Find stopping times $S$ and $T$ of Brownian motion $B$ such that $S\le T$, $E[S]$ finite, and $E[B(S)^2]>E[B(T)^2]$ This is exercise 2.12 of Peter Morters and Yuval Peres' book Brownian Motion:


  Find two stopping times $S\le T$ with $E[S]&lt;\infty$ such that $E[B(S)^2]&gt;E[B(T)^2]$.


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?
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.
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.