# 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....

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
$$E[W]=(E[T \mid T \leq \xi] + E[W])P[T \leq \xi] + E[T \mid T>\xi] P[T>\xi].$$
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.)