I'm trying to solve the following exercise from Resnick's book:

For any sequence of random variables {$X_n$} set $S_n=\sum_{i=1}^{n}X_i$

Show $X_n \xrightarrow{a.s.} 0$ implies $S_n/n \xrightarrow{a.s.} 0.$

I know that if $X_n$ converges almost surely, then its Cesaro sum must also converge. But how to formalize the proof?

Note that I can't apply Strong Law of Large Numbers, since it applies for any sequence, so that the theorem may not be applicable.


This can be solved pointwise.

Let $(x_n)_n$ be a sequence in $\mathbb R$ with $\lim_{n\to\infty}x_n=0$, let $s_n:=\sum_{i=1}^n x_i$ and let $\bar x_n:=s_n/n$.

$\lim_{n\to\infty}x_n=0$ implies the existence of some $c>0$ such that $|x_n|\leq c$ for every $n\in\mathbb N$.

If $\epsilon>0$ then some $N\in\mathbb N$ exists with $n>N\implies |x_n|<\epsilon$.

Then for $n>N$ we have: $$|\bar x_n|\leq\frac1n\sum_{i=1}^n|x_n|\leq\frac{N}{n}c+\frac{n-N}{n}\epsilon\leq \frac{N}{n}c+\epsilon$$

Then consequently $$\limsup|\bar x_n|\leq\epsilon$$

This for every $\epsilon>0$ so$$\limsup|\bar x_n|=0\text{ or equivalently }\lim_{n\to\infty}\bar x_n=0$$

This proves that: $$\{X_n\to0\}\subseteq\{S_n/n\to0\}$$ so that $$\mathsf P(X_n\to0)=1\implies\mathsf P(S_n/n\to0)=1$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.