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Let $\gamma_t$ be the last zero of brownian motion before $t$ and $\beta_t$ be the first zero after $t$. I need to calculate the joint distribution of $\gamma_t$ and $\beta_t$, i.e. $P(\gamma_t<x, \beta_t<y)$. I know about arcsine low for these zeroes:

$P(\gamma_t<x)=\frac{2}{\pi}\arcsin\sqrt\frac{x}{t}$

$P(\beta_t<y)=\frac{2}{\pi}\arcsin\sqrt\frac{t}{y}$

But how can I get the joint distribution?

Thanks!

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Got something from my answer below? –  Did Jul 25 '12 at 16:52

1 Answer 1

Let me assume for a moment that one knows the distribution of the first hitting time $\tau$ of level $1$ by a Brownian motion starting from $0$. By scaling and translation invariance, the first hitting time of $0$ for a Brownian motion starting from $x$ is distributed like $x^2\tau$.

For every $0\lt u\lt t\lt v$, $[\gamma_t\lt u,v\lt\beta_t]=[\gamma_v\lt u]$. The scale invariance of Brownian motion implies that $[\gamma_v\lt u]$ has the same probability as $[\gamma_1\lt u/v]$. The last event depends on $\gamma_1$ only, and one knows that the distribution of $\gamma_1$ is the Arcsine distribution. Let us briefly show this.

Conditionally on $B_{s}=x$, the event $[\gamma_1\lt s]$ corresponds to the fact that a Brownian motion starting from $x$ hits $0$ after time $1-s$, hence $$ \mathrm P(\gamma_1\lt s\mid B_{s}=x)=\mathrm P(x^2\tau\gt 1-s). $$ Introducing $h(s)=\mathrm P(\tau\gt s)$ for every $s\gt0$ and writing $(g_s)_{s\gt0}$ for the transition semi-group of Brownian motion $g_s(x)=\mathrm e^{-x^2/(2s)}/\sqrt{2\pi s}$, this means that $$ \mathrm P(\gamma_1\lt s)=\int_{-\infty}^{+\infty}h\left(\frac{1-s}{x^2}\right)g_s(x)\mathrm dx, $$ from which the distribution of $\gamma_1$, thus also the joint distribution of $(\gamma_t,\beta_t)$, follow by differentiation.

To complete this, recall that Désiré André's reflexion principle shows that the probability of $[\tau\lt s]$ is twice the probability of $[B_s\gt1]$, hence $$ h(s)=1-2\mathrm P(B_s\gt 1)=2\int_0^1g_s(x)\mathrm dx. $$ After some simplifications and change of variables in the double integral, this leads to $$ \mathrm P(\gamma_1\lt s)=\frac2\pi\arctan\left(\sqrt{\frac{s}{1-s}}\right)=\frac2\pi\arcsin\left(\sqrt{s}\right), $$ hence $$ \mathrm P(\gamma_t\lt u,v\lt\beta_t)=\frac2\pi\arcsin\left(\sqrt{\frac{u}{v}}\right), $$ a formula which is all over the place, and probably in Rick Durrett's Probability Theory with Applications.

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Thanks! But we calculate the probability $P(\gamma_t<u,\beta_t>v)$. Can we get it in the form $P(\gamma_t<u,\beta_t<v)$? –  Leo May 27 '12 at 14:20
    
$[\gamma_t\lt u,\beta_t\lt v]=[\gamma_t\lt u,t\lt\beta_t]\setminus[\gamma_t\lt u,v\lt\beta_t]$. –  Did May 27 '12 at 15:28
    
So in this case formula looks like:$\frac{2}{\pi}\arcsin\sqrt\frac{u}{t}-\frac{2}{\pi}\arcsin\sqrt\frac{u}{v}$ and depends on $t$? –  Leo May 27 '12 at 15:39
    
The event $[\gamma_t\lt u,\beta_t\lt v]$ corresponds to no zero in $(u,t)$ and at least one zero in $(t,v)$ hence indeed its probability depends on $t$ (and on $u$ and $v$). –  Did May 27 '12 at 16:07

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