Let $B_t$ be standard Brownian motion and $a < 0 < b$. Define stopping time $T$ as follows. $$T = \min \{t \geq 0: B_t \in \{a, b\} \}.$$ The expectation of $T$ is $\mathbb ET = |a|b$ and can be found here. The question now is how to find the expectation of the square of $T$, i.e., $\mathbb E T^2$. Following the hint, one also needs the iterated law of expectations.

  • $\begingroup$ @20824: There must be some sign mistake because you find a negative value for $\Bbb E[T^2]$. $\endgroup$
    – Siméon
    Sep 2 '14 at 9:25
  • $\begingroup$ @Siméon Yes, you are right. I think I fixed it. BTW, I would like to delete my answers so that other people can have a try. $\endgroup$
    – LaTeXFan
    Sep 2 '14 at 12:52

Let us start from the answer of this question: Expectation of Stopping Time w.r.t a Brownian Motion.

  1. Use the martingale $B_t^3 - 3tB_t$ to compute $E[T \mid B_T = a]$ and $E[T \mid B_t = b]$. Deduce the value of $E[TB_t^2]$.
  2. Use the martingale $B_t^4 - 6tB_t^2 + 3t^2$ to compute $E[T^2]$.
  • $\begingroup$ How were you able to come up with these martingales? $\endgroup$ May 2 '20 at 1:47

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