In order to show that a $X_n$ converges to $X$ almost surely, one can often use the Borel-Cantelli lemma. I suppose it is a sufficient but not necessary condition for almost sure convergence. (?)
Let's look at the following example:
Let $X_1,X_2,…$ be a sequence of independent random variables with: $$P(X_n=n^2)=\frac{1}{n^2}=1-P(X_n=-1)$$ I want to prove that $X_n\xrightarrow{\text{a. s.}} -1$ . My argumentation goes like this (using Borel-Cantelli):
$\{X_n=n^2\}_n$ are independent events and $$\sum \limits_{n=1}^{\infty}P(X_n=n^2)=\sum \limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}<\infty$$ Thus, we have $\mathbb{P}(\limsup\limits_{n\to\infty}\{X_n=n^2\}_n)=0$. I read that as saying "The probability that the event $\{X_n=n^2\}$ will happen infinitely many times is $0$". Hence: For sufficiently large $n$, we have $X_n=-1$, which means that $X_n$ converges almost surely to $-1$.
Is this argumentation method correct and how can I make it more formal?
Best Answer
I agree with @Surb that this looks good as is, although whether this is sufficient is ultimately a question for your instructor.
If you really want to go down the analysis rabbit hole, you can do something like this: Let $\epsilon > 0$. Since $\sum_1^{\infty} \frac 1 {n^2} < \infty$, it follows that there is some $N$ such that $\sum_{n=N}^{\infty} \frac 1 {n^2} < \epsilon$. Correspondingly, $\mathbb P \left( \bigcup_{n=N}^{\infty} \{X_n \neq -1\} \right) \leq \sum_{n=N}^{\infty} \frac 1 {n^2} < \epsilon$; that is, the probability that $X_n$ will ever again (after the $N^{\text{th}}$ term) be something other than $-1$ is less than $\epsilon$. Since $\epsilon$ was arbitrary, we have no proved that the probability of $X_n$ being something other than $-1$ infinitely often is zero.
But... that argument is really just re-proving the Borel-Cantelli Lemma, whose proof is just a general version of exactly that. I prefer your argument instead.