Chapter 5.2, Problem 12E

Let $W_1,W_2$ be independent, each with a Cauchy distribution. In this exercise we find the pdf of the sample mean $\bar{W}=\frac{\left(W_1+W_2\right)}{2}$.

(a) Show that the pdf of $X_1=\left(\frac{1}{2}\right)W_1$ is $f\left(x_1\right)=\frac{2}{\pi \left(1+4x_1^2\right)},-\infty <x_1<\infty$

(b) Let $Y_1=X_1+X_2=\bar{W}$ and $Y_2=X_1$, where $X_2=\left(\frac{1}{2}\right)W_2$.

Show that the joint pdf of $Y_1$ and $Y_2$ is

$g\left(y_1,y_2\right)=f\left(y_1-y_2\right)f\left(y_2\right),-\infty <y_1<\infty ,-\infty <y_2<\infty$

(c) Show that the pdf of $Y_1=\bar{W}$ is given by the convolution formula.

$g_1\left(y_1\right)={\displaystyle {\int }_{-\infty }^{\infty }f\left(y_1-y_2\right)f\left(y_2\right)d{y}_2.}$

(d) Show that $g_1\left(y_1\right)=\frac{1}{\pi \left(1+y_1^2\right)},-\infty <y_1<\infty .$

That is, $\bar{W}$, the pdf of W is the same as that of an individual W.