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Example Problems on Sequences of Random Variables

ST779 Homework 6 on Sequences of Random Variables

1

Let X1,X2,, be i.i.d. with P(X1=1)=p=1P(X1=1) and Snni=1Xi.

Is {Sn=0 infinitely often} a tail event? Supply arguments.

Show that if p1/2, P(Sn=0 infinitely often)=0. (Hint: Use Stirling’s approximation)

{Sn=0 infinitely often} is a tail event. If we remove finitely many Sn from the, there will still be infinitely many Sn’s equal to 0. Thus, it is a tail event.

Notice that P(Sn=0 infinitely often)=0 only when the number of Xi that equal to 1 is the same as the number equal to 0.

P(Sn=0)=(nn/2)pn/2(1p)n/2=n!(n/2)!2pn/2(1p)n/22πnnnen2π(n/2)(n/2)n/2en/2pn/2(1p)n/2=2πnnnen2π(n/2)(n/2)nenpn/2(1p)n/2=2πn[4p(1p)]n/2.

Then for p1/2,

i=1P(Sn=0)=ni=12πn[4p(1p)]n/2ni=12π[4p(1p)]n/2<.

Thus, by the Borel-Cantelli Lemma, P(Sn=0 infinitely often)=0.

2

Let An be events such that P(An)0 and i=1P(AnACn+1)<. Show that P(lim supAn)=0.

Take X=lim supAn and Y=lim supACn. Then,

P(X)=P(XY)+P(XYC)P(XY)+P(YC).

So if we show P(XY)+P(YC)=0 then P(X)=0.

Let’s start by showing P(XY). Notice that by the Borel-Cantelli Lemma, since i=1AnACn+1<, then P(AnACn+1 i.o.)=0.

For sake of contradiction, assume that P(XY)0. Then both P(An i.o.)0 and P(ACn i.o.)0. However, then P(AnACn+1 i.o.)0. Thus, P(XY)=0.

Now we will show that P(YC)=0.

P(YC)=limnP(m=nAm)limnP(An)0

Thus,

P(X)P(XY)+P(YC)0

so P(lim supAn)=0.

3

Let X1,X2, be independent with common distribution P(Xx)=1exα, x0 for some α>0 (Weibull distribution). Show that with probability 1, lim sup[Xn/log(n)1/α]=1.

Take ε>0 and

An={Xnlog(n)>1+ε} and Bn={Xnlog(n)>1ε}.

To show lim sup[Xn/log(n)1/α]=1 with probability 1, we need to show that P(lim supAn)=0 and P(lim supBn)=1.

P(An)=1(1exp{[(1+ε)log(n)1/α]α})=exp{(1+ε)αlog(n)}=exp{log(n(1+ε)α)}=n(1+ε)α

Then,

i=1P(An)=i=1n[(1+ε)α]<

because 1+ε>1. By the Borel-Cantelli Lemma P(lim supAn)=0.

Similary,

i=1P(Bn)=i=1n[(1+ε)α]=

because 1+ε<1. By the Borel-Cantelli Lemma P(lim supBn)=1.

Thus, lim sup[Xn/log(n)1/α]=1.

4

For a sequence of numbers {an}, the radius of convergence of the power series i=1anzn is the largest number r such that for any z∣<r, the series i=1anzn is absolutely convergent. It is known that if the series absolutely converges for some z with z∣=c, then it also converges for nay w with w∣<c. If the series for all z we say that r=. If the series does not converge for an z0, we say that r=0.

Now if {Xn} is a sequence of independent random variables, then show that the radius of convergence of the random series i=1Xnzn is a degenerate random variable (possibly infinite valued).

Take R to be the random variable that describes the radius of convergence. Also notice that convergence is a tail event. Then by Kolmogorov’s zero-one law, since X1,X2, are independent all tail events are P-trivial. So, R is a P-trivial event and P(R=r0)=1 for some r0.

5

Suppose that Xn, n=1,2, are nonneative integrable random variables such that the sequence {E(Xn)} is bounded and XnX for some random variable X. Show that X is also integrable and E(X)=limnE(Xn).

Since Xn0 and XnX, by the Monotone Convergence Theorem, E(Xn)E(X).

To show that X is integrable, we have to show that E(X),E(X+)<.

Since 0XnX, then X=0. Thus, E(X)=0<.

Since E(X+n) is bounded and integrable, limnE(X+n=E(X+)=E(X)<.

Thus, X is integrable.