2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$Z_1 + Z_2, Z_1 - Z_2$ 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 $(1,1)$

$\endgroup$
  • $\begingroup$ yes! I see it now, thank you very much $\endgroup$ – otupygak Nov 18 '14 at 12:07
  • $\begingroup$ @Kimo Good, you are welcome! $\endgroup$ – Petite Etincelle Nov 18 '14 at 12:08
1
$\begingroup$

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 $\frac{1}{\pi\sqrt{t}(1+t)}$.

$\endgroup$
-1
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

Also, \begin{align} &\color{#66f}{\large\pp\pars{X}}={\Theta\pars{x} \over 2\pi}\int_{-\pi}^{\pi} \delta\pars{% X - {\cos^{2}\pars{\theta - \pi/4} \over \cos^{2}\pars{\theta + \pi/4}}}\,\dd\theta \\[5mm]&={\Theta\pars{x} \over 2\pi}\bracks{\int_{0}^{\pi}\delta\pars{% X - {\cos^{2}\pars{\theta - \pi/4}\over\cos^{2}\pars{\theta + \pi/4}}}\,\dd\theta +\int_{0}^{\pi}\delta\pars{% X - {\cos^{2}\pars{\theta + \pi/4}\over\cos^{2}\pars{\theta - \pi/4}}}\,\dd\theta} \\[5mm]&={\Theta\pars{x} \over 2\pi}\bracks{\int_{-\pi/2}^{\pi/2}\!\delta\pars{% X - {\sin^{2}\pars{\theta - \pi/4}\over\sin^{2}\pars{\theta + \pi/4}}}\,\dd\theta +\int_{-\pi/2}^{\pi/2}\!\delta\pars{% X - {\sin^{2}\pars{\theta + \pi/4}\over\sin^{2}\pars{\theta - \pi/4}}}\,\dd\theta} \\[5mm]&={\Theta\pars{x} \over \pi}\bracks{\int_{0}^{\pi/2}\!\delta\pars{% X - {\sin^{2}\pars{\theta - \pi/4}\over\sin^{2}\pars{\theta + \pi/4}}}\,\dd\theta +\int_{0}^{\pi/2}\!\delta\pars{% X - {\sin^{2}\pars{\theta + \pi/4}\over\sin^{2}\pars{\theta - \pi/4}}}\,\dd\theta} \\[5mm]&={\Theta\pars{x} \over \pi}\bracks{\int_{-\pi/4}^{\pi/4}\!\delta\pars{% X - {\sin^{2}\pars{\theta}\over\cos^{2}\pars{\theta}}}\,\dd\theta +\int_{-\pi/4}^{\pi/4}\!\delta\pars{% X - {\cos^{2}\pars{\theta}\over\sin^{2}\pars{\theta}}}\,\dd\theta} \\[5mm]&={2\Theta\pars{x} \over \pi}\bracks{\int_{0}^{\pi/4}\!\delta\pars{% X - \tan^{2}\pars{\theta}}\,\dd\theta +\int_{0}^{\pi/4}\!\delta\pars{X - \cot^{2}\pars{\theta}}\,\dd\theta} \\[5mm]&={2\Theta\pars{x} \over \pi}\left\{\Theta\pars{1 - X}\int_{0}^{\pi/4}{\delta\pars{\theta - \arctan\pars{\root{X}}} \over \verts{2\tan\pars{\theta}\sec^{2}\pars{\theta}}}\,\dd\theta\right. \\&\left.\phantom{{2\Theta\pars{x} \over \pi}\left[\right.} +\Theta\pars{X - 1}\int_{0}^{\pi/4} {\delta\pars{\theta - {\rm arccot}\pars{\root{X}}} \over \verts{2\cot\pars{\theta}\bracks{-\csc^{2}\pars{\theta}}}}\,\dd\theta\right\} \\[5mm]&={2\Theta\pars{x} \over \pi}\bracks{% {\Theta\pars{1 - X} \over 2\root{X}\pars{X + 1}} +{\Theta\pars{X - 1} \over 2\root{X}\pars{X + 1}}} =\color{#66f}{\large{\Theta\pars{X} \over \pi}\,{1 \over \root{X}\pars{X + 1}}} \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.