# A stopped canonical Markov process is a Markov process

I came across the following question while studying for a Stochastic Processes exam:

Consider the space $(\Omega,\mathcal F)$, where $\Omega$ is the space of real valued continuous functions, and $\mathcal F$ the associated path $\sigma$- algebra. Let $X_t=X_0+W_t$ be the canonical process on this space, where $W$ is a standard BM. For each distribution $\nu$ on $(\mathbb R,\mathcal B(\mathbb R))$ there exists a corresponding measure $\mathbb P_\nu$ on $(\Omega,\mathcal F)$ under which $X$ is a Markov process w.r.t the natural filtration $(\mathcal F_t^X):=(\mathcal F_t)$ with $X_0$ independent of $W$. Define $T=\inf\{t\geq 0: X_t\notin (a,b)\}$.

Show that the stopped process $X^T$ is Markov w.r.t the filtration $(F_{t\land T})$. It is not necessary to compute the transition function.

I think I have the vague outline of a proof, but it is missing some subtleties. First let us remember the shift operator, $\theta_t:\Omega\to \Omega$, defined for every $t\geq 0$ by $(\theta_t\omega)_s=\omega_{t+s}$ for every $s\geq 0$. It is possible to prove that $$\theta_{T\land (t+s)}=\theta_{T\land t}\circ\theta_{T\land s}\label{1}\tag{1}$$. For any $s,t\geq 0$, initial distribution $\nu$ and bounded measurable $f$ we need to prove that \begin{equation}\label{2}\tag{2} E_\nu[f(X_{t+s})|\mathcal F_{T\land s}]=E_{X_s}f(X_t)\quad \mathbb P_\nu ~ \text{a.s}. \end{equation} Intuitively it makes sense to say this is equivalent to proving that for $\mathbb P_\nu$ a.e. $\omega\in \Omega$ \begin{equation}\label{3}\tag{3} E_\nu[f(X_{T(\omega)\land(t+s)})|\mathcal F_{T(\omega)\land s}](\omega)=E_{X_{s\land T(\omega)}}f(X_{t\land T})(\omega). \end{equation} The definition of the canonical process along with the identity \eqref{1} mentioned above gives $$E_\nu[f(X_{T(\omega)\land(t+s)})|\mathcal F_{T(\omega)\land s}](\omega)=E_\nu[(f\circ X_{T\land t})\circ \theta_{T(\omega)\land s}|\mathcal F_{T(\omega)\land s}](\omega).$$ Now $f\circ X_{T\land t}$ is a bounded $(\mathcal F^X_\infty)$ measurable r.v, so by the generalized Markov property for the canonical process (we know $X$ is Markov) we have $$E_\nu[(f\circ X_{T\land t})\circ \theta_{T(\omega)\land s}|\mathcal F_{T(\omega)\land s}](\omega)=E_{X_{T(\omega)\land s}}f(X_{T\land t})(\omega).$$

Now I believe this gives the result if the claimed equivalence between statements \eqref{2} and \eqref{3} is actually correct. I am unable to convince myself that it is, as I do not believe that the relationship between a stopped filtration and conditional expectation is this simplistic. I'd thus appreciate any and all comments on my proof, and indeed any alternative proofs if my "proof" is junk.

we need to prove that $$E_\nu[f(X_{t+s})|\mathcal F_{T\land s}]=E_{X_s}f(X_t)\quad \mathbb P_\nu ~ \text{a.s}.$$

That's not correct, but perhaps it's just a typo... you actually need to prove that

\begin{equation} E_\nu[f(X_{\color{red}{T \wedge (t+s)}})|\mathcal F_{T\land s}]=E_{X_\color{red}{T \wedge s}}f(X_{\color{red}{T \wedge t}})\quad \mathbb P_\nu ~ \text{a.s}. \end{equation}

Regarding your proof: No, it doesn't work that way - you cannot simply plug in $\omega$ (... and even if you could, you would ran into trouble because of exceptional null sets which are showing up). However, you can fix the gap in your proof by invoking the strong Markov property of $(X_t)_{t \geq 0}$ which states that

$$\mathbb{E}_{\nu}(f(X_{\sigma}) \circ \theta_{\tau} \mid \mathcal{F}_{\tau}) = \mathbb{E}_{X_{\tau}}f(X_{\sigma}) \quad \text{\mathbb{P}^{\nu}-a.s.}$$

for stopping times $\tau$ and $\sigma$. If you use this identity for $\sigma := T \wedge t$ and $\tau := \sigma \wedge s$, you will obtain the desired Markov property of $(X_{t \wedge T})_{t \geq 0}$.

• Thanks for your answer. I should have trusted my instincts about the junk. And yes that was a typo. I was tentative using the strong Markov property, because as far as I'm aware we can only use it because $X$ is Feller-Dynkin, so $x\mapsto E_{x}f(X_s)$ is continuous. However the very next question in the paper involves proving that $X^T$ is Feller-Dynkin, and the one after that is using the Feller-Dynkin property to show that $X^T$ is strong Markov. Thus I was unsure if the markers would think I was "jumping the gun". – K.Power Jun 23 '18 at 13:11
• However, this subject is known for setting papers where the mark allocation and what you are allowed to use when is not very clear, so I am sure your solution is what they would be looking for. (It definitely is the simplest I believe) – K.Power Jun 23 '18 at 13:13
• @K.Power I doubt that there is a proof which does not involve the strong Markov property of $(X_t)_{t \geq 0}$ (... well, obviously, we can "reprove" the strong Markov property but that's not the point...) – saz Jun 23 '18 at 14:14