BM and interpretation of stopping time sigma algebra

Question

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.

Answer 1

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.

Answer 2

I am not sure what kind of interpretation you want, but heuristically, $\mathcal{F}_{t\wedge\tau_H}$ is the information we have from observing the Brownian motion up until the time $t\wedge\tau_H$. Then $E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]=P(t\le\tau_K\mid\mathcal{F}_{t\wedge\tau_H})$ is the probability that $B_t$ is still in $K$, given this information.

As for $E[1_EE[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]]$, one thing you might observe (and again this may not be the kind of thing you are looking for) is the following: \begin{align*} E[1_EE[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]] &= E[E[1_EE[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}] \mid\mathcal{F}_{t\wedge\tau_H}]]\\\\ &= E[E[1_E\mid\mathcal{F}_{t\wedge\tau_H}] E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]]\\\\ &= E[P(E\mid\mathcal{F}_{t\wedge\tau_H}) P(t\le\tau_K\mid\mathcal{F}_{t\wedge\tau_H})]. \end{align*} So informally, you observe the Brownian path until time $t\wedge\tau_H$, compute the conditional probabilities of $E$ and $\{t\le\tau_K\}$, multiply them together, and then average the result over all such observations.

Edit:

Perhaps I have misunderstood your question. I wanted to address your comment, "By $E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]$ I mean the $\mathcal{F}_{t\wedge\tau_H}$-measurable r.v. It is not a probability anymore." I am not yet permitted by the software to post comments of my own, so I must address it as an edit to my answer.

The r.v. $E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]$ is a [0,1]-valued, measurable function of the Brownian path up until time $t\wedge\tau_H$. Given such a path, say $\omega$, the number $E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}] (\omega)\in[0,1]$ is generally interpreted (albeit informally) as the probability that $t\le\tau_K$, given that the Brownian path was $\omega$. This was the genesis of the final comment in my original answer.