$X$ and $Y$ are independent rv having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$ determine pdf of $Z=\frac{X+Y}{3}$

Question

Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.

So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$.

We shall find the CDF of $Z$ and then differentiate it.

$F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \ dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.

At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.

How do I do this? Or otherwise is there different more elegant way to get the required pdf?

Answer 1

$X$ and $Y$ are standard Cauchy hence $\frac{X+Y}{2}$ is standard Cauchy (what easily can be seen using characteristic functions)

So the pdf of $\frac{3}{2}Z$ is also $$f(t) = \frac{1}{\pi}\frac{1}{1+t^2}$$

Hence the distribution of $Z$ is: $$f(z) = \frac{d}{dz}P(Z \le z) = \frac{d}{dz}F\left(\frac{3}{2}Z \le \frac{3}{2}z\right) = \frac{3}{2}f\left(\frac{3}{2}z\right)$$

Answer 2

If you are comfortable with performing a slightly hefty partial fraction decomposition while evaluating an integral, you could do the following standard procedure:

Denote the joint density of $(X,Y)$ as $f_{X,Y}(x,y)$ where the support of the distribution is $x,y\in\mathbb{R}$

Transform $(X,Y)\to (U,V)$ such that $U=\frac{X+Y}{3}, V=Y$

$\Rightarrow x=3u-v,\quad y=v$ where $u,v\in\mathbb R$

The Jacobian of the transformation is

$\det\left(J\left(\dfrac{x,y}{u,v}\right)\right)=\det\begin{pmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\\\end{pmatrix}=\det\begin{pmatrix}3&-1\\ 0& 1\end{pmatrix}=3$

The joint density of $(U,V)$ is then given by

$f_{U,V}(u,v)=f_{X,Y}(3u-v,v)|\det(J)|=\dfrac{3}{\pi^2(1+(3u-v)^2)(1+v^2)},\qquad u,v\in\mathbb{R}$

So the marginal density of $U$ is

$f_U(u)=\displaystyle\int_\mathbb Rf_{U,V}(u,v)\,\mathrm{d}v$

$\qquad\quad=\dfrac{3}{\pi^2}\displaystyle\int_\mathbb R\frac{\mathrm{d}v}{(1+(3u-v)^2)(1+v^2)}$

$\qquad\quad=\dfrac{3}{\pi^2}\cdot\dfrac{2\pi}{9u^2+4}=\dfrac{6}{\pi(9u^2+4)}\quad,u\in\mathbb{R}$

You can refer to this post for evaluating the above integral.