solutions probability

8/3/2019 Solutions Probability

1/12

Bocconi University

PhD in Economics and Finance

2010-2011

ProbabilitySandra Fortini, Caterina May

SOLUTIONS

1)

a) An is an infinite sequence of heads (H) and tails (T) having one tail at (n 1)-th toss, n heads fromn-th toss to (2n 1)-th and again one tail at 2n-th toss. IfP(T) = P(H) = 1/2, then P(A1) = (1/2)2and, for any n 2, P(An) = (1/2)n+2.

b) The event An has a H at n-th toss, while An+1 has a T at n-th toss. Hence, An and An+1 are disjoint,and it follows that the sequence An cant be increasing (An An+1).

c) For the same reason then in (b), An is not decreasing (An An+1).

d) Since An and An+1 are disjoint (for any n), then: lim infAn = (An,ult.) =m

nmAn = .

On the other site, lim sup An = (An, i.o.) =m

nmAn is non empty; for instance, the sequence

having a T at (22n 1)-th toss, all H from 22n-th toss to (22n+1 1)-th toss and a T at 22n+1-th tossis contained in lim sup An. It follows that lim infAn = lim sup An, and hence An is not convergent.

e) Sincen P(An) < , then, from Borel-Cantelli Lemma, P(An, i.o.) = 0.

f) P(An,ult.) = 0, because (An,ult.) = .

2)

a) Tn is not measurable with respect to Gn, for instance: the event (T1 = 3) is the event (X1 = 6, X2 =6, X3 = 6), and then it is not contained in (X1).

b) By definition, Tj is a stopping time with respect to Gn if and only if (Tj = n) Gn for any n.We prove it by induction on j.For j = 1 this is true, in fact:{T1 = 1} = {X1 = 6} G1, and, for any n > 1:{T1 = n} = {X1 = 6, . . . , X n1 = 6, Xn = 6} Gn.Suppose that (Tj = n) Gn for any fixed j and let us prove that (Tj+1 = n) Gn:{Tj+1 = n} =

i=1,...,n1{Tj = i, Xi+1 = 6, . . . , X n1 = 6, Xn = 6} Gn.

c) TX1 is not

G1-measurable; for instance, if X1 = 1 then TX1 can assume all values 1, 2, 3,... and this

implies that it doesnt exist any measurable function g such that TX1 = g(X1).

d) TX1 is a stopping time with respect to Gn, in fact:{TX1 = n} =

j=1,...,6{X1 = i, Tj = n} and this belongs to Gn from (b).

e) We can write Y =T2i=1 Xi. Y is not FT2-measurable because, for instance, if T2 = 3 then Y can

assume all values between 13 and 17; this implies that there is not any measurable function f suchthat Y = f(T2).

1


2/12

f) Y is measurable with respect to GT2 since we can show that the event {Y = y} GT2 for any y.Let us write:

{Y = y

}= m{

Y = y, T2 = m

}= m{

mi=1 Xi = y, T2 = m

};

we have that each event {mi=1 Xi = y, T2 = m} of the countable union belongs to GT2 ;in fact, by calling Am = {

mi=1 Xi = y, T2 = m},

Am{T2 = n} is the null set ( Gn) when n = m, and Am{T2 = n} is the event {ni=1 Xi = y, T2 =

n} Gn when n = m.

3)

a) FY1 FZ and FZ FY1 . In fact: suppose, for instance, that X1, X2, X3 take values in {1, 0, 1}; inthis case, if Z = 1 then Y1 can be either 1 or 1. On the other side, if Y1 = 1 then Z can be 1, 1 or0. This implies that there is not any function g such that Y1 = g(Z) and there is not any function hsuch that Z = h(Y1).

b)

FY1,Y2,Y3

FX1,X2,X3 since (Y1, Y2, Y3) = (X1X2, X2X3, X3X1), which is a measurable function of

(X1, X2, X3).FX1,X2,X3 FY1,Y2,Y3 ; in fact, as in (a), if for instance X1, X2, X3 take values in {1, 0, 1}, then when(Y1, Y2, Y3) = (1, 1, 1) we can have (X1, X2, X3) equal to (1, 1, 1) or (1, 1, 1).

c) FY1,Y2,Z FX1,X2,X3 , since (Y1, Y2, Z) = (X1X2, X2X3, X1X2X3), which is a measurable function of(X1, X2, X3).Also FX1,X2,X3 FY1,Y2,Z, since (X1, X2, X3) = (Z/Y2, Y1Y2/Z, Z/Y1), which is a measurable functionof (Y1, Y2, Z).

d) FW,Z FX1,X2,X3 , since (W, Z) = (X1 + X2 + X3, X1X2X3), which is a measurable function of(X1, X2, X3).FX1,X2,X3 FW,Z; in fact, if for instance X1, X2, X3 take values in {1, 0, 1}, then when (W, Z) =(1, 1) we can have (X1, X2, X3) equal to (1, 1, 1) or (1, 1, 1) etc.

e) FX1,W,Z FX1,X2,X3 , since (X1, W, Z) = (X1, X1+X2+X3, X1X2X3), which is a measurable functionof (X1, X2, X3).FX1,X2,X3 FX1,W,Z ; in fact if for instance X1, X2, X3 take values in {1, 0, 1}, then when (X1, W, Z) =(1, 1, 1) we can have (X1, X2, X3) equal to (1, 1, 1) or (1, 1, 1).

4)

a) The moment generation of X is MX(s) = E(esX) =k=0 eskke

k!= e

k=0

(es)k

k!;

by indicating = es, we have:

k=0

(es)k

k!=

k=0

k

k!= e = ee

s

;

we can conclude that MX(s) = e ees = e(es1).b) Since MX

(k)(0) = E(Xk), in particular we have E(X2) = MX(2)(0). Since, calculating from (a),

MX(2)(s) = e(e

s1)es(1 + es),

we obtain MX(2)(0) = (1 + ), and then E(X2) = (1 + ).

2


3/12

c) By definition, M(X,Y)(s, t) = E(esX+tY); since X and Y are independent, E(esX+tY) = E(esX)E(etY),

and, from (a), this is equal to

e(es1)e(e

t1);

we can conclude that M(X,Y)(s, t) = ees+et.

d) MX+Y(s) = E(es(X+Y)) = E(esX+sY); from (c), this is equal to ees+es = e(+)(e

s1).

e) IfZ is a Poisson random variable with parameter +, from (a) we have that Z has moment generatingfunction MZ(s) = e

(+)(es1); on the other side, we have from (d) that also X + Y has momentgenerating function equal to e(+)(e

s1). It follows that Z and X+ Y have the same distribution.

5)

a)

MX(s) = E(esX) =+

esx 12

e(x

)2

22 dx =+

12

e1

22 (x2

+2

2x22

sx)dx

= e2s2/2+s

+

12

e

1

22(x(+2s))2

dx,

where the last equality follows from: (x2 + 2 2x 22sx) = (x ( + 2s))2 (2s)2 22s.Since we are integrating between and + the density of a Gaussian random variable with para-meters ( + 2s) and 2 (and this is equal to one), we obtain:

MX(s) = e2s2/2+s.

b) MY(s) = E(esY) = E(esaX+sb) = esb

E(esaX) = esb+

2a2s2/2+sa = e(2a2)s2/2+(b+a)s.

From (a), it follows that Y is a Gaussian random variable with parameters (b + a) and 2a2.

c) Z has moment generating function MZ(s) = E(esZ) = E(esX

) = E(es/Xs/), and this is,from (a), equal to: es

2/2+s/s/ = es2/2.

By the Taylor expansion of the exponential function we find: MZ(s) = es2/2 =

k=0

s2k

2kk!;

on the other side, MZ(s) = E(esZ) = E(k=0

skZk

k!) =

k=0

skE(Zk)k!

;

it follows that

- if k is odd, then E(Zk) = 0;

- if k is even, thenE(Zk)

k!=E(Z2m)

(2m)!=

1

m!2m, and hence

E(Z2m) =

(2m)!

m!2m.

d) E(Ws) = E((eX)s) = E(esX) = MX(s); it follows that:

E(W) = MX(1) = e2/2+ (from (a));

E(W2) = MX(2) = e22+2 (similarly),

and V ar(W) = E(W2) E(W)2 = e2+2(e2 1).

3


4/12

6)

a) Since, for k, > 0, the density function of X

Gamma(, k) is f(x; k, ) = xk1kex

(k)I(0,)

(x),

then its moment generating is

MX(s) = E(esX) =

0

esxxk1kex

(k)dx

=k

( s)k0

xk1( s)k e(s)x

(k)dx

=

sk

0

f(x; k, s) dx =

sk

,

(1)

where f(x; k, s) is a density function for s < and then the last equality holds for s < ; hence,for k = 1, 2, 3,

MXk(s) = 2

2 sk

, for s < 2.

b) Since, for a R+, MaX(s) = MX(as) (in its domain), then, by (1), if X Gamma(, k)

MaX(s) = MX(as) =

ask

= aa s

k= MY(s) (2)

where Y Gamma(a , k). Hence, aXk Gamma(2/a,k).c) From (1), if Xk Gamma(, k) for k = 1, 2, 3 and all Xk are independent, then

MX1+X2+X3(s) =

3k=1

MXk(s) =

31

s

k=

s

1+2+3=

s

6,

then X1 + X2 + X3 Gamma(, 6). Thus, with = 6, X1 + X2 + X3 Gamma(2, 6).d) From (c), X1 + X2 + X3 Gamma(2, 6). By (2), (X1 + X2 + X3)/3 Gamma(6, 6).e) Using the density of X Gamma(, k) in (a), we have

E(X2) =

0

x2xk1kex

(k)dx

=(k + 2)

(k)2

0

x(k+2)1k+2ex

(k + 2)dx

=(k + 2)

(k)2

0

f(x; k + 2, ) dx =(k + 2)

(k)2=

k(k + 1)

2.

(The same result can be gained by using the m.g.f. MX(s) obtained in (a), and calculating

E(X2) = MX(0)).

From (c), Y = (X1 + X2 + X3) Gamma(2, 6), and hence E(Y2) = 674 = 424 .

4


5/12

7)

a) E(Yn|Ym) = E(ni=1 Xi|

mi=1 Xi) = E(

mi=1 Xi+

ni=m+1 Xi|

mi=1 Xi) =

mi=1 Xi+E(

ni=m+1 Xi),

where the last equality follows from the linearity of conditional expectation, the measurability ofmi=1 Xi with respect to (

mi=1 Xi), and the independence between Xm+1,...,Xn and X1,...,Xm.

Since E(ni=m+1 Xi) =

ni=m+1 E(Xi) = 0, we can conclude that E(Yn|Ym) = Ym.

b) By symmetry, we know that E(X1|Yn) = E(X2|Yn) = = E(Xn|Yn); moreover,E(nj=1 Xj|Yn) = E(Yn|Yn) = Yn; it follows that, for any j n,

E(Xj |Yn) = 1n

ni=j

E(Xi|Yn) = 1n

Yn.

c) E(Ym|Yn) = E(m

i=1 Xi|Yn) = m

i=1 E(Xi|Yn) =m

n Yn,

where the last equality follows from (b).

d) E(Sn2|Sm2) = E(

ni=1 Xi

2

n|Sm2) = E(

mi=1 Xi

2

n+

ni=m+1 Xi

2

n|Sm2)

= E(m

nSm

2|Sm2) + E( 1n

ni=m+1 Xi

2|Sm2) = mn

Sm2 +

1

n

ni=m+1 E(Xi

2),

where the the last equality follows from the independence between Xm+1,...,Xn and X1,...,Xm.

We can conclude that E(Sn2|Sm2) = m

nSm

2 +n m

n2.

e) Since, by symmetry, E(X1|Yn, Sn2) = E(X2|Yn, Sn2) = = E(Xn|Yn, Sn2), it follows thatE(Xj|Yn, Sn2) = 1

n

ni=j E(Xi|Yn, Sn2) =

1

nE(Yn|Yn, Sn2) = 1

nYn.

f) Let us compute: E(N SN2|N) = E(Ni=1 Xi2|N) = NE(Xi2) = N 2, where the last but one equality

follows from the independence between N and the Xis.Since E(N SN

2) = E(E(N SN2|N)), we obtain that E(NSN2) = E(N)2 = 2.

8)

a) Given Bn and Yn, we have that Bn+1 = Bn + 1 with probability Yn, and Bn+1 = Bn with probability(1 Yn); it follows that

E(Bn+1|Bn, Yn) = Bn + Yn 1 + (1 Yn) 0 = Bn + Yn.

b) E(Bn+1|Bn) = E(E(Bn+1|Bn, Yn)|Bn) = E(Bn + Yn|Bn), from (a),

= E(Bn + Bn/(r + b + n)|Bn) = r + b + n + 1r + b + n

Bn.

c) From (b) we have that E(Bn+1|Bn) = r + b + n + 1r + b + n

Bn > Bn, and hence Bn is a submartingale.

5


6/12

d) By definition, E(Yn+1|Yn) = E( Bn+1r + b + n + 1

| Bnr + b + n

) =1

r + b + n + 1E(Bn+1|Bn);

from (c), this is equal toBn

r + b + n= Yn, and hence Yn is a martingale.

e) E(Yn) = E(Y0) because Yn is a martingale, and moreover E(Y0) = Y0 =b

r + b.

f) E(Bn) = E((r + b + n)Yn) = (r + b + n)b

r + b(from (e)).

9)

a) Gn Fn, since Yn = f(X1, . . . , X n). Fn Gn, consider for instance F2 and G2: if Y1 = 1, Y2 = 2, wecan have either X1 = 1, X2 = 1 or X1 = 1, X2 = 1.

b) E(Yn|Yn1) = E(XnI(Yn1=0) + nYn1|Xn||Yn1) = I(Yn1=0)E(Xn) + nYn1E(|Xn|) = Yn1.

c) Yn is a martingale with respect to Gn because (b) and because E(|Yn|) < .d) E(Yn|Fn1) = E(XnI(Yn1=0)+nYn1|Xn||(Xn1, ...X1)) = I(Yn1=0)E(Xn)+ nYn1E(|Xn|) = Yn1,

because Yn1 is Fn1-measurable and X1, X2, ...Xn,... are independent random variables. Hence Ynis a martingale with respect to Fn

e) By martingale properties, E(Yn) = E(Y1); E(Y1) = E(X1) = 0.

f) E(|Yn|) = E(|Xn|I(Yn1=0)+n|Yn1||Xn|) = E(|Xn|(I(Yn1=0)+n|Yn1|)) =1

nP(Yn1 = 0)+E(|Yn1|),

where the last equality follows from E(|Xn|) = 1/n and from the independence between Xn and Yn1.

Hence, E(

|Yn

|)

E(

|Yn1

|) =

P(Yn1 = 0)

n

1 1/(n 1)

n

=1

n

1

n(n 1),

and E(|Yn|) = E(|Y1|) +ni=2(E(|Yi|) E(|Yi1|)) 1 +

ni=1(

1

i 1

i(i 1) ),which is a divergent series. We conclude that E(|Yn|) is not uniformly bounded.

g) Following the calculations in (f), we can see that limn E(|Yn|) = +.

10)

a) E(X1) = 1/2; E(X2) = E(E(X2|X1)) = E(X1/2) = 1/4; E(X3) = E(E(X3|X1, X2)) = E(X2/2) = 1/8.b) P(X2 1/2) = E(P(X2 1/2|X1)) = E(I(X11/2)P(X2 1/2|X1)) + E(I(X1>1/2)P(X2 1/2|X1))

= P(X1

1/2) + E(I(X1>1/2)P(X2

1/2|X1)) = 1/2 + E(I(X1>1/2)

1

2X1)

=1

2+

11/2

1

2xdx =

1

2+

log2

2.

c) P(X1 1/2, X2 1/4) = E(P(X1 1/2, X2 1/4|X1)) = E(I(X11/2)P(X2 1/4|X1))= E(I(X11/4)P(X2 1/4|X1)) + E(I(1/4


7/12

d) P(X3 1/4|X1) = E(P(X3 1/4|X1)|X1, X2) = E(P(X3 1/4|X1, X2)|X1)= E(IX21/4P(X3 1/4|X1, X2)|X1) + E(IX2>1/4P(X3 1/4|X1, X2)|X1)

= E(IX21/4|X1) + E(IX2>1/41

4X2|X1)

= E(IX21/4IX11/4|X1)+E(IX21/4IX1>1/4|X1)+E(IX2>1/4IX11/41

4X2|X1)+E(IX2>1/4IX1>1/4

1

4X2|X1)

= IX11/4+IX1>1/4P(X2 1/4|X1)+1

4

X11/4

1

x2

1

X1dx2 = IX11/4+IX1>1/4

1

4X1+IX1>1/4

1

4X1log(4X1).

e) fX1|X2(t|x) =fX1,X2(t, x)

fX2(x), where:

fX1,X2(t, x) = fX2|X1(x|t)fX1(t) =1

tI[0,t](x)I[0,1](t),

and fX2(x) =

10

1

tI[0,t](x)I[0,1](t)dt = log(x)I[0,1](x),

and then fX1|X2(t|x) = 1

log x

1

tI[0,1](x)I[x,1](t).

Hence we have:

P(X1 x1|X2 = x) =1x1

fX1|X2(t|x)dt = I[x,1](x1)I[0,1](x)(1)log x

1x1

1

tdt+I[0,x)(x1)I[0,1](x)

(1)log x

1x

1

tdt

= I[x,1](x1)I[0,1](x)log x1log x

+ I[0,x)(x1)I[0,1](x)log x

log x,

and we can conclude that

P(X1 x1|X2) = I[0,1](X2)I[X2,1](x1)log x1log X2

+ I[0,1](X2)I[0,X2)(x1).

11)

a) = E(E(Xj|)) = E() = 1/2.V ar(Xj) = E(Xj

2) E(Xj)2 = E(E(Xj2|)) 1/4 = E() 1/4 = 1/2 1/4 = 1/4.b) Cov(Xi, Xj) = E((Xi 1/2)(Xj 1/2)) = E(XiXj 1/2 Xi 1/2 Xj + 1/4) = E(2) 1/4

=10

x2dx 1/4 = 1/12.c) Since Cov(Xi, Xj) = 0 from (b), they cant be stochastically independent.

d) P( p|X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1) = P( p, X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1)P(X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1)

=

p0

P(X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1| = s)ds10

P(X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1| = s)ds=

p0

s4(1 s)ds10

s4(1 s)ds=

p5/5 p6/61/30

;

hence,

f|X1=1,X2=0,X3=1,X4=1,X5=1(p) =d

dpP( p|X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1) = 30p4(1 p).

7


8/12

e) From (d) we have that: E(|X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1) =

=

1

0 pf|X1=1,X2=0,X3=1,X4=1,X5=1(p)dp = 30

1

0 p

5

(1 p)dp = 5/7.

f) From (d) we have that:

P( < 1/2|X1 = 1, X2 = 0, X3 = 1, X4 = 1, X5 = 1) = 301/20

p4(1 p)dp = 7/64.

12)

a) We know that, if pij(2) = P(Xn+2 = j|Xn = i) is the (i, j)-th element of the matrix P(2), then

P(2) = P2 =

1/2 1/4 1/41/4 1/2 1/4

1/4 1/4 1/2

b) The chain is not periodic since, from (a), the 2-step transition probabilities are all positive.

c) The chain is irreducible since, from (a), the 2-step transition probabilities are all positive.

d) By writing the equations: (1, 2, 3)P = (1, 2, 3) and 1 + 2 + 3 = 1, we find the unique:(1, 2, 3) = (1/3, 1/3, 1/3).

e) No because (1, 0, 0)P = (1, 0, 0). Alternatively, from (d), the unique stationary distribution is =(1/3, 1/3, 1/3) and therefore the initial distribution must be for the chain to be stationary.

f) From (b), (c) and (d) and the properties of stationary distributions, limn P(Xn = 1) = 1/3 if thechain starts in 1.

g) No because of the properties of stationary distributions, as in (f ).

13)

a) E(X22) = E(E(X2

2|X1)) = E(X1 + X12) = E(X1) + E(X12) < .b) E(X2|X1) = X1 and V(X2|X1) = X1, because X2|X1 Poisson(X1).c) E(X2) = E(E(X2|X1)) = E(X1) = 1.d) From theorem of decomposition of variance we have:

V(X2) = E(V(X2|X1)) + V(E(X2|X1)) = E(X1) + V(X1) = 1 + 1 = 2.e) ||X2 X1||2 = E[(X2 X1)2]1/2 = [E(X22) + E(X12) 2E(X1X2)]1/2

= [E(X22

) + E(X12

) 2E(E(X1X2|X1))]1/2

= [3 + 2 2E(X12

)]

1/2

= (3 + 2 4)1/2

= 1.

14)

a) Xt is an AR(1) process: Xt = c+Xt1+Ut. From the autocovariance function of the AR(1) processeswe obtain:

(k) = E(Ut2)

k

1 2 =0.5k

1 0.52 =0.5k

0.75.

8


9/12

b) By backward substitution we obtain:

Xt = 1 + 0.5Xt1 + Ut = 1 + 0.5(1 + 0.5Xt2 + Ut1) + Ut = 1 + 0.5 + 0.52Xt2 + 0.5Ut1 + Ut

= 1 + 0.5 + 0.52

(1 + 0.5Xt3 + Ut2) + 0.5Ut1 + Ut = ...... =

k1j=0 0.5

j + 0.5kXtk +k1j=0 0.5

jUtj , (by iterating k times),

which converges, as k , to: 2 +j=0 12j Utj . We have showed that Xt = 2 +j=0

1

2jUtj .

c) E(Xt|Xt1) = E(1 + 0.5Xt1 + Ut|Xt1) = 1 + 0.5Xt1.d) V(Xt|Xt1) = E((XtE(Xt|Xt1))2|Xt1) = E((1+0.5Xt1+Ut10.5Xt1)2|Xt1) = E(Ut2) = 1.e) By decomposing the variance we have:

V(Xt) = E(V(Xt|Xt1)) + V(E(Xt|Xt1)) = 1 + V(1 + 0.5Xt1) = 1 + 0.52V(Xt1) = 1 + 0.52

1 0.52 .

In fact: V(Xt) =1

1 0.52

= 1 +0.52

1 0.52

.

15)

a) Ut is integrable, it satisfies E(Ut|Ut1,...,U1) = E(Ut) = 0, hence it is a martingale difference. Bydefinition it is also a white noise, since it is a sequence of independent random variables with zeromean.

b) Xt is stationary: it is a M A(1) process: Xt = + Ut + Ut1 and its autocorrelation function is:(1) = /(1 + 2) = 0.2/1.04, (k) = 0 for any k > 1.

c) Yt = Xt Xt1 = Ut 0.8Ut1 0.2Ut2.It follows that: E(Yt) = 0, and, indicating by (k) the autocovariance function and remembering thatUt, Ut1, Ut2,... are independent, we get:

(0) = V(Yt) = V(Ut 0.8Ut1 0.2Ut2) = 1 + 0.82

+ 0.22

= 1.68;(1) = E((Ut 0.8Ut1 0.2Ut2)(Ut1 0.8Ut2 0.2Ut3)) = 0.8 + 0.2 0.8 = 0.64;(2) = E((Ut 0.8Ut1 0.2Ut2)(Ut2 0.8Ut3 0.2Ut4)) = 0.2;(k) = 0 for any k > 2.

d) Zt = XtUt = Ut+ 0.2UtUt1 + Ut2. It follows that: E(Zt) = 1, and, indicating by (k) the autocovari-

ance function we get:

(0) = V(Zt) = E(Ut + 0.2UtUt1 + Ut2)2 1

= E(Ut2 + 0.22Ut

2Ut12 + Ut

4 + 2 0.2Ut2Ut1 + 2Ut3 + 2 0.2Ut3Ut1) 1 = 1 + 0.22 + 4 1 = 4.04;(1) = Cov(Zt, Zt1) = E((Ut + 0.2UtUt1 + Ut

2 1)(Ut1 + 0.2Ut1Ut2 + Ut12 1)) = 0.e) Wt is a MA(1) as Xt (see (b)), and its autocorrelation function is:

(1) = /(1 + 2) = 5/26 = 0.2/1.04, (k) = 0 for any k > 1.

The autocovariance functions are not identical because V(Xt) = 0.22 + 1, while V(Wt) = 52 + 1.

16)

a) For any > 0, P(|Xn| > ) = 1n

0 as n . Hence, by definition, Xn converges to 0 in probability.

b) IfXn converges almost surely, then the limit must be 0 for (a). Since

n=0

P(Xn = 1) =n=0

1

n= ,

9


10/12

from Borel-Cantelli second lemma this implies that P(lim sup(Xn = 1)) = 1, and then Xn cantconverge almost surely to 0.

c) IfXn converges in L2

, then the limit must be 0 for (a).||Xn||2 = E(|Xn|2)1/2 = P(Xn = 1)1/2 = 1

n1/2 0 as n . Hence, by definition, Xn converges to

0 in L2.

d) Yn = nXn can not converge almost surely because Xn dont converge almost surely as proved in (b).

For any > 0, P(|Yn| > ) = 1n

0 as n . Hence, by definition, Yn converges to 0 in probability.If Yn converges also in L

2, then the limit must be 0:||Yn||2 = E(|Yn|2)1/2 = E(|nXn|2)1/2 = nP(Xn = 1)1/2 = n1/2 + as n . Hence, Yn cantconverge in L2.

e) P(Zn = 0) = P(X1 = 0,...,Xn = 0) =ni=1 P(Xi = 0) = (1

1

2)(1 1

3)...(1 1

n) =

1

nand then

P(Zn = 1) = 1 1

n ;

as in (a), for any > 0, P(|Zn 1| > ) = 1n

0 as n , and hence Zn converges to 1 inprobability.Similarly to the results for Xn in (c), Zn converges to 1 also in L

2; moreover, Zn is an increasingsequence, and then it converges also almost surely to 1.

f) P(Wn = 1) = P(X1 = 1,...,Xn = 1) =ni=1 P(Xi = 1) = 1

1

2 1

3 ... 1

nand

P(Wn = 0) = 1 1 12

13

... 1n

; it follows that, for any > 0,

P(|Wn| > ) = 1 12

13

... 1n

0, as n ,

and then Wn converges to 0 in probability.Since |Wn| |Xn|, we have that Wn converges to 0 also in L2 from (c).Finally we have

n=0

P(Wn = 1) =n=0

1 12

13

... 1n

=n=0

1

n!< ,

and then, from Borel-Cantelli first lemma, P(lim supXn = 1) = 0; hence Xn converge also almostsurely to 0.

17)

a) MTn(s) = E[exp(sTn)] = E[exp(snj=1 Xnj]) =

nj=1 E[exp(sXnj]) = E[exp(sXn1)]

n

= (1

2pn +pn exp(s) +pn exp(

s))n = (1 +pn(exp(s) + exp(

s)

2))n.

b) In this case, MTn(s) = (1+1

n2(exp(s)+exp(s)2))n = and this converges to 1 as n (remember

that:(1 +

c

n2)n

2/c exp(1) as n , and then (1 + cn2

)n = [(1 +c

n2)n

2/c]c/n exp(1)0 = 1 as n ).This implies that Tn converges to 0 in distribution, and then it converges to 0 also in probability.

c) In this case, MTn(s) = (1 +1

n(exp(s)+exp(s) 2))n exp(es+ es 2) = exp(es 1)exp(es 1),

which is equal to MXY(s) = E[exp(s(X Y)] = E[exp(sX)]E[exp(sY)], where X and Y are twoindependent Poisson(1).

10


11/12

d) M|Tn|(s) = (1 +1

n(2 exp(s) 2))n exp(2(es 1)) as n , which is the m.g.f. of a Poisson(2).

Thus,

|Tn

|converges in distribution to a Poisson(2).

e) Since E(Xnj) = 0 and V(Xnj) = 2p, from Levy Central Limit TheoremTn2np

converges in distribution

to a standard normal distribution; since g(x) = x2 is a continuous function, it follows thatTn

2

2np=

(Tn2np

)2 converges in distribution to a chi-square distribution with one degree of freedom.

18)

a) Since, for k, > 0, the density function of X Gamma(, k) is f(x; k, ) = xk1k ex

(k)I(0,+)(x),

then its moment generating function is, for s < ,

MX(s) = E(esX) =

0

esxxk1kex

(k)dx

=k

( s)k0

xk1( s)k e(s)x

(k)dx

=

sk

0

f(x; k, s) dx =

sk

;

(3)

hence if Xk Gamma(1, k), we get MXk(s) = (1 s)k for s < 1.b) Since Ma(X+b)(s) = e

absMX(as) (for as in the domain of MX), we get

MYk(s) = esk

1 skk

.

c) From (b), and by using the Taylor expansion of the logarithm,

MYk(s) = exp

s

k k log 1 sk

= exp

s

k k s

k s

2

2k+ o(1/k)

= exps2

2+ o(1)

exp

s22

as k ,

which is the m.g.f. of a standard normal distribution; this shows that the limit distribution of Yk is astandard gaussian distribution.

d) Since g(x) = x2 is a continuous function, then Y2k converges to the square of the limit ofYk (ContinuousMapping Theorem) . By (c), Y2k converges to

21 (a chisquare distribution with 1 degree of freedom).

e) Y1, Y2, . . . are independent, since X1, X2, . . . are. Then, from (d),

MY2k+Yk+12+Y2k+2

(s) = MY2k

(s) MYk+12(s) MY2k+2(s) k(M21(s))3.

Note that 2m (a chisquare distribution with m degrees of freedom) is the sum of m independent 21,

which means that M2m(s) = (M21(s))m. We get

M23

(s) = (M21

(s))3 = limk

MY2k+Yk+12+Yk+22(s),

and hence the limit of Yk2 + Yk+1

2 + Yk+22 is a 23 distribution.

11


12/12

f) Y1, Y2, . . . are independent, since X1, X2, . . . are. Then, from (c) and (d),

M(Yk,Yk+12)(s1, s2) = MYk(s1) MYk+12(s2)

MN(0,1)(s1)M2

1(s2) as k

,

which means that the joint vector (Yk, Yk+12) converges to (Z, D), where Z N(0, 1) and D 21

are independent. Since g(x, y) = x/

y is continuous for y > 0, then, from the Continuous Mapping

Theorem,YkY2k+1

=Yk

|Yk+1| converges to Z/

D, a Students tdistribution with one degree of freedom.

12

solutions probability

Documents