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

• Jan 29th 2009, 11:04 PM
alakazam
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?
• Jan 30th 2009, 02:11 AM
mr fantastic
Quote:

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?

$\displaystyle Z = \frac{Y - \rho X}{\sqrt{1 - \rho^2}} \Rightarrow Y = \rho X + Z \sqrt{1 - \rho^2}$.

Therefore $\displaystyle Y > 0 \Rightarrow \rho X + Z \sqrt{1 - \rho^2} > 0 \Rightarrow Z > \frac{- \rho X}{\sqrt{1 - \rho^2}}$.

Therefore $\displaystyle \Pr(X > 0, Y > 0) = \Pr \left( X > 0, Z > \frac{- \rho X}{\sqrt{1 - \rho^2}} \right)$ $\displaystyle = \frac{1}{2 \pi} \int_{x = 0}^{+\infty} \, \int_{z = -\rho x/\sqrt{1 - \rho^2}}^{+\infty} e^{-x^2/2} \cdot e^{-z^2/2} \, dz \, dx$

using the independence of X and Z.

Now switch to polar coordinates (be sure to sketch the region of integration) and note that $\displaystyle 0 \leq r < +\infty$ and $\displaystyle \alpha \leq \theta \leq \frac{\pi}{2}$ where $\displaystyle \alpha = - \tan^{-1} \frac{\rho}{\sqrt{1 - \rho^2}} = - \sin^{-1} \rho$.
• Jan 30th 2009, 08:35 AM
alakazam
I had completely forgotten about the possibility of switching to polars to evaluate the integral. This helps enormously - thank you very much indeed!