If $\{X_t: t\in \mathbb{R_{\geq 0}}\}$ convergences to $X_1$ in probability, then $\{X_t\}$ admits a sequence converging to $X_1$ almost surely.

Question

I want to prove the following claim:

If $\{X_t: t\in \mathbb{R_{\geq 0}}\}$ convergences to $X_1$ in probability when $t\to 1$, then there is a sequence $\{t_n\}\subset \mathbb{R_{\geq 0}}$ converging to 1 such that $\{X_{t_n}\}$ converges to $X_1$ almost surely.

Since $\{X_t: t\in \mathbb{R_{\geq 0}}\}$ convergences to $X_1$ in probability, we have $$ \forall\ m\in \mathbb N,\ \exists\ \{t_{m,n}\}_n \text{ such that } \lim_{n\to \infty}\mathbb{P}(|X_{t_{m,n}}-X_1|>\frac{1}{m})=0 $$ I guess that $t_{n,n}$ is the desired sequence, but I don't know how to prove it. Any ideas?

Have a look here (especially claim 2). – drhab Feb 28 '20 at 13:11 — drhab, Feb 28 '20 at 13:11

score 1 · Answer 1 · answered Feb 28 '20 at 13:09

1

Let $\varepsilon \gt0$. Choose $t_n $ so that $$ \mathbb{P}(|X_{t_{n}}-X_1|>\varepsilon)\leq \frac{1}{n^2}. $$ Apply the first Borel-Cantelli lemma to conclude.

answered Feb 28 '20 at 13:09

cqfd

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If $\{X_t: t\in \mathbb{R_{\geq 0}}\}$ convergences to $X_1$ in probability, then $\{X_t\}$ admits a sequence converging to $X_1$ almost surely.

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