# Question on the proof of strong Markov property

I am studying strong Markov property and the following question came up.

Let $$Y_{ s}( \omega)$$ be a bounded jointly measurable stochastic process and $$X_{ s}( \omega)$$ be a standard Brownian motion for $$s \geq 0$$, $$\omega \in \Omega$$ . Let $$P^ { x}$$ be the probability associated with the Browman motion starting at $$x \in \mathbb R$$ and $$E^{ x}$$ be the corresponding expectation. Finally, let $$\tau$$ be a stopping time and $$\mathcal F_{\tau}$$ be the associated sigma-algebra.

I want to show:

$$\phi (X _{\tau},\tau) \ \text{ is} \ \mathcal F_{\tau } \ \text{ measurable on }\{\tau < \infty \}\ \text{ where } \ \phi ( y, s):= E ^{ y} [ Y_{ s}]$$ for $$s \geq 0$$ and $$y \in \mathbb R$$.

I know that $$\tau$$ and $$X_{\tau} 1_{\tau < \infty}$$ are $$\mathcal F_{\tau}$$ measurable and want to use this fact, but I cannot proceed on.

Any help is appreciated.

• When you say that $Y$ is (jointly) measurable process, is it with respect to the filtration $\mathcal{F}$ ? Sep 5, 2019 at 9:00
• @TheBridge Yes, the big one Sep 5, 2019 at 9:13
• $\phi(X_{\tau},\tau) = E^{X_{\tau}}[Y_{\tau}]$ is the composition of $y\mapsto E^yY_{\tau}$ (which is non-random, so its measurability is not an issue) and $\omega \mapsto X_{\tau}(\omega)$. So you only have to show that $X_{\tau}$ is $\mathcal{F}_{\tau}$ measurable. Sep 5, 2019 at 10:54
• @Sayantan We are not considering such composition; we take expectaion of each section of $Y$ and then insert randomness again in two different places. I edited question to be more clear. Sep 5, 2019 at 13:53

I came up with the following solution.

It suffices to show $$\phi$$ is jointly Borel m'ble.

Let a random variable $$Y$$ be called special if $$Y(\omega)=\prod_{i=1} ^m f_i(\omega(t_i))$$ for some $$f_i \in C_b(\mathbb R)$$ and $$t_i \geq 0$$. In other words, $$Y=\prod _{i=1} ^m f_i \circ \pi_{t_i}$$ where $$\pi_{t_i}$$ is projection onto time $$t_i$$ i.e., $$\pi_{t_i} (\omega) := \omega(t_i)$$.

$$B \subseteq \Omega$$ is a finite dimensional set if $$B=\bigcap_{i=1}^m \pi_{t_i}^{-1}(A_i)$$ for some $$t_i \geq 0$$ and open sets $$A_i \subseteq \mathbb R$$.

STEP 1

Consider the case $$Y_s(\omega)=f(s)Y(\omega)$$ for some $$f \in C_b(\mathbb R)$$ and $$Y$$ special.

Using DCT, it can be shown that $$y \mapsto E^y(Y)$$ is continuous since $$Y$$ is special .

Then $$\phi(y,t)=f(s)E^y(Y)$$ is continuous, hence measurable.

STEP2

Consider the case $$Y_s(\omega)= 1 _U Y$$ for some open set $$U \subset \mathbb R$$.

It is a fact that $$1_U$$ is a monotone pointwise limit of continuous bounded functions. (generally holds in LCH spaces using Urysohn lemma, but can be explicitly constructed in metric spaces)

Then, using STEP1, $$\phi(y,t)=1_U E^y(Y)$$ is pointwise limit of measurable functions, hence measurable.

STEP3

Consider the case $$Y_s(\omega)=1_U(s) 1_B(\omega)$$ for some open $$U\subseteq \mathbb R$$ and finite dimensional $$B=\bigcap_{i=1}^m \pi_{t_i}^{-1}(A_i)$$.

Claim: $$1_B$$ is monotone pointwise limit of some special random variables.

proof. \begin{align} 1_B &=1_{\bigcap_{i=1}^m \pi_{t_i}^{-1}(A_i)} = \prod_{i=1}^m 1_{A_i} \circ \pi_{t_i} \\&= \prod_{i=1}^m \lim_{n \to \infty} f_n^{(i)} \circ \pi_{t_i} = \lim_{n \to \infty} \prod_{i=1}^m f_n^{(i)} \circ \pi_{t_i} \end{align} for some $$f_n^{(i)} \in C_b(\mathbb R)$$ such that $$f_n^{(i)} \uparrow 1_{A_i}$$ as $$n \to \infty$$ for each $$i$$. ////

Then, using STEP2 and MCT, $$\phi(y,t)=1_U(t) E^y 1_B$$ is monotone pointwise limit of measurable functions, hence measurable.

STEP4

Now, we can use the monotone class theorem. Note that $$\mathcal P = \{U\times B: U \text{ open in } \mathbb R , B \text{ is finite dimensional}\}$$ is a $$\pi$$-system generating the product $$\sigma$$-algebra on $$\mathbb R \times \Omega$$. Let $$\mathcal H:=\{Y_{s}(\omega): Y_{s}(\omega) \text{ is bounded jointly measurable and } \phi(y,s) = E^y(Y_s) \text{ is jointly Borel measurable}\}$$ Firstly, constant $$1 \in \mathcal H$$ is obvious. Secondly, $$U\times B \in \mathcal P \implies 1_{U\times B} \in \mathcal H$$ follows from STEP3. Lastly, $$0\leq Y_n \uparrow Y$$ for some $$Y_n \in \mathcal H$$ and $$Y$$ bounded implies $$Y\in \mathcal H$$ by MCT and the fact that pointwise limit of measurable is measurable.

Thus, by monotone class theorem, $$\mathcal H$$ contain all bounded $$\sigma(\mathcal P)$$ measurable function, which yields the result.