4
$\begingroup$

I would like to show the following:

For $(X_n)_{n \geq1}$ independent RVs, $$ X_n \rightarrow X \ \text{ a.s.} \Rightarrow\ \forall \varepsilon \gt0, \ \sum _{n\geq1} P(|X_n -X| \gt \varepsilon) \lt \infty$$

We don't know that $|X_n -X| \gt \varepsilon$ are independent so second Borel-Cantelli cannot be applied. Any hint is appreciated.

$\endgroup$
2
$\begingroup$

You can show that the events $|X_n-X|>\epsilon$ are independent. This is because you can use Kolmogorov's $0-1$ law to show that $X$ is constant almost surely.

$X$ is measurable with respect to the tail $\sigma$-algebra $$\bigcap_{n\ge 1}\sigma(X_n,X_{n+1},X_{n+2}\dots).$$

(Intuitively, in order to know the limit of $X_n$, it suffices to know the values of all but any finite subset of the $X_n$.)

Kolmogorov's $0-1$ law says that all events in the tail $\sigma$-algebra of series of independent random variables have probability $0$ or $1$. Therefore, for any $c\in \mathbb R$, $P(X\le c)=0\text{ or }1$, so $X$ is constant almost surely.

$\endgroup$

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.