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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.

little o
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  • Do you know contour integration? – Shashi Nov 30 '18 at 20:12
  • Yeah. I know. Is it related to contour integration? – little o Nov 30 '18 at 20:19
  • Yes! Maybe you can try to do that, or partial fraction decomposition. The latter needs longer calculations I guess. – Shashi Nov 30 '18 at 20:25
  • Yeah. I have finally found $g$ which is given by $g(x)= \frac {2} {\pi (x^2+4)},x\in \Bbb R$. – little o Nov 30 '18 at 20:31
  • So we can say that $\frac {X+Y} {2} \sim \text {Cauchy distribution}$. Isn't it so? – little o Nov 30 '18 at 20:34
  • Yes, you can say that! – Shashi Nov 30 '18 at 20:38
  • Very interesting problem indeed! – little o Nov 30 '18 at 20:40
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    The usual approach is to use characteristic functions. If $X$ and $Y$ are Cauchy with parameters $a$ and $b$, $$E(e^{itX})=e^{-a|t|}\qquad E(e^{itY})=e^{-b|t|}$$ If furthermore $X$ and $Y$ are independent, $$E(e^{it(X+Y)})=E(e^{itX})E(e^{itY})=e^{-a|t|}e^{-b|t|}=e^{-(a+b)|t|}$$ hence $X+Y$ is Cauchy with parameter $a+b$. Thus, it suffices to know how to prove that $$\int_\mathbb Re^{itx}\frac{dx}{1+x^2}=e^{-|t|}$$ Do you? – Did Nov 30 '18 at 20:47
  • @Did O, after having finished my answer, I saw that you already did the same in a comment – Shashi Nov 30 '18 at 20:55

2 Answers2

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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}

mechanodroid
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  • I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid. – little o Nov 30 '18 at 21:14
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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.

Shashi
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  • The process you have mentioned is somewhat related to Fourier transform. Isn't it so? – little o Nov 30 '18 at 21:11
  • @Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful. – Shashi Nov 30 '18 at 21:15
  • Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus. – little o Nov 30 '18 at 21:24