Does anyone know how to calculate $P(Z(3)>Z(2), Z(2)>0)$ if $Z(3)$ and $Z(2)$ are on the same sample path, i.e. not independent?
I found a solution for the case $P(Z(2)<0, Z(1)<0)$ in Fima C Klebaner, Introduction to Stochastic Calculus with Applications, Imperial College Press, 1998. Example 3.1. (check Google Books)
The solution suggested there equals 0.375 and not 0.25 as under independence.
Therefore, for the problem above I conclude that it is wrong to argue as follows $P(Z(3)>Z(2), Z(2)>0)=P(Z(3)-Z(2)>0, Z(2)>0)=P(Z(3)-Z(2)>0)P(Z(2)>0)=(0.5)(0.5)=0.25$.
For every $0\lt s\lt t$, the probability of $[Z(s)\gt0,Z(t)\gt Z(s)]$ is $1/4$ since the increments of $Z$ are independent and symmetric.
To compute the probability of $A=[Z(s)\lt0,Z(t)\lt0]$, note that $$(Z(s),Z(t))=(\sqrt{s}U,\sqrt{s}U+\sqrt{t-s}V),$$ where $(U,V)$ is i.i.d. standard normal hence $A=[(U,V)\in D]$ where $$D=\{(u,v)\mid u\lt0,\sin(\alpha) u+\cos(\alpha) v\lt0\}, $$ and the angle $\alpha$ in $(0,\pi/2)$ is defined by $$ \sin(\alpha)=\sqrt{s/t},\qquad\cos(\alpha)=\sqrt{(t-s)/t}. $$ The domain $D$ is the angular sector $$D=\{(r\cos(\theta),r\sin(\theta))\mid r\gt0,\cos(\theta)\lt0,\sin(\alpha+\theta)\lt0\}, $$ that is, in the interval $(0,2\pi)$, $D$ corresponds to the angles in $$(\pi/2,3\pi/2)\cap(\pi-\alpha,2\pi-\alpha)=(\pi-\alpha,3\pi/2). $$ The distribution of $(U,V)$ is invariant by the rotations hence the measure of $D$ is $$P[A]=\frac{3\pi/2-(\pi-\alpha)}{2\pi}=\frac14+\frac\alpha{2\pi}=\frac14+\frac1{2\pi}\arctan\sqrt{\frac{s}{t-s}}. $$ For example, if $s=1$ and $t=2$, then $P[A]=3/8$.