I'm given (X,Y) ~ standard bivariate normal (p) -(assume this is the greek letter ro).
I'm asked to find the P(XY>0) as functions of p and other values as indicated.
I know by definition, two random variables X and Y are said to be bivariate normal if and only if aX+bY has a normal distribution. Though, I'm not certain I'm able to satisfy this axiom. Is the product of two normal distributions univariate normal? How should I approach this question?
Assuming the Bivariate Normal has mean $0$. We can do the following:
First, let $Z:=\frac{X}{Y}$ denote the ratio between the two components of the bivariate normal then $Z \sim Cauchy (\rho\frac{\sigma_x}{\sigma_y},\frac{\sigma_x}{\sigma_y}\sqrt{1-\rho^2})$. Therefore for any number $z$ the distribution function is: $F_z=\frac{1}{2}+\frac{1}{\pi}\tan^{-1}\left(\frac{\sigma_yx-\rho\sigma_x}{\sigma_x\sqrt{1-\rho^2}}\right)$
You should note that $P(XY>0)=1-P(XY<0)$ and further $P(XY<0)=P(Z<0)$, now by taking $F_z(0)$ you get $\frac{1}{2}+\frac{1}{\pi}\tan^{-1}\left(-\frac{\rho}{\sqrt{1-\rho^2}}\right)$, now, $\tan^{-1}$ is and odd function and the trigonometric identity $\tan(\sin^{-1}(x))=\frac{x}{\sqrt{1-x^2}}$ imply that $P(Z<0)=\frac{1}{2}-\frac{1}{\pi}\sin^{-1}(\rho)$ and in your case $P(XY>0)=\frac{1}{2}+\frac{1}{\pi}\sin^{-1}(\rho)$