Q:

Calculating expectations in terms of quantile functions?

I have a well behaved random variable, $X$, where I can solve for the quantile in closed form, but in general cannot invert it to get the pdf/cdf.  Assume whatever you need on the properties of $X$ and that the quantile function is $F^{-1}(p)$.
Then through standard results, we know that the expectation can be calculated as:
$$
E(X) = \int_{-\infty}^{\infty}x F'(x)dx = \int_{0}^{1} F^{-1}(p)dp
$$
My question is whether this generalizes for expectations of functions of $X$.  i.e. for  strictly increasing $h(X)$ assuming whatever measureability necessary, can $E(h(X))$ be written in terms of quantiles?
$$
E(h(X)) = \int_{0}^{1}h(F^{-1}(p))dp ???
$$

A:

We require that $h$ is right-continuous and we will first assume that $h$ is strictly increasing so that $h^{-1}$ exists and later we will relax this requirement. Define the random variable $Y=h(X)$ and assume that $Y$ is integrable.
In what follows $F_X^{-1}$ is the quantile function of $X$. We have that
$$
(F_X\circ h^{-1})^{-1}=h \circ F_X^{-1}.
$$ 
Notice that we used the fact that $(h^{-1})^{-1}$ and
$$
\begin{align}
(F_X\circ h^{-1})(y) &= \mathrm{P}[X\leq h^{-1}(y)]\\
&=\mathrm{P}[h(X)\leq y]\\
&=F_{Y}(y),
\end{align}
$$
therefore, $F_Y^{-1}(p)=h(F_X^{-1}(p))$ and as a result
$$
\mathbb{E}[Y] = \int_0^1 h(F_X^{-1}(p))\mathrm{d}p.
$$
Further, we need to assume that the underlying probability space is nonatomic. Hereafter, we assume that $h$ is simply nondecreasing and $h^{-1}$ will therefore stand for its generalised inverse.
$$
\begin{align}
(F_X\circ h^{-1})(y) &= 1-\mathrm{P}[X>h^{-1}(y)]\\
                     &= 1-\mathrm{P}[X\geq h^{-1}(y)]\\
                     &= 1-\mathrm{P}[h(X)\geq y]\\
                     &= \mathrm{P}[h(X)< y]\\
                     &= \mathrm{P}[h(X)\leq y]
\end{align}
$$
We have used the fact that $X\geq h^{-1}(y)$ implies $h(X)\geq y$ which holds provided that provided that $h$ is right-continuous.
We then have:
$$
(F_X\circ h^{-1})^{-1}=(h^{-1})^{-1} \circ F_X^{-1},
$$
where $(h^{-1})^{-1}$ is not necessarily equal to $h$. In that case, the expectation becomes
$$
\mathbb{E}[Y] = \int_0^1 (h^{-1})^{-1}(F_X^{-1}(p))\mathrm{d}p.
$$

