MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
By $\mathbb{E}[1_{t\wedge\tau_K}|\mathcal{F}_{t\wedge\tau_H}]$ I mean the $\mathcal{F}_{t\wedge\tau_H}$-measurable r.v. It is not a probability anymore. When you take the expectation of it times an indicator function, then maybe it can be interpreted in terms of probabilities (the reason for the question that starts "Specifically.."). If $E$ is $\mathcal{F}_{t\wedge\tau_H}$-measurable, then that expectation really is the probability of $E$ intersect $t\leq \tau_K$. Not quite the case when $E$ is not $\mathcal{F}_{t\wedge\tau_H}$-measurable. It is this interpretation I am looking for. – mcecil Jun 21 '11 at 1:50

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.

share|cite|improve this answer

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.


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.