We have a point process where the interarrival time is described by a random variable, $T$. Further, $E(T)=t$ and $T=t+\epsilon$, where $E(\epsilon)=0$ and $\epsilon \in (-t, \infty)$. The $\epsilon$ terms of successive inter-arrival samples are denoted $\epsilon_i$ and they can be correlated with each other.
We take some arbitrary interval of size $u$ and observe the number of events, $N$ that fall into this interval. Note that the interval is started at some random time, $J$ which is completely un-correlated with the point process itself. This makes the interval $(J,J+u)$. Here by definition, we pick $J$ in a way that conditional on the arrivals, $T_i$ for the $i$th event and the fact that $J$ lies between $T_i$ and $T_{i+1}$, $J-T_i$ is uniform over the interval $(T_i, T_{i+1})$. This can be achieved for example, by requiring $J$ be uniform over a very large interval, much larger than $t$.
I conjecture (regardless of the distribution of $T$):
$$E(N) = \frac{u}{t}$$
How do I prove or disprove this in general? Here are some special cases.
Case-1: when $T \sim Exp(\lambda)$, this becomes a Poisson process and it is easy to show this.
Case-2: when $\epsilon = 0$, we get a deterministic process where events happen every $t$ interval. Please pay careful attention to this case since it is crucial in understanding the premise of the question.
Let $X$ be the time from the start of the interval to the very first event. By definition, $X$ is a uniform random variable between $0$ and $t$. In this case we get:
$$N = 1 + \left[\frac{u-X}{t}\right]=1+\left[\frac{u}{t}-\frac{X}{t}\right]=1+\left[\frac{u}{t}-U\right]$$
Where $U$ is a uniform random variable between $0$ and $1$. Taking expectation, we get:
$$E(N) = \frac{u}{t}$$
using the result here: Prove that $E([c-U]) = c-1$.
One approach to prove this for the general case could be to somehow take all the $\epsilon_i$'s and fold them into $u$.
Here is a python simulation that demonstrates this for a particular point process that has inter-arrival time: $10$ with $50\%$ probability and $20$ with $50\%$ probability. This makes the average inter-arrival time, $t=15$. And the number of events within a window of size $1$ unit is $\frac{1}{15}$. You can plug other inter-arrival distributions into the code below and see that the conjecture continues to hold.
import numpy as np
catches = 0
for _ in range(50000):
j = np.random.uniform()*1000
#j = np.random.exponential(500)
t_i = 0
while t_i < j+100:
if np.random.uniform() < 0.5:
t_i += 10
else:
t_i += 20
if j < t_i and t_i < j+1:
catches += 1
print(catches/50000)
See here: https://gist.github.com/ryu577/662f9cb593d40920e161cfef3eba0244 for a variety of inter-arrival distributions all satisfying this conjecture.
EDIT: In case anyone is looking for inspiration, here is something I was trying: Need help validating a proof that for any point process with MTBF $t$, the events in an interval sized $u$ will be $\frac{u}{t}$. It's an attempt at a proof, but I now know the proof doesn't work. In the end of it, $\eta$ and $U$ are not independent and that's what kills it.
EDIT: I missed the requirement that the interval was to start at a random point in time (very ill defined concept). The counter example proves that this cannot happen for any arbitrary interval. I have update the prove to reflect that.
I'd offer an attempt at a counter example. I will make many assumptions showing there is a very special case where the formula doesn't hold. Let the inter arrival time $T$ be a discrete random variable that takes the value $0$ with probability $\frac 12$ and $2t$ with probability $\frac 12$.
Let $T_i, i\in\mathbb{N}$ denote the arrival times. And assume that $T_{i+1}-T_i$ are also independent. Then look at the intervall $[0,u]$. Assume further that the first arrival $T_0$ is distributed like $T$ as well and independent of all other waiting/arrival times.
$$\mathbb{E}[N]=\sum_{i\geq 1} \mathbb{P}(T_i\leq u, T_{i+1} > u) \cdot i$$
Let us compute the probability that at least $N$ events happen in $[0,u]$ that is $\mathbb{P}(T_{N-1}\leq u)=\mathbb{P}(T_{N-1}-T_0 + T_0\leq u)=\mathbb{P}\left(\sum\limits_{i = 1}^{N-1} T_i-T_{i-1} + T_0 - 0 \leq u\right)$.
By the i.i.d. assumptions this is basically $\mathbb{P}(NT\leq u)$. Assume $u<t$. Then for any $N\geq 1$ this implies that all waiting times must be $0$, which happens with probability $2^{-N}$. The probability that exactly $N$ events occur is half that (the next arrival time must have length $2t$). Thus the expectation:
$$\mathbb{E}(N)=\sum\limits_{N\geq 1} \frac 12 2^{-N}N=\sum\limits_{N\geq 1} \sum\limits_{i=1}^N \frac 12 2^{-N}=\sum\limits_{i\geq 1} \sum\limits_{N=i+1}^\infty \frac 12 2^{-N}=\sum_{i\geq 1} \frac 12 2^{-i}=\frac 12\neq \frac ut$$.
Now do not consider a stationary interval, but the interval $[J, J+u[$ with $u$ fixed and $J$ a non-negative real valued random variable. Let $J$ take two discrete values $0,a$ such that $\mathbb{P}(J=0)=\mathbb{P}(J=a)=\frac 12$ and $0<a<\frac t{1000}$. Also assume that $(J,T_1,\ldots)$ are jointly independent.
As before we have seen, that $\mathbb{E}(N|J=0)=\frac 12 \neq \frac ut$. Now consider the case $J=a$. What is the probability that at least $N$ arrivals occur in the interval? We have that the first $k$ arrivals occur at $0$ and the next events occur at time $s>a>0$. Since $s$ must be $2t$ these occure outside the interval alltogether (since $s = 2t > u + a = \frac{1001}{1000} t$). Thus the probability of any event occuring in the interval is $0$ and thus $\mathbb{E}(N|J=a)=0$. Together this yields $\mathbb{E}(N)=\frac 14\neq \frac ut$.