The last argument shows that if $X_n\to X_\infty$ a.s. and $N(n)\to\infty$ a.s., then $X_{N(n)}\to X_\infty$. We have written this out with care because the analogous result for convergence in probability is false. If $X_n\in\{0,1\}$ are independent with $P(X=1)=a_n\to0$ and $\sum_n a_n=\infty$, then $X_n\to0$ in probability, but if we let $N(n)=\inf\{m\ge n; X_m=1\}$ then $X_{N(n)}=1$ a.s.

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It is from Durrett 5th Edition To be consistent to the Subsequence Criterion, N(n) should have further subsequence that converges to 0 almost surely. But it is not possible. So, what's wrong with my understanding to subsequence criterion?


The sequence $(X_{N(n)})_{n=1}^\infty$ is not a subsequence of $(X_n)_{n=1}^\infty$, which I believe is how you are reading it, since you refer to a "further" subsequence. $(X_{N(n)})_{n=1}^\infty$ is not a subsequence because the $N(n)$ are random variables. So $X_{N(n)}$ is a Frankenstein monster cobbled together from different terms of the sequence, depending on the various values $N(n)$ takes over the probability space.

With the understanding that $(X_{N(n)})_{n=1}^\infty$ is not a subsequence of $(X_n)_{n=1}^\infty$, but a sequence of Frankensteins, I think this should clear things up. Any subsequence of $(X_n)_{n=1}^\infty$ has a further subsequence which converges to $0$ a.s., but since $(X_{N(n)})_{n=1}^\infty$ is not a subsequence, this does not apply to $(X_{N(n)})_{n=1}^\infty$. $(X_{N(n)})_{n=1}^\infty$ does not converge to zero in probability or a.s.

  • $\begingroup$ Yes. This is about using a stopping time as the subscript. Something that is often done in probability theory. The "last argument" result quoted shows that it is OK for a.s. convergence. $\endgroup$ – GEdgar Mar 22 '19 at 10:44
  • $\begingroup$ Of course, $X_{N(n)}$ does converge to 0 a.s. under the stated hypotheses. $\endgroup$ – ofer zeitouni Mar 22 '19 at 18:46
  • $\begingroup$ @ofer "Of course, $X_{N(n)}$ does converge to $0$ a.s. under the stated hypotheses." The specific choice of $X_{N(n)}$ from the end of the text is $1$ a.s, and does not converge to $0$ a.s. $\endgroup$ – bangs Mar 24 '19 at 13:11

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