Proving the sum of two independent Cauchy Random Variables is Cauchy

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.

• 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 3 '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 3 '15 at 14:25
• Read about the partial fractions method in a calculus text or on Wikipedia. – Dilip Sarwate May 3 '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 3 '15 at 17:40
• 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 4 '15 at 3:27 2 Answers 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 ($$K(k)$$). • How do you get equation (5)?$u/\sqrt{4m + u^2} = 0$? – Alain Nov 5 '17 at 23:55 • @Alain: that function is odd and integrable when multiplied by$\frac{1}{1+u^2}$, so it does not really contribute to the integral. – Jack D'Aurizio Nov 6 '17 at 0:13 • And that term comes from$(1 + \frac{m}{x^2}) dx = du$? – Alain Nov 6 '17 at 0:53 • @Alain: exactly. – Jack D'Aurizio Nov 6 '17 at 0:54 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.