Can a probability density funtion have a negative domain?

Question

I've only ever seen domains stretching from $X\ge0$. I have a question where:

$F(x)=cx^2$, Domain: $− 1 ≤ x ≤ 1$

When finding value of $c$, would I only need to integrate $0$ to $1$. And does that go the same for $E(x)/$mean?

Answer 1

Yes, a random variable can be negative. If you have an associated density function, you have to integrate over the whole image of the random variable, which means in your case, since $f$ is a probabilty density function, that $$1 = \int_{-1}^1 f(x) dx = \frac{2}{3}c,$$ hence, $c = \frac{3}{2}$. Note that because of this we have $c>0$, which is in accordance with the fact that the image of a PDF must almost surely never be negative. In this case regardless of $c$ holds $$E[X] = \int_{-1}^1 xf(x) dx = 0,$$ since the $f$ is an even function. Because of this, it would be sufficient to solve for $c$ via $$\frac{1}{2} = \int_{0}^1 f(x) dx,$$ as you suggested.

Answer 2

It can have a negative domain. And you will need to integrate over the whole domain (from $-1$ to $1$ in this case).