TITLE: $P(X>0,Y>0)$ for a bivariate normal distribution with correlation $\rho$
QUESTION [8 upvotes]: $X$ and $Y$ have a bivariate normal distribution with $\rho$ as covariance. $X$ and $Y$ are standard normal variables.
I showed that $X$ and $Z= \dfrac{Y-\rho X}{\sqrt{1-\rho^2}}$ are independent standard normal variables.
Using this I need to show that 
$$P(X >0,Y>0) = \frac14 + \frac{1}{2\pi} \cdot\arcsin(\rho).$$

REPLY [5 votes]: Using the OP's hint: The event $\{X>0,Y>0\}$ is the same as the event $\{X>0,Z>\frac{-\rho}{\sqrt{1-\rho^2}}X\}$, where now $X$ and $Z$ are independent standard normal variables.  Writing $a:=\frac{-\rho}{\sqrt{1-\rho^2}}$ for brevity, the desired probability is expressible as a double integral involving the joint density of $(X,Z)$:
$$
P(X>0,Y>0)=P(X>0,Z>aX)=\int_{x=0}^\infty\int_{z=ax}^\infty\frac1{\sqrt{2\pi}}e^{-x^2/2}\frac1{\sqrt{2\pi}}e^{-z^2/2}\,dz\,dx.
$$
Switching to polar coordinates, this equals
$$
\begin{align}
\int_{\theta=\arctan(a)}^{\pi/2}\int_{r=0}^\infty\frac1{2\pi}e^{-r^2/2}\,rdrd\theta&=\int_{\theta=\arctan(a)}^{\pi/2}\frac1{2\pi}\,d\theta\\
&=\frac1{2\pi}\left(\frac\pi2-\arctan a\right)\\
&=\frac14+\frac1{2\pi}\arctan\frac\rho{\sqrt{1-\rho^2}};
\end{align}
$$
for the last equality we substitute $a:=\frac{-\rho}{\sqrt{1-\rho^2}}$ and use the fact that the arctan function is odd. To finish off, remember that if $\theta\in[-\pi/2,\pi/2]$ then $$\theta=\arctan\frac{\rho}{\sqrt{1-\rho^2}}\ \Longleftrightarrow\ \tan\theta=\frac{\rho}{\sqrt{1-\rho^2}}\ \Longleftrightarrow\ \sin\theta=\rho\ \Longleftrightarrow\ \theta=\arcsin \rho.$$