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?
$\begingroup$
$\endgroup$
1
-
7$\begingroup$ Right.${}{}{}{}$ $\endgroup$– Michael GreineckerOct 27, 2012 at 20:14
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
2
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 $X$.
answered Oct 27, 2012 at 20:38
-
2$\begingroup$ 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? $\endgroup$ Jul 13, 2019 at 12:52
-
1$\begingroup$ 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. $\endgroup$– rimsSep 26, 2022 at 2:36
$\begingroup$
$\endgroup$
3
Also, you can directly apply Borel Cantelli to the sequence of events $|X_{n_k}-X|>2^{-k}$.
-
7$\begingroup$ Would you mind expanding upon this? $\endgroup$ Dec 27, 2015 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 $$P\left( \bigcup_{k \in \mathbb{N}} \bigcap_{\ell \ge k}\left \{ \left| X_{n_{\ell}} - X \right| \ge 2^{-\ell} \right\}\right) = 1$$. $\endgroup$ Jul 12, 2019 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. $\endgroup$ Jul 12, 2019 at 2:32
$\begingroup$
$\endgroup$
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 $|X_{n_k}-X|\to 0$.
