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=1−P(X1=−1) and Sn∑ni=1Xi.
Is {Sn=0 infinitely often} a tail event? Supply arguments.
Show that if p≠1/2, P(Sn=0 infinitely often)=0. (Hint: Use Stirling’s approximation)
{Sn=0 infinitely often} is a tail event. If we remove finitely many S′n 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(1−p)n/2=n!(n/2)!2pn/2(1−p)n/2≈√2πnnne−n√2π(n/2)(n/2)n/2e−n/2pn/2(1−p)n/2=√2πnnne−n2π(n/2)(n/2)ne−npn/2(1−p)n/2=√2πn[4p(1−p)]n/2.Then for p≠1/2,
∞∑i=1P(Sn=0)=n∑i=1√2πn[4p(1−p)]n/2≤n∑i=1√2π[4p(1−p)]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(An∩ACn+1)<∞. Show that P(lim supAn)=0.
Take X=lim supAn and Y=lim supACn. Then,
P(X)=P(X∩Y)+P(X∩YC)≤P(X∩Y)+P(YC).So if we show P(X∩Y)+P(YC)=0 then P(X)=0.
Let’s start by showing P(X∩Y). Notice that by the Borel-Cantelli Lemma, since ∑∞i=1An∩ACn+1<∞, then P(An∩ACn+1 i.o.)=0.
For sake of contradiction, assume that P(X∩Y)≠0. Then both P(An i.o.)≠0 and P(ACn i.o.)≠0. However, then P(An∩ACn+1 i.o.)≠0. Thus, P(X∩Y)=0.
Now we will show that P(YC)=0.
P(YC)=limn→∞P(∞⋃m=nAm)≤limn→∞P(An)→0Thus,
P(X)≤P(X∩Y)+P(YC)→0so P(lim supAn)=0.
3
Let X1,X2,… be independent with common distribution P(X≤x)=1−e−xα, x≥0 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)>1−ε}.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−(1−exp{[−(1+ε)log(n)1/α]α})=exp{−(1+ε)αlog(n)}=exp{log(n−(1+ε)α)}=n−(1+ε)αThen,
∞∑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 z≠0, 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 Xn↑X for some random variable X. Show that X is also integrable and E(X)=limn→∞E(Xn).
Since Xn≥0 and Xn↑X, 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 0≤Xn≤X, then X−=0. Thus, E(X−)=0<∞.
Since E(X+n) is bounded and integrable, limn→∞E(X+n=E(X+)=E(X)<∞.
Thus, X is integrable.