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Let $(X_t)_{t\ge 0}$ be an $\mathscr{F}_t$ adapted, $d$-dimensional stochastic process with (right-)continuous paths and $\sigma$ be a stopping time.

The usual Markov process is the following relation: We have for all $t \ge 0$, $u \in \mathscr{B}_b(\mathbb{R}^d)$ and $P$ almost all $\omega \in \{\sigma < \infty\}$

$$E[u(X_{t+ \sigma})|\mathscr{F}_{\sigma+}](\omega) = E[u(X_t+x)]|_{x=X_\sigma(\omega)}=E^{X_\sigma(\omega)}u(X_t).$$

I would like to derive from this, the more general property that for all bounded $\mathscr{B}(C)/\mathscr{B}(\mathbb{R})$ measurable functionals $\Psi:C[0,\infty) \to \mathbb{R}$ which may depend on a whole path and $P$ almost all $\omega \in \{\sigma < \infty\}$ this becomes

$$E[\Psi(X_{\bullet+ \sigma})|\mathscr{F}_{\sigma+}]=E[\Psi(X_\bullet+ x)]|_{x=X_\sigma}=E^{X_\sigma}[\Psi(X_\bullet)].$$

I think the way to show this is using the monotone class theorem since $\mathscr{B}(C)$ is the intersection of $\mathscr{C}[0,\infty)$ and $\pi_t^{-1}(A)$ for $A \in \mathbb{R}^d$. Clearly, linearity and monotonicity conditions are satisfied, hence we need only show that for any finite $t_1< t_2 < \cdots <t_n$, and $A_1, \cdots , A_n$, we have the second relation for $\Psi = I(\pi_{t_1, \dots , t_n}^{-1}(A_1 \times \cdots \times A_n))$, where $I$ is the indicator. I think we can show this using the first relation, but I am having difficulty extending this. How could we extend this for $n>1$? I would greatly appreciate any help.

Arctic Char
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nomadicmathematician
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1 Answers1

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For fixed $t_1 < \ldots < t_n$ and measurable sets $A_i$, the functional $\Psi=1_{\pi_{t_1,\ldots,t_n}^{-1}(A_1 \times \ldots \times A_n)}$ satisfies $$\Psi(X_{\bullet+\sigma}) = \prod_{j=1}^n 1_{A_j}(X_{t_j+\sigma}).$$

By the tower property, we have

\begin{align*} \mathbb{E}(\Psi(X_{\bullet+\sigma}) \mid \mathcal{F}_{\sigma+}) &= \mathbb{E} \bigg[ \mathbb{E} \bigg( \prod_{j=1}^n 1_{A_j}(X_{t_j+\sigma}) \mid \mathcal{F}_{(t_{n-1}+\sigma)+} \bigg) \mid \mathcal{F}_{\sigma+} \bigg]. \end{align*}

Since the random variables $X_{t_j+\sigma}$ are $\mathcal{F}_{(t_{n-1}+\sigma)+}$-measurable for $j=1,\ldots,n-1$, it follows using the Markov property (from your first display) that \begin{align*} \mathbb{E}(\Psi(X_{\bullet+\sigma}) \mid \mathcal{F}_{\sigma+}) &= \mathbb{E} \bigg[ \prod_{j=1}^{n-1} 1_{A_j}(X_{t_j+\sigma}) \mathbb{E}(1_{A_n}(X_{t_n+\sigma}) \mid \mathcal{F}_{(t_{n-1}+\sigma)+}) \mid \mathcal{F}_{\sigma+} \bigg] \\ &=\mathbb{E} \bigg[ \prod_{j=1}^{n-1} 1_{A_j}(X_{t_j+\sigma}) \mathbb{E}^{X_{t_{n-1}+\sigma}}(1_{A_n}(X_{t_n-t_{n-1}})) \mid \mathcal{F}_{\sigma+} \bigg]. \end{align*}

Now we repeat the procedure: next we condition on $\mathcal{F}_{(t_{n-2}+\sigma)+}$ and obtain that

\begin{align*} \mathbb{E}(\Psi(X_{\bullet+\sigma}) \mid \mathcal{F}_{\sigma+}) = \mathbb{E} \bigg[ \prod_{j=1}^{n-2} 1_{A_j}(X_{t_j+\sigma}) \mathbb{E}^{X_{t_{n-2}+\sigma}}\bigg(1_{A_{n-1}}(X_{t_{n-1}-t_{n-2}}) \mathbb{E}^{X_{t_{n-1}-t_{n-2}}}(1_{A_n}(X_{t_n-t_{n-1}})) \bigg) \mid \mathcal{F}_{\sigma+} \bigg]. \end{align*}

By iteration, this gives

$$\mathbb{E}(\Psi(X_{\bullet+\sigma}) \mid \mathcal{F}_{\sigma+})=\mathbb{E}^{X_\sigma}\bigg(1_{A_1}(X_{t_1}) \mathbb{E}^{X_{t_1}}\big(1_{A_2}(X_{t_2-t_1}) \mathbb{E}^{X_{t_2-t_1}}( \ldots \mathbb{E}^{X_{t_{n-1}-t_{n-2}}}(1_{A_n}(X_{t_n-t_{n-1}})) \big) \bigg)$$

The right-hand side is nothing but

$$\mathbb{E}^{X_{\sigma}} \left( \prod_{j=1}^n 1_{A_j}(X_{t_j}) \right);$$

this follows using a very similar reasoning to that in the above calculation (essentially you can put $\sigma=0$ and replace $\mathbb{E}$ by $\mathbb{E}^{X_\sigma}$; if you prefer to do it by hands, then use the tower property to condition first on $\mathcal{F}_{t_{n-1}}$ and use Markov, then condition on $\mathcal{F}_{t_{n-2}}$ and so on.)

saz
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