# What is the expectation of $E(\min_{1\le s\le 2}B_S)$?

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*}
Why $\Pr[\min_{1\le s\le 2}(B_s - B_1) \leq t] = 2 \Pr[B_2 - B_1 \leq t].$? The version i learnt about the reflection principle didn't involved any min RV –  Mathematics Dec 18 '12 at 2:45
@Yury Let $T_a\le t$ be the first time the brownian motion process hits a. $P(X_t\ge a|T_a\le t)=\frac{1}{2}$ –  Mathematics Dec 18 '12 at 2:57
This is an equivalent formulation: You get for $X_t = B_{1+t} - B_1$, $\Pr[\min_{0\leq t\leq 1} X_t \leq a] =\Pr[T_a \leq 1] = 2\Pr[X_1 \leq a \text{ and } T_a \leq 1] = 2\Pr[X_1 \leq a]$. –  Yury Dec 18 '12 at 3:07