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This question already has an answer here:

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.

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marked as duplicate by StubbornAtom, Community Nov 3 '18 at 16:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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 $Unif(0,2\pi)$.

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