# Is this a stopping time?

Consider a random variable $X$ such that $X \sim \mathcal{E}(\lambda)$. Define the probability space $(\Omega,\mathcal{F},\mathbb{P})$ such that $X$ is $\mathcal{F}$-measurable and the filtration $(\mathcal{F}_t)_{t\geq 0}$ such that, for all $s<t$, $\mathcal{F}_s \subset \mathcal{F}_t$ and $\mathcal{F}_t \subset \mathcal{F}$. Now, let $(X_t)_{t \geq 0}$ be a continuous stochastic process such that for all $t$, $X_t = X$. Finally, define the following random variable:

$$\tau = \{ t : X_t = t \}$$

Is the r.v. $\tau$ a stopping time?

Edit: I am slightly familiar with stopping times, however I am not familiar with the approach used to prove that a certain r.v. is a stopping time. How is this problem tackled?

• I am not sure what you mean by the definition of $\tau$. It appears to be a stochastic set, which is not a random variable. Perhaps you meant $\tau=X$, but then why would it be a stopping time? You have assumed no relation between $X$ and the filtration. Mar 10, 2017 at 16:12
• Informally, the idea is that all $X_t$ are equal to the realization of $X$, hence the set $\{t:X_t=t\}$ has a unique element. I have also forget to mention that $X$ is measurable with respect to $\mathcal{F}$ $-$ I am fixing this. Mar 10, 2017 at 16:18

A stopping time $T$ is a $\mathcal{F}$-measurable function $\Omega\rightarrow[0,\infty]$ such that $\{T\leq t\}\in\mathcal{F}_{t}$ for all $t\geq0.$ As it stands, you haven't specified the filtration, but perhaps the natural one to consider is just $\mathcal{F}=\mathcal{F}_{t}=X^{-1}(\mathcal{B}([0,\infty))$ for all $t\geq0,$ since this makes $X_{t}$ adapted. (This could be seen as natural, because the filtration is in some sense a quantification of the "information available at time $t$." Since for all $t\geq0,$ we know the value of $X_{t}=X,$ we have the same information at all times, so it makes sense that our filtration would be constant.)
$\tau$ is measurable, since for $A=[a,b],$ $0\leq a<b<\infty$, $\tau^{-1}(A)=\{\omega:\tau(\omega)\in A\}=\{\omega:X(\omega)\in A\}=X^{-1}(A)\in\mathcal{F}$, and $\tau^{-1}(\{\infty\})=\varnothing.$ Also, $\{\tau\leq t\}=\tau^{-1}([0,t])\in\mathcal{F}_{t}$ for all $t\geq0,$ so $\tau$ is a stopping time for this filtration.