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Consider a sequence of real-valued random variables $\{X_n\}_n$ almost surely converging to a real-valued random variable $X$ along a subsequence $\{n_k\}_k \subseteq \mathbb{N}$: $X_{n_k} \rightarrow_{a.s.}X$ as $k \rightarrow \infty$.

Does this imply convergence in probability along the whole sequence, i.e. $X_n \rightarrow _p X$ as $n \rightarrow \infty$? Hint for the proof?

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    $\begingroup$ Assuming something about a subsequence (and assuming nothing else) cannot possibly ever prove anything about the whole sequence! You're given exactly zero information about $X_n$ for $n\ne n_k$, so you cannot possibly prove anything about it. (Now, in whatever problem this came from there may be more information given...) $\endgroup$ – David C. Ullrich Feb 15 '16 at 15:19
  • $\begingroup$ Yes, thanks a lot. $\endgroup$ – user299158 Feb 15 '16 at 16:06
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What is true is this: If every subsequence of $\{X_n\}$ has a further subsequence that converges almost surely to $X$, then $\{X_n\}$ converges to $X$ in probability. The proof is straightforward.

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  • $\begingroup$ do you have a reference for this? $\endgroup$ – user61038 Apr 23 '16 at 6:12
  • $\begingroup$ Theorem 6.3.1 in S. Resnick's A Probability Path. $\endgroup$ – John Dawkins Apr 23 '16 at 15:20
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No.

Formal definition of the subsequence convergence has constraints on $X_{n_k}$. But for any ${n}>{n_{k_0}}$ there can be an ${X_n}$ s.t. $|X_n - X|> \epsilon$

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