# Weird Density function

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• December 7th 2007, 02:58 PM
paolopiace
Weird Density function
Hello.

Could someone help me with this, please? Thanks!

Given the following probability density function

p(x) = (3/4)*(1-x^2) for |x|<1

p(x) = 0 otherwise

I need to find the probability that the random variable X falls between 0 and 1/2 (i.e. 0<X<1/2)

Integrating from -1 to 1 the density function, I find its area equals -(1/2).

Am I doing sometihng wrong?
• December 7th 2007, 03:49 PM
Plato
Quote:

Originally Posted by paolopiace
Integrating from -1 to 1 the density function, I find its area equals -(1/2).
Am I doing sometihng wrong?

Yes you are. Look at the integral again.
$\int\limits_{ - 1}^1 {\left( {1 - x^2 } \right)dx} = \frac{4}{3}$
• December 7th 2007, 04:16 PM
paolopiace
Thanks Plato...
I really don't know how I went wrong on that trivial integral...
• December 7th 2007, 05:00 PM
Krizalid
Always take advantage of the even functions.

$f(x)=(1-x^2)$ is even, since $f(x)=f(-x).$ (You may confirm this easily.)

Now $\int_{ - 1}^1 {\left( {1 - x^2 } \right)\,dx} = 2\int_0^1 {\left( {1 - x^2 } \right)\,dx} = 2\left( {\left. {x - \frac{1}
{3}x^3 } \right|_0^1 } \right) = \frac{4}
{3}.$

This leads easy computations and prevent to get mistaken.