Convergence Properties of Sums of iid Random Variables"

Question

I want to check my solutions for this problem. Can someone help me?

Let $(X_n)_{n\geq1}$ be independent and identically distributed real-valued random variables.

Suppose $E[|X_n|] = ∞$.

a. Show that $P(|X_n| > an \text{ for infinitely many } n) = 1$ for any $a > 0$. Hint: Justify $\int_{0}^{\infty} P(|X_1| > at) dt = ∞$ and then use $P(|X_n| > an) > P(|X_n| > at)$ for $t > n$.

b. For $S_n = \sum_{i=1}^n X_i$, follow that $P(|S_n| > an \text{ infinitely often}) = 1$ for every $a > 0$, and therefore $\lim \sup_{n→∞}|S_n|/n = ∞$ almost surely. What does it imply for $\lim \sup_{n→∞}|S_n|/n$ in the case where $X_n \in L^2$, that is, $E[X_n^2] < ∞$?

I have done a) but for b) I am having some problems:

First we want to show that from $S_n = \sum_{i=1}^n X_i$, follows that $P(|S_n| > an \text{ infinitely often}) = 1$ for every $a > 0$

So $P(|S_n| > an \text{ infinitely often}) = \\ P(|X_1+...+X_n| > an \text{ infinitely often})\overset{\text{since iid}}{=} \\ P(|X_1|+...+|X_n| > an \text{ infinitely often})=\\ P((|X_1|+...+|X_n|)n > a \text{ infinitely often})=\\P(|X_1| > a \text{ infinitely often})\overset{\text{from a.}}{=}1 $

For the second part $\lim \sup_{n→∞}|S_n|/n = ∞$ almost surely, I have used Borel Cantelli Since the $X_n$ are independen we have that also $S_n$ are independent.

We know from the first part that $P(|S_n|/n > a)=1$ and therefore that $\sum_{i=1}^n=P(|S_n|/n > a)= \infty$ we can use BCand say that $P(\lim \sup_{n→∞}|S_n|/n>a)=1$ therefore $\lim \sup_{n→∞}|S_n|/n=\infty$ almost surely.

I think that I have done a lot of mistakes, can someone help me?

For the question "What does it imply for $\lim \sup_{n→∞}|S_n|/n$ in the case where $X_n \in L^2$, that is, $E[X_n^2] < ∞$?" I have really no idea what to write.

Answer 1

A link to the solution of a part of your problem has been pointed out in the comments by Snoop.

Let us look at the steps

$P(|S_n| > an \text{ infinitely often}) = \\ P(|X_1+...+X_n| > an \text{ infinitely often})\overset{\text{since iid}}{=} \\ P(|X_1|+...+|X_n| > an \text{ infinitely often})=\\ P((|X_1|+...+|X_n|)n > a \text{ infinitely often})=\\P(|X_1| > a \text{ infinitely often})\overset{\text{from a.}}{=}1$

The second equal sign should be a $\leq$ and this is not due to the fact that the random variables are i.i.d.

For the third equal sign, one should replace $(|X_1|+...+|X_n|)n$ by $(|X_1|+...+|X_n|)/n$.

For the last equal sign, what does $\{\lvert X_1\rvert>a\mbox{ infinitely often}\}$ mean?

The last question is a bit weird, since usually, we deduce results under infinite mean/variance from those under finite mean/variance. We know in any case by the strong law of large numbers that $\limsup_{n\to\infty}\lvert S_n\rvert/n$ is finite, but the answer depends a lot on what was covered before this exercise.