How do I prove that $\Bbb{P}(|X_n|>n~~i.o.)=0$?

Question

I have the following problem:

We have given a probability space $(\Omega, F,\Bbb{P})$ and $X_i:\Omega \rightarrow \Bbb{R}$ an i.i.d. sequence of random variables. We assume that there is a r.v. $Y$ such that $$\lim_{n\rightarrow \infty} \frac{\sum_{i=1}^n X_i(\omega)}{n}=Y(\omega)$$ I need to show that $$\Bbb{P}(|X_n|>n~~ i.o.)=0$$

I somehow thought about using B.C. lemma, but I don't see how. Because in part b) of the exercise I need to show that $\sum_{n=0}^\infty \Bbb{P}(|X_n|>n|)<\infty$, and thus I don't think B.C. is used, since for this we should know from the beginning that the sum is finite. Therefore I wanted to ask if someone can help me with other ideas. Because also the use of the Law of large numbers does not work since I don't know that $X_i$ is in $L^1$ this I need to deduce in part c) of the exercise.

I would appreciate your help.

Answer 1

In fact, one should not use Borel-Cantelli here. Write $S_n=\sum_{i=1}^n X_i$. (As usual in probability theory, we suppress the argument $\omega$.) The hypothesis is $$\lim_{n\rightarrow \infty} \frac{S_n}{n}=Y \quad (*) \,.$$ Rewriting this with the index $n-1$ replacing $n$ gives $$\lim_{n\rightarrow \infty} \frac{S_{n-1}}{n-1}=Y \,.$$ Multiplying this by $ (n-1)/n$ gives $$\lim_{n\rightarrow \infty} \frac{S_{n-1}}{n}=Y \,.$$ Subtracting this from (*), we obtain $$\lim_{n\rightarrow \infty} \frac{X_n}{n}=0 \,,$$ which implies $$\Bbb{P}(|X_n|>n~~ i.o.)=0 \,,$$ by the definition of convergence.

Answer 2

Suppose $P[|X_n|>n \text{ io.}] > 0$. There is some $\omega$ and a subsequence such that $|X_{n_k}(\omega)| > n_k$ for all $k$.

Let $S_n = X_1+\cdots X_n$, then ${S_n(\omega) \over n} \to Y(\omega)$ and so ${| {S_n(\omega) \over n} - {S_{n-1}(\omega) \over n-1} | } \to 0$.

\begin{eqnarray} | {S_n(\omega) \over n} - {S_{n-1}(\omega) \over n-1} | &=& | { (n-1) S_n(\omega)- n S_{n-1}(\omega) \over n (n-1)} | \\ &=& | { n( S_n(\omega)- S_{n-1}(\omega)) - S_n(\omega) \over n (n-1)} | \\ &=& | { n X_n(\omega) - S_n(\omega) \over n (n-1)} | \\ &\ge& {n \over n-1}| { X_n(\omega)\over n} | -| { S_n(\omega) \over n (n-1)} | \end{eqnarray} This gives $\limsup _n | {S_n(\omega) \over n} - {S_{n-1}(\omega) \over n-1} | \ge 1$, which is a contradiction.

Answer 3

Assume that $A:=\lim_n\frac{1}{n}S_n$ converges $\Pr$--a.s. Then \begin{align} \frac{X_n}{n}=\frac{S_n}{n}- \Big(\frac{n-1}{n}\Big)\frac{S_{n-1}}{(n-1)}\rightarrow0 \end{align} Consequently, $\Pr[|X_n|>n,\,\text{i.o}]=0$.

Bonus: By the reversed Borel--Cantelli lemma and Fubini's theorem \begin{align} \mathbb{E}[|X_1|]=\int^\infty_0\Pr[|X_1|>t]\,dt\leq 1+\sum_{n\geq1}\Pr[|X_1|>n]<\infty. \end{align}