How can I prove this staments?
Let $X$ and $Y$ independent random variables with common distribution $N(0,1)$. Prove that $R=\sqrt{X^{2}+Y^{2}}$ and $\Theta=\tan^{-1}(Y/X)$ also independent random variables, $\Theta \sim U(0,2\pi)$ and $R$ has Rayleigh's distribution, this's has density $$f_{R}(r)=re^{-r^{2}/2}, r>0$$
My approach: By hypothesis, we have that $$f_{X,Y}(x,y)=f_{X}(x)f_{Y}(y)$$and since that $$f_{X}(x)=\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{x^{2}}{2} \right)$$ and $$f_{Y}(y)=\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{y^{2}}{2} \right)$$ So, $$f_{X,Y}(x,y)=\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{x^{2}}{2} \right)\cdot \frac{1}{\sqrt{2\pi}}\exp \left(-\frac{y^{2}}{2} \right)$$ we need to find the PDF of $R=\sqrt{X^{2}+Y^{2}}$.
and we need to prove that $$f_{R,\Theta}(r,\theta)=f_{r}(r)f_{\Theta}(\theta)$$ where $f$ is de PDF.
But, I don't know how can I write here $f_{R,\Theta}(r,\theta)$ and the marginal functions $f_{R}(r)$ and $f_{\Theta}(\theta)$.
In the previuous problem I proved that $X^{2}+Y^{2}\sim \chi^{2}(2)$. I think that I can use that result, since that $R^{2}=\chi^{2}(2)$. But, I don't sure about my approach.
You have $$ f_{X,Y}(x,y) = \frac{1}{2\pi}\exp\left ( -\frac{1}{2} (x^2+y^2)\right ) $$ Note further that $X = R \sin \Theta, Y=R \cos \Theta$
Now, we can use the density transform formula to write $$ f_{R,\Theta}(r,\theta) = f_{X,Y}(r \sin \theta, r\cos\theta)|J|^{-1} $$ where $|J|^{-1}$ is the inverse Jacobian of the transformation, you should be able to show $|J|^{-1} = r$. Putting this together we have
$$ f_{R,\Theta}(r,\theta) = \frac{1}{2 \pi} r e^{-r^2/2} \mathbb{1}\{r >0\} \mathbb{1}\{0\le\theta \le 2\pi\} = \left ( \frac{1}{2 \pi} \mathbb{1}\{0\le\theta \le 2\pi\} \right ) \left (r e^{-r^2/2} \mathbb{1}\{r >0\} \right) $$ and the result follows immediately (independence holds due to the fact that the density factorizes)