# $\frac{N_1}{\sqrt{N_{1}^{2} + N_2^2}} \perp \frac{N_2}{\sqrt{N_{1}^{2} + N_2^2}}$ 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?

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.

• Thanks, this is really helpful !
– Bozu
Dec 2 '21 at 21:39

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" :)

• Thank you Thomas ! I do not have enough reputation to upvote your answer but it was helpful ! :)
– Bozu
Dec 2 '21 at 21:40