This is part of a practice exam, to prepare for an upcoming final -- I honestly have no idea how to approach this. I tried various things and nothing seemed to make sense. Even a hint on how to approach this would be appreciated.
Let $Z_1$ and $Z_2$ be independent standard normal r.v's. Let $Y_1 = \frac{Z_1}{(Z_1^2 + Z_2^2)}$ and $Y_2 = \frac{Z_2}{(Z_1^2 + Z_2^2)}$. Find the joint density of $Y_1$ and $Y_2$, the marginal density of $Y_1$, the conditional density of $Y_2$ given $Y_1$, and then calculate the conditional expectation of $Y_2$ given $Y_1$
Dude, I had to warn you, whoever designed this exam had a blood feud with you guys :]. We have to find the Jacobean of transformation
$$\det \begin{bmatrix}\frac{\partial Y_1}{\partial Z_1} & \frac{\partial Y_1}{\partial Z_1} \\ \frac{\partial Y_2}{\partial Z_1} & \frac{\partial Y_2}{\partial Z_1} \end{bmatrix}=\det \begin{bmatrix}\frac{Z_2^2-Z_1^2}{(Z_1^2+Z_1^2)^2} & \frac{-2Z_2Z_1}{(Z_1^2+Z_1^2)^2}\\ \frac{-2Z_2Z_1}{(Z_1^2+Z_1^2)^2} & \frac{Z_1^2-Z_2^2}{(Z_1^2+Z_1^2)^2}\end{bmatrix}=-1$$
Therefore $f(Y_1,Y_2)=f(Z_1,Z_2)$ and we only need to replace $Z_1,Z_2$ with $Y_1,Y_2$ in the density of $f(Z_1,Z_2)$. $f(Z_1,Z_2)=f(Z_1)f(Z_2)=\frac{1}{2\pi\sigma_1\sigma_2}\exp\left( -\frac{1}{2} \left( \frac{Z_1-\mu_1}{\sigma_1} \right)^2 -\frac{1}{2} \left( \frac{Z_2-\mu_2}{\sigma_2} \right)^2 \right)$.
$$Y_1^2+Y_2^2=\frac{1}{Z_1^2+Z_2^2}\Rightarrow Z_1=\frac{Y_1}{Y_1^2+Y_2^2},Z_2=\frac{Y_2}{Y_1^2+Y_2^2} \\ \Rightarrow f(Y_1,Y_2)=\frac{1}{2\pi\sigma_1\sigma_2}\exp\left( -\frac{1}{2} \left( \frac{\frac{Y_1}{Y_1^2+Y_2^2}-\mu_1}{\sigma_1} \right)^2 -\frac{1}{2} \left( \frac{\frac{Y_2}{Y_1^2+Y_2^2}-\mu_2}{\sigma_2} \right)^2 \right)$$ Since $Z_1,Z_2$ are standard normal $\mu_1=\mu_2=0,\sigma_1=\sigma_2=1$ therefore
$$f(Y_1,Y_2)=\frac{1}{2\pi}\exp\left( -\frac{1}{2} \left( \frac{Y_1}{Y_1^2+Y_2^2} \right)^2 -\frac{1}{2} \left( \frac{Y_2}{Y_1^2+Y_2^2} \right)^2 \right) \\ = \frac{1}{2\pi}\exp\left( -\left( \frac{1}{2(Y_1^2+Y_2^2)} \right) \right)$$
$$f(Y_1)=\int_{-\infty}^{\infty} \frac{1}{2\pi}\exp \left( -\frac{1}{2(Y_1^2+Y_2^2)} \right) dY_2=\frac{1}{\pi} \int_{0}^{\infty}\exp \left(- \frac{1}{2(Y_1^2+Y_2^2)} \right)dY_2$$ Unfortunately this integral is very hard (if possible ?!) to evaluate. So we go back to definition of $Y_1$ to find the marginal distribution.
$$F_{Y_1}(y)=\text{Pr}\left( \frac{Z_1}{Z_1^2 + Z_2^2}\le y \right) = \text{Pr}\left( Z_1^2 + Z_2^2 - \frac{Z_1}{y} \ge 0 \right)=1-\int_{\left( Z_1-\frac{1}{2y} \right)^2+Z_2^2\le \frac{1}{4y^2}} f(Z_1,Z_2)dZ_1dZ_2$$
Next
$$\int_{\left( Z_1-\frac{1}{2y} \right)^2+Z_2^2\le \frac{1}{4y^2}} f(Z_1,Z_2)dZ_1dZ_2 = \frac{2}{2\pi}\int_{0}^{\frac{1}{2y}}\exp\left(-\frac{Z_1^2}{2}\right)dZ_1\int_{0}^{\sqrt{\frac{Z_1}{y}-Z_1^2}}\exp\left(-\frac{Z_2^2}{2}\right)dZ_2$$
Therefore the density of $Y_1$ is ($g(Z_1,y)=\int_{0}^{\sqrt{\frac{Z_1}{y}-Z_1^2}}\exp\left(-\frac{Z_2^2}{2}\right)dZ_2 , \frac{\partial g}{\partial y} = -\frac{Z_1}{2y^2\sqrt{\frac{Z_1}{y}-Z_1^2}}\exp\left(-\frac{\frac{Z_1}{y}-Z_1^2}{2}\right)$
$$f_{Y_1}(y)=-\frac{d}{dy}\left( \frac{1}{\pi}\int_{0}^{\frac{1}{2y}}\exp(-\frac{Z_1^2}{2})g(Z_1,y)dZ_1\right)=\frac{1}{2\pi y^2}\exp{\left(-\frac{1}{8y^2}\right)}g(y,y)\\- \frac{1}{\pi}\int_{0}^{\frac{1}{2y}}\exp(-\frac{Z_1^2}{2})\frac{\partial g(Z_1,y)}{\partial y}dZ_1 \\ = \frac{1}{2\pi y^2}\exp{\left(-\frac{1}{8y^2}\right)}\int_{0}^{\sqrt{1-y^2}}\exp\left(-\frac{Z_2^2}{2}\right)dZ_2- \frac{1}{2\pi y^2}\int_{0}^{\frac{1}{2y}}\frac{Z_1}{\sqrt{\frac{Z_1}{y}-Z_1^2}}\exp\left(-\frac{Z_1}{2y}\right)dZ_1$$
And as we expected, this cannot be evaluated and we hit the deadend. The conditional density is $f(Y_1|Y_2)=\frac{f(Y_1,Y_2)}{f(Y_1)}$ which is useless without marginal of $Y_1$. To improvise another derivation path, we notice that $\frac{Y_1}{Y_2}=\frac{Z_1}{Z_2}$, therefore $$F_{Y_1 | Y_2}\left( y|x \right)=\text{Pr}(Y_1\le y|Y_2=x)=\text{Pr}\left(\frac{Z_1}{Z_2}Y_2\le y|Y_2=x\right)=\text{Pr}\left(Z_1\le \frac{y}{x}Z_2|Y_2=x\right) \\ =\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\frac{y}{x}Z_2}\exp\left(-\frac{Z_2^2}{2}\right)dZ_2\exp\left(-\frac{Z_1^2}{2}\right)dZ_1$$
Again we use Leibniz rule to find the density
$$f_{Y_1 | Y_2}\left( y|x \right)=\frac{dF_{Y_1 | Y_2}( y|x )}{dy}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp\left(-\frac{Z_2^2}{2}\left(1+\frac{y^2}{x^2}\right)\right)dZ_2=\frac{1}{\sqrt{2\pi\left( 1+\frac{y^2}{x^2} \right)}}$$