arctan of ratio of two normal variables is uniform

Question

Say $X, Y$ are independent standard normals, and $\theta = \arctan(Y/X)$. Prove that $\theta$ is uniformly distributed over it's range.

It is pretty intuitive that the distribution of $\theta$ would be uniform given a scatter plot of $X,Y$, but how can I mathematically show it?

Answer 1

Let $R=\sqrt{X^2+Y^2}$. We want to calculate the distribution of $(R,\theta)$, i.e. the polar cordinates of $(X,Y)$. For $r\ge 0$ and $\alpha\in [-\pi,\pi]$ we have $\Bbb P[R\le r,\, \theta\le\alpha]=\Bbb P[(X,Y)\in\{(x,y)\in \Bbb R^2|\; x$ and $y$ have polar cordinates $(\rho , \beta)$ with $\rho \le r$ and $\beta\le\alpha\}=\Bbb P[(X,Y)\in \Omega]$.

So, since $X$ and $Y$ are indipendents, density of $(X,Y)$ is the product of the densities of $X$ and $Y$. So, passing in polar cordinates:$$\Bbb P[R\le r,\, \theta\le\alpha]=\int\int_{\Omega} \frac 1{2\pi} e^{-\frac {x^2+y^2}2}dxdy=\int_{-\pi}^{\alpha}\int_0^r \frac 1{2\pi} \rho e^{-\frac{\rho}2}d\rho d\theta=\frac {\alpha}{2\pi}(1-e^{-\frac{r^2}2})$$ Now we have that $(R,\theta)$ has density $f(r,\alpha)=\frac r{2\pi} e^{-\frac{r^2}2}\chi _{[0,+\infty)\times [-\pi,\pi]} (r,\alpha)$, from this, finally we obtain the density of $\theta$: $\int_0^{+\infty} \frac r{2\pi} e^{-\frac{r^2}2} \chi _{[0,+\infty)\times [-\pi,\pi]} (r,\alpha) dr=\frac 1{2\pi}\chi _{[-\pi,\pi]} (\alpha)$.

Answer 2

Some ideas that may or may not be relevant/useful.

I would use the Transfer Theorem for this. Let $f$ be a regular enough function (e.g. continuous bounded on $\mathbb{R}$). You need to show : $$\mathbb{E}[f(\theta)]=\int_\alpha^\beta\frac{f(t)}{\beta-\alpha}\text{d}t.$$

Now, some change of variables may help :

$\begin{align} \mathbb{E}[f(\theta)]&=\int_{-\infty}^{+\infty}\left\{\int_{-\infty}^{+\infty}f\circ\arctan\left(\frac{y}{x}\right)e^{-y^2/2}\text{d}y\right\}\frac{e^{-x^2/2}}{2\pi}\text{d}x\\ &=\int_{-\infty}^{+\infty}\left\{\int_{-\pi/2}^{\pi/2}f(t)\frac{xe^{-(x\tan(t))^2/2}}{1+t^2}\text{d}t\right\}\frac{e^{-x^2/2}}{2\pi}\text{d}x. \end{align}$

I'm not 100% sure about my calculations. Now, by justifying it, I suppose you'd need to swap the two integrals using a Fubini theorem, and you'd be left with something that hopefully cancels out and leaves you with $\theta\sim\mathcal{U}([-\pi/2,\pi/2])$.

Edit: there may be a mistake, as $\theta$ should be uniform on $[-\pi,\pi]$...

Answer 3

Here's another way to do it. You can check that the random variables $(\cos \theta) X - (\sin \theta) Y$ and $(\sin \theta) X + (\cos \theta) Y$ are uncorrelated standard normals, and since they are jointly normal they are independent. These are the real and imaginary parts of $e^{i \theta} (X + iY)$ so we conclude that this distribution is rotation invariant. $\tan^{-1}(Y/X)$ is the argument of $X+iY$, so it must be a rotation invariant distribution on the circle, and the only such one is the uniform one.