Expectation of $X$ given a Cumulative Function I've (hard) to find the expectation of $X$:
CDF (Cumulative Function) = $1 - x^{-a}$, $1 \leqslant x < \infty$
$E[X]$ = ??
 A: First you have to establish the proper domain for the random variable $X$. That can be done by using $F(x_{\min}) = 0$. You random variable will be supported on $[x_{\min}, \infty)$. 


*

*Find the probability density function, $f_X(x)$ by differentiating the cumulative function, $f_X(x) = F_X^\prime(x)$.

*Apply the law of the unconscious statistician: $$ \mathbb{E}(X) = \int_{x_\min}^\infty x f_X(x) \mathrm{d}x $$ 

*For what values of parameter $a$ does the integral exist?

*Once you found the answer, you have found the mean of the Pareto distribution.
A: To compute the expectation of a nonnegative random variable $X$ when one is given its cumulative distribution function $F_X$, one does not need to deduce the probability density function $f_X$ from $F_X$ and to plug $f_X$ into the well known integral formula. Rather, one can rely on the more direct
$$
\mathbb E(X)=\int_0^{+\infty}(1-F_X(x)) \, \mathrm dx.
$$
In the present case, $F_X(x)=0$ if $x\leqslant1$ and $F_X(x)=1-x^{-a}$ if $x\geqslant1$ hence
$$
\mathbb E(X)=1+\int_1^{+\infty}x^{-a} \, \mathrm dx=\ldots
$$
Similar direct formulas exist when $X$ is real-valued, not necessarily almost surely nonnegative, and still others are available when one needs to compute higher moments $E(X^\alpha)$ for $\alpha>1$, all using only the CDF $F_X$ and not the (possibly nonexistent) PDF of $X$.
