# Another question on almost sure and convergence in probability

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?

• Right.${}{}{}{}$ Oct 27, 2012 at 20:14

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 $$n_1 such that for all $$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$$.
• How does $P\left(\limsup_{k\to+\infty}\left\{|X_{n_k}-X|>2^{-k}\right\}\right)=0$. imply almost sure convergence? I commented on the other answer to this question on how I think on can do it. Is that correct? Jul 13, 2019 at 12:52
Also, you can directly apply Borel Cantelli to the sequence of events $|X_{n_k}-X|>2^{-k}$.
• Here's a beginning on how one could expand, I'd love if someone could help me finish this: Because of $$\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$$. Jul 12, 2019 at 2:16
• I'm not sure but I think the second line in the above comment also implies (we used that in the proof) $$\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. Jul 12, 2019 at 2:32