Relationship between Exponential and Poisson Distributions

I have been studying queueing theory. In many basic models, each entity’s service time (death/lifetime) is modeled as an exponential random variable with rate $\mu$, independent of how arrivals occur. Queueing theory analyzes quantities such as the average number of entities in the system and the waiting time until service begins.

A key fact we will use is this: if service times are exponential with rate $\mu$, then the number of service completions in any time interval of length $t$ follows a Poisson distribution with mean $\mu t$. This post explains the bridge between exponential service times (lifetime until completion) and the Poisson distribution for service‑completion counts.

Poisson Distribution (Service‑Completion Counts)

The Poisson distribution models the number of service completions $N_s(t)$ occurring in a time interval of length $t$ when completions happen at a constant average rate $\mu$: \begin{equation} \mathbb{P}\big(N_s(t)=k\big) = \frac{(\mu t)^k e^{-\mu t}}{k!}. \end{equation}

Exponential Distribution (Service Times)

The exponential distribution describes positive waiting times whose density decays exponentially. Here, we use it for service time (lifetime until completion) with rate $\mu$: \begin{equation} f(t) = \mu e^{-\mu t},\quad t\ge 0. \end{equation} It is memoryless: \begin{equation} \mathbb{P}(X > t + s \mid X > t) = \mathbb{P}(X > s),\quad s,t\ge 0. \end{equation}

Renewal Process for Service Completions

A renewal process is one in which each renewal starts after the previous one completes. Focusing on the server, let service times $s_1,s_2,\dots$ be i.i.d. exponential with rate $\mu$. Let $S_k = \sum_{i=1}^k s_i$ be the time of the $k$-th service completion, and let $N_s(t)=\max\qty{k: S_k \le t}$ be the number of completions by time $t$. Then \begin{equation} \mathbb{P}\big(N_s(t) \ge k\big) = \mathbb{P}\big(S_k \le t\big) \equiv g_k(t). \end{equation}

These distribution functions satisfy the convolution recursion (for $k \ge 2$) \begin{align} g_k(t) &= \int_0^t f(t’)\, g_{k-1}(t - t’)\, dt’ = (f * g_{k-1})(t), \newline g_1(t) &= \int_0^t f(t’)\,dt’ = 1-e^{-\mu t}, \end{align} where $f(t)=\mu e^{-\mu t}$ is the exponential service‑time density. Solving yields the closed form (Erlang/Gamma CDF) \begin{equation} g_k(t) = 1 - e^{-\mu t} \sum_{i=0}^{k-1} \frac{(\mu t)^i}{i!}. \end{equation}

Therefore the service‑completion count distribution is \begin{equation} \mathbb{P}\big(N_s(t) = k\big) = g_k(t) - g_{k+1}(t) = \frac{(\mu t)^k e^{-\mu t}}{k!}, \end{equation} which is precisely the Poisson distribution with mean $\mu t$.

Intuitively, exponential service times are memoryless, which implies independent increments for the completion count process—exactly the defining property of a Poisson process. This service‑centric view aligns with classical queueing models such as $M/M/1$, $M/M/c$, etc.