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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?

BCLC
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Lost1
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3 Answers3

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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 $n\geqslant N$, $\Pr\left(|X_{n}-X|>2^{-k}\right)\leqslant 2^{-k}$. 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 $X$.

rims
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Davide Giraudo
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    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? – ViktorStein Jul 13 '19 at 12:52
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    The sequence of RVs ${X_{n_k} }$ converges almost surely to $X$ if $P({\limsup_{k} |X_{n_k} - X| \neq 0}) = 0$. Clearly, $P({\limsup_{k} |X_{n_k} - X| \neq 0})$ is upper bounded by $P({\limsup_{k} |X_{n_k} - X| > 2^{-k}})$ which is shown to be equal to zero. – rims Sep 26 '22 at 02:36
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Also, you can directly apply Borel Cantelli to the sequence of events $|X_{n_k}-X|>2^{-k}$.

patrick12345
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    Would you mind expanding upon this? – Michael Albanese Dec 27 '15 at 05:21
  • 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 $$P\left( \bigcup_{k \in \mathbb{N}} \bigcap_{\ell \ge k}\left { \left| X_{n_{\ell}} - X \right| \ge 2^{-\ell} \right}\right) = 1$$. – ViktorStein Jul 12 '19 at 02:16
  • 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. – ViktorStein Jul 12 '19 at 02:32
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If $n_k$ is a subsequence with $P(|X_{n_k}-X|>\frac{1}{2^k}) \le \frac{1}{3^k}$ then $$ E[\sum_{k\ge 0} \min(1,|X_{n_k}-X|) ] \le \sum_{k\ge 0} \frac{1}{2^k} + \frac{1}{3^k} <+\infty. $$ The non-negative random variable $\sum_{k\ge 0} \min(1,|X_{n_k}-X|)$ is thus finite with probability 1 and in this event the summand must converge to 0, that is, $\min(1,|X_{n_k}-X|)\to 0$, which implies $|X_{n_k}-X|\to 0$.

jlewk
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