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Math Help - Bivariate normal distribution: what is P(X>0, Y>0)?

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    Bivariate normal distribution: what is P(X>0, Y>0)?

    Let X and Y have the standard bivariate normal pdf

    f(x,y) = 1 / [2*pi*sqrt(1-p^2) ] exp { -(x^2 - 2pxy + y^2)/(2(1-p^2)) }

    where |p| < 1. Show that X and Z = (Y - pX) / sqrt(1-p^2) are independent N(0,1) variables, and deduce that

    P(X>0, Y>0) = 1/4 + arcsin(p) / (2*pi).

    I can do everything apart from the final deduction. I can't even see where this might be coming from. Where does the relevance to Z and X being independent come into it?
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  2. #2
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    Quote Originally Posted by alakazam View Post
    Let X and Y have the standard bivariate normal pdf

    f(x,y) = 1 / [2*pi*sqrt(1-p^2) ] exp { -(x^2 - 2pxy + y^2)/(2(1-p^2)) }

    where |p| < 1. Show that X and Z = (Y - pX) / sqrt(1-p^2) are independent N(0,1) variables, and deduce that

    P(X>0, Y>0) = 1/4 + arcsin(p) / (2*pi).

    I can do everything apart from the final deduction. I can't even see where this might be coming from. Where does the relevance to Z and X being independent come into it?
    Z = \frac{Y - \rho X}{\sqrt{1 - \rho^2}} \Rightarrow Y = \rho X + Z \sqrt{1 - \rho^2}.


    Therefore Y > 0 \Rightarrow \rho X + Z \sqrt{1 - \rho^2} > 0 \Rightarrow Z > \frac{- \rho X}{\sqrt{1 - \rho^2}}.


    Therefore \Pr(X > 0, Y > 0) = \Pr \left( X > 0, Z > \frac{- \rho X}{\sqrt{1 - \rho^2}} \right) = \frac{1}{2 \pi} \int_{x = 0}^{+\infty} \, \int_{z = -\rho x/\sqrt{1 - \rho^2}}^{+\infty} e^{-x^2/2} \cdot e^{-z^2/2} \, dz \, dx

    using the independence of X and Z.

    Now switch to polar coordinates (be sure to sketch the region of integration) and note that 0 \leq r < +\infty and \alpha \leq \theta \leq \frac{\pi}{2} where \alpha = - \tan^{-1} \frac{\rho}{\sqrt{1 - \rho^2}} = - \sin^{-1} \rho.
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  3. #3
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    I had completely forgotten about the possibility of switching to polars to evaluate the integral. This helps enormously - thank you very much indeed!
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