Finding the pdf of $(X+Y)^2/(X^2+Y^2)$ where $X$ and $Y$ are independent and normal

Question

$X$ and $Y$ are iid standard normal random variables. What is the pdf of $(X+Y)^2/(X^2+Y^2)$?

I am guessing you would transform into polar coordinates and go from there, but I am getting lost. Do we need two variable transformations here?

Answer 1

Hint. Write $(X, Y) = (R\cos\Theta, R\sin\Theta)$. Can you conclude that $\Theta$ is uniform on $[0,2\pi]$? And if this true, how $(X+Y)^2/(X^2+Y^2)$ is written in terms of $\Theta$?

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ In order to keep it simple enough, I'll assume normal distributions like $\ds{\expo{-\xi^{2}/2} \over \root{2\pi}}$. Lets $\ds{z \equiv {\pars{x + y}^{2} \over x^{2} + y^{2}}}$:

\begin{align} \color{#66f}{\large\pp\pars{z}}&=\int_{-\infty}^{\infty}{\expo{-x^{2}/2} \over \root{2\pi}} \int_{-\infty}^{\infty}{\expo{-y^{2}/2} \over \root{2\pi}}\, \delta\pars{z - {\bracks{x + y}^{2} \over x^{2} + y^{2}}}\,\dd x\,\dd y \\[5mm]&={1 \over 2\pi}\ \overbrace{\int_{0}^{\infty}\dd r\,r\expo{-r^{2}/2}} ^{\ds{=}\ \dsc{1}}\ \int_{0}^{2\pi}\delta\pars{z - 1 - \sin\pars{2\theta}}\,\dd\theta \end{align}

It's clear that $\ds{\pp\pars{z} = 0}$ whenever $\ds{z < 0}$ or $\ds{z > 2}$. Hereafter, I'll assume that $\ds{z \in \pars{0,2}}$:

\begin{align} \left.\color{#66f}{\large\pp\pars{z}}\right\vert_{z\ \in\ \pars{0,2}}& ={1 \over \pi}\int_{0}^{\pi}\delta\pars{z - 1 - \sin\pars{2\theta}}\,\dd\theta ={1 \over \pi}\int_{-\pi/2}^{\pi/2} \delta\pars{z - 1 + \sin\pars{2\theta}}\,\dd\theta \\[5mm]&={1 \over \pi}\int_{0}^{\pi/2}\bracks{% \delta\pars{z - 1 + \sin\pars{2\theta}} + \delta\pars{z - 1 - \sin\pars{2\theta}}} \,\dd\theta \\[5mm]&={1 \over \pi}\int_{-\pi/4}^{\pi/4}\bracks{% \delta\pars{z - 1 + \cos\pars{2\theta}} + \delta\pars{z - 1 - \cos\pars{2\theta}}} \,\dd\theta \\[5mm]&={2 \over \pi}\int_{0}^{\pi/4}\bracks{% \delta\pars{z - 1 + \cos\pars{2\theta}} + \delta\pars{z - 1 - \cos\pars{2\theta}}} \,\dd\theta \\[5mm]&={1 \over \pi}\int_{0}^{\pi/2}\bracks{% \delta\pars{z - 1 + \cos\pars{\theta}} + \delta\pars{z - 1 - \cos\pars{\theta}}} \,\dd\theta \\[5mm]&={1 \over \pi}\int_{0}^{\pi/2} \delta\pars{\verts{z - 1} - \cos\pars{\theta}}\,\dd\theta \\[5mm]&={1 \over \pi}\int_{0}^{\pi/2} {\delta\pars{\theta - \arccos\pars{\verts{z - 1}}} \over \verts{\sin\pars{\theta}}}\,\dd\theta \\[5mm]&={1 \over \pi}\,{1 \over \root{1 - \verts{z - 1}^{2}}} ={1 \over \pi}\,{1 \over \root{z\pars{2 - z}}}\,,\qquad z \in \pars{0,2} \end{align}

$$ \color{#66f}{\large\pp\pars{z}} =\color{#66f}{\large\left\{\begin{array}{lcl} {1 \over \pi}\,{1 \over \root{z\pars{2 - z}}} & \color{#000}{\mbox{if}} & z \in \pars{0,2} \\ 0 && \mbox{otherwise} \end{array}\right.} $$