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I am taking a course in Probability Theory and in a proof, the professor made the following claim:

If for all $\epsilon>0$, the sum $\sum_{n\geq 1}P(|X_n-E[X_n]|> n\epsilon)<\infty$ then by Borel Cantelli Lemma $$\limsup_n \left|\frac{X_n}{n}-\frac{E[X_n]}{n}\right|\leq \epsilon\quad \text{P a.s.}$$

Here $X_n$ is some sequence of random variables assumed that are positive.

I am trying to understand how the inequality can be deduced from the Borel Cantelli Lemma. Here is my attempt.

If the sum is finite then $P(\limsup_n A_n) =0$ where $A_n$ is the event $A_n=\{\omega \in \Omega: |X_n(w)-E[X_n]| > n\epsilon \}.$ This implies that $P(\liminf A_n^c) = 1$ where $A_n^c= \{\omega \in \Omega: |X_n(w)-E[X_n]| \leq n\epsilon \}.$ If we take $\omega \in \liminf A_n^c = \bigcup_{m\geq 1}\bigcap_{n\geq m}A_n^c$ then there is an $m\geq 1$ such that for all $n\geq m$ the $|\frac{X_n(\omega)}{n}-\frac{E[X_n]}{n}|\leq \epsilon$ and thus we can conclude that,
$$\limsup_n \left|\frac{X_n}{n}-\frac{E[X_n]}{n}\right|\leq \epsilon\quad \text{P a.s.}$$

Is this reasoning correct?

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  • $\begingroup$ Yes, it is correct. But if the series converges for every $\epsilon >0$ then we can show that $$\frac{X_n}{n}-\frac{E[X_n]}{n} \to 0$$ a.s. and this requires some additional work. $\endgroup$ Commented Jan 7, 2021 at 23:17

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Yes your reasoning is correct, though you may be using more words than necessary. For fixed $\epsilon>0$, Borel-Cantelli tells us that since the sum converges, a.s. only finitely many $n$ have $|X_n-E[X_n]|/n > \epsilon$. On this event, it is clear that the limsup is at most $\epsilon$. Taking a sequence $\epsilon_k \to 0$, on the intersection of all the a.s. events for $\epsilon_k$, it follows that $|X_n-E[X_n]|/n \to 0$, and hence the limit is $0$ a.s.

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