Events, occurring according to a Poisson process with rate $\lambda$, are registered by a counter. However, each time an event is registered the counter becomes inoperative for the next $b$ units of time and does not register any new events that might occur during that interval. Let $R(t)$ denote the number of events that occur by time $t$ and are registered.
For $t\ge (n-1)b$, find $\mathbb{P}(R(t)\ge n)$
I found a comment from another post, saying that the distribution of $R(t)$ is the identical to $N(t-(n-1)b)$. I don't see the equivalence between these two distributions. Any help would be appreciated!
Best Answer
By the strong Markov property and time homogeneity of $N$ (alternatively, independent and identically distributed interarrival times$^\star$), the jump times $(T_n)_{n\in\mathbb{N}}$ of $R$ are given by \begin{align*} T_1 &= S_1&& \\ T_2 &= T_1 + b + S_2&& \\ T_3 &= T_2 + b + S_3&& \\ &\,\,\,\vdots&& \end{align*} where $(S_n)_{n\in\mathbb{N}}$ are independent and identically distributed with $S_1 \sim \text{Exp}(\lambda)$.
Note that $T_n = S_1 + \sum_{i=2}^n (S_i + b)$ for $n\geq2$. Now, with $n\in\mathbb{N}$ and $t\geq0$, \begin{align*} \mathbb{P}(R(t) \geq n) &= \mathbb{P}\!\left(T_n \leq t\right) \\ &= \mathbb{P}\!\left(\sum_{i=1}^n S_i \leq t - (n-1)b\right)\!. \end{align*} Let $(\tau_n)_{n\in\mathbb{N}}$ denote the jump times of $N$. Since $(S_n)_{n\in\mathbb{N}}$ follow the same distribution as the interarrival times $\tau_{n} - \tau_{n-1}$ of $N$ (under the convention $\tau_0 := 0$), we find that \begin{align*} \mathbb{P}\!\left(\sum_{i=1}^n S_i \leq t - (n-1)b\right) = \mathbb{P}(\tau_n \leq t - (n-1)b) = \mathbb{P}(N(t-(n-1)b)\geq n) \end{align*} for $t\geq (n-1)b$. Collecting results, we find that \begin{align*} \mathbb{P}(R(t) \geq n) = \mathbb{P}(N(t-(n-1)b) \geq n) \end{align*} for $n\in\mathbb{N}$ and $t\geq (n-1)b$.
$^\star$Let me know if additional details or explanations are needed; heuristically, you may think of the process starting anew at times $T_n + b$.