Exact probability distribution for an alternating renewal counting process

Question

Consider a scenario where a detector counts the number of photons incident on it's surface over a time interval of $[0,\tau]$. We suppose that the photons arrive at the detector's surface with independent exponentially distributed arrival times $T\sim\operatorname{exp}(\Phi)$ where $\Phi$ denotes the photon flux in units of photons/s. Thus, $f_T(t)=\Phi e^{-\Phi t}$, $t\geq 0$. When a photon arrives at the detector's surface, the detector shuts down for a deterministic amount of time $\tau_D$ $(0\leq\tau_D\leq\tau)$ to count the photon. While the detector is shut down, no other photons arriving at its surface can be counted. If we let $N(\tau)$ denote the number of photons counted at time $\tau$, what is the probability distribution of $N(\tau)$?

This problem can be modeled as an alternating renewal process. For the special case $\tau_D=0$, we have the classic result $$ \mathsf P(N(\tau)=n)=\frac{(\Phi\tau)^n}{n!}e^{-\Phi\tau},\quad n=0,1,2,\dots $$ Since the time the detector shuts down is $\tau_D=0$, the probability distribution has support on the nonnegative integers. We know for the case $\tau_D>0$ that the support of $N(\tau)$ must be finite on $n=0,1,\dots,\lfloor\tau/\tau_D\rfloor$.

Answer 1

We assume that the counter is not shut down at time $t = 0$. Let $T_1, T_2, \ldots$ be independent, $\exp(\Phi)$-distributed random variables. Then $S_1 := T_1$ is the time when photon 1 is detected, $S_2 := T_1 + \tau_D + T_2$ the time when photon 2 is detected, generally $$S_n := T_1 + \tau_D + T_2 + \tau_D + \ldots + \tau_D + T_n = \sum_{i=1}^n T_i + (n-1)\tau_D$$ the time when photon $n$ is detected. (Not detected photons are not counted.) Then $$\mathbb{P}(N(\tau) < n) = \mathbb{P}(S_n > \tau) = \mathbb{P}\left(\sum_{i=1}^n T_i > \tau - (n-1)\tau_D\right) =: f(n),$$ hence $$\mathbb{P}(N(\tau) = n) = \mathbb{P}(N(\tau) < n+1) - \mathbb{P}(N(\tau) < n) = f(n+1) - f(n)$$ Now $\sum_{i=1}^n T_i \sim \Gamma(n,\Phi)$. Let $F_n$ be the distribution function of $\Gamma(n,\Phi)$. Then $f(0) = 0$ and $f(n) = 1 - F_n(\tau - (n-1)\tau_D)$ for $n > 0$.