Find the following expectation: $E\Big[g\big(f(x)-c\big)\Big]$

Question

I want to find the expectation of the following function: $$ E\Big[g\big(f(x)-c\big)\Big] \tag{1}$$ where $c$ is a constant.

1. First consider the case when $g(y)=-e^{-y}$, $f(x)=ax^b\; (a,b>0)$, and $x\sim U(i,j) \;(j>i>0)$. Is it possible to find explicitly the above expectation?
2. Is it possible to find the expectation term for an arbitrary function $f$, while function $g$ and random variable $x$ are as specified in 1? If it is then how? If not, then under what conditions would this be feasible?

With respect to 1, I was trying to write $(1)$ as $E[-e^{-{(ax^b-c)}}]$ but now I am unsure how to proceed. If we take a function $h(x)=-e^{-(ax^b-c)}$, would $E[h(x)]=h(E[x])$?

Answer 1

If $\ X\sim U(i,j)\ $, then for any (sufficiently regular) function $\ h\ $ $$ E[h(x)]=\frac{\displaystyle\int_i^jh(y)dy}{j-i}\ . $$ In particular, if $\ h(x)={-}e^{-(ax^b-c)}\ $, then \begin{align} E\big[{-}e^{-(ax^b-c)}\big]&=\frac{\displaystyle\int_i^j{-}e^{-(ay^b-c)}dy}{j-i}\\ &=\frac{{-}e^c}{j-i}\int_i^je^{-ay^b}dy\ . \end{align} The integral can be evaluated in terms of elementary functions if $\ b=1\ $, but not, as far as I know, for any other value of $\ b\ $. If $\ b=2\ $, it can be evaluated in terms of Gauss's error function, erf. Specifically, for those values of $\ b\ $ we have \begin{align} E\big[e^{-(ax^1-c)}\big]&=\frac{e^{c-aj}-e^{c-ai}}{a(j-i)}\\ E\big[e^{-(ax^2-c)}\big]&=\frac{\sqrt{\pi}\,\big(\text{erf}(i\sqrt{a})-\text{erf}(j\sqrt{a})\big)}{2\sqrt{a}(j-i)}\ . \end{align} For other values of $\ b\ $ you can express the integral as a power series, by using the power series expansion $\ e^z=\sum_\limits{n=0}^\infty \frac{z^n}{n!}\ $: $$ E\big[e^{-(ax^b-c)}\big]=\Big(\frac{e^c}{j-i}\Big)\Big(\sum_{n=0}^\infty\frac{(-a)^ni^{bn+1}}{(bn+1)n!}-\sum_{n=0}^\infty\frac{(-a)^nj^{bn+1}}{(bn+1)n!}\Big)\ . $$

For an arbitrary function $\ f\ $ it's not possible to give any simpler expression than $$ \frac{-1}{j-i}\int_i^je^{-f(y)}dy $$ for $\ E\big[{-}e^{-f(x)}\big]\ $ unless you have much more specific information about the form of $\ f\ $.

The identity $\ E[h(x)]=h\big(E[x]\big)\ $ is only likely to hold if $\ h\ $ is a linear function of the form $\ h(x)=cx+d\ $. For non-linear functions such as $\ h(x)={-}e^{-(x^b-c)}\ $ it's only going to hold as the result of a very unlikely coincidence. In fact, for $\ b>1 $ and $\ y>1\ $, $\ {-}e^{-(y^b-c)}\ $ is a strictly concave function function of $\ y\ $, so if $\ i>1\ $ and $\ b>1\ $ then $\ E[h(x)]<h\big(E[x]\big)\ $.