Thread: Bivariate normal distribution: what is P(X>0, Y>0)?

Let X and Y have the standard bivariate normal pdf

f(x,y) = 1 / [2*pi*sqrt(1-p^2) ] exp { -(x^2 - 2pxy + y^2)/(2(1-p^2)) }

where |p| < 1. Show that X and Z = (Y - pX) / sqrt(1-p^2) are independent N(0,1) variables, and deduce that

P(X>0, Y>0) = 1/4 + arcsin(p) / (2*pi).

I can do everything apart from the final deduction. I can't even see where this might be coming from. Where does the relevance to Z and X being independent come into it?

Originally Posted by alakazam

Let X and Y have the standard bivariate normal pdf

f(x,y) = 1 / [2*pi*sqrt(1-p^2) ] exp { -(x^2 - 2pxy + y^2)/(2(1-p^2)) }

where |p| < 1. Show that X and Z = (Y - pX) / sqrt(1-p^2) are independent N(0,1) variables, and deduce that

P(X>0, Y>0) = 1/4 + arcsin(p) / (2*pi).

I can do everything apart from the final deduction. I can't even see where this might be coming from. Where does the relevance to Z and X being independent come into it?

.


Therefore .


Therefore

using the independence of X and Z.

Now switch to polar coordinates (be sure to sketch the region of integration) and note that and where .

I had completely forgotten about the possibility of switching to polars to evaluate the integral. This helps enormously - thank you very much indeed!