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Is there any method to show that the sum of two independent Cauchy random variables is Cauchy? I know that it can be derived using Characteristic Functions, but the point is, I have not yet learnt Characteristic Functions. I do not know anything about Complex Analysis, Residue Theorem, etc.

I would want to prove the statement only using Real Calculus. Feel free to use Double Integrals if you please.

On searching, I found this. However, I was wondering if I could get some help directly on the convolution formula:

$F_Z(t)=P(X+Y<t)=\int\int\limits_{x+y<t}f_{X}(x)f_{Y}(y)dxdy=\int\limits_{-\infty}^{t}\int\limits_{-\infty}^{+\infty} f_X(x)f_Y(x-y)dxdz=\int\limits_{-\infty}^{t}\int\limits_{-\infty}^{+\infty}\dfrac{1}{\pi^2}.\dfrac{1}{(1+x^2)}.\dfrac{1}{[1+(x-y)^2]}dxdz\tag{1}$

Here I have supposed that $X,Y$ are Independent Standard Cauchy. But I think the general formula can be derived easily after some substitutions. I need some help on how to proceed from $(1)$.

EDIT: Just as what the hint in the hyperlink said, I got the answer using that hint. However, I am not quite sure that the hint is algebraically correct. Maybe there has been some typing mistake in the book.

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  • $\begingroup$ Can you write the integrand in the form $\displaystyle \frac{a(x)}{1+x^2} + \frac{b(x)}{1+(z-x)^2}$ and break up that integral into the sum of two integrals that might be computable more easily? $\endgroup$ – Dilip Sarwate May 3 '15 at 13:27
  • $\begingroup$ Could you provide some more insight into finding $a$ and $b$? I am not really able to think of such functions whose integral I can compute easily. $\endgroup$ – Landon Carter May 3 '15 at 14:25
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    $\begingroup$ Read about the partial fractions method in a calculus text or on Wikipedia. $\endgroup$ – Dilip Sarwate May 3 '15 at 15:20
  • $\begingroup$ I know what partial fractions are, of course. I am saying that I do not understand how we can select $a(x)$ and $b(x)$. $\endgroup$ – Landon Carter May 3 '15 at 17:40
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    $\begingroup$ Actually, if you observe my Partial Fraction Decomposition, it follows that upon integration, the first and third terms together yield $log1% as $x$ approaches $\infty$ or $-\infty$. Hence that will not be problem. $\endgroup$ – Landon Carter May 4 '15 at 3:27
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We may exploit the Lagrange identity: $$ (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2 \tag{1}$$ to state: $$ I_z=\int_{-\infty}^{+\infty}\frac{dx}{(1+x^2)(1+(z-x)^2)}=\int_{-\infty}^{+\infty}\frac{dx}{(1+x(z-x))^2+(z-2x)^2}\tag{2}$$ and by replacing $x$ with $x+\frac{z}{2}$ in the last integral, we get: $$ I_z = \int_{-\infty}^{+\infty}\frac{dx}{(1+\frac{z^2}{4}-x^2)^2+4x^2}\tag{3}$$ hence $I_z$ just depends on $\left(1+\frac{z^2}{4}\right)$. For the sake of brevity, let: $$ J(m)=\int_{0}^{+\infty}\frac{dx}{(x^2-m)^2+4x^2}\tag{4} $$ for any $m\geq 1$. With the change of variable $x-\frac{m}{x}=u$ we have: $$ J(m) = \int_{-\infty}^{+\infty}\frac{1-\frac{u}{\sqrt{4m+u^2}}}{8m+2m \,u^2}\,du = \frac{1}{2m}\int_{-\infty}^{+\infty}\frac{du}{4+u^2}=\frac{\pi}{4m}\tag{5}$$ from which it follows that:

$$ I_z = \frac{\pi}{2+\frac{z^2}{2}}=\frac{2\pi}{4+z^2}.\tag{6}$$

The interesting thing is that this proof is just a variation of the proof of the relation between the arithmetic-geometric mean (AGM) and the complete elliptic integral of the first kind ($K(k)$).

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    $\begingroup$ How do you get equation (5)? $u/\sqrt{4m + u^2} = 0$? $\endgroup$ – Alain Nov 5 '17 at 23:55
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    $\begingroup$ @Alain: that function is odd and integrable when multiplied by $\frac{1}{1+u^2}$, so it does not really contribute to the integral. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 0:13
  • $\begingroup$ And that term comes from $(1 + \frac{m}{x^2}) dx = du$? $\endgroup$ – Alain Nov 6 '17 at 0:53
  • $\begingroup$ @Alain: exactly. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 0:54
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So after no satisfactory answer to this question, here I am posting the ultimate hint which I found after a long hard search and from which the problem becomes immediately obvious.

Decompose $\dfrac{1}{(1+x^2)(1+(z-x)^2)}=\dfrac{1}{z^2(z^2+4)}\big[\dfrac{2zx}{1+x^2}+\dfrac{z^2}{1+x^2}+\dfrac{2z^2-2zx}{1+(z-x)^2}+\dfrac{z^2}{1+(z-x)^2}\big]$

I post this keeping in mind that there must be an online record which I may also use for my personal computations at a later stage.

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