Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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*}

share|improve this answer
    
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
1  
That's multiple steps. Here is an example step by step (7); galton.uchicago.edu/~lalley/Courses/390/Lecture5.pdf –  gnometorule Dec 18 '12 at 2:48
    
@Mathematics, what version of the reflection principle do you know? –  Yury Dec 18 '12 at 2:50
    
@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
1  
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.