Suppose $H$ and $K$ are open subsets of $\mathbb{R}^d$ containing the origin with $H\subset K$, $B_t$ a standard Brownian motion starting at the origin, $\mathcal{F}_t$ its canonical filtration, and $\tau_H$ and $\tau_K$ the first exit times.
For some fixed time $t>0$, I'm having trouble interpreting the r.v. $\mathbb{E}[1_{t\leq \tau_K}|\mathcal{F}_{t\wedge\tau_H}]$. Specifically, if I'm given some event $E$ (not necessarily $\mathcal{F}_{t\wedge\tau_H}$-measurable), is there a way to interpret the expectation $\mathbb{E}[1_E\mathbb{E}[1_{t\leq \tau_K}|\mathcal{F}_{t\wedge\tau_H}]]$?
I understand the abstract definition of conditional expectation, and I apologize for the somewhat imprecise phrasing, but I'm looking for some intuition on how to interpret conditioning on such a stopping time $\sigma$-algebra. Thank you.
Best Answer
You may think of the conditional expectation as follows: $\mathbb{E}(X|\mathcal{F})$ for a r.v. $X$ is the average value of $X$ based on our knowledge of "information" that is given by the sigma-algebra $\mathcal{F}$. In particular, if $\mathcal{F}$ is the trivial sigma-algebra (consisting of $\varnothing,\Omega$), then the expectation is just $\mathbb{E}(X)$, i.e., we know very little "information".
Back to your example: since $H\subset K$, you must first exit $H$ and then $K$. So if $t<\tau_H$, then $t<\tau_K$. The sigma-algebra $\mathcal{F}_{t\wedge \tau_H}$ consists of the "information" you know up till moment $t$ or up till you exit $H$ (e.g., rigorously: the point at which you exit $H$ is $\mathcal{F}_{t\wedge \tau_H}$-measurable). The expectation $\mathbb{E}(1_{t\le \tau_K}|\mathcal{F}_{t\wedge \tau_H})$ is the average of the probability that you've exited $K$ by time $t$, but in calculating this probability you use the "information" from $\mathcal{F}_{t\wedge \tau_H}$, and thus, the probability that you've exited $K$ depends on this "information". Informally, if you exit $H$ from one point, you're more likely to exit $K$ sooner than if you exit $H$ from another point.
As for the second expectation, I can't think of something equally illustrative (as it seems to me) as above, than just say that you've given two events: $E$ and the "event" $\mathbb{E}(1_{t\le \tau_K}|\mathcal{F}_{t\wedge \tau_H})$, which depends on the "information" from $\mathcal{F}_{t\wedge \tau_H}$, and you compute the probability of their intersection.