# Finding the Probability for a Bivariate Normal Distribution [duplicate]

Given $$(x_1,x_2)' \sim N_2 \left(\bf{0},\Sigma\right) = N_2 \left(\left(\begin{array}{l}0\\0\end{array}\right), \left(\begin{array}{l}1&\rho\\\rho&1\end{array}\right)\right)$$, find $$Pr(x_1>0,x_2>0)$$.

I have been struggling with solving this problem using multivariate algebra. I think the final answer is $$\frac{1}{4}+\frac{sin^{-1}(\rho)}{2\pi}$$, but not sure how to get to it. I am considering the trick is to notice that $$Pr(x_1>0,x_2>0)=\frac{1}{2}Pr(x_1x_2>0)$$; but from here I am stuck.

Any help and/or advice is appreciated.

• – StubbornAtom Nov 3 '18 at 6:54

Hint: Let $$Z=\dfrac{Y-\rho X}{\sqrt{1-\rho^2}}$$ then $$Z$$ and $$X$$ are independent $$N(0,1)$$. Make the change of variables $$(X,Z)\mapsto (R\cos\theta,R\sin\theta)$$ where you know that $$\theta\sim Unif(0,2\pi)$$. Next write the event $$\{X_1>0,X_2>0\}$$ in terms of $$R$$ and $$\theta$$ and note that it gets free of $$R$$. So the resulting probability can be computed solely using the distribution of $$Unif(0,2\pi)$$.