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.

• You can try approximate $\tau$ by $\tau_n = \frac{k+1}{2^n}$ if $\tau \in [\frac{k}{2^n},\frac{k+1}{2^n})$ which is a discrete stopping time. Show that for discrete stopping time $\eta$, $X_{\eta}$ is a random variable (hint: $X_{\eta} = \sum_{k=1}^\infty X_{\eta_k} 1_{\{\eta = \eta_k\}}$, where $\{\eta_k\}_{k \in \mathbb N_+}$ are values that $\eta$ can take). Then by continuity of process (right continuity is enough here), show that $X_{\tau_n} \to X_{\tau}$ almost surely, hence $X_{\tau}$ is a random variable as a limit of random variables. Commented Aug 4, 2020 at 14:31
• Why doesn't $X_{\tau_n}\to X_{\tau}$ everywhere? Otherwise, the underlying prob. space needs to be complete. Commented Aug 4, 2020 at 17:05

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 $$\mathscr{B}([0,\infty))\otimes\mathscr{F}$$.

1. 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.

1. 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.