calculate some integral of sum of two variables

Question

Let $X_1$ and $X_2$ be two i.i.d. continuous random variables which have probability density function $$f(x) = \frac{1}{\pi}\frac{1}{1+x^2}.$$ Now I want to calculate the density of $X_1 + X_2$.
$$ f_{X_1+X_2}(z) = \int_{-\infty}^{\infty} f_{X_1}(x_1)f_{X_2}(z-x_1)dx_1 = \int_{-\infty}^{\infty} \frac{1}{\pi} \frac{1}{1+x_1^2} \frac{1}{\pi} \frac{1}{1+(z-x_1)^2} dx_1 $$ Is there any efficient way to calculate this integral?

Does this integral exist? What happens when $x_1 = z$? – Ryan Goulden Mar 29 '19 at 01:35 — Ryan Goulden, Mar 29 '19 at 01:35
@RyanGoulden this one does not but the correct one does – gt6989b Mar 29 '19 at 01:56 — gt6989b, Mar 29 '19 at 01:56

score 1 · Accepted Answer · answered Mar 29 '19 at 01:51

1

You made a mistake, the correct integral is $$ \frac{1}{\pi^2} \int_\mathbb{R} \frac{1}{1+x^2} \frac{1}{1 + (z-x)^2} dx $$ and I would look around trig substitution or partial fractions...

answered Mar 29 '19 at 01:51

gt6989b

54,930

thanks for pointing it out – Jonny Mar 29 '19 at 02:07
I think it is not possible to get the original one. – Jonny Mar 29 '19 at 02:40

score 0 · Answer 2 · answered Mar 29 '19 at 01:31

0

You could try partial fractions. An integral of (1+(x1)^2)^-1 on its own is fine to work out, it is an arctan integral. Likewise, an integral of (1+(z-x1)^2)^-1 is also very doable.

answered Mar 29 '19 at 01:31

fGDu94

4,072

Also double check your expression for fx2(z-x1) – fGDu94 Mar 29 '19 at 01:34

calculate some integral of sum of two variables

2 Answers2