F, IP) one has thatfp= 0 if andonly if IP(f= 0) = 0. The relationf g if IP(f =g) = 1 defines anequivalence class relation.

Remark 3.1.4

(i) Because of assertion (ii) of Proposition 3.1.3 the expression p is asemi-normonLp. Moreover, (ii) and (iv) imply that Lp is a normon Lp, that means [Lp, Lp ] is a normed space.

(ii) Items (iii) and (iv) say that everyCauchy-sequencein Lp converges toa limit inLp. Hence [Lp, p] is acompletenormed space. A completenormed space is called Banach3 space.

(iii) Ifp= 2, then [L2, 2] is a Hilbert 4 spacewhere the inner product isgiven by

f , g:=

f gdIP for f f and gg.

Conditional expectation. There are different ways to approach condi-tional expectations. Let us explain the approach of the best approximation.Assume = [0, 1),F =B([0, 1)) and IP = to be the Lebesgue measure.Define the sub--algebras

Fdyadn = 0, 12n , 12n , 22n , . . . , 2n12n , 1 , n= 0, 1, 2, ...,which is the system of all possible unions of sets of type

k12n

, k2n

for k =

1, 2, 3, . . . , 2n where n 1 is fixed. ThenFdyadn FdyadN F for 1 nN


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Proposition 3.1.5 (i) The best approximationg is of form

g(x) = 2n

k2n

k12n

f(y)dy forx k12n , k2n .(ii) For allA Fdyadn one has that

A

f(x)dx=

A

g(x)dx.

Proof. (i) Take anotherFdyadn -measurable functionh: [0, 1)IR. Then

1

0

|f(x) h(x)|2dx=2n

k=1 k2n

k12n

|f(x) h(x)|2dx.

We will prove that k2n

k12n

|f(x) h(x)|2dx k

2n

k12n

|f(x) g(x)|2dx

=

k2n

k12n

f(x) 2n k

2n

k12n

f(y)dy

2

dx

Consider

(y) : = k2nk12n

|f(x) y|2dx

=

k2n

k12n

|f(x)|2dx 2y k

2n

k12n

f(x)dx + 1

2ny2.

Then the minimum is attained for y = 2n k

2nk12n

f(x)dxand we are done.

(ii) A Fdyadn means A =

kIk12n

, k2n

. Then

A

f(x)dx=kI

k2nk12n

f(x)dx

=kI

k2n

k12n

g(x)dx

=

A

g(x)dx.

Behind this example there is the general notion of the expected value weintroduce now.


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Proposition 3.1.6 (Conditional expectation) Let(, F, IP) be a proba-bility space andG Fbe a sub--algebra. Letf L1(, F, IP).

(i) There exists ag L1(, G, IP) such thatB

f dIP =

B

gdIP for allB G.

(ii) Ifg andg are like in(i), thenIP (g=g ) = 0.

The main point of the theorem above is that g isG-measurable. This leadsto the following definition:

Definition 3.1.7 (Conditional expectation) TheG

measurable and in-tegrable random variable g from Proposition 3.1.6 is called conditional ex-pectationoffwith respect toG and is denoted by

g= IE(f| G).

One has to keep in mind that the conditional expectation is only unique upto null sets fromG.

We will prove Proposition 3.1.6 later. First we continue with some basicproperties of conditional expectations.

Proposition 3.1.8 Letf , g , f 1, f2 L1(, F, IP) and letH G Fbe sub-algebras ofF. Then the following holds true:

(i) Linearity: if, IR, thenIE(f+ g| G) =IE(f| G) + IE(g| G) a.s.

(ii) Monotonicity: iff1 f2 a.s., thenIE(f1| G) IE(f2| G) a.s.(iii) Positivity: Iff 0 a.s., thenIE(f| G) 0 a.s.

(iv) Convexity: one has that|IE(f| G)| IE(|f| | G) a.s.(v) Projection property: iff isG-measurable, thenIE(f| G) =f a.s.

(vi) Advanced projection property:

IE(IE(f| G)| H) = IE(IE(f| H)| G) = IE(f| H)a.s.

(vii) Ifh: IR isG-measurable andf h L1(, F, IP), thenIE(hf| G) =hIE(f| G)a.s.

(viii) IfG ={, }, thenIE(f|G) = IEf a.s.


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(ix) The case that f is independent fromG : if for all B B(IR) and allA Gone has that

IP ({f B} A) = IP(f B)IP(A),

thenIE(f|G) = IEf a.s.(x) Monotone convergence: assumef 0 a.s. and random variables0

hnf a.s. Thenlimn

IE(hn|G) = IE(f|G)a.s..

Proof. (i) is an exercise.

(ii) Assume that (ii) does not hold. We find an A G

, IP(A) > 0 and < such that

IE(f2| G) < IE(f1| G) on A.

Hence A

f2dIPIP(A)< IP(A)A

f1dIP

which is a contradiction.

(iii) Apply (ii) to 0 =f1f2 = f.

(iv) The inequality f |f| gives IE(f|G) IE(|f||G) a.s. andf |f|givesIE(f|G) IE(|f||G) a.s., so that we are done.

(v) follows directly from the definition.

(vi) Since IE(f|H) isH-measurable and henceG-measurable, item (v)implies that IE(IE(f|H)|G) = IE(f|H) a.s. so that one equality is shown. Forthe other equality we have to show that

A

IE(IE(f|G)|H)dIP =A

f dIP

for A H. Letting h:= IE(f|G) this follows fromA

IE(IE(f|G)|H)dIP =A

IE(h|H)dIP

=

A

hdIP

=

A

IE(f|G)dIP

=

A

f dIP

since A H G.


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(vii) Assume first that h =N

n=1 nAn , whereN

n=1

An = is a disjoint

partition with An G

. For A

Gwe get, a.s., that

A

hf dIP =N

n=1

n

A

Anf dIP

=N

n=1

n

AAn

f dIP

=N

n=1

n

AAn

IE(f|G)dIP

= A N

n=1nAn IE(f|G)dIP

=

A

hIE(f|G)dIP.

Hence IE(hf|G) = hIE(f|G) a.s. For the general case we can assume thatf, h 0 since we can decompose f = f+ f and h = h+ h withf+ := max {f, 0} and f := max {f, 0} (and in the same way we proceedwith h). We find step-functions 0hn h such that hn()h(). Then,by our fist step, we get that

hnIE(f|G) = IE(hnf|G) a.s.By n the left-hand side follows. The right-hand side is a consequenceof the monotone convergence given in (x).

(viii) Clearly, we have that

f dIP = 0 =

(IEf)dIP and

f dIP = IEf=

(IEf)dIP

so that (viii) follows.

(ix) is an exercise.(x) is an exercise.

Proof of Proposition 3.1.6. For this purpose we need the theorem ofRadon 5 -Nikodym6 .

5Johann Radon, 16/12/1887 (Tetschen, Bohemia)- 25/05/1956 (Vienna, Austria),worked on the calculus of variations, differential geometry and measure theory.

6Otton Marcin Nikodym, 13/8/1887 (Zablotow, Austria-Hungary ; now Ukraine)-4/05/1974 (Utica, USA), worked in measure theory, functional analysis, projections onto

convex sets with applications to Dirichlet problem, generalized solutions of differentialequations, descriptive set theory and the foundations of quantum mechanics.


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Definition 3.1.9 (Signed measures) Let (, F) be a measurable space.(i) A map :F IR is called (finite) signed measureif and only if

= + ,where , 0 and+ and are probability measures onF.

(ii) Assume that (, F, IP) is a probability space and that is a signedmeasure on (, F). Then IP ( is absolutely continuous withrespect to IP) if and only if

IP(A) = 0 implies (A) = 0.

Example 3.1.10 Let L L1(, F, IP) and

(A) := A

LdIP.

Then is a signed measure and IP.Proof. We letL+ := max {L, 0} andL:= max {L, 0} so thatL= L+L.Assume that

LdIP> 0 and define

(A) =

ALdIP

LdIP

.

Now we check that are probability measures. First we have that

() = LdIP

LdIP = 1.

Then assume An F to be disjoint sets such that A =

n=1

An. Set :=

L+dIP. Then

+ n=1

An

=

1

n=1

AnL+dIP

= 1

n=1 An()L+dIP

= 1

limN

Nn=1

An

L+dIP

= 1

limN

Nn=1

An

L+dIP

=

n=1

+(An)

where we have usedLebesgues dominated convergence theorem. The same

can be done for L.


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Theorem 3.1.11 (Radon-Nikodym) Let(, F, IP) be a probability spaceand a signed measure withIP. Then there exists anL L1(, F, IP)such that

(A) = A

L()dIP(), A F. (3.1)The random variable L is unique in the following sense. If L and L arerandom variables satisfying (3.1), then

IP(L=L) = 0.Definition 3.1.12 Lis called Radon-Nikodymderivative. We shall write

L= d

dIP.

We should keep in mind the rule

(A) =

Ad=

ALdIP,

so that d= LdIP.

Proof of Proposition 3.1.6. Define

(A) :=

A

f dIP for A G

so that is a signed measure onG. Applying the Theorem of Radon-Nikodym gives an g L1(, G, IP) such that

(A) =

A

gdIP

so that A

gdIP =(A) =

A

f dIP.

Assume now another g L1(, G, IP) withA

gdIP =

A

gdIP

for all A G and assume that IP(g= g) > 0. Hence we find a set A Gwith IP(A)> 0 and real numbers < such that

g() < g () for Aor

g() < g() for A.Consequently (for example in the first case)

A

gdIPIP(A)< IP(A)A

gdIP

which is a contradiction.


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3.2 Definition and examples of martingales

Looking up the word martingale from an encyclopedia (for example in

www.dict.org/bin/Dict) gives the following:

(i) A strap fastened to a horses girth, passing between his fore legs, andfastened to the bit, or now more commonly ending in two rings, throughwhich the reins pass. It is intended to hold down the head of the horse,and prevent him from rearing.

(ii) (Naut.) A lower stay of rope or chain for the jib boom or flying jibboom, fastened to, or reeved through, the dolphin striker. Also, thedolphin striker itself.

(iii) (Gambling) The act of doubling, at each stake, that which has beenlost on the preceding stake; also, the sum so risked; metaphoricallyderived from the bifurcation of the martingale of a harness. [Cant]Thackeray.

We start with the notation of a filtration which describes the information wehave at a certain time-point.

Definition 3.2.1 (Filtration and adapted process) (i) Let (, F) bea measurable space. An increasing sequence (

Fn)n=0 of -algebras

F0 F1 . . . Fn . . . F is called filtration. If (, F, IP) isa probability space, then (, F, IP, (Fn)n=0) is called filteredprobabil-ity space.

(ii) A stochastic process (Xn)n=0, whereXn: IR are random variables,

is called adapted with respect to the filtration (Fn)n=0 (or (Fn)n=0-adapted) provided that Xn isFn-measurable for all n= 0, 1,...

(iii) Given a stochastic processX= (Xn)n=0of random variablesXn :

IR, then the filtration (FXn )n=0 given by

FXn :=(X0,...,Xn)is called natural filtrationof the process X.

It is clear that a process Xis adapted with respect to (FXn )n=0.

Definition 3.2.2 (Martingale) LetM= (Mn)n=0withMn L1(, F, IP)

for n = 0, 1,...and let (Fn)n=0 be a filtration on (, F, IP).(i) The process M is called martingaleifM is (Fn)n=0-adapted and

IE(Mn+1| Fn) =Mn a.s. for n= 0, 1,...


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3.2. DEFINITION AND EXAMPLES OF MARTINGALES 47

(ii) The process M is called sub-martingale or super-martingale, if M is(Fn)n=0-adapted and

IE(Mn+1| Fn) Mn a.s. or IE(Mn+1| Fn) Mn a.s.,for n= 0, 1,..., respectively.

The definition of a (sub-, super-) martingale depends on the filtrationand the measure. To emphasize this, one sometimes uses the phrases(Fn)n=0-martingaleor IP-martingale(and the same for the sub- and super-martingales). Now we consider some examples.

Random walk. Let 0 < p, q < 1, p+q= 1 and (p)1 ,

(p)2 , . . .: IR be

independent random variables such that

IP((p)n =1) =p and IP((p)n = 1) =q(see Definition 2.1.1). As filtration we useF0 :={, } and

Fn := ((p)1 , . . . , (p)n )for n1 so thatFn consists of all possible unions of sets of type

{(p)1 =1, . . . , (p)n =n}with 1, . . . , n {1, 1}. In dependence on pwe get that our random walkconsidered in Chapter 2 is a (sub-, super-) martingale.

Proposition 3.2.3 LetM= (Mn)n=0 be given byMn :=

(p)1 + . . . +

(p)n for

n1 andM0:= 0. Then one has the following:(i) Ifq=p = 1

2, thenM is a martingale.

(ii) Ifq > 12

, thenMis a sub-martingale.

(iii) Ifq < 12

, thenM is a super-martingale.

Proof. Using the rules for conditional expectations we get that

IE(Mn+1| Fn) = IE(Mn+ n+1| Fn)= IE(Mn| Fn) + IE(n+1| Fn)= Mn+ IEn+1.

It remains to observe that IEn+1 = qp is negative, zero, or positive independence on p.

The random walk from Proposition 3.2.3 has an additivestructure since,for 0kn,

Mn Mk is independent fromFk, Mn Mk has the same distribution as Mnk.


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Exponential random walk. In applications, for example in StochasticFinance, one needs instead of the additive structure a multiplicative struc-ture. For instance, changes in share price processes are always relative tothe current price. Hence we are looking for a process Mn >0 such that, for0kn,Mn/Mk is independent fromFk and has the same distribution asMnk. One process of this type is given by

Proposition 3.2.4 Letp(0, 1), >0, cIR, M0:= 1,

Mn:=ePn

i=1(p)i +cn

forn = 1, 2,...,F0 ={, }, andFn = ((p)1 , . . . , (p)n ). Then one has thefollowing:

(i) Ifp(e

e

) =ec

e

, thenM= (Mn)n=0 is a martingale.(ii) Ifp(e e)ec e, thenM= (Mn)n=0 is a sub-martingale.

(iii) Ifp(e e)ec e, thenM= (Mn)n=0 is a super-martingale.

Proof. We have that

IE(Mn+1| Fn) = IE(e(p)n+1+cMn| Fn) =MnIEe

(p)n+1+c =Mn(pe

c + qec+)

and obtain a martingale if and only ifpe + (1 p)e =ec or

p(e e) =ec e.If we have an inequality, then we get a sub- or super-martingale.

Dyadic martingales. Let = [0, 1),F =B([0, 1)), f L1([0, 1)), andIP = be the Lebesguemeasure. As filtration we define

Fdyadicn =

0, 12n

,

12n

, 22n

, . . . ,

2n1

2n , 1

for n = 0, 1, . . .Finally, we let

Mn(t) := IE(f|Fdyad

n )(t).The process (Mn)

n=0is calleddyadicmartingale. This process is a martingale

as shown by the more general

Proposition 3.2.5 For f L1(, F, IP) and a filtration (Fn)n=0 the se-quence(Mn)

n=0 withMn:= IE(f|Fn) is a martingale.

Proof. One has that, a.s.,

IE(Mn+1| Fn) = IE(IE(f| Fn+1)| Fn) = IE(f| Fn) =Mn.


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3.2. DEFINITION AND EXAMPLES OF MARTINGALES 49

Remark 3.2.6 The function fin Proposition 3.2.5 is called the closureofthe martingale (IE(f|Fn))n=0.

Remark 3.2.7 Assume a continuous f : [0, 1] IR and Mn(t) :=IE(f|Fdyadn ) as above. Then one can easily show that

limn

Mn(t) =f(t)

for all t [0, 1) and obtain our first limit theorem. To check this, fix t0[0, 1), find k0, k1,...such that

kn

1

2n t0 0, f0 := 1,

and

fn+1() :=X(n+1)0 () + + X(n+1)fn()() and Mn:=

fnn

.

Then M = (Mn)n=0 is a martingale with respect to the filtration (Fn)n=0

given byFn:=(M0, . . . , M n) forn0.

Proof. Clearly, Mn isFn-measurable by definition ofFn. Next we showMn L1(, F, IP). We proceed by induction and assume that IEMn1


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or equivalently, that IEfn1


51/99

3.3. SOME ELEMENTARY PROPERTIES OF MARTINGALES 51

3.3 Some elementary properties of martin-gales

Here we collect some very basic properties of martingales. It will be use-ful to introduce for a martingale M = (Mn)

n=0 the sequence ofmartingale

differencesdM0:= 0, and

dMn:=Mn Mn1 for n= 1, 2,...

We start with

Proposition 3.3.1 Let M = (Mn)n=0 be a martingale on a filtered proba-bility space(, F, IP, (Fn)n=0). Then one has

(i) IE(Mm| Fn) =Mn a.s. for0n < m


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(iii) Assume that 0m < n. Then

dMndMmdIP=

(Mn Mn1)(Mm Mm1)dIP

=

MnMmdIP

MnMm1dIP

Mn1MmdIP +

Mn1Mm1dIP

=

IE(MnMm|Fm)dIP

IE(MnMm1| Fm1)dIP

IE(Mn1Mm|Fm)dIP +

IE(Mn1Mm1|Fm1)dIP

=

MmIE(Mn| Fm)dIP

Mm1IE(Mn| Fm1)dIP

MmIE(Mn1|Fm)dIP +

Mm1IE(Mn1|Fm1)dIP

=

M2mdIP

M2m1dIP

M2mdIP +

M2m1dIP

= 0

where we have used Proposition 3.1.8.

Lemma 3.3.2 (Jensens inequality) 7

Let : IRIR be a convex func-tion, f L1(, F, IP) such that(f) L1(, F, IP), andG F be a sub-algebra. Then

(IE(f|G))(IE((f)|G)) a.s.

Proof. Any convex function : IRIR is continuous, hence measurable. Tocheck the continuity we fixxIR and observe first that for xnx we havethat

lim supn

(xn)(x)

(if this is not true we would get, by drawing a picture, a contradiction tothe convexity of). Assume now that an x and bn x and that eitherlim infn (an) or lim infn (bn) (or both) is (are) strictly less than (x). Pick-ing n(0, 1) such that x= (1 n)an+ nbn we would get that

(1 n)(an) + n(bn)< (x)for nn0 which is a contradiction to the convexity of. Next we observethat there exists a countable set D of linear functions L : IRIR such that,

7Johan Ludwig William Valdemar Jensen, 08/05/1859 (Nakskov, Denmark)- 05/03/1925 (Copenhagen, Denmark), studied infinite series, the gamma function and in-

equalities for convex functions. Only did mathematics in his spare time, his actual jobwas in a telephone company.


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3.3. SOME ELEMENTARY PROPERTIES OF MARTINGALES 53

for allxIR, one has that(x) = sup

LDL(x).

To get that D we proceed as follows. For q Q we let

aq := limxq

(q) (x)q x and bq := limxq

(q) (x)q x .

By convexity of we have that < aq bq


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Definition 3.3.5 Assume a filtered probability space (, F, IP, (Fn)n=0).(i) A sequence of random variables (vn)

n=1, vn :

IR, is called pre-

dictableifvn isFn1-measurable forn = 1, 2,...(ii) Given a martingale M = (Mn)

n=0 we define its martingale transform

Mv = (Mvn)n=0 byM

v0 :=M0 and

Mvn :=M0+n

k=1

vk(Mk Mk1) for n= 1, 2,...

Proposition 3.3.6 If M = (Mn)n=0 is a martingale and v = (vn)

n=1 is

predictable such that IE|vndMn|


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3.4. STOPPING TIMES 55

Let us start with some examples. The most easiest one is

Example 3.4.2 Consider :

{0, 1,...

}with ()

n0. Then is a

stopping time, since

{=n}=

: n= n0

: n=n0.

One typical and often used class of stopping times is the class ofhitting times.

Example 3.4.3 (Hitting time) Let B B(IR), the stochastic processX= (Xn)

n=0 be adapted, and

B() := inf

{n

0 :Xn()

B

}with inf:=. Then B is a stopping time and called hitting timeofB .

Proof. That B is a stopping time follows from

{B =n}={X0B} . . . {Xn1B} {XnB} Fnfor n1 and{B = 0}={X0B} F0.

Example 3.4.4 Let X = (Xn)n=0 and Y = (Yn)

n=0 be adapted processes

and

() := inf{n0 : Xn() =Yn()}with inf=:. Then is a stopping time. In fact, if we let Zn:=Xn Ynand B :={0}, then is the hitting time ofB with respect to the adaptedprocessZ.

Example 3.4.5 Let X= (Xn)n=0 be adapted. Then

() := inf{n0 : Xn+1()> 1}with inf:=is, in general, nota stopping time.

Now we continue with some general properties of stopping times.

Proposition 3.4.6 Let , : {0, 1, 2, . . .} {} be stopping times.Then one has the following:

(i) The map : IR{}is a random variable, that means1()F and1(B) F for allB B(IR).

(ii) The timemax{, } is a stopping time.(iii) The timemin{, } is a stopping time.


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(iv) The time+ is a stopping time.

Proof. Item (i) follows from

1(B) =

nB,n01({n})

n0Fn F

and

1() = \

n=0

1({n}) F.

Items (ii) and (iii) are consequences of

{max{, }= n}={= n, n} {n, =n}= (

{=n

} {

n

})

(

{

n

} {=n

})

Fnand

{min{, }= n}= ({=n} { n}) ({n} {=n})= [{=n} ( \ { < n})] [{=n} ( \ { < n})] Fn.

(iv) is an exercise.

Now we introduce the -algebraFfor a stopping time . The-algebrawill contain all events one can decide until the stopping time occurs. For

example one can decideA={=n0}

until the stopping time: We take and wait untilis going to happen,that means on{ = 0} the stopping time occurs at time 0, on{ = 1} attime 1, and so on. If() is going to happen, and ()=n0, then A, if() =n0, then A. The abstract definition is as follows:Definition 3.4.7 Let : {0, 1, 2, . . .}{}be a stopping time, thena set A Fbelongs toFif and only if

A {

=n} F

n

for all n= 0, 1, 2, . . ..

Proposition 3.4.8F is a-algebra.Proof. We have that { = n} = Fn and { = n} ={ =n} Fn for all n {0, 1, 2,...} so that, F. Now let A F. ThenAc {=n}={ =n} \ (A {=n}) Fn so that Ac F. Finally, letA1, A2, . . . F. ThenAk {=n} Fn and (

k=1

Ak) { = n} Fn and

k=1Ak F.


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3.4. STOPPING TIMES 57

Example 3.4.9 If n0, thenF =Fn0. In fact, by definitionA F ifand only ifA { = n} Fn for n = 0, 1, 2,.... Sincen= n0 immediatelygives that A

{ = n

}=

Fn the condition reduces to n = n0 which

means that A F if and only ifA { = n0} Fn0 which simply meansthat A Fn0.

Proposition 3.4.10 Let be a stopping time. Then A F if and onlyif A = A

n=0

An with An Fn and An {=n}, and A F andA {=}.

Let us illustrate this by an example.

Example 3.4.11 Let Xn=1+ . . . + n, X0= 0, and

() = inf{n0 : Xn()2}.

Then one has

A = A k{,0,1,...}

{=k}=k{,0,1,...}

[A {=k}]

=:

k{,0,1,...}Ak

with{= 0}={= 1}=(after 0 or 1 steps one cannot have the value 2)and

{= 2} = {1=2 = 1},{= 3} = ,{= 4} = {1=1, 2=3 =4 = 1} {2 =1, 1=3 = 4= 1}

{3=1, 1=2 =4 = 1} {4 =1, 1=2 = 3= 1},

We continue with some general properties.

Proposition 3.4.12 LetX = (Xn)n=0 be an adapted process and :

{0, 1, . . .} a stopping time. Then

Z() :=X()() : IR

isF-measurable.


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Proof. We have to show that

{

: Z()

B

} F for all B

B(IR).

By definition this is equivalent to

{ : Z()B} { : () =n} Fn,or

{ : X()()B, () =n} Fn,or

{ : Xn()B, () =n} Fn,or

{XnB} {=n} Fn,which is true.

Proposition 3.4.13 Let0S T be stopping times. ThenFS FT.

Proof. We have thatA FSif and only ifA{S=n} Fnfor n= 0, 1, 2, . . .Hence

A {

T n}

= A {

S n} {

T n} F

n

since A {S n} Fn and{T n} Fn. But this implies that A {T =n} Fn.

3.5 Doob-decomposition

TheDoob8 -decomposition of a sub-martingale consists in a martingale andin a predictable increasing process. In this way, we get two components whichcan be treated separately.

Proposition 3.5.1 LetY = (Yn)n=0 be a sub-martingale. Then there exists

a processA= (An)n=0 such that

(i) 0 =A0()A1(). . . a.s.,(ii) An isFn1-measurable forn= 1, 2, . . .,

(iii) Mn := Yn An is a martingale.8Joseph Leo Doob, 27/02/1910 (Cincinnati, USA)- 7/06/2004 (Clark-Lindsey Village,

USA),Doobs work was in probability and measure theory. Made major contributions instochastic processes, martingales, optimal stopping, potential theory.


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3.5. DOOB-DECOMPOSITION 59

The processA = (An)n=0 is unique in the following sense: if there are two

such processesA andA, thenIP(An=An) = 1 forn= 0, 1, 2, . . ..

Proof. In order to deduce the formula for An let us assume that we alreadyhave the decomposition Yn = Mn+ An. We get that, a.s.,

IE(Yn+1 An+1| Fn) = IE(Mn+1| Fn) =Mn=Yn Anand

An+1= IE(Yn+1| Fn) Yn+ An.Now let us go the other way round. Define

A0:= 0 and An+1:= IE(Yn+1| Fn) Yn+ An.The properties ofA = (An)

n=0 are that

An+1 isFn-measurable forn0, that An+1 An = IE(Yn+1| Fn) Yn 0 a.s. since Y = (Yn)n=0 is a

sub-martingale, and that

the process M is a martingale because, a.s.,IE(Mn+1

| Fn) = IE(Yn+1

An+1

| Fn)

= IE(Yn+1 IE(Yn+1| Fn) + Yn An| Fn)=Yn An=Mn.

The uniqueness can be seen as follows: if (An)n=0 and (A

n)n=0 are such

processes, then, a.s.,

An+1= IE(Yn+1| Fn) Yn+ AnAn+1= IE(Yn+1| Fn) Yn+ An

with A0=A00, which implies by induction that An = An.

Remark 3.5.2 In fact, we have an equivalence: ifYn=Mn+ An, where Mis a martingale and A an increasing, predictable, and integrable process asabove, then Y is a sub-martingale.

Definition 3.5.3 The decomposition

Yn = Mn+ An

from Proposition 3.5.1 is called Doob-decomposition.


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One might think that the process (An)n=0measures how far a sub-martingale

is from a martingale. Let us consider now some examples.

Example 3.5.4 Let 0 = Y0 Y1 Y2 . . . be real numbers. For anyfiltration we get a sub-martingale since IE(Yn+1| Fn) =Yn+1Yn a.s. TheDoob-decomposition is then Mn= 0 a.s. and An= Yn a.s.

Example 3.5.5 IfY = (Yn)n=0 is a martingale, then An = 0 a.s.

Example 3.5.6 We consider independent random variables (p)1 ,

(p)2 ,... :

IR such that IP((p)n =1) =p and IP((p)n = 1) =q, where 0< p, q 1

2, then the process Y = (Yn)

n=0 with Y0= 0 and

Yn=(p)1 + . . . + (p)n

is a sub-martingale with respect toFn := ((p)1 ,...,(p)n ) for n 1 andF0 :={, }, and

An+1 = IE(Yn+1| Fn)Yn + An = IEn+1 + An= (n + 1)IE1 = (n +1)(qp).

Example 3.5.7 Using the same filtration and random variables as inthe previous example we consider the exponential random walk Sn =

ePn

i=1(p)i +cn withS0:= 1. Ifp(e

e)ec e we get a sub-martingaleand

An+1= IE(Sn+1| Fn) Sn+ An=SnIE

e

(p)n+1+c

Sn+ An

=Sn[ 1] + An=Sn[ 1] + Sn1[ 1] + An1= ( 1)(Sn+ Sn1+ + S0).

Moreover, we have that 1 if and only ifp(e e)ec e.

3.6 Optional stopping theorem

IfM= (Mn)n=0 is a martingale and 0ST 1, we would like to win by the following game: if you have at


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3.6. OPTIONAL STOPPING THEOREM 61

time n an amount ofNn, we toss a fair coin for 2n + |Nn| Euro, that means

thatN0

1 and Nn+1 := Nn+ n+1(2

n +|Nn

|)

where 1, 2,... are independent symmetric Bernoulli random variables.Letting also 0 = 1, we takeFn:=(0,...,n). We obtain

Nn+1 2n ifn+1= 1,

this is what we want, and that the process (Nn)n=0 is a martingale transform

of (0+ + n)n=0 sinceNn+1 Nn= (2n + |Nn|)n+1.

Now we introduce our stopping strategy : {0, 1, 2, . . .}{} by() := inf{n0 : Nn()c}.

The properties ofare summarized in

Proposition 3.6.2 The random time is a stopping time with

(i) IP({ :() T})> 0.Proof. The process (Nn)

n=0 is adapted and can be considered as the

hitting time of the set B= [c, ). Hence we have a stopping time.(i) We observe that one has, for n0,

IP(Nn+1 2n) IP(n+1 = 1) =

1

2.

Assuming now an n00 with 2n0 c gives that

IP( supnn0 Nn+1 c) IP(n0+1 = 1) + IP(n0+1=1, n0+2= 1)

+IP(n0+1 =1, n0+2=1, n0+3 = 1) + = 1

2+ 1

2 1

2+ 1

2 1

2 1

2+

= 1.

(ii) follows by the very definition of.

(iii) We can take 1=. . .= T =1 and have for this path thatN0()< c, . . . , N T()< c

so that IP( > T)2T.


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From that we get

N()()

c a.s. and IE(N{ T0 sothat

IE(YT YS) =m

n=1

IE(Hn(Yn Yn1))

=m

n=1

IE(IE(Hn(Yn Yn1)|Fn1))

=m

n=1IE(HnIE((Yn Yn1)|Fn1))

= 0


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3.6. OPTIONAL STOPPING THEOREM 63

because IE((Yn Yn1)|Fn1) = 0 a.s. so that IEYT = IEYS. Let nowB FSand define the new random times

SB :=S B+ mBc,

TB :=T B+ mBc.

The random times SB and TB are stopping times since, for example for SB,one has

{SB =n}= (B {S=n}) (Bc {m= n})where the first term belongs toFn by the definition of the -algebraFSandthe second term belongs toFn since it is equal the empty set ifm=n andBc FS Fm=Fn ifm= n. Hence we have stopping times with

0 SB TB m

so that

IE(YTB+ YmBc) = IEYTB = IEYSB = IE(YSB+ YmBc)

and B

YTdIP =

B

YSdIP.

Together with Proposition 3.4.12 this implies that IE(YT| FS) =YS a.s.(b) Now we assume that we have a sub-martingale: we first apply the

Doob-decomposition and get that

Yn=Mn+ An.

By step (a) we haveIE(MT| FS) =MSa.s.,

so that, a.s.,

IE(YT| FS) = IE(MT+ AT| FS)= MS+ IE(AT| FS) MS+ IE(AS

| FS)

= YS.

Example 3.6.1(continued) Assume now that we have to stop our game atsome time T0>0. Our new strategy is then

T() := min{(), T0}which is as the minimum of two stopping times again a stopping time. More-overT T0. Letting S

0, we have

0 S T T0


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and, by the optional stopping theorem, IE(MT|FS) =MSa.s. so thatIEMT= IE(IE(MT

|FS)) = IEMS= 1.

Hence our unfair doubling strategy does not work anymore as before, wecannot become rich, if we only have finite time at our disposal.

3.7 Doobs maximal inequalities

Doobs maximal inequality is a replacement ofKolmogorovs inequality inLemma 2.2.11. In the following we also work with (sub-, super-) martingalesX = (Xn)

Nn=0 over the finite time scale{0,...,N}. Moreover, given X =

(Xn)Nn=0 or X= (Xn)

n=0, we need the maximal functions

Xk() := sup0nk

|Xn()| and X() := supn=0,1,...

|Xn()|,

where the latter is used for X= (Xn)n=0 only.

Proposition 3.7.1 (Doobs maximal inequalities) LetX= (Xn)Nn=0 be

a martingale or non-negative sub-martingale. Then the following holds true:

(i) For0 one hasIP (X

N)

IE

{X

N}|X

N|.

(ii) Forp(1, ) one has

XNLp p

p 1XNLp .

Proof. If X = (Xn)Nn=0 is a martingale, then (|Xn|)Nn=0 is a non-negative

sub-martingale, so that it remains to consider this case only.

(i) For 0 we define the stopping times() := inf{n : Xn} N and () :=N

so that 0N. By the Optional Stopping Theorem we may deduceXIE(X| F) = IE(XN| F) a.s.

and

IEXN IEX = IE(X{XN}) + IE(X{XN


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3.7. DOOBS MAXIMAL INEQUALITIES 65

(ii) Let c >0 be fixed. Then

IE((XN c)p

) = IE (XNc)

0 pp

1

d

= IE

c0

pp1{XN}d

=

c0

pp1IP(XN)d.

By step (i) we continue with

IE((XN c)p) c

0

pp2IE(|XN|{XN})d

= pIE[|XN| XNc0

p2d]

= p

p 1 IE[|XN|(XN c)p1]

pp 1 (IE|XN|

p)1p (IE((XN c)p

(p1))1

p

with 1 = 1p

+ 1p

. Because ofp(p 1) =p this implies that

IE((XN c)p)11p p

p

1(IE|XN|p)

1p .

By c we conclude the proof.

Now we reprove Lemma 2.2.11.

Corollary 3.7.2 (Inequality ofKolmogorov) Let1, 2,...: IR bea sequence of independent mean zero and quadratic integrable random vari-ables. IfMn:=1+ + n forn1 andM0 := 0, then one has that

IP (Mn ) IEM2n

2

for all >0.

Proof. Applying Proposition 3.3.3 we get that (M2n)n=0 is a sub-martingale.

Hence we may apply Proposition 3.7.1 and get that

IP

(Mn)2 IEM2n

or2IP (Mn)IEM2n.


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Corollary 3.7.3 (Exit probability of a random walk) Let 1, 2, . . . : IR be independent Bernoulli random variables, M0 : 0, andMn=1+ . . . + n forn= 1, 2,...Then, for

0 andn= 1, 2,...,

IP

1

nMn

2e

2

2 .

Proof. We can assume >0. We define the stochastic exponential

Zn:=eMn

2

2 n

where , > 0 are fixed later. The process Z = (Zn)n=0 is a positive

martingale if 12

(e +e) = e2

2 under the natural filtration (

Fn)n=0 given

byF0 ={, } andFn=(1, . . . , n) for n= 1, 2,...In fact, we haveIE(Zn+1| Fn) =Zn

if and only if

IE(en+12

2 ) = 1,

or12

(e + e) =e2

2 .

Hence we are in a position to apply Doobs maximal inequality and obtain

IP sup0kn

Mk = IPe sup0knMk22 n e22 n IP

e

sup0kn

Mk

2

2 k

e

2

2 n

= IP

Zne

2

2 n

e+

2

2 nIEZn

= e+2

2 n

and

IP

1n

sup0kn

Mk

en+

2

2 n =e

n

e + e

2

ne

ne

2

2 n =e

2

2 nn,

where the inequalitye+e 2e22 can be verified by a Taylor9 -expansionof both sides and the comparison of the expansions. Finally, we have to find

9Brook Taylor, 18/08/1685 (Edmonton, England) - 29/12/1731 (Somerset House, Eng-

land), calculus of finite differences, integration by parts, discovered the formula knownas Taylors expansion.


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3.8. UNIFORMLY INTEGRABLE MARTINGALES 67

the minimum of

f() := 2

2n n.

Since f() =n n, we get 0 = n and

IP

1n

sup0kn

Mk

e 2

22 =e2

2 .

We finish the proof by

IP

Mnn

IP

1n

sup0kn

Mk

+IP

1n

sup0kn

(Mk)

2e2

2 .

3.8 Uniformly integrable martingales

In this section we consider uniformly integrable martingales. Let us recallthe definition ofuniformly integrable.

Definition 3.8.1 A family (fi)iIof random variables fi : IR is calleduniformly integrableprovided that for all >0 there exists a c >0 such that

supiI

|fi|c

|fi|dIP.

An equivalent condition to the above one is

limc

supiI

|fi|c

|fi|dIP = 0.

Moreover, we need the following criteria for uniform integrability:

Lemma 3.8.2 LetG: [0, )[0, ) be non-negative and increasing suchthat

limt

G(t)

t =

and(fi)iIbe a family of random variablesfi: IR such that

supiI

IEG(|fi|)


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Proof. We let >0 and M:= supiIIEG(|fi|) and find a c >0 such thatM

G(t)

t

for tc. Then |fi|c

|fi|dIP M

|fi|c

G(|fi|)dIP.

Corollary 3.8.3 Let p (1, ) and X = (Xn)n=0 be a martingale or anon-negative sub-martingale with supn=0,1,... XnLp


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Proposition 3.8.6 Let(, F, IP, (Fn)n=0) be a stochastic basis with

F= n=0

Fn .andM= (Mn)

n=0 be a martingale. Then the following assertions are equiv-

alent:

(i) The sequence(Mn)n=0 is a Cauchy-sequence inL1, that means for all

>0 there exists ann00 such that for allm, nn0 one has thatMm MnL1 < .

(ii) There exists an Z L1 such that limn Mn = Z inL1 that meanslimn Mn ZL1 = 0.(iii) There exists anM L1 such thatMn= IE(M| Fn) a.s. forn0.(iv) The familyM= (Mn)n0 is uniformly integrable.

If these equivalent conditions hold, then

(a) Z=M a.s.,

(b) M= limn Mn a.s.

(c) IfsupnIE|Mn|p


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Since limm

|Z Mm|dIP = 0 we conclude that IE|Mn IE(Z| Fn)| = 0

and Mn = IE(Z| Fn) a.s.(iii)=

(iv) For c >0 we get that

{|Mn|>c}|Mn|dIP =

{|Mn|>c}

|IE(M| Fn)|dIP

{|Mn|>c}

IE(|M| | Fn)dIP

=

{|Mn|>c}

|M|dIP

{supm |Mm|>c}

|M|dIP.

Since for c >0 Doobs inequality gives

IP

supm0

|Mm|> c

= supN1

IP

sup

0mN|Mm|> c

sup

N1

1

cIE|MN|

= supN1

1

cIE|IE(M|FN)|

supN1

1

cIEIE(|M||FN)

= 1c

IE|M| 0as c , we are done.

Now we interrupt the proof to prepare the implication (iv) (ii). Weneed the notion ofdown-crossings of a function: let I IR, f : I IR {, }, < a < b b},s2 : = inf

{ti> s1 : f(ti)< a

},

...

s2n+1 : = inf{ti> s2n : f(ti)> b},s2n+2 : = inf{ti> s2n+1 : f(ti)< a},

where we use inf:= td, andD0(f,F, [a, b]) := #{n : s2n< td}.

Definition 3.8.7

D(f , I, [a, b]) := sup

{D0(f,F, [a, b]) : F

I finite

}is the number of down-crossingsoffover [a, b].


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Proposition 3.8.8 Let X = (Xn)n=0 be a sub-martingale and let I

{0, 1, 2,...} be non-empty. Then one has for all< a < b b and on A2n we have that Xs2n < a.Hence

(b a)IP(A2n)A2n1\A2n

(Xs2n b)dIP.

Since s2n=N on Ac2n this implies that

(b a)IP(A2n) A2n1\A2n

(XN b)dIP.


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Summing over none gets

(b a)

n=1 IP(D0(X,F, [a, b]) n) = (b a)

n=1 IP(A2n) (XN b)+dIP.

Corollary 3.8.9 Let(Xn)n=0 be a sub-martingale such that

supn

IEX+n


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Proposition 3.8.10 Let(fn)n=0 L1(, F, IP) be uniformly integrable and

f : IR be a random variable such thatf = limn fn a.s. Then onehas that

(i) fL1(, F, IP),(ii) limnIE|f fn|= 0.

Proof. (i) Let >0 and find c >0 such that|fn|c

|fn|dIP

for n = 0, 1, 2,... Then IE|fn| c+ and supnIE|fn|


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ifA, B A with AB , then B\ A A,

ifA1

A2

and A1, A2,...

A, then

n=1 An

A.

From the implication (ii)(iii) we get thatIE(M|Fn) =Mn= IE(Z|Fn) a.s.

so that B:= n=1 Fn A. Since Bis a-system (closed under intersections)which generates F, the--Theorem implies(B) A. Hence A=FwhichgivesZ=M a.s.

(b) From Corollary 3.8.9 we know that limn Mn exists almost surely. Atsame time we know that there is a subsequence such that limkMnk =Za.s.Hence limn Mn = Za.s. Applying (a) gives the assertion of (b).

(c) ByDoobs maximal inequality we have

IE(M)p

p

p 1p

supn

IE|Mn|p


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3.9. APPLICATIONS 75

3.9.2 The branching process example

The key-point is that we use Corollary 3.8.9 to get a limit of our re-normalized

branching process.

Proof of Proposition 1.2.1: Corollary 3.8.9 says that there exists a limitM = limn Mn a.s. Let

n:= IP(Mn= 0)

for n= 0, 1,... Since by construction Mn() = 0 implies that Mn+1() = 0,we get a bounded increasing sequence

0 =0< q0=123 1with the limit := limn n. We shall investigate the value ofby a functionalmethod. For that purpose we let the generating function offn, n= 0, 1,...,be

n() :=k=0

kIP(fn=k) = IEfn, [0, 1],

and, in particular,

() := 1() =N

k=0

qkk

the generating function of the off-spring distribution. Then

n+1= n 1 = 1 1 1since

n+1() = IEfn+1

=k=0

IP(fn=k)IEX

(n+1)1 ++X(n+1)k

=

k=0 IP(fn=k)IEX(n+1)1

k

= n

IEX

(n+1)1

= n 1()

with the convention that 00 := 1 and empty sums are treated as zero. Weget that

(i) 0< q0=(0)< (1) = 1,

(ii) n+1 = (n),

(iii) = (1),


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(iv) () =,

Assertion (i) is clear by our assumptions. Part (ii) follows from n = n(0)

andn+1= n+1(0) = (n(0)) =(n).

To check (iii) we see that

() =N

k=1

qkkk1 and =

Nk=1

qkk= (1).

Finally, (iv) follows from (ii), limn n =, and the continuity of. Now weconsider the different cases for the expected number of sons .

(a) 0 <

1: The only fix-point of is = 1 so that = 1. Hence

limnIP(fn= 0) = 1 and

IP(M(, )) = limn

IP(Mn(, ))limn

IP(fn= 0) = 1

asMn= fnn

. Since this is true for all >0 we conclude that IP(M = 0) = 1.

(b) >1: Here we have exactly one fix-point 0 in (0, 1). Moreover, bydrawing a picture we check that

= limn

n = 0(0, 1).

The equality IEM= 1 we do not prove here. To understand the different cases better, here are two figures:

Figure 3.1: On the left 0 < 1, on the right >1 and 0= 0.25.

Remark 3.9.1 In dependence on the parameter one distinguishes the fol-lowing cases for our branching process:

0< 1 : supercritical case


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3.9.3 The Theorem of Radon-Nikodym.

The following proof is according to P.-A. Meyer [4] (see also D. Williams

[7]). Assume a measurable space (, F) and two finite measures IP andon(, F). We recall that is absolutely continuous with respect to IP providedthat IP(A) = 0 implies that (A) = 0. The aim of this section is to provethe following fundamental theorem by martingale methods.

Proposition 3.9.2 (Theorem of Radon-Nikodym) Assume a probabil-ity space(, F, IP) and a finite measure on(, F). If is absolutely con-tinuous with respect to IP, then there exists a non-negative random variableL L1(, F, IP) such that

(A) = A LdIP for all A F.First we show that our assumption implies some quantitative uniformity.

Lemma 3.9.3 Assume a probability space(, F, IP) and a finite measureon(, F)such that is absolutely continuous with respect to IP. Then, given >0 there is some(0, 1) such thatIP(A)implies that(A).

Proof. Assume the contrary, that means that there is an 0 > 0 such that

for all(0, 1) there is a set A() such thatIP(A()) but (A())> 0.

In particular, we get a sequence Bn := A(2n) for n = 1, 2, 3,... From the

Lemma ofFatou we know that

0lim supn

(Bn)

n=1

k=n

Bk

.

On the other hand, for all n01,

IP

n=1

k=n

Bk

IP

k=n0

Bk

k=n0

IP(Bk) 22n0

so that

IP

n=1

k=n

Bk

= 0 but

n=1

k=n

Bk

0

which is a contradiction to the absolute continuity of with respect to IP.

Proofof Proposition 3.9.2. (a) First we assume a filtration (Fn)n=0 suchthat


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Fn=

A(n)1 ,...,A

(n)Ln

,

the A(n)

1 ,...,A

(n)

Ln are pair-wise disjoint andLnl=1 A(n)l = , every A(n)l is a union of elements from

A(n+1)1 ,...,A

(n+1)Ln+1

,

F=

A(n)l :n = 0, 1,...and l= 1,...,Ln

.

We define the random variables Mn : IR by

Mn() :=

(A(n)l

)

IP(A(n)l )

: IP(A(n)l )> 0

1 : IP(A(n)l ) = 0

whenever A(n)l . The process M= (Mn)n=0 is a martingale with respectto the filtration (Fn)n=0. Clearly IE|Mn| 0. By assumption we can express

A(n)l as disjoint union

A(n)l = mI

A(n+1)m

for some index set I {1,...,Ln+1}. HenceA(n)l

Mn+1dIP =mI

A(n+1)m

Mn+1dIP

=mI

IP(A(n+1)m )(A(n+1)m )

IP(A(n+1)m )

= mI(A(n+1)m )= (A

(n)l )

=

A(n)l

MndIP

with the convention that 0/0 := 1 (note that our assumption of absolute con-

tinuity is used for the equality IP(A(n+1)m )((A

(n+1)m )/IP(A

(n+1)m )) =(A

(n+1)m )).

In particular, the above computation shows that Mn is the density of withrespect to IP onFn, i.e.

(A) = A

MndIP for all A Fn.


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Next we show that the martingaleM= (Mn)n=0 is uniformly integrable. We

know that, for c >0, one has that

IP(Mnc) IEMnc

= ()c

.

Applying Lemma 3.9.3, for any > 0 we find a c > 0 such that IP(A)()/cimplies that (A). Hence

MncMndIP =(Mnc)

and the uniform integrability is proved. Applying Proposition 3.8.6 we havea limit L

L1(,

F, IP) such that

Mn= IE (L|Fn) a.s.

The random variableL is the density we were looking for: defining a measureonF by

(A) :=

A

LdIP

we get = on

n=0 Fn which is a generating algebra ofF. Hence = onFaccording to Caratheodory10s extension theorem.

(b) Next we assume the general case. Let

Abe the collection of all sub-

-algebrasG ofFsuch thatG =(B1,...,BL)

for L = 1, 2,...andBl F. As we saw in step (a) for everyG Athere is anon-negative LG L1(, G, IP) such that

d

dIP|G =LG.

Fact 3.9.4 For all >0 there is an

G

Asuch that

IE|LG1 LG2|<

for allG1 G andG2 G.

Proof. Assuming the contrary gives an 0 > 0 and a sequenceG1 G2G3 fromAsuch that

IE|LGn+1 LGn| 010Constantin Caratheodory, 13/09/1873 (Berlin, Germany)- 2/02/1950 (Munich, Ger-

many), significant contributions to the calculus of variations, the theory of measure, andthe theory of functions of a real variable.


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forn = 1, 2,...However, using the argument from step (a) says that (LGn)n=1

should be a uniformly integrable martingale with respect to the filtration(Gn)n=1

which yields to a contradiction so that the fact is proved.

Now we define our candidate for the density L: We takeG(n) A suchthat

IE|LG1 LG2|< 1

2n+1

for allG1 G(n) andG2 G(n). LettingHn := (G(1), ...,G(n)) we obtain amartingale (LHn)

n=1 with respect to the filtration (Hn)n=1 which converges

inL1 and almost surely to a limit L. Now, for B Fwe get

(B) B LdIP = B L(Hn,B)dIP B LdIP IE L(Hn,B) L L(Hn,B) LHnL1+ LHn LL1

1

2n+1+ LHn LL1n0

so that (B) =B

LdIP.

3.9.4 On a theorem ofKakutani

Martingales that are generated by products of independent random vari-ables play an important role in stochastic modeling and in understanding

equivalent probabilities on a given measurable space. Let us start with theKakutanis theorem

Proposition 3.9.5 Let X0, X1, X2,... : IR be independent and non-negative random variables withX01 andIEXn = 1. Define

Mn:=X0X1X2 XnandFn:=(X0,...,Xn) forn= 0, 1, 2,...

(i) Then M = (Mn)n=0 is a non-negative martingale with respect to

(Fn)n=0 so thatM := lim Mn exists almost surely.(ii) The following assertions are equivalent:

(a)

n=0(1 IE

Xn) 0.

(c) IEM


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Before we prove the proposition we consider a simple example.

Example 3.9.6 Let 1, 2,...:

IR be independent Bernoulli random

variables, i.e. IP(n =1) = IP(n = 1) = 1/2. Define M0:= 1 and

Mn:=e1++ncn such that IEe1c = 1 with c= log

e + 1

e

2

.

Then Xn = enc for n1 and

n=1

(1 IEe nc2 ) =

because we have that IEenc

2 = IEe1c2 0 (note thatn= 0 would imply thatXn = 0 a.s. so that IEXn= 1would not be possible).

(a)(b) is a simple fact from analysis.(b)= (c) Defining

Nn:=

X0

0

Xn

n=

Mn

0 nwe get again a martingale N= (Nn)

n=0. Since

IEN2n = 1

20

2n

k=01

2k


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almost surely, so that IP(M = 0) = 1. This proves (e)(b) and at sametime item (iii).

Now let = IRIN the space of all real valued sequences = (n)n=1IR

and

Yn: IR given by Yn() :=n.As filtration we useF0 :={, } and

Fn:=(Y1,...,Yn).

Finally, we letF :=n=0Fn. Assume Borel-measurable fn, gn : IR IRsuch that fn(x)> 0 and gn(x)> 0 for all xIR and such that

IR

fn(x)dx=

IR

gn(x)dx= 1

for all n = 1, 2,... which means that fn and gn are densities of probabilitymeasures n and n, respectively, on IR. Define the measures

IP :=n=1n and Q:=n=1non the measurable space (, F) and

Xn() := gn(Yn())fn(Yn())= gn(n)fn(n)

.

The random variablesXn form aLikelihood-Ratio Test. We get the following

Proposition 3.9.7 The following assertions are equivalent:

(i) The measureQ is absolutely continuous with respect to IP, that meansthatIP(A) = 0 implies thatQ(A) = 0.

(ii)

n=1IEIP

Xn=

n=1 IR

fn(x)gn(x)dx >0.

(iii)n=1 IR(fn(x) gn(x))2dx


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for n1. Moreover,

IEIPXn = IR gn(x)fn(x) fn(x)dx= IR gn(x)fn(x)dxand

n=0

(1 IEIP

Xn) =

n=1

1

IR

gn(x)

fn(x)fn(x)dx

=

n=1

1

IR

fn(x)gn(x)dx

= 1

2

n=1 IR(fn(x) gn(x))2dx.Applying Kakutani11 s theorem Proposition 3.9.5 gives that conditions(ii) and (iii) are equivalent to the uniform integrability of the martingaleM= (Mn)

n=0 given by

Mn:= X0 Xn.Moreover, we have that

Mn= dQ

dIP|Fn

since

nk=1k(B) = B

g1(x1) gn(xn)dx1...dxn

=

B

g1(x1) gn(xn)f1(x1) fn(xn)d(

nk=1k)(x).

(ii) = (i) Here we use the proof of the Radon-Nikodym theorem wherewe exploit that Mis uniformly integrable.

(i)=(ii) Here we get a density L = dQdIP

and

IE(L|Fn) =Mn a.s.

But then Proposition 3.8.6 gives the uniform integrability ofM.Finally, ifQ is not absolutely continuous with respect to IP, then M isnot uniformly integrable and Kakutanis theorem says that in this caseMn 0 a.s. The other way round: ifMn 0 a.s. but, IEMn = 1 for alln = 1, 2,..., then (Mn)

n=0 cannot be uniformly integrable and Q cannot be

absolutely continuous with respect to IP.

Looking at the symmetry in conditions (ii) and (iii) of Proposition 3.9.7with respect to the densities fn and gn one gets immediately the following

11Shizuo Kakutani, 28/08/1911 (Osaka, Japan)- 17/08/2004 (New Haven, USA), con-tributed to several areas of mathematics: complex analysis, topological groups, fixed point

theorems, Banach spaces and Hilbert spaces, Markov processes, measure theory, flows,Brownian motion, and ergodic theory.


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