# Find the Distribution of the following;

Give that $Z_1,Z_2,...,Z_n$ are independent identically distributed standard Gaussian random variables with mean 0 and variance 1.
Find the Distribution of $$X=\dfrac{(Z_1+Z_2)^2}{(Z_1-Z_2)^2}$$

Now first thing I did was open it up, but no common distributions I know are looking familiar. Any hints would be appreciated.

$Z_1 + Z_2, Z_1 - Z_2$ are jointly Gaussian and luckily their convariance is $0$, so they are also independent, then it's easy to see the quotient of their squares follows a F-distribution with parameter $(1,1)$
It is useful to recall that, if $Z_1$ and $Z_2$ are independent normal variables with unit variance, then $W=Z_2/Z_1$ has a Cauchy distribution. Obviously $U=\left(\frac{Z_1+Z_2}{Z_1-Z_2}\right)^2=\left(\frac{1+W}{1-W}\right)^2$ is a non-negative random variable, and, for every $t\in[0,1)$: $$\mathbb{P}[U\leq t]=\mathbb{P}\left[\frac{1+\sqrt{t}}{-1+\sqrt{t}}\leq W\leq\frac{-1+\sqrt{t}}{1+\sqrt{t}}\right]=\frac{2}{\pi}\arctan\sqrt{t}$$ while for $t\in[1,+\infty)$ we have: $$\mathbb{P}[U\leq t]=\mathbb{P}\left[W\leq\frac{-1+\sqrt{t}}{1+\sqrt{t}}\right]+\mathbb{P}\left[W\geq\frac{1+\sqrt{t}}{-1+\sqrt{t}}\right]=\frac{2}{\pi}\arctan\sqrt{t}$$ hence the pdf of $U$ is supported on $\mathbb{R}^+$ and given by $\frac{1}{\pi\sqrt{t}(1+t)}$.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#66f}{\large\pp\pars{X}}&\equiv \int_{-\infty}^{\infty}\dd Z_{1}\,{\exp\pars{-Z_{1}^{2}/2}\over \root{2\pi}} \int_{-\infty}^{\infty}\dd Z_{2}\,{\exp\pars{-Z_{2}^{2}/2}\over \root{2\pi}}\, \delta\pars{X - {\pars{Z_{1} + Z_{2}}^{2} \over \pars{Z_{1} - Z_{2}}^{2}}} \\[5mm]&={1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{Z_{1}^{2} + Z_{2}^{2} \over 2}} \delta\pars{X - {\bracks{Z_{1} + Z_{2}}^{2} \over \bracks{Z_{1} - Z_{2}}^{2}}} \,\dd Z_{1}\,\dd Z_{2} \\[5mm]&={1 \over 2\pi}\int_{0}^{2\pi}\int_{0}^{\infty} \exp\pars{-\,{\rho^{2} \over 2}} \delta\pars{X - {\bracks{\cos\pars{\theta} + \sin\pars{\theta}}^{2} \over \bracks{\cos\pars{\theta} - \sin\pars{\theta}}^{2}}}\rho\,\dd\rho\,\dd\theta \\[5mm]&=\color{#66f}{\large{\Theta\pars{X} \over 2\pi}\int_{-\pi}^{\pi} \delta\pars{% X - {\bracks{\cos\pars{\theta} + \sin\pars{\theta}}^{2} \over \bracks{\cos\pars{\theta} - \sin\pars{\theta}}^{2}}} \,\dd\theta} \\[5mm]&\mbox{because}\quad \int_{0}^{\infty}\exp\pars{-\,{\rho^{2} \over 2}}\rho\,\dd\rho = 1 \end{align}