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Let $X,X_1,X_2,...$ be random variables on the probability space $(\Omega, \mathcal{F}, P)$. Let $S$ denoted the set of all bounded stopping times on $(\Omega, \mathcal{F}, (\mathcal{F}_n))$, where $\mathcal{F}_n = \sigma(X_1,...,X_n)$.

$S$ is a directed set under the pointwise partial order $\leq$, so we can speak of stopping time directed nets of random variables.

We write $X_\tau \xrightarrow{p} X$ to mean: for all $\epsilon, \delta > 0$, there exists $\sigma \in S$ such that $\tau \geq \sigma$ implies $$P\Big( \Big\{\omega: |X_{\tau(\omega)}(\omega) - X(\omega)| > \delta \Big\} \Big) < \epsilon. $$

It's clear that if $X_n \xrightarrow{a.s.} X$, then $X_\tau \xrightarrow{p} X$.

A paper I was reading seemed to claim that the converse is also true and obvious, but I cannot see it. The argument given is that if $X_\tau \xrightarrow{p} X$, then there exists $\tau_1 < \tau_2 < ... \in S$ such that $$X_{\tau_n} \xrightarrow{a.s.} \limsup_n X_{\tau_n} = \limsup_n X_n.$$ I don't understand how this proves the claim, and I guess I'm just confused about something simple. Any pointers or hints would be appreciated.

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For fixed $\delta,\epsilon>0$ let $\sigma$ be a bounded stopping time such that

$$\forall \tau \in S, \tau \geq \sigma: \quad \mathbb{P}(|X_{\tau}-X|>\delta) < \epsilon. \tag{1}$$

Without loss of generality, we may assume $\sigma \geq 1$; otherwise we replace $\sigma$ by $\sigma+1$. Note that $(1)$ implies, in particular,

$$\mathbb{P}(|X_{\sigma}-X|>\delta) < \epsilon. \tag{2}$$

For fixed $k \in \mathbb{N}$ we define a bounded stopping time $\tau_k$ by

$$\tau_k(\omega) := \inf\{n \geq \sigma(\omega); |X_n(\omega)-X_{\sigma}(\omega)| \geq 2 \delta\} \wedge (k \sigma(\omega)).$$

Clearly,

$$\begin{align*} \mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) &\leq \mathbb{P}(|X_{\sigma}-X|>\delta) + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_k-X|>3 \delta \, \, \text{i.o.}) \\ &\stackrel{\text{(2)}}{\leq} \epsilon + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_k-X_{\sigma}|>2 \delta \, \, \text{i.o.}). \end{align*}$$

By the definition of $\tau_k$ this implies

$$\mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) \leq \epsilon + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_{\tau_k}-X_{\sigma}|> 2\delta \, \, \text{i.o.}).$$

As $k \sigma \to \infty$ as $k \to \infty$ it follows (from the definition of $\tau_k$) that we can choose $K \gg 1$ sufficiently large such that

$$\mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) \leq 2\epsilon + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_{\tau_K}-X_{\sigma}|>2\delta).$$

Hence, $$\mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) \leq 2\epsilon+ \mathbb{P}(|X_{\tau_K}-X|>\delta) \stackrel{\text{(1)}}{\leq} 3 \epsilon.$$

Since $\epsilon>0$ and $\delta>0$ are arbitrary, this proves $X_k \to X$ almost surely.

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  • $\begingroup$ Nice, thanks. I'm still mystified by the argument that I quoted from the paper. Any idea why that's supposed to make the result obvious? $\endgroup$
    – aduh
    Feb 16, 2018 at 20:51
  • $\begingroup$ @aduh Not exactly, no. Intuitively it was kind of obvious to me, but it took me some time to write down a rigorous proof. $\endgroup$
    – saz
    Feb 16, 2018 at 20:56
  • $\begingroup$ Care to share your intuition about this? $\endgroup$
    – aduh
    Feb 17, 2018 at 15:01
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    $\begingroup$ @aduh By the very definition, $X_{\sigma}$ is close to the limit $X$ with a large probability. If we consider the first time $\tau \geq \sigma$ where the process moves away from a neighborhood of $X_{\sigma}$ (and hence of $X$), then the convergence $X_{\tau} \to X$ tells us that $X_{\tau}$ is still close to $X_{\sigma}$ (and hence to $X$) with a high probability. The annoying thing is that $\tau$ is not bounded, and therefore we have to use some truncation argument $\endgroup$
    – saz
    Feb 17, 2018 at 15:27

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