# Almost sure convergence along a subsequence implies convergence in probability over the whole sequence

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?

• 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...) – David C. Ullrich Feb 15 '16 at 15:19
• Yes, thanks a lot. – user299158 Feb 15 '16 at 16:06

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.
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. $|X_n - X|> \epsilon$