Conditioned Maximum of Brownian Motion 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.
 A: 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:
$$
  \mathbb{P}\left(M_t > a |B_t=M_t\right) = \mathbb{P}\left(M_t > a |M_t - B_t=0\right) = \lim_{y \to 0^+} \int_{a}^\infty f_{M_t|M_t-B_t}(x,y) \mathrm{d}x = \lim_{y \to 0^+} \exp\left(-\frac{a(a+2y)}{2t} \right) = \mathrm{e}^{-\frac{a^2}{2t}} 
$$
A: 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?
