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Let $W_t$ be a Brownian motion and

$$ M_t = \max_{o

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.

2 Answers 2

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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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    @MichaelChernick Thank you for the edits, and sorry for being sloppy in my grammar.2012-09-12
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    I have a follow up question: Is there a simple proof for the joint density of $(M_t, B_t)$ that Sasha gives above?2012-09-12
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    @Ben See sect. 3.8 of Fima Klebaner's "[Introduction to Stochastic Calculus with Applications](http://www.amazon.com/Introduction-Stochastic-Calculus-With-Applications/dp/1848168314)". 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.2012-09-12
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    @Sasha. Thanks.2012-09-13
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I have the following question related with the same problem.

I was trying to solve this problem with the following reasoning:

  • I use the join distribution of $M_t=x$ and $X_t=y$ $f_{M_t,X_t}(x,y)=\frac{2}{t\sqrt{2\pi t}}(2x-y)\exp\{ -\frac{1}{2t}(2x-y)^2\}$

  • I use the distribution of $X_t=y$ $f_{X_t}(y)=\frac{1}{\sqrt{2\pi t}}\exp\{-y^2/2t\}$

  • The and use $f_{M_t|X_t=y}(x|y)=\frac{f_{M_t,X_t}(x,y)}{f_{X_t}(y)}$

So I get the following result $f_{M_t|X_t=y}(x|y)=\frac{2}{t}(2x-y)\exp\{-\frac{2}{t}x(x-y)\}$

Finally I sustitute $y=x-h$ and get that $f_{M_t|X_t=y}(x|y)=\frac{2}{t}(x+h)\exp\{-\frac{2}{t}x(h)\}$

And when I try to compute $\lim_{h\rightarrow0}$ I dont get the same solution.

Why?

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    These kind of things should be put in a seperate question.2018-10-26
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    How can I put it in another question?2018-10-26
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    By asking a new question?2018-10-26
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    Ok, I added a new question [here](https://math.stackexchange.com/questions/2973030/conditioned-maximum-of-brownian-motion-alternative-metthod)2018-10-27