$\frac{N_1}{\sqrt{N_{1}^{2} + N_2^2}} \perp \frac{N_2}{\sqrt{N_{1}^{2} + N_2^2}}$ where $N_1, N_2 \sim \mathcal{N}(0,1)$ are independent? I have the following situation
Let $N_1, N_2 \sim \mathcal{N}(0,1)$ two independent r.v. Let $X = \frac{N_1}{\sqrt{N_{1}^{2} + N_2^2}}$ and $Y = \frac{N_2}{\sqrt{N_{1}^{2} + N_2^2}}$.
Now I know how show to that $X$ and $Y$ are not independent, but I don't know how to show $X$ and $Y$ are uncorrelated. Can anybody helps me?
Thanks in advance
 A: Hint:
To prove that $X$ and $Y$ are uncorrelated, you need to prove that the covariance is null, in other words $$Cov(X,Y) = \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y) = 0 \tag{1}$$
If we can prove that $\mathbb{E}(X) = 0$ and $\mathbb{E}(XY) = 0$ then $(1)$ holds true.
We have
$$
\begin{align}
\mathbb{E}(X) &= \iint_{(s,t)\in \Bbb R^2} \frac{e^{-\frac{s^2+t^2}{2}}}{2\pi}\frac{s}{\sqrt{s^2+t^2}}dsdt \tag{2}
\end{align}
$$
Make a change of variable $s = r \cos(\theta)$ and $t = r\sin(\theta)$,
$$\mathbb{E}(X) = \int_{0}^{+\infty} \frac{e^{-\frac{r^2}{2}}}{2\pi}  \int_0^{2\pi}(\cos(\theta)d\theta) rdr = 0$$
Same methode for $\mathbb{E}(XY)$, by making change of variables to polar coordinates, you can prove also that $\mathbb{E}(XY) = 0$.
So, $(1)$ holds true.
A: Additional method looking at the parity of the integrands:
$E[XY]=\frac{1}{2\pi} \int dn_1 dn_2e^{-(n_1^2+n_2^2)/2}\frac{n_1n_2}{n_1^2+n_2^2}$
The integral is odd in (say) $n_1$, therefore the result is 0. More formally after a change of variables $n'_1=-n_1,n'_2=n_2$ the integral is minus itself, therefore vanishes.
Analogously:
$E[X]=\frac{1}{2\pi} \int dn_1 dn_2e^{-(n_1^2+n_2^2)/2}\frac{n_1}{\sqrt{n_1^2+n_2^2}}$
The integrand is odd in $n_1$, therefore again the result is 0.
Probably this is a good example where to use mutual information rather than correlation as a measure of "dependence" :)
