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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:

$$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx=\int_{-\infty}^\infty\frac{1}{\pi^2}.\frac{1}{1+x^2}.\frac{1}{1+(z-x)^2}dx\tag{1}$$

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.

StubbornAtom
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Landon Carter
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  • 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? – Dilip Sarwate May 03 '15 at 13:27
  • 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. – Landon Carter May 03 '15 at 14:25
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    Read about the partial fractions method in a calculus text or on Wikipedia. – Dilip Sarwate May 03 '15 at 15:20
  • I know what partial fractions are, of course. I am saying that I do not understand how we can select $a(x)$ and $b(x)$. – Landon Carter May 03 '15 at 17:40
  • The main problem is that by using a partial fraction decomposition we introduce an eliminable discontinuity, then integration is not straighforward if we want to stay on the real line. However, there is a way to circumvent this issue, just see below. – Jack D'Aurizio May 03 '15 at 19:57
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    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. – Landon Carter May 04 '15 at 03:27
  • Apparently someone misunderstood the convolution formula and edited the original work; unfortunately that edit was accepted. – StubbornAtom Feb 08 '20 at 08:03

3 Answers3

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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 ($K(k)$).

Jack D'Aurizio
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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.

Landon Carter
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I will try to show a detailed proof of the above $(4)$ to $(5)$ (for those who does not understand of course). I am not able to comment the above reply so I will reply from here. $$J(m)=\int_{0}^\infty \frac{dx}{(x^2-m)^2+4x^2}$$ for any $m≥1$. With the change of variable $x−\frac{m}{x}=u$ we have:

$$x^2-m=(xu) \; ;\; (x+\frac{m}{x})^2-(x-\frac{m}{x})^2=4m \; ; \; \left(1+\frac{m}{x^2}\right)=du$$ Then, in the above integral : $$\frac{dx}{(x^2-m)^2+4x^2}=\frac{dx}{x^2u^2+4x^2}=\frac{2m(1+\frac{m}{x^2})dx}{2m(1+\frac{m}{x^2})x^2(u^2+4)}=\frac{(2m)du}{(x^2+m)(2m)(u^2+4)}$$ But $$\frac{2m}{x^2+m}=\frac{x^2+m-(x^2-m)}{x^2+m}=1-\frac{xu}{x^2+m}=1-\frac{u}{x+\frac{m}{x}}=1-\frac{u}{\sqrt{2m+(x-m/x)^2}}=1-\frac{u}{\sqrt{u^2+4m}}$$ Hence we obtain the implication $(4)⇒(5)$

arofenitra
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