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Here is the question:

$(B_t,t\ge 0)$ is a standard brwonian motion, starting at $0$. $S_t=\sup_{0\le s\le t} B_s$. $T=\inf\{t\ge 0: B_t=S_1\}$. Show that $T$ follows the arcsinus law with density $g(t)=\frac{1}{\pi\sqrt{t(1-t)}}1_{]0,1[}(t)$.

I used Markov property to get the following equality:


where $\hat{S}_{1-t}$ is defined for the brownian motion $\hat{B}_s=B_{t+s}-B_t$, which is independant of $F_t$.

However the reflexion principle tells us that $S_t-B_t$ has the same law as $S_t$, so we can also write that $P(T<t)=P(\hat{S}_{1-t}<S_t)$.

To this point, we can calculate $P(T<t)$ because we know the joint density of $(\hat{S}_{1-t},S_t)$, but this calculation leads to a complicated form of integral and I can not get the density $g$ at the end.

Do you know how to get the arcsinus law? Thank you.

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I corrected the title but I do not feel like correcting the post itself. – Did Oct 13 '12 at 18:12
up vote 2 down vote accepted

Let us start from the formula $\mathbb P(T\lt t)=\mathbb P(\hat S_{1-t}\lt S_t)$, where $0\leqslant t\leqslant 1$, and $\hat S_{1-t}$ and $S_t$ are the maxima at times $1-t$ and $t$ of two independent Brownian motions.

Let $X$ and $Y$ denote two i.i.d. standard normal random variables, then $(\hat S_{1-t},S_t)$ coincides in distribution with $(\sqrt{1-t}|X|,\sqrt{t}|Y|)$ hence $$ \mathbb P(T\lt t)=\mathbb P(\sqrt{1-t}|X|\lt\sqrt{t}|Y|)=\mathbb P(|Z|\lt\sqrt{t}), $$ where $Z=X/\sqrt{X^2+Y^2}$. Now, $Z=\sin\Theta$, where the random variable $\Theta$ is the argument of the two-dimensional random vector $(X,Y)$ whose density is $\mathrm e^{-(x^2+y^2)/2}/(2\pi)$, which is invariant by the rotations of center $(0,0)$. Hence $\Theta$ is uniformly distributed on $[-\pi,\pi]$ and $$ \mathbb P(T\lt t)=\mathbb P(|\sin\Theta|\lt\sqrt{t})=2\,\mathbb P(|\Theta|\lt\arcsin\sqrt{t})=\tfrac2\pi\,\arcsin\sqrt{t}. $$ The density of the distribution of $T$ follows by differentiation.

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What a great idea to write $Z=\sin\theta$. Thanks. – Iew Oct 13 '12 at 15:26

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