$X_1$ and $X_2$ are independent from Unif(-1,1) and have joint $f(x_1,x_2) = 1/4$ for $-1<x_1<1$ and $-1 < x_2 < 1$. Let $Z_1 = X_1$ and $Z_2 = X_2/(1+X_1^2)$. Find the conditional density of $Z_1$ given $Z_2 = 0$.
I have tried to get the joint density of $f(z_1,z_2) = 1/4*z_1^2 - 1/2*z_1z_2 + 1/4$, so I only need the $f(z_2 = 0)$ to complete this question. But I think the position $Z_2 = 0$ is not existed. How can I get the density for $Z_2$ at $Z_2 = 0$?
Why? $Z_2 \in(-1;1)$ thus $f_{Z_2}(0)$ exists!
The joint density you calculated doesn't look right to me.
The Jacobian is $1+z_1^2$ thus
$$f_{Z_1Z_2}(z_1,z_2)=\frac{1+z_1^2}{4}$$
It is understood that you have to understand which is the bivariate support of $(Z_1,Z_2)$
It results to me the following
$$\mathbb{1}_{(-1;1)}(z_1)\cdot\mathbb{1}_{\Big(-\frac{1}{1+z_1^2};\frac{1}{1+z_1^2}\Big)}(z_2)$$
That is the area between these two curves
To get the density of $Z_2$ you have only to integrate in $Z_1$