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.
Best Answer
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.