0
$\begingroup$

In Karatzas and Shreve pg. 244 Chapter 4.2, it states the following equation:

$$u(a) = E^{a}f(W_{\tau_D}) = E^a\{E^a[f(W_{\tau_D})|\mathcal{F}_{\tau_{a+B_r}}]\} = E^a\{u(W_{\tau_{a+B_r}})\}$$

where he claims the last equality holds by Strong Markov Property. Note that $\tau_A := \inf\{t\geq 0 : W_t \in A^c\}$ and $u(x) := E^xf(W_{\tau_D})$, $x\in \bar{D}$. However, when I try to apply SMP, I arrive at:

$$E^a[f(W_{\tau_D})|\mathcal{F}_{\tau_{a+B_r}}](\omega) = E^{W_{\tau_{a+B_r}(\omega)}(\omega)}[f(W_{\tau_D(\omega) - \tau_{a+B_r}(\omega)}(\cdot))]$$

How can I get from here to $u(W_{\tau_{a+B_r}(\omega)}(\omega))$? I'm stuck because the expectation depends on $\omega$ inside but definition of $u(x)$ does not depend on $\omega$ inside.

$\endgroup$
2
  • 1
    $\begingroup$ Seems like this is the one you asked in the help room. I realized that Proposition 6.17 is not actually applicable in this case since $\tau_D$ is definitely not measurable with respect to $\mathcal F_{\tau_{a+B_r}}$, so we were looking at the wrong statement entirely. The actual statement of what's usually referred to as the SMP can be found in the previous section, see for instance Problems 6.9 and 6.11 on pages 83-84. $\endgroup$
    – shalop
    Commented Apr 13, 2018 at 3:52
  • $\begingroup$ You are absolutely right. Thanks @Shalop! $\endgroup$
    – James Yang
    Commented Apr 13, 2018 at 18:25

0

You must log in to answer this question.

Browse other questions tagged .