# How is Borel Cantelli Lemma being applied in this case?

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?

• 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. Commented Jan 7, 2021 at 23:17

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.