Proof of strong Markov property in Oksendal

Question

I am reading Oksendal's SDE book on the strong Markov property for Ito diffusions. There is a step in the proof that I cannot follow. Here is some notation:

$\mathcal{F}_t$ is the sigma-field generated by the Brownian motion $B_r, r \le t$. $E^{x}[\cdot]$ denotes the expected value assuming that the diffusion process $X_t$ satisfies $X_0=x$.

Let $f_1,f_2$ be bounded Borel functions and $\tau$ be a $\mathcal{F}_t$-stopping time. Let $0 \le h_1 \le h_2$.

Then $E^{x}[f_1(X_{\tau + h_1}) f_2(X_{\tau+h_2}) \, | \, \mathcal{F}_{\tau}] = E^{X_{\tau}}[f_1(X_{h_1}) f_2(X_{h_2})]$.

The last line of the proof goes as follows: $E^{X_{\tau}}[f_1(X_{h_1}) E^{x}[f_2(X_{h_2}) \, | \, \mathcal{F}_{h_1}]] = E^{X_{\tau}}[f_1(X_{h_1}) f_2(X_{h_2})]$.

My question is: it seems that the proof is saying that $ E^{x}[f_2(X_{h_2}) \, | \, \mathcal{F}_{h_1}] = f_2(X_{h_2})$. This can happen if $X_{h_2}$ is $\mathcal{F}_{h_1}$ measurable which I don't understand since $h_2 \ge h_1$. Am I missing something? I would appreciate some clarification.

Answer 1

Note that $E^x[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]$ does not depend on $x$. This is apparent from the previous line of Øksendal's derivations or, equivalently, from the ordinary Markov property. (It equals $E^{X_{h_1}}[f_2(X_{h_2-h_1})]$, where $E^y[f_2(X_{h_2-h_1})] = E[f_2(X^y_{h_2-h_1})]$.)

Thus for any $z\in\mathbb{R}$, we find that \begin{align*} E^z\!\left[f_1(X_{h_1})E^x[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]\right] &= E^z\!\left[f_1(X_{h_1})E^z[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]\right] \\ &= E^z\!\left[E^z[f_1(X_{h_1})f_2(X_{h_2}) \, | \, \mathcal{F}_{h_1}]\right] = E^z[f_1(X_{h_1})f_2(X_{h_2})], \end{align*} where the last equation follows from the law of iterated expectations. Consequently, $$ E^{X_{\tau}}\!\left[f_1(X_{h_1})E^x[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]\right] = E^{X_{\tau}}[f_1(X_{h_1})f_2(X_{h_2})], $$ as desired.

I am very happy about your question; it might at first seem to simply be resolved by the law of iterated expectations, but as my answer attempts to highlight, there is a bit more to it. (Good understanding of the ordinary Markov property is a must.)