12
$\begingroup$

Convergence in probability implies convergence on a subsequence almost surely.

But this means we fix a subsequence, such that $X_{n_k}$ converges for almost every $\omega$, right? The subsequence we pick does not depend on the $\omega$ right?

$\endgroup$
16
$\begingroup$

Yes, we can take a sequence $\{n_k\}$ which works for almost all $\omega$. To see that, fix a subsequence $\{n_k\}$ such that for each $k$, $$P(|X_{n_k}-X|>2^{-k})\leq 2^{-k}.$$ This one can be constructed by induction. We have $$P(\limsup_{k\to+\infty}\{|X_{n_k}-X|>2^{-k}\})=0,$$ by a similar argument of Borel-Cantelli theorem proof, using the fact that $\sum_{j\geq k}2^{—j}\to 0$ when $k\to +\infty$. This proves convergence almost everywhere of $\{X_{n_k}\}$ to $X$.

$\endgroup$
3
$\begingroup$

Also, you can directly apply Borel Cantelli to the sequence of events $|X_{n_k}-X|>2^{-k}$.

$\endgroup$
  • 6
    $\begingroup$ Would you mind expanding upon this? $\endgroup$ – Michael Albanese Dec 27 '15 at 5:21

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.