renewal process with underlying distribution Consider a renewal process with underlying distribution function $F(x)$.
Let $W$ be the time when the interval duration from the preceding renewal event first exceeds $\xi > 0$ (a fixed constant). Determine an integral equation satisfied by
$$ V(t)= Pr(W \leq t)$$
Assume an event occurs at time $t = 0$.
So, I'm studying this renewal theory, and I would like to know if someone could help me to compute $E[W]$.
Could someone give me hints to solve this problem, pls?
Thanks for your time and help everyone. Some help pls....
 A: The expectation is easier to calculate. Either the first jump occurs before time $\xi$ or it doesn't. The case when it does restart tells you $E[W \mid T \leq \xi]=E[T \mid T \leq \xi] + E[W]$, where $T$ is the time of the first jump. When it doesn't, $W=T$, so you have $E[T \mid T>\xi]$. So 
$$E[W]=(E[T \mid T \leq \xi] + E[W])P[T \leq \xi] + E[T \mid T>\xi] P[T>\xi].$$
This can be quite dramatically simplified (and the final expression is quite intuitively appealing).
The CDF of $W$ can be calculated using a similar "renewal" approach: either the first jump occurs before time $\xi$ or it doesn't. If it does, there is a contribution to the probability of $\int_0^\xi V(t-s) dF(s)$, corresponding to jumping at time $s$ and then having time $t-s$ left over. If it doesn't, there is a contribution to the probability of $\int_\xi^t dF(s)=F(t)-F(\xi)$, corresponding to jumping at time $\xi<s \leq t$. In any case $V(t)=0$ if $t<\xi$. 
This gives the desired integral equation. When $F$ is absolutely continuous, it is a Fredholm integral equation of the second kind, of the form $V=K*V+g$, where the kernel $K(x,y)=\begin{cases} f(x-y) & x-y \leq \xi \\ 0 & x-y>\xi \end{cases}$, the inhomogeneity $g(x)=\max \{ 0,F(x)-F(\xi) \}$ and $*$ denotes causal convolution.
(The above might be slightly wrong if $F$ is not continuous at $\xi$ i.e. $P(T=\xi)>0$, so be careful about that case.)
