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Let $W_t$ be a Brownian motion and

$$ M_t = \max_{o<s<t} W_s $$

Can anyone give me some insights on how to prove:

$$ P[M_t >a \mid W_t=M_t]= \exp(-a^2/2t) \quad;\quad a>0 $$

Many thanks in advance.

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1 Answer 1

up vote 3 down vote accepted

The joint distribution of $(B_t, M_t)$ is well-known. The probability density reads: $$ f_{(B_t, M_t)}(x,y) = \sqrt{\frac{2}{\pi}} \frac{2y-x}{t^{3/2}} \exp\left(-\frac{(2y-x)^2}{2t} \right) [ y>0, x \leqslant y ] $$ where $[ y>0 ]$ denotes Iverson bracket.

From here the distribution of $(M_t, M_t-B_t)$ is easy to read off: $$ f_{(M_t, M_t-B_t)}(x,y) = \sqrt{\frac{2}{\pi}} \frac{x+y}{t^{3/2}} \exp\left(-\frac{(x+y)^2}{2t} \right) [ y>0, x > 0 ] $$ Now finding the conditional probability density is also straightforward. Assuming $y>0$: $$ f_{M_t|M_t-B_t}(x|y) = \frac{f_{M_t,M_t-B_t}(x,y)}{f_{M_t-B_t}(y)}= \frac{x+y}{t} \exp\left(-\frac{x(x+2y)}{t} \right) [x >0] $$ This will now allow you to find the quantity of interest: $$ \mathbb{P}\left(M_t > a |B_t=M_t\right) = \mathbb{P}\left(M_t > a |M_t - B_t=0\right) = \lim_{y \to 0^+} \int_{a}^\infty f_{M_t|M_t-B_t}(x,y) \mathrm{d}x = \lim_{y \to 0^+} \exp\left(-\frac{a(a+2y)}{2t} \right) = \mathrm{e}^{-\frac{a^2}{2t}} $$

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Thank you very much Sasha! Very clear answer! The question is indeed homework. Proper tag already added. –  RCA Sep 12 '12 at 13:54
    
@MichaelChernick Thank you for the edits, and sorry for being sloppy in my grammar. –  Sasha Sep 12 '12 at 17:04
    
I have a follow up question: Is there a simple proof for the joint density of $(M_t, B_t)$ that Sasha gives above? –  Ben Sep 12 '12 at 21:33
    
@Ben See sect. 3.8 of Fima Klebaner's "Introduction to Stochastic Calculus with Applications". Reflection principle is used to evaluate $\mathbb{P}\left(B_t \leq x, M_t \geq y\right)$ and is found to be $1-\Phi\left((2y-x)/\sqrt{t}\right)$. The density follows by differentiation. –  Sasha Sep 12 '12 at 21:38
    
@Sasha. Thanks. –  Ben Sep 13 '12 at 10:20

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