# Probability Brownian Motion - dependence

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 $P(Z(3)>Z(2), Z(2)>0)=P(Z(3)-Z(2)>0, Z(2)>0)=P(Z(3)-Z(2)>0)P(Z(2)>0)=(0.5)(0.5)=0.25$.

• Didn't you ask exactly the same question quite recently? Was it not indicated to you that Klebaner's computes the probability of a different event? This is not a "special case". Once again, $[Z_{t+s}\gt Z_s\gt0]\ne[Z_{t+s}\gt0, Z_s\gt0]$. Are you going to continue asking the same question ad infinitum without reading their answers?
– Did
Nov 21, 2013 at 9:49
• Apologies if I broke a rule. I am new here. My original post was deleted because I added it into an answer. Hence the repost. That's why I did not seeany answers. Nov 21, 2013 at 12:40
– Did
Nov 21, 2013 at 13:14
• So would anybody know the answer to my little problem as formulated above? Nov 22, 2013 at 20:45
– Did
Nov 22, 2013 at 21:13

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 $P[A]=3/8$.

• Wow! Thanks a lot! In my simple excel simulation model your answer works great. Can you recommend some background references, if I want to understand the basics better, in particular the part concerning the domain and the angular sector? Dec 13, 2013 at 8:03
• I understand now that the solution involves a switch fro cartesian to polar coordinates. Jan 27, 2014 at 20:22
• I understand now that the solution involves a switch from cartesian to polar coordinates. Also I learned after some research that Did's answer is known in some literature since already some time (Sheppard 1900). I did not find any generalization of the result provided by Did to the case of Prob[Z(s)<a, Z(t)<b]. Is there a general solution to this problem? I tried to extend Did's approach. However, the radius r does no longer cancel out. Hence, I end up with a distribution for theta and r. As U and V are non-zero correlated I was not able to find a general solution. Can you confirm? Jan 28, 2014 at 7:29
• New problem? Then new question on the site, not add-ons to an old one.
– Did
Jan 28, 2014 at 9:30