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This is exercise 2.12 of Peter Morters and Yuval Peres' book Brownian Motion:

Find two stopping times $S\le T$ with $E[S]<\infty$ such that $E[B(S)^2]>E[B(T)^2]$.

I considered about deterministic stopping times, but it does not work. And by Wald's Lemma, it seems we need to find a stopping time $T$ with $E[T]=\infty$. While Wald's second lemma says $E[B(S)^2]=E[S]$. May I get some hint about it?

Did
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Connor
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  • $$S=\inf{t\mid B_t\in{-1,2}}\quad T=\inf{t\mid B_t=-1}$$ – Did Apr 15 '17 at 22:12
  • I checked these stopping times satisfying all requirements except $E[S]<\infty.$ I know $E[T]=\infty$ and $P(S<\infty)=1$. But I don't know if $E[S]<\infty.$ I try to use $E[S]=\int_0^\infty P(S>x)dx$ and use $P(S>x)\le P(B_s\in (-1,2))$ but it seems does't work. I wonder if there is a way to show $E[S]<\infty$. – Connor Apr 15 '17 at 23:59
  • I find this question solving it. https://math.stackexchange.com/questions/261889/expectation-of-stopping-time-w-r-t-a-brownian-motion – Connor Apr 16 '17 at 00:07
  • $E(S)=(0-(-1))(2-0)=2$. – Did Apr 16 '17 at 07:51

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Your answer is correct. In general if $S=\inf\{t:|B(t)|=a\}$ and $B(0)=x$, then $\mathbb{E}[S]=a^2-x^2$. You can find the expectation of your $S$ by shifting but maybe just make $S=\inf\{t:B(t)\in\{-1,1\}$ instead.

Pepe Silvia
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