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Suppose we have a discrete-time stochastic process $\{X_t\}$ defined on a space $(\Omega,\mathcal{F})$ equipped with the probability measure $\mathbb{P}$. Suppose we know that $X_t \rightarrow 0$ almost surely. Let $T(\epsilon)$ be defined as $T(\epsilon)=sup\{t\in\mathbb{N}: |X_t| \geq \epsilon\}$. My question is whether $T(\epsilon)$ is measurable w.r.t. the $\sigma$-algebra $\mathcal{F}$. This appears to be true, but a proof or a reference to a proof would be very helpful.

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  • $\begingroup$ $T(\epsilon)$ is an $\mathbb{N}$-valued random variable, not a subset of $\Omega$. What do you mean by "$T(\epsilon)$ is measurable w.r.t. the $\sigma$-algebra $\mathcal{F}$"? $\endgroup$
    – kccu
    Commented Apr 8, 2019 at 22:28
  • $\begingroup$ @kccu en.m.wikipedia.org/wiki/Measurable_function#Formal_definition $\endgroup$ Commented Apr 8, 2019 at 22:30
  • $\begingroup$ Can you clarify whether each $X_t$ is defined on $(\Omega,\mathcal{F})$ or whether the entire sequence $(X_0,X_1,X_2,\dots)$ is? $\endgroup$
    – kccu
    Commented Apr 8, 2019 at 22:33
  • $\begingroup$ Ah, the entire sequence is defined on $\mathcal{F}$. $\endgroup$
    – Brian
    Commented Apr 8, 2019 at 22:37
  • $\begingroup$ @kccu : Basically, given that the entire discrete stochastic process $\mathcal{X}_t$ is measurable w.r.t. the sigma algebra $\mathcal{F}$, I am interested in knowing whether the quantity $T(\epsilon)$ qualifies as a $\mathcal{F}$-measurable random variable. $\endgroup$
    – Brian
    Commented Apr 8, 2019 at 22:41

1 Answer 1

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To verify a positive integer valued random variable $N$ is measurable, it suffices to show the sets $\{N=n\}$ are measurable, for $n\in \mathbb N$.

$$ \{T(\epsilon)=n\}=\{|X_n|\ge \epsilon\}\cap \bigcap_{t=n+1}^{\infty} \{|X_t|< \epsilon\} $$ The RHS is measurable because it is a countable intersection of measurable events.

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  • $\begingroup$ Thanks, that makes sense. $\endgroup$
    – Brian
    Commented Apr 8, 2019 at 22:57
  • $\begingroup$ I have an unrelated (perhaps trivial) question. Suppose $\mathbb{X},\mathbb{Y}$ are random variables on a common space, and $a,\epsilon$ are positive constants with $\epsilon < a$. Is the following true: $\mathbb{P}(\mathbb{X} \leq a-\epsilon) \leq \mathbb{P}(\mathbb{X} \leq a-\mathbb{Y} | |\mathbb{Y}| \leq \epsilon)$? This seems to be trivially true because of the obvious implication of events, but am I overlooking any subtlety induced by the conditional operator? $\endgroup$
    – Brian
    Commented Apr 8, 2019 at 23:25
  • $\begingroup$ You're welcome. If an answer is helpful, you should mark it as accepted by clicking the check mark. Also, your second comment would be better asked as a separate question. $\endgroup$ Commented Apr 9, 2019 at 0:03
  • $\begingroup$ Okay, let me ask it as a separate question then: any comments on the same would be appreciated, thanks. $\endgroup$
    – Brian
    Commented Apr 9, 2019 at 0:30

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