Distribution of $\arctan(X/Y)$

Question

Given $X$ and $Y$ independent random variables with a standard normal distribution, I've been asked to calculate thee distribution of $Z=\arctan(X/Y)$. For that I thought about calculating the cumulative distribution function of $Z$ and then take its derivative. So $P(Z \leq t)$ would be:

$$ \iint_{R}^{ } \frac{1}{2\pi } e^{\frac{-(x^{2}+y^{2})}{2}} $$

where $R=\{(x,y) \in \Bbb R : \arctan(x/y) \leq t)\}$

Then I thought about solving the integral by using polar coordinates, but I'm not sure what the boundaries for $r$ and $\theta$ should be.

Answer 1

You can use a very standard approach, as follows.

First perform a change of variables $$ \begin{cases} U = \arctan\left(\frac{X}{Y}\right)\\ V = X. \end{cases} $$ If you invert the relationships, they become $$ \begin{cases} X = V\\ Y = V\cot U. \end{cases} $$

The Jacobian of this transformation is $$J = \left| \begin{array}{cc} 0 & 1\\ -\frac{V}{\sin^2 U} & \cot U \end{array} \right| = \frac{V}{\sin^2 U}.$$

The joint distribution of $(U,V)$ is

$$f_{U,V}(u,v) = f_{X,Y}(v,v\cot u) |J|,$$ limited to the correct domain, which we can find by considering the codomain of the functions involved in the original change of variables. In our case, the domain is $$ D_{U,V}=\left(-\frac{\pi}2; \frac{\pi}2\right) \times \Bbb R.$$

So we have

$$ f_{U,V}(u,v) = \begin{cases} \frac{1}{2\pi} e^{-\frac{v^2+v^2\cot^2 u}2} \cdot \left|\frac{v}{\sin^2 u}\right| &((u,v) \in D_{U,V})\\ 0 & ((u,v) \not\in D_{U,V}), \end{cases} $$ that is, with some simplifications, $$ f_{U,V}(u,v) = \begin{cases} \frac{1}{2\pi} e^{-\frac{v^2}{2\sin^2 u}} \cdot \frac{|v|}{\sin^2 u} &((u,v) \in D_{U,V})\\ 0 & ((u,v) \not\in D_{U,V}). \end{cases} $$ Finally, by integration we obtain the marginal distribution you need.

$$f_U(u)= \begin{cases} \int_{-\infty}^{+\infty} f_{U,V}(u,v) dv & -\frac{\pi}2 \leq u \leq \frac{\pi}2 \\ 0 & \mbox{otherwise}.\end{cases}\tag{1}\label{1}$$

With the change of variables $t = v^2/2\sin^2 u$ the integral becomes \begin{eqnarray} \int_{-\infty}^{+\infty} f_{U,V}(u,v) dv &=& \frac1{2\pi}\left[\int_0^{+\infty}e^{-t}dt-\int_{-\infty}^0 e^{-t}dt\right]=\\ &=&\frac1{\pi} \int_0^{+\infty}e^{-t}dt = \frac1{\pi}. \end{eqnarray}

Using \eqref{1} leads to the conclusion that $U$ has uniform distribution between $-\frac{\pi}2$ and $\frac{\pi}2$.

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If, instead, you want to use the approach you mention (in this case it's faster), then consider the inequality

$$U = \arctan\left(\frac{X}{Y}\right) < \alpha.$$

If $0\leq \alpha \leq \frac{\pi}2$, the condition is satisfied by the points $(X,Y)$ in the red shaded region (Figure below).

So in this case, using the symmetry of the original distribution, we get, for the CDF of $U$, \begin{eqnarray} F_U(\alpha) &=& P(U < \alpha) = \frac1{\pi}\int_0^{+\infty}\int_{\pi/2-\alpha}^{\pi} e^{-\frac{\rho^2}2}\rho d\theta d\rho=\\ &=& \left(\frac12+\frac{\alpha}{\pi}\right)\int_0^{+\infty}e^{-\frac{\rho^2}2}\rho d\rho=\\ &=&\left(\frac12+\frac{\alpha}{\pi}\right). \end{eqnarray}

If $-\frac{\pi}2 \leq \alpha \leq 0$, the points $(X,Y)$ that satisfy the relationship are in the blue shaded region below.

This gives again

\begin{eqnarray} F_U(\alpha) &=& P(U < \alpha) = \frac1{\pi}\int_0^{+\infty}\int_{\pi/2-\alpha}^{\pi} e^{-\frac{\rho^2}2}\rho d\theta d\rho=\\ &=&\left(\frac12+\frac{\alpha}{\pi}\right). \end{eqnarray}

In conclusion, $$ F_U(\alpha) = \begin{cases} \left(\frac12+\frac{\alpha}{\pi}\right) & \left(|\alpha| \leq \frac{\pi}2\right)\\ 0 & \mbox{otherwise}, \end{cases} $$ which clearly is a CDF of a uniform distribution between $-\frac{\pi}2$ and $\frac{\pi}2$.

Answer 2

Hint: Try to use $\pi/2 - \arctan(y/x)=\arctan(x/y)$.

If $(x,y) = (r \cos\theta, r\sin\theta)$ where $r>0$ and $\pi/2>\theta>0$ then for any real number $t$, the following are equivalent:

1. $\arctan(x/y) \leq t$
2. $\pi/2 - \arctan(y/x) \leq t$
3. $\pi/2 - \theta \leq t$
4. $\pi/2 - t \leq \theta$

This should help you to get the right limits for the integration.

Also, notice that changing $r$ does not affect whether $\pi/2 - t \leq \theta$.

Answer 3

Firstly, we'll find distribution of $W= \frac{X}{Y}$. Consider vector $V = (X,Y)$ with density $f_V(x,y) = \frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}$.

$F_W(t)= \mathbb P(W \le t) = \mathbb P( \frac{X}{Y} \le t) = \mathbb P( (X \le tY, Y\ge 0) \cup (X \ge tY, Y < 0)) = \mathbb P(X \le tY, Y\ge 0) + \mathbb P( (X \ge tY, Y < 0)= \mathbb P( V \in A_t) + \mathbb P( V \in B_t) = \mu_V(A_t) + \mu_V(B_t)$, where:

$\mu_v(C) = \int_C f_V(v) d\lambda_2(v)$, $A_t = \{ (x,y) \in \mathbb R^2 : y\ge 0, x \le yt\}, B_t = \{ (x,y) \in \mathbb R^2 : y< 0, x \ge yt\}$.

That is:

$\mu_V(A_t) = \frac{1}{2\pi}\int_0^{+\infty} \int_{-\infty}^{yt} e^{-\frac{x^2+y^2}{2}} dxdy$

$ \mu_V(B_t) = \frac{1}{2\pi}\int_{-\infty}^{0} \int_{yt}^{+\infty} e^{-\frac{x^2+y^2}{2}} dxdy $

Note, that functions $t \to \mu_V(A_t), t \to \mu_V(B_t)$ are continuosly differentiable, so $F_w$ as a function of $t$ is, too.

We get: $f_W(t) = \frac{dF_W}{dt}(t) = \frac{1}{2\pi}\frac{d}{dt}(\int_0^{+\infty} \int_{-\infty}^{yt} e^{-\frac{x^2+y^2}{2}} dxdy + \int_{-\infty}^{0} \int_{yt}^{+\infty} e^{-\frac{x^2+y^2}{2}} dxdy) $

$ =\frac{1}{2\pi}(\int_0^{+\infty} \frac{d}{dt} \int_{-\infty}^{yt} e^{-\frac{x^2+y^2}{2}}dxdy + \int_{-\infty}^{0} \frac{d}{dt}\int_{yt}^{+\infty} e^{-\frac{x^2+y^2}{2}} dxdy) = $

$= \frac{1}{2\pi}(\int_0^{+\infty}e^{-\frac{y^2}{2}}( \frac{d}{dt}(G(yt) - G(-\infty)) )dy + \int_{-\infty}^0 e^{-\frac{y^2}{2}}( \frac{d}{dt}( G(+\infty) - G(yt)))dy)$, where $G(x) = \int e^\frac{-x^2}{2}dx$. So $\frac{d}{dt}G(yt) = ye^\frac{-(yt)^2}{2}$.

So that we get:

$f_W(t) = \frac{1}{2\pi}(\int_0^{+\infty} ye^{-\frac{y^2(t^2+1)}{2}}dy - \int_{-\infty}^0 ye^{-\frac{y^2(t^2+1)}{2}}dy)$. The same substitution $ w = \frac{y^2}{2} $, gives us:

$f_W(t) = \frac{1}{2\pi}(\int_0^\infty e^{-w(t^2+1)} dw + \int_0^\infty e^{-w(t^2+1)} dw ) = \frac{1}{\pi} \int_0^\infty e^{-w(t^2+1)}dw = \frac{1}{(t^2+1)\pi}$. Luckily it integrates to one ;).

So now, as we have $W$, we can look at the distribution of $Z=\arctan(W)$

$F_Z(t) = \mathbb P( Z \le t ) = \mathbb P( \arctan(W) \le t) $. So at this point, for $t\le -\frac{\pi}{2}$ we have $F_Z(t) = 0$, and for $t \ge \frac{\pi}{2}$, we have $F_Z(t) = 1$. Let's take any $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

$F_Z(t) = \frac{1}{\pi}\int_{-\infty}^{\tan(t)} \frac{1}{(t^2+1)} = \frac{1}{\pi}(\arctan(\tan(t)) - \arctan(-\infty)) = \frac{t}{\pi} + \frac{1}{2}.$

So, taking derivative of $F_Z$ (as we see, it is piecewise continuously differentiable), we have, that the density of $Z$ is $f_Z(z) = \frac{1}{\pi} \chi_{(-\frac{\pi}{2},\frac{\pi}{2})}$, so to conclude $Z \sim \mathcal U((-\frac{\pi}{2},\frac{\pi}{2}))$