# Thread: PDF of a random variable

1. ## PDF of a random variable

The pdf of a random variable X is given by f(x)=c(1-x^2) for x element of (-1,1) (the element of symbol that looks like a curved E). For which values of c is this function actually a pdf? (c being a constant)

I know that in order for something to be a pdf it must meet the requirements that
p(x)>or= 0 for all x
and
the probabilities must sum to 1.

Where I am confused is by the element of (-1,1) I am not sure how to solve this with an interval.
Thanks to anyone who can help.
(I attached the equation in a word doc just in case my notation above is confusing)

2. ## Re: PDF of a random variable

Originally Posted by acasas4
The pdf of a random variable X is given by f(x)=c(1-x^2) for x element of (-1,1) (the element of symbol that looks like a curved E). For which values of c is this function actually a pdf? (c being a constant)

I know that in order for something to be a pdf it must meet the requirements that
p(x)>or= 0 for all x
and
the probabilities must sum to 1.

Where I am confused is by the element of (-1,1) I am not sure how to solve this with an interval.
Thanks to anyone who can help.
(I attached the equation in a word doc just in case my notation above is confusing)
Since this is a continuous random variable the way that you check that it "sums" to 1 is to use an integral. That is what you need an interval!

This must hold.

$\displaystyle \int_{-1}^{1}f(x)dx=1$

Evaluate the integral and solve for c.