What will be the pdf of $X+Y$ if $X$ and $Y$ are iid from Cauchy?

Question

Suppose $X$ and $Y$ follow Cauchy distribution independent of each other. What will be the pdf of $X+Y$?

What I got by using convolution theorem is that the density $g$ of $X+Y$ is $:$

$$g(x) = \int_{-\infty}^{\infty} f(y) f(x-y)\ \mathrm {dy}$$ where $f$ is the density of the Cauchy distribution given by $f(x)=\frac {1} {\pi ({1+ x^2})},x \in \Bbb R$. Then the whole integration becomes $$\frac {1} {\pi^2} \int_{-\infty}^{\infty} \frac {\mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.$$ Now how do I solve this integral? Please help me in this regard.

Thank you very much.

Answer 1

You can try partial fraction decomposition:

\begin{align}\frac1{(1+y^2)(1+(x-y)^2)} &= \frac{x+2y}{x(x^2+4)(1+y^2)} + \frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \\ &= \frac{1}{(x^2+4)(1+y^2)}+ \frac{2y}{x(x^2+4)(1+y^2)} + \frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + \frac{1}{(x^2+4)(1+(x-y)^2)}\\ &= \frac1{x^2+4}\left(\frac1{1+y^2} + \frac1{1+(x-y)^2}\right) + \frac1{x(x^2+4)}\left(\frac{y}{1+y^2} + \frac{x-y}{1+(x-y)^2}\right) \end{align} so we have \begin{align} \frac1{\pi^2}\int_{-\infty}^\infty \frac{dy}{(1+y^2)(1+(x-y)^2)} &= \frac1{\pi^2(x^2+4)}\left[\int_{-\infty}^\infty\left(\frac1{1+y^2} + \frac1{1+(x-y)^2}\right)dy + \frac1x\int_{-\infty}^\infty\left(\frac{y}{1+y^2} + \frac{2(x-y)}{1+(x-y)^2}\right)dy \right]\\ &= \frac1{\pi^2(x^2+4)}\left[\Big(\arctan(1+y^2) + \arctan(1+(x-y)^2)\Big)\Big|_{-\infty}^\infty + \frac1x\Big(\ln(1+y^2) - \ln(1+(x-y)^2)\Big)\Big|_{-\infty}^\infty \right]\\ &= \frac1{\pi^2(x^2+4)}\left[2\pi + \frac2x \lim_{y \to \infty} \ln \left(\frac{1+y^2}{1+(x-y)^2}\right)\right]\\ &= \frac{2}{\pi(x^2+4)} \end{align}

Answer 2

What about using characteristic functions? Let $\phi$ be the characteristic function of a Cauchy distribution. We know that $$\phi(t) =\frac 1\pi\int_\mathbb{R} \frac{e^{itx}} {1+x^2}\, dx$$ Such integrals are many times solved on this site, for instance here, the result is $$\phi(t) = e^{-|t|} $$ So the characteristic function of $X+Y$ is $\phi$ squared, due to the independence of $X$ and $Y$. We conclude $$\phi_{X+Y} (t) =\phi(t) \phi(t) =e^{-2|t|}=e^{-|2t|}$$ Indeed, we may also conclude from this that $(X+Y) /2$ is Cauchy distributed.