I've been struggling on a problem for a while if anyonw could point me to the right direction I'd be so happy
Problem Let $X_1 = \sqrt{2Z} \cos(U)$ and $X_2 = \sqrt{2Z} \sin(U)$ where $U$ has uniform distribution $(0,2\pi)$ and $Z$ follows a Exp($\lambda = 1)$.
Suppose $Z$ and $U$ are independant. Show that $X_1, X_2$ follows $N(0,1)$ and are independant.
Solution
We have that $\sqrt{2Z}$, $\cos(U)$ and $\sin(U)$ are derivable.
Moreover, let
$x_1 = g_1(z,u) = \sqrt{2z} \cos(u)$
$x_2 = g_2(z,u) = \sqrt{2z} \sin(u)$
we can solve for $z,u$, where
$\frac{x_1}{x_2} = tan(u) \Rightarrow$ $u = \arctan(\frac{x_1}{x_2})$
and
$(x_1)^2 + (x_2)^2 = 2z(cos^2(u) + sin^2(u)) \Rightarrow z =\frac{x_1^2 +x_2^2}{2}$
Then the Jacobien $J(z,u)$ = $\frac{2\sqrt{z}(cos^2(u))}{2\sqrt{z}}$ + $\frac{2\sqrt{z}(\sin^2(u))}{2\sqrt{z}}$ = $\cos^2(u) + \sin^2(u) = 1$
So from the manual (Ross),
$f_{X1X2}(x_1,x_2) =|J(z,u)|^{-1}f_{zu}(z,u)$, and since $Z,U$ are independant, we can write
$f_{X1X2}(x_1,x_2) = f_z(z) f_u(u)$ =$ \frac{1}{2\pi}$$e^{-\frac{x_1^2 +x_2^2}{2}}$
where $e^{-\frac{x_1^2 +x_2^2}{2}}$ = $e^{-\frac{(x_1^2 + 2x_1 x_2 +x_2^2) - 2xy}{2}}$ = $e^{-\frac{(x_1 + x_2)^2}{2}}$$e^{x_1x_2}$
and this is where I'm lost, i'm stuck with
$f_{X1X2}(x_1,x_2) = f_z(z) f_u(u)$ =$ \frac{e^{x_1 x_2}}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}$$e^{-\frac{(x_1 +x_2)^2}{2}}$
If anyone could help me or point me to the right directions I would appreciate it very much.
Thank you
You are done. Now, all you need to note is that the PDF can be decomposed into a product. (That is the definition of independence), and the PDFs tell you that these have to be Gaussian. (This is essentially a form of uniqueness; when you decompose the density into a product form, the constants come for free since you know that the density has to integrate to 1.)
Here is a useful theorem you might be able to use to clean it up:
Joint p.d.f and independence
It also shows you that the PDFs you get are (a version of) the densities of $X_1$ and $X_2$.