Is there a closed form solution for $P[Z(s)<-a, Z(t)<-b]$ where $Z$ is a Brownian motion and $0<a<b$ are constants?
In the post Probability Brownian Motion - dependence the user Did gave the following answer to $P[Z(s)<0, Z(t)<0]$=$1/4$$+$$(1/2π)$arctan$(\sqrt s/(t-s))$
Following Did's approach I have tried as follows:
With $U$ and $V$ being standard normally distributed we have $P[Z(s)<-a, Z(t)<-b]=P[U\sqrt(s)<-a, U\sqrt(s)+V\sqrt(t-s)<-b]<=>$ $P[Z(s)<-a, Z(t)<-b]=P[U<-a/\sqrt(s), U\sqrt(s/t)+V\sqrt((t-s)/t)<-b/\sqrt(t)]$
As $sin[m]^2+cos[m]^2=1=s/t+(t-s)/s$ we can rewrite $P[Z(s)<-a, Z(t)<-b]=P[U<-a/\sqrt(s), U sin[m]+V cos[m]<-b/\sqrt(t)]$
Note that as $0<\sqrt(s/t)$, $\sqrt((t-s)/t) <1$ implies that $0<m<\pi/2$.
Switching to polar coordinates, we define:
$U=r cos(\theta)$, $V=r sin[\theta]$,
this gives:
$P[Z(s)<-a, Z(t)<-b]$=$P[$$r cos[\theta]<-a/\sqrt(s)$, $r cos[\theta] sin[m]+r sin[\theta] cos[m]$$<$$-b/\sqrt(t)$]
$P[Z(s)<-a, Z(t)<-b]=P[rcos[\theta]<-a/\sqrt(s), r sin[\theta+m]<-b/\sqrt(t)]$
As $a$, $b>0$ per assumption and $r>0$ as it is a radius, a necessary condition for both inequalities is $cos[\theta]<0$ and $sin[\theta+m]<0$. This implies
$P[Z(s)<-a, Z(t)<-b]$$=$$P[r>-a/(\sqrt(s)/cos[\theta]), r>-b/(\sqrt(t)/sin[\theta+m])]$
These inequalities imply the following domains for $r$ and $\theta$
$b/\sqrt(t)<r<\infty$ and $\pi$$-$$m$$<$$\theta$$<$$3*\pi/2$
The question now is, which distribution can I assume for $\theta$ and r. Initially, I thought that $\theta$ is uniformly distributed (that's what I learned from Did's post) and that r is Rayleigh distributed. However, Excel simulations show me that this is wrong.
Does anyone know whether this problem can be solved analytically, or can it only be solved numerically, i.e. with simulations?
$$ P[Z(s)<-a, Z(t)<-b]=\int_a^{+\infty}\Phi\left(\frac{x-b}{\sqrt{t-s}}\right)\,\varphi\left(\frac{x}{\sqrt{s}}\right)\,\frac{\mathrm dx}{\sqrt{s}} $$