6
$\begingroup$

Let $B_t,\;t\geq0$ be a standard Brownian motion. Define the stopping time $$\tau = \min_t\{B_t^2\geq t+1\}$$

Is the expected value $E(\tau)$ finite?

Actually, my raw problem as following: $$\gamma = \min_t\{ B_t = \sqrt{t+1}\quad \text{or}\quad B_t=-1\}$$ I want to prove $E(\gamma) < \infty$ by $E(\tau) < \infty$.


As commented by @zhoraster, since $B_t^2-t$ is a martingale, using stopping time theorem we have $$E(B_\tau^2-\tau) = E(B_0^2) = 0$$ However, $E(B_\tau^2-\tau) = E(1) = 1$, is this imply $E(\tau) = \infty$ ?


So I can't prove $E(\gamma) < \infty$ by $\tau$. I have simulated using compute program which shows that $E(\gamma) < 10$.

$\endgroup$
5
  • 2
    $\begingroup$ Any own thoughts on the problem? Hint: $B^2_t - t$ is a martingale. $\endgroup$
    – zhoraster
    Commented Oct 12, 2015 at 14:00
  • $\begingroup$ If the hypotheses of the optional stopping theorem hold, $E(B_\tau^2-\tau)$ would have to be zero. But it is $1$. So the hypotheses can't hold. What do you conclude about $\tau$? $\endgroup$
    – Ian
    Commented Oct 12, 2015 at 15:11
  • $\begingroup$ Note that your original problem is rather different: when $\tau$ is large, you will probably have $B_\gamma=-1$. $\endgroup$
    – Ian
    Commented Oct 12, 2015 at 15:15
  • $\begingroup$ Seems that the expectation is finite, but there is no great margin. I bet that with upper bound being $1.2\sqrt{t+1}$ it is infinite. Still no time at the moment to think about this properly... $\endgroup$
    – zhoraster
    Commented Oct 13, 2015 at 11:09
  • $\begingroup$ I've updated the answer. $\endgroup$
    – zhoraster
    Commented Feb 5, 2016 at 7:18

1 Answer 1

6
$\begingroup$

Your guess is correct but is missing some important details. In order to use the optional sampling theorem, you need some uniform integrability or boundedness of $\tau$.

But it is possible to complete this argument. Assume that $E[\tau]<\infty$ (in particular, $\tau<\infty$ a.s.). Then, by the optional sampling theorem, for each $n\ge 1$ $$ E[B^2_{\tau\wedge n}] = E[\tau\wedge n]. $$ Therefore, letting $n\to\infty$, we get by the Fatou lemma (in the lhs) and the monotone/dominated convergence theorem (in the rhs), $$ E[B^2_\tau]\le E[\tau], $$ whence, as you noted, it follows that $1\le 0$, which is absurd.

This answers the question as originally posted.


For a modified question (with lower bound being $-1$), the expectation of the stopping time is finite.

Idea First, it can be shown that the stopping time $$ \sigma = \inf\left\{s\ge 0: |B_s| = \frac12(1+\sqrt{1+s})\right\} $$ is integrable.

Indeed, for any $t\ge 0$ $$ 0=E\left[B^2_{\sigma\wedge t}-\sigma\wedge t\right]\le E\left[\frac14\left(1+\sqrt{1+\sigma\wedge t}\right)^2-\sigma\wedge t\right]\\= \frac12 + E\left[\frac12\sqrt{1+\sigma\wedge t}-\frac34\sigma\wedge t\right] \le 1 - \frac12 E[\sigma\wedge t], $$ whence $E[\sigma\wedge t]\le 2$. Letting $t\to+\infty$, we get $E[\sigma]\le 2$ by virtue of the Fatou lemma.

Now for each $t>0$ $$ P(\gamma>t) = P(W \in G_t) $$ with $$ G_t = \big\{f: \forall s\in[0,t]\ f(s)\in (-1,\sqrt{1+s})\big\}. $$ Similarly, for each $t>0$ $$ P(\sigma>t) = P(W \in S_t). $$ It is easy to see that $S_t = \frac12(G_t - G_t)$. Then, by the log-convexity of the distribution of $W$, $$ P(\sigma>t) = P(W \in S_t) \ge \left(P(W \in G_t)P(W \in -G_t)\right)^{1/2} = P(W\in G_t) = P(\gamma>t), $$ where we have used the symmetry of $W$. Therefore, $\gamma$ also has finite expectation, moreover, $E[\gamma]\le E[\sigma]<2$.

$\endgroup$
2
  • $\begingroup$ Could you explain why stopping time $\sigma$ is integrable? $\endgroup$ Commented Apr 2, 2017 at 12:52
  • $\begingroup$ @Cofibration, sorry for late response; see updated answer. $\endgroup$
    – zhoraster
    Commented Apr 6, 2017 at 15:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .