I am trying to do the following prove: Let $X$ be a non-negative random variable. Show that if for some $a \in (-1,\infty), b>0$, it holds that the mean excess function $\epsilon_x(t)$ is:
$$\epsilon_X(t) = at + b, t\in[0,x^*] $$
Then $X$ follows a generalized Pareto distribution and give it's parameters. Here we set $X^* = \infty$ for $a \geq 0$ and $x^* = -b/a$ for $a < 0$ (which, as it turns out , is in fact the right endpoint of $F$).
So far I have tried using the following theorem :
$$ε_X(t) = \frac{1}{\bar{F(t)}}\int_t^{X^*}\bar{F(u)} du$$
$$\bar{F(t)}=\frac{\bar{F(0)}}{ε_X(t)}ε_X(0)\exp\left(- \int_0^{t}$ $\frac{1}{εX(u)} du \right) $$
I tried to derive an ordinary differencial equation for $\bar{F}$ to try to get the expression of f and see if it would match the generalized pareto distributio but without sucess. If anyone can help me would be really great!