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?

$\endgroup$
17
$\begingroup$

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 $k$, $$ \Pr\left(|X_{n_k}-X|>2^{-k}\right)\leqslant 2^{-k}.$$ This one can be constructed by induction. Indeed, we first use the definition of convergence in probability with $\varepsilon=1/2$. We know that $\Pr\left(|X_{n}-X|>2^{-1}\right)$ goes to zero as $n$ goes to infinity, hence we are sure that for some $n_1$, $\Pr\left(|X_{n_1}-X|>2^{-1}\right)\leqslant 2^{-1}$. Now assume that we constructed $n_1<n_2<\dots<n_{k-1}$ such that for all $j\in\{1,\dots,k-1\}$, $$ \Pr\left(|X_{n_j}-X|>2^{-j}\right)\leqslant 2^{-j}.$$ We now use the definition of convergence in probability with $\varepsilon=2^{-k}$. We know that $\Pr\left(|X_{n}-X|>2^{-k}\right)$ goes to zero as $n$ goes to infinity, hence we are sure that there is some $N$ such that for all $n\geqslant N$, $\Pr\left(|X_{n}-X|>2^{-1}\right)\leqslant 2^{-k}$. Consequently, a $n_k$ bigger than $n_{k-1}$ and $N$ does the job.

By the Borel-Cantelli lemma applied to $A_k:=\left\{|X_{n_k}-X|>2^{-k}\right\}$, $$P\left(\limsup_{k\to+\infty}\left\{|X_{n_k}-X|>2^{-k}\right\}\right)=0.$$ This proves convergence almost everywhere of $\{X_{n_k}\}$ to $X$.

$\endgroup$
  • $\begingroup$ How does $P\left(\limsup_{k\to+\infty}\left\{|X_{n_k}-X|>2^{-k}\right\}\right)=0$. imply almost sure convergence? I commented on the other answer to this question on how I think on can do it. Is that correct? $\endgroup$ – Viktor Glombik Jul 13 at 12:52
3
$\begingroup$

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

$\endgroup$
  • 7
    $\begingroup$ Would you mind expanding upon this? $\endgroup$ – Michael Albanese Dec 27 '15 at 5:21
  • $\begingroup$ Here's a beginning on how one could expand, I'd love if someone could help me finish this: Because of $$\sum_{k \in \mathbb{N}} P( \left| X_{n_k} - X \right| > 2^{-k})\le \sum_{k \in \mathbb{N}} 2^{-k}= 1 < \infty$$ Borel-Cantelli implies $$P\left( \limsup_{k \to \infty} \left| X_{n_k} - X \right| > 2^{-k} \right)=P\left( \bigcap_{k \in \mathbb{N}} \bigcup_{\ell \ge k}\left \{ \left| X_{n_{\ell}} - X \right| > 2^{-\ell} \right\} \right)=0,$$ in turn yielding $$P\left( \bigcup_{k \in \mathbb{N}} \bigcap_{\ell \ge k}\left \{ \left| X_{n_{\ell}} - X \right| \ge 2^{-\ell} \right\}\right) = 1$$. $\endgroup$ – Viktor Glombik Jul 12 at 2:16
  • $\begingroup$ I'm not sure but I think the second line in the above comment also implies (we used that in the proof) $$\lim_{k \to \infty} P\left( \bigcup_{\ell \ge k}\left \{ \left| X_{n_{\ell}} - X \right| > 2^{-\ell} \right\} \right)= \lim_{k \to \infty} P\left( \left| X_{n_{k}} - X \right| > 0 \right) = 0$$ and this (I'm not sure about this part) yields $$\lim_{k \to \infty} P\left( \left| X_{n_{k}} - X \right| = 0 \right)= 1.$$ This can be completely right since this doesn't show convergence almost surely on a subsequence but on the whole sequence. $\endgroup$ – Viktor Glombik Jul 12 at 2:32

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.