continuous time stochastic process prove $X_{\tau}(\omega):=X_{\tau(\omega)}(\omega)$ is a random variable Let {$X_t, t\geq0$} be a continuous time stochastic process, and $\tau:\Omega \rightarrow [0,\infty)$ be a stopping time. Let now define the stopped process at $\tau$ as $X_{\tau}:\Omega \rightarrow [0,\infty)$, as: $$X_{\tau}(\omega):=X_{\tau(\omega)}(\omega),$$ for $\omega \in \Omega$. Prove that $X_{\tau}$ is a random variable.
I would simply say that $\tau(\omega)$=t, for some t, and we know that $X_t$ is a RV, for every t in $\mathbb{R^+}$. On the other hand, I'm not sure we can say it is a composite function (the composition of two RV is again a RV so we would be done).
Is there anyone that might tell me if it is just as such or is missing something?
By the way, I'm studing chapter 3 of Resnick - Probability Path, quite far from deepful explanations of stochastic processes (which come just at the end of the book!).
Many thanks for the help.
 A: Here is a proof that shows that in fact, $X_T$ is $\mathscr{F}_T$-measurable.
I will make use of a couple of technical facts, but which are not difficult to motivate and prove
Suppose $X$ is a continuous stochastic process adapted to the filtration $\{\mathscr{F}_t:t\geq0\}$, and let  $\mathscr{F}=\sigma\Big(\bigcup_t\mathscr{F}_t\Big)$.
Recall that $T$ is a topping time (with respect to the filtration $\mathscr{F}_t$) iff $\{T\leq t\}\in\mathscr{F}_t$ for all $t\geq0$. Also, recall that a stopping time induces a $\sigma$--algebra
$$\mathscr{F}_T=\{A\subset\Omega:A\cap\{T\leq t\}\in\mathscr{F}_t,\,\text{for all},\,t\geq0\}\subset\mathscr{F}$$

The process $X$ may be viewed  as the  map $X:[0,\infty)\times\Omega\mapsto\mathbb{R}$ given by $(t,\omega)\mapsto X(t,\omega)=X_t(\omega)$.
Equip $[0,\infty)\times\Omega$ with the product $\sigma$ algebra $\mathscr{B}([0,\infty))\otimes\mathscr{F}$.

*

*The assumptions on $X$ imply that $X$ is progressively measurable, that is, for any $t>0$ fixed, the restriction of $X$ to $[0,t]\times\Omega$ is $\mathscr{B}([0,t])\otimes\mathscr{F}_t$-measurable; equivalently, the map $X^t(s,\omega):=X(s\wedge t,\omega)$ is $\mathscr{B}([0,\infty))\otimes\mathscr{F}_t$-measurable.

The check this, one may consider
$$
Y_n(s,\omega)=\left\{
\begin{array}{rcl}
X\big(\tfrac{k}{2^n}t,\omega\big)&\text{if}&\tfrac{k}{2^n}t\leq
  s< \tfrac{k+1}{2^n}t, \quad k=0,\ldots,2^n-1\\
X(t, \omega) &\text{if}& s\geq t
\end{array}
\right.
$$
that is $Y_n(s,\omega) = \sum^{2^n-1}_{k=0}X(\tfrac{k}{2^n}t,\omega)\mathbb{1}_{\big[\tfrac{k}{2^n}t,\frac{k+1}{2^n}t\big)}(s) + X(t,\omega)\mathbb{1}_{[t,\infty)}(s)$. Clearly $Y_n$ is right-continuous with left limits, and $\mathscr{B}([0,\infty))\otimes\mathscr{F}_t$-measurable.
The continuity of $X$ implies that  $\lim_{n\rightarrow}Y_n(s,\omega)=X(s\wedge t,\omega)=X^t(s,\omega)$ for all $(s,\omega)\in[0,\infty)\times\Omega$. Thus, $X$ is $\mathscr{B}([0,\infty))\otimes\mathscr{F}_t$ measurable.


*Any progressively measurable map $Y:[0,\infty)\times\Omega\rightarrow\mathbb{R}$ is adapted.

To check this, let $t>0$. Then $S=\big(X^t\big)^{-1}(A)=\{(s,\omega):X(s\wedge t,\omega)\in A\}$ is $\mathscr{B}([0,\infty))\otimes\mathscr{F}_t$-measurable for any Borel set $A\subset\mathbb{R}$. Since $(X_t)^{-1}(A)=\{\omega: X(t,\omega)\in A\}=\{X_t\in A\}$ is the $t$-cross section of $S$, that is $S_t=\{\omega\in\Omega:(t,\omega)\in S\}$, $\{X_t\in A\}\in\mathscr{F}_t$.

Going back to the OP, consider the stoping time $T$, and fix $t>0$.

*

*Consider the process $X^T:[0,\infty)\times\Omega\rightarrow\mathbb{R}$ defined by $X^T(s,\omega):=X(T(\omega)\wedge s,\omega)$. For fixed $\omega$, $X^T$ is continuous as a function of $s$. Define the map
$G_{T,t}:[0,\infty)\times\Omega\rightarrow[0,\infty)\times\Omega$ by
$$
G_{T,t}(s,\omega)=(T(\omega)\wedge t\wedge s,\omega)
$$
Since
$$
\{T\wedge t\leq u\}=\left\{\begin{array}{lcr}
\Omega& \text{if} & t\leq u\\
\{T\leq u\} &\text{if}& u< t
\end{array}\right.
$$
it follows that $\{T\wedge t\leq u\}\in\mathscr{F}_t$ for all $u\geq0$. In particular
$$
G^{-1}_{T,t}([0,u]\times A)=\big([0,u]\times A\big)\cup\big((u,\infty)\times(A\cap \{T\wedge t\leq u\})\big)\in\mathscr{B}([0,\infty)\otimes\mathscr{F}_t$$
Hence, $(X^T)^t=X^{T\wedge t}=X^t\circ G_{T,t}$ is progressively measurable. By (2), $X^T$ is also adapted to the filtration $\{\mathscr{F}_t:t\geq0\}$.


*To conclude, notice that for any Borel set $B\subset\mathbb{R}$,
$$
\{X_T\in B\}\cap\{T\leq t\}=\{X^{T\wedge t}\in B\}\cap\{T\leq t\}\in\mathscr{F}_t
$$
for all $t\geq0$ since $\{X^{T\wedge t}\in B\}\in\mathscr{F}_t$, and $T$ is a stoping time. Therefore, $X_T$ is $\mathscr{F}_T$-measurable.

Notes:

*

*Continuity of $X$ is used to shoe progressive measurability. In fact, we only used right-continuity.


*Statement (1) can be extended to left-right continuous and right-continuous processes.


*In the proof of measurability of $X_T$, only progressive measurability of $X$ plays a role. That is, if $X$ is a progressive measurable process and $T$ is a stoping time, then $X_T(\omega)=X(T(\omega),\omega)=X_{T(\omega)}(\omega)$ is $\mathscr{F}$--measurable.


*Many stochastic processes used in the theory of Markov processes, Stochastic analysis, and Stochastic differential equations are progressively measurable. In particular, the so called càdlàg (French: "continue à droite, limite à gauche") (English: "right continuous with left limits") which appear everywhere.
