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If $X_1$ and $X_2$ are jointly normal with $ 0<\rho < 1$ , and both individually follow the standard normal distribution, how do I get $ \Pr(X_2\leq x_2 \mid X_1=x_1)$ from the Copula function? I know that it's a partial derivative of the Copula but I am not able to get an expression for it. Kindly help!

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My attempt. We have that:

$$(X_{1},X_{2})$$

has a bivariate normal distribution with standard marginals and correlation $\rho$.

Now we want:

$$\text{Pr}(X_{2}\leq x_{2}\,|\, X_{1}=x_{1})$$

We know that:

$$X_{2}\,|\,X_{1}=x_{1}\sim N(\rho x_{1},1-\rho^{2})$$

Therefore:

$$\text{Pr}(X_{2}\leq x_{2}\,|\, X_{1}=x_{1})=\Phi\Bigg(\frac{x_{2}-\rho x_{1}}{\sqrt{1-\rho^{2}}}\Bigg)$$

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Let $U_1 = \Phi(X_1)$ and $U_2 = \Phi(X_2)$, then both $U_1$ and $U_2$ are uniformly distributed. We know that the conditional distribution of bivariate Gaussian copula is $$P(U_2 \leq v|U_1 = u) = C_{2|1}(v|u; \rho)=\Phi(\frac{\Phi^{-1}(v) -\rho \Phi^{-1}(u)}{\sqrt{1-\rho^2}}).$$ Therefore \begin{multline*} P(X_2 \leq x_2|X_1 = x_1) = P(U_2 \leq \Phi(x_2)|U_1 = \Phi(x_1)) \\= \Phi(\frac{\Phi^{-1}(\Phi(x_2)) -\rho \Phi^{-1}(\Phi(x_2))}{\sqrt{1-\rho^2}}) = \Phi(\frac{x_2 -\rho x_1}{\sqrt{1-\rho^2}}). \end{multline*}

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