Let $X$ be a positive random variable whose moments are all finite. Define the stationary excess $X_{e}$ to be the random variable with distribution
$$P\{X_{e}\leq t\} = \frac{1}{\mathbb{E}X} \int_0^t P\{X>s\} ds$$
Show that if $f$ is a differentiable function where $\mathbb{E}[f(X)]$ and $\mathbb{E}[f'(X_e)]$ are well defined, then
$$\frac{\mathbb{E}[f(X)-f(0)]}{\mathbb{E}X} = \mathbb{E}[f'(X_e)]$$.
My attempt: By the fundamental theorem of calculus,
$$ \begin{eqnarray} \mathbb{E}[f(X)-f(0)] &=& \mathbb{E}\left[\int_0^X f'(\xi) \ d\xi \right] \\ &=& \mathbb{E}\left[\int_0^\infty \mathbb{1}_{X\geq \xi} f'(\xi) \ d\xi \right] \\ &=& \int_0^\infty P\{X\geq \xi\} f'(\xi) \ d\xi \end{eqnarray}$$
But I'm not sure how to proceed.
Many thanks in advance!
By the definition of expectation we have: $$E(g(X))=\int_{\mathbb{R}}g(x)h(x)dx, $$ where $h(x)$ is a pdf of $X$.
Now $$E(f'(X_e))=\int_0^\infty f'(x)h_e(x)dx $$ and $h_e(x)$ pdf of $X_e$. What does $h_e(x)$ equal?
Let calculate it directly from the $h_e(t)=(P\{X_e\leq t\})_t'=(\frac{1}{E(X)}\int_0^tP\{X> s\}ds)_t'=\frac{1}{E(X)}P\{X>t\}.$
Now just plug-in this $h_e(x)$ into the $E(f'(X_e))$ formula and get the result you need:
$$E(f'(X_e))=\frac{1}{E(X)}\int_0^\infty f'(x)P\{X>x\}dx, $$ does it look familiar now?