I'm going to start out by saying this is a homework problem straight out of the book. I have spent a couple hours looking up how to find expected values, and have determined I understand nothing.
Let $X$ have the CDF $F(x) = 1 – x^{-\alpha}, x\ge1$.
Find $E(X)$ for those values of $\alpha$ for which $E(X)$ exists.
I have no idea how to even start this. How can I determine which values of $\alpha$ exist? I also don't know what to do with the CDF (I'm assuming this means Cumulative Distribution Function). There are formulas for finding the expected value when you have a frequency function or density function. Wikipedia says the CDF of $X$ can be defined in terms of the probability density function $f$ as follows:
$F(x) = \int_{-\infty}^x f(t)\,dt$
This is as far as I got. Where do I go from here?
EDIT: I meant to put $x\ge1$.
Best Answer
Edited for the comment from probabilityislogic
Note that $F(1)=0$ in this case so the distribution has probability $0$ of being less than $1$, so $x \ge 1$, and you will also need $\alpha > 0$ for an increasing cdf.
If you have the cdf then you want the anti-integral or derivative which with a continuous distribution like this
$$f(x) = \frac{dF(x)}{dx}$$
and in reverse $F(x) = \int_{1}^x f(t)\,dt$ for $x \ge 1$.
Then to find the expectation you need to find
$$E[X] = \int_{1}^{\infty} x f(x)\,dx$$
providing that this exists. I will leave the calculus to you.