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?
Best Answer
With direct computation I get (similar to yours but I calcualated also the normalizing constant)
$$f_Y(y)=\sqrt{\frac{\pi}{1+y^2}}e^{-(1+y^2)}$$
Thus the conditional density is, by definition,
$$f_{X|Y}(x|y)=\frac{f_{XY}(x,y)}{f_Y(y)}=\sqrt{\frac{1+y^2}{\pi}}e^{-x^2(1+y^2)}$$
Thus immediately we recognize that
$$(X|Y=y)\sim N\Bigg(0;\frac{1}{2(1+y^2)}\Bigg)$$