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Let $\{B_t:t\ge0\}$ be a standard brownian process. What is the expectation of $E(\min_{1\le s\le 2}B_S)$?

I think the problem is i am not sure how $\min_{1\le s\le 2}B_S$ is distributed. I try that $\min_{1\le s\le 2}B_S=X_s$, $P(X_s\le x)=P(x\le B_s)=1-P(B_s\le x) \text{for}\ s\in[1,2]$ but i am not sure how to proceed or is there any other easier method to find the expectation?

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We have, ${\mathbb E}[\min_{1\le s\le 2}B_s] = {\mathbb E}[B_1 + \min_{1\le s\le 2}(B_s-B_1)]={\mathbb E}[B_1] + {\mathbb E}[\min_{1\le s\le 2}(B_s-B_1)]={\mathbb E}[\min_{1\le s\le 2}(B_s-B_1)].$

By the Reflection Principle, for $t\leq 0$, we have $\Pr[\min_{1\le s\le 2}(B_s - B_1) \leq t] = 2 \Pr[B_2 - B_1 \leq t].$ Note that $B_2 - B_1 \sim\cal{N}(0,1)$. Thus \begin{align*} \mathbb{E}[\min_{1\le s\le 2}(B_s - B_1)] &= -\int_0^\infty \Pr[\min_{1\le s\le 2}(B_s - B_1) \leq -t]dt \\ &= -2\int_0^\infty \Pr[B_s - B_1 \leq -t]dt=-\mathbb{E}[|B_s-B_1|] = -\sqrt{\frac{2}{\pi}}. \end{align*}

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    o i see, i didn't notice $Pr(\min X_t\le a)=Pr(T_a\le 1)$2012-12-18