Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution.

Question

Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution.

MY SOLUTION

To begin with, let us remember that the Cauchy probability density function is given by \begin{align*} f_{X}(x) = \frac{1}{\pi(1+x^{2})}\quad\text{for}\quad x\in\textbf{R} \end{align*}

Consequently, the sought distribution is described as \begin{align*} F_{Y}(Y\leq y) &= \textbf{P}(Y\leq y) = \textbf{P}\left(\frac{a}{1+X^{2}} \leq y\right) = \textbf{P}\left(1+X^{2}\geq \frac{a}{y}\right)\\\\ & = \textbf{P}\left(|X|\geq \sqrt{\frac{a-y}{y}}\right) = \textbf{P}\left(X\geq \sqrt{\frac{a-y}{y}}\right) + \textbf{P}\left(X \leq -\sqrt{\frac{a-y}{y}}\right)\\\\ & = 1 - F_{X}\left(\sqrt{\frac{a-y}{y}}\right) + F_{X}\left(-\sqrt{\frac{a-y}{y}}\right) \end{align*}

where $y\in(0,a]$. Finally, we have \begin{align*} F_{X}(x) = \int_{-\infty}^{x}f_{X}(u)\mathrm{d}u = \int_{-\infty}^{x}\frac{\mathrm{d}x}{\pi(1+x^{2})} = \frac{\arctan(x)}{\pi} + \frac{1}{2}\end{align*}

I have two questions. Firstly, I would like to know if this approach is correct. Secondly, I would like to know if there is another way to solve this problem. Thanks in advance!

Answer 1

You can use the usual change of variables method.

Density of $X$ is

$$f_X(x)=\frac{1}{\pi(1+x^2)}\qquad,\,x\in\mathbb R$$

Note that $Y=\frac{a}{1+X^2}$ will have separate densities for $a>0$ and $a<0$.

Consider the case $a>0$.

The preimages $x_1,x_2$ for the 2-to-1 transformation are $$Y=\frac{a}{1+X^2}\implies \begin{cases}x=x_1(y)=\sqrt{(a-y)/y}&,\text{ if }x>0 \\\\ x=x_2(y)=-\sqrt{(a-y)/y}&,\text{ if }x<0 \end{cases}$$

And $$x\in\mathbb R\implies 0<y< a $$

Absolute value of the jacobian is

$$\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|=\frac{a}{2y^{3/2}\sqrt{(a-y)}}$$

Hence the density of $Y$ is

\begin{align} f_Y(y)&=f_X(x_1(y))\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|+f_X(x_2(y))\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right| \\&=2\times\frac{1}{2\pi\sqrt {y(a-y)}}\mathbf1_{0<y< a} \end{align}

That is, pdf of $Y$ when $a>0$ is $$f_Y(y)=\frac{1}{\pi\sqrt {y(a-y)}}\mathbf1_{0<y< a}\tag{1}$$

Proceeding similarly for $a<0$, keeping in mind there are no negative quantities inside square roots yields the pdf

$$f_Y(y)=\frac{1}{\pi\sqrt {y(a-y)}}\mathbf1_{a< y<0}\tag{2}$$

The densities $(1)$ and $(2)$ make sense because $Y$ is of the form $Y=aZ$ where $Z\sim\text{Beta}(1/2,1/2)$, as suggested in the answer by @kimchilover using the fact that a standard Cauchy variable can be expressed as the ratio of two independent standard normal variables.

Answer 2

A different way to approach this is to represent a Cauchy r.v. $X$ as the ratio of two independent $N(0,1)$ r.v.s. Once you write $X=S/T$, with $S,T\sim N(0,1)$ the representation $Y=aT^2/(S^2+T^2)$ drops out. And so on...

This method relies on "pattern matching" (you have to `know the ratio of Gaussians fact, you have to know what the distribution of $T^2/(S^2+T^2)$ is). But if you have these facts in your working tool kit, the calculus details are less intricate.

Answer 3

first notice that if $\alpha =0$ then $Y=0$ (So $P_Y=\delta_0$)

Let's suppose that $\alpha \in \mathbb{R^*}$, take any bounded function $f$, we have: $$E[f(Y)]=\frac{2}{\pi}\int_{]0;+\infty[}f(\frac{\alpha}{x^2+1})\frac{1}{x^2+1}dx=\frac{1}{\pi}\int_{]\min(\alpha,0),\max(\alpha,0)[}f(u)\frac{1}{\sqrt{u(\alpha-u)}}du$$

by the substitution $u=\frac{\alpha}{x^2+1}$ and in these case Y has a density

Answer 4

Your approach is fine.

Perhaps you are aware of this result (very useful to generate a Cauchy variable): if $Z \sim U(-\frac{\pi}{2},\frac{\pi}{2})$ then $X = \tan(Z)$ follows a canonical Cauchy distribution.

Now, we have $Y = \frac{a}{1+X^2}= a \cos^2(Z)$ which is slightly easier.

Then, assuming $a>0$ $$\begin{align} f_Y(y) &= 2 \frac{f_Z(z)}{|2 a \cos(z) \sin(z)|} \\ &=\frac{1}{a \pi} \frac{1}{|\cos^2(z)\tan(z)|}\\ &=\frac{1}{a \pi} \frac{1}{|\frac{y}{a}x|}\\ &=\frac{1}{ \pi} \frac{1}{y\sqrt{a/y-1}} \\ &=\frac{1}{ \pi} \frac{1}{\sqrt{y(a-y)}} \end{align} $$

for $0<y<a$