Let $(X,Y)$ have joint density $ f(x,y)=c\exp(-(1+x^2)(1+y^2)) $, $(x,y)\in\mathbb R^2$, $c$ is a constant such that $f$ is indeed a probability density function. Then what is the conditional density of $X$ given $\{Y=y\}$.
We compute the conditional distribution of $X$ given $\{Y=y\}$. \begin{align} \Pr(X<x_0|Y=y)&=\frac{\int\int_{-\infty}^{x_0}f(x,y)dxdy}{\int f(x,y) dx}. \end{align}
Applying the Fubini-Tonelli theorem and taking the derivative in terms of $x_0$ gives us $$\frac{\int f(x_0,y)dy}{\int f(x,y) dx}.$$
Direct computation shows that $$ \frac{\int f(x_0,y)dy}{\int f(x,y) dx}=\frac{\exp(-(1+x_0^2))\sqrt{1+y^2}}{\exp(-(1+y^2))\sqrt{1+x_0^2}} $$
But I think there should be a better way to compute the density other than taking the derivative of the distribution function. Can we do that?
