Expectation of function of random variable? How to calculate the expectation of function of random variable without using probability density function? Note:- only cumulative distribution function is available.
For example $E[g(X)]$=? where X is  nonnegative r.v. with CDF $F_{X}(x)$.
 A: If $X\geqslant0$ almost surely and if $g$ is regular,
$$
\mathrm E(g(X))=g(0)+\int_0^{+\infty}g'(x)\cdot(1-F_X(x))\cdot\mathrm dx.
$$
Proof: integrate with respect to $\mathrm P$ both sides of the almost sure relation
$$
g(X)=g(0)+\int_0^{+\infty}g'(x)\cdot[x\lt X]\cdot\mathrm dx,
$$
where $[\ \ ]$ denotes Iverson bracket.
A: I'm going to hazard a guess about what the original poster might mean.  Suppose wone knows the probability density function of $X$ but not of $g(X)$.  My guess is that it is the latter that is not to be used.  In that case, one can write
$$
\mathbb{E}(g(X)) = \int_{-\infty}^\infty g(x)f(x)\;dx
$$
where $f$ is the probability density function of $f$.  In other words, one doesn't need to find the probability density function of $g(X)$.
A: At least formally, the probability density function is $f(X) = dF/dX$, so 

$ E[g(X)] = \int dX f(X) g(X) = \int dX g(X) dF/dX = Fg(X_f) - Fg(X_i) - \int dX F(X) dg/dX $.

So if $g(X)$ is differentiable, and $g$ is finite at the endpoints of the domain of $X$ ($X_f$ and $X_i$), then you can try to evaluate the integral above.
