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Suppose $x$ and $y$ are jointly normal with known parameters. I need to find the probability of the event in the parenthesis: $\Pr\left(x>f(y; a),y>g(x;b)\right)$. The function $f(\cdot)$ is increasing, differentiable and invertible and $a$ is a parameter. $g(\cdot)$ is decreasing, differentiable and invertible while $b$ is a parameter. Thank you

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Since $f(\cdot)$ is increasing and invertible $X > f(Y;a)$ implies $f^{(-1)}(X;a) > Y$, thus: $$ \begin{eqnarray} \Pr\left(X>f(Y; a),Y>g(X;b)\right) &=& \Pr\left(f^{(-1)}(X;a)>Y>g(X;b)\right) \\ &=& \mathbb{E}\left(\Pr\left(f^{(-1)}(X;a)>Y>g(X;b) \mid X \right)\right) \\ &=& \mathbb{E}\left( \Phi_{Y|X}\left(f^{(-1)}(X;a)\right) - \Phi_{Y|X} \left( g(X;b) \right) \right) \end{eqnarray} $$ Now for $(X,Y)$ jointly normal with $\mathbb{E}(X) = m_1$, $\mathbb{E}(Y)=m_2$, $\mathbb{Var}(X) = \sigma_1^2$, $\mathbb{Var}(Y) = \sigma_2^2$, $\mathbb{Cor}\left(X, Y\right)=\rho$, $$ Y \mid X \sim \mathcal{N}\left(m_2 + \frac{\sigma_2}{\sigma_1} \rho \left(X-m_1\right), \sigma_2 \sqrt{1-\rho^2}\right) $$ Thus $$\Phi_{Y|X}\left(Z\right) = \Phi\left(\frac{1}{ \sqrt{1-\rho}} \left( \frac{1}{\sigma_2} \left(Z-m_2\right) + \frac{\rho}{\sigma_1} \left(X-m_1\right) \right) \right) $$ Now substitute back and cross your fingers that $f^{(-1)}$ and $g$ are simple enough to permit analytic evaluation. Otherwise numeric methods can be used.

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