I have a question about the following problem:

Let $2$ independent r.v.’s be $Z_{1}, Z_{2} \sim Normal(0,1)$. Now, let:

$$Y_{1}=\frac{Z_{1}}{Z_{1}^{2}+Z_{2}^{2}}, Y_{2}=\frac{Z_{2}}{Z_{1}^{2}+Z_{2}^{2}}.$$

Find the joint density of $Y_{1}$ and $Y_{2}$, the marginal density of $Y_{1}$.

My attempt:

Let $Z_{1}=Rcos(\Theta), Z_{2}=Rsin(\Theta), R \geq 0, \Theta \in [0, 2\pi]$. Then, the joint p.d.f of $R, \Theta$ is: $$f_{R,\Theta}(r,\theta)=\frac{1}{2\pi}re^{\frac{-r^2}{2}}.$$

Now, we have: $$Y_{1}^2+Y_{2}^2=\frac{1}{Z_{1}^2+Z_{2}^2}=\frac{1}{R^{2}}, Y_{1}=\frac{1}{R}cos(\Theta), Y_{2}=\frac{1}{R}sin(\Theta).$$ Therefore, using transformation of r.v.’s, we get: $$R=\frac{1}{\sqrt{Y_{1}^2+Y_{2}^2}}, \Theta=tan^{-1}(\frac{Y_{2}}{Y_{1}}).$$ Then, the determinant of Jacobian matrix is: $$|J(y_{1},y_{2})|=\frac{1}{(y_{1}^2+y_{2}^2)^{3/2}}=(r^2)^{\frac{3}{2}}=r^3.$$ Thus, the joint density of $Y_{1}, Y_{2}$ is: $$f_{Y_{1},Y_{2}}(y_{1},y_{2}) =f_{R,\Theta}(r(y_{1},y_{2}),\theta(y_{1},y_{2}))|J(y_{1},y_{2}|=\frac{1}{2\pi}re^{-\frac{r^2}{2}}r^3.$$ So, $$f_{Y_{1},Y_{2}}(y_{1},y_{2}) =\frac{1}{2\pi(y_{1}^2+y_{2}^2)^2}e^{-\frac{1}{2(y_{1}^2+y_{2}^2)}}, -\infty < y_{1}, y_{2} < \infty.$$

My question: I don’t know my work is correct since I am not sure about the domain of $Y_{1},Y_{2}$ and I tried to integrate to get the marginal density $Y_{1}$, but it didn’t seem right. Also, is there any other way to do this problem ? Because I feel like there is an easier way to obtain the joint density.

  • 2
    $\begingroup$ There is no restriction on the values of $Y_1$ and $Y_2$. They take all real values. $\endgroup$ – Kavi Rama Murthy Feb 14 '18 at 7:19
  • $\begingroup$ @KaviRamaMurthy oh, ok. Can you let me know whether I’m doing it’s right? Because I can’t ingrate the joint density to get the marginal. $\endgroup$ – Harry Feb 14 '18 at 7:21

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