# Probability Problem

• Dec 11th 2006, 08:22 AM
WeeG
Probability Problem
Hello,

I need to prove that the next function is a density function, meaning, to prove that the integral equals 1. I know that I need to use Cauchy's distribution , but I can't figure out how exactly, you can say it's a calculus problem....

thanks

the function is:

f(x)= (1/pi) * ( beta / beta^2+(x-a)^2)

beta does not refers to the beta function, just the greek letter
• Dec 13th 2006, 02:38 PM
CaptainBlack
Quote:

Originally Posted by WeeG
Hello,

I need to prove that the next function is a density function, meaning, to prove that the integral equals 1. I know that I need to use Cauchy's distribution , but I can't figure out how exactly, you can say it's a calculus problem....

thanks

the function is:

f(x)= (1/pi) * ( beta / beta^2+(x-a)^2)

beta does not refers to the beta function, just the greek letter

$
I=\int_{-\infty}^{\infty}\,f(x)dx=\frac{\beta}{\pi}\,\int_{-\infty}^{\infty}\,\frac{1}{\beta^2+(x-a)^2}dx
$

...... $
=\frac{1}{\beta\, \pi}\,\int_{-\infty}^{\infty}\,\frac{1}{1+((x-a)/\beta)^2}dx
$

Now put $x'=(x-a)/\beta$, then:

$
I=\frac{1}{\pi}\,\int_{-\infty}^{\infty}\,\frac{1}{1+x'^2}dx'
$

Now the integral is a standard integral and is equal to $\pi$,
so:

$I=1$, and so $f$ is a density.

RonL
• Dec 15th 2006, 12:16 PM
WeeG
Thanks, superb solution
:)