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I wish to calculate $\mathbb{E}(\tau)$ and $\mathbb{E}(W_\tau)$, where $\tau=\sup\{t\leq1:W_t=1\}$ (i.e. the last hitting time).

In a previous question, we determined that $\tau$ is not a stopping time, and so, unlike the case for the first hitting time $\inf\{t\leq1:W_t=1\}$, the optional stopping theorem cannot be used. How would $\mathbb{E}(\tau)$ or $\mathbb{E}(W_\tau)$ then be calculated? I'm guessing the reflection principle should be used, but I can't think of how to bring it in (particularly to capture the distribution of $\tau$).

Any ideas would be greatly appreciated, thanks!

Ice Tea
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  • Hint: With the conventions in your previous question, $W_\tau$ is a Bernoulli random variable. – Chris Janjigian Nov 01 '21 at 01:16
  • @ChrisJanjigian Hi, thanks for the comment. How is it so? Isn't it Bernoulli only at $t=0$? – Ice Tea Nov 01 '21 at 22:10
  • Brownian motion is continuous, so $W_\tau = 1$ if there exists $t \in [0,1]$ with $W_t = 1$. If no such $t$ exists, then $W_\tau = W_0 = 0$. This is a Bernoulli(p) random variable, with $p=P(W_t = 1 \text{ for some }t \in [0,1]).$ E: Since the question has now changed, computing $\mathbb{E}[\tau]$ is harder. For that, try thinking about the time reversal $\tilde{B}{t} = B{1-t}$, for which $\tau$ is a stopping time. – Chris Janjigian Nov 01 '21 at 22:24

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