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For $\left(U_1, V_1\right) \sim \mathcal{N}(0,0,1,1, \rho)$. In a standard bivariate normal with zero means and unit variances, the joint probability of both variables being positive equals:

$$ P\left[U_1>0, V_1>0\right]=\frac{1}{4}+\frac{\arcsin (\rho)}{2 \pi} $$

How to show the above result?

Chao
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  • In a sense, the ellipses associated with $(U_1,V_1)$ are oriented with the diagonals so rotate them so they are oriented with the axes, and then turn these into circles and look at the angle formed by the constraints. So consider $A=a(U_1+V_1)$ and $B=b(U_1-V_1)$ for suitable $a$ and $b$ which make $\left(A, B\right) \sim \mathcal{N}(0,0,1,1, 0)$ and see what that means in terms of $P(U_1>0,V_1>0)$. – Henry Apr 07 '25 at 17:05
  • $P[U>0,V>0]=\int_0^\infty \int_0^\infty f(u,v) du dv$ where $f(u,x)$ is the pdf of your bi-variate normal distribution. – Fisherman's Friend Apr 07 '25 at 17:06

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