The Poisson Process

The Poisson process is the most fundamental continuous-time stochastic process, modeling the occurrence of random events over time. It is characterized by independent, stationary increments and plays a central role in queueing theory, insurance mathematics, and the theory of point processes.

Definition and basic properties

Definition2.1Poisson process

A stochastic process $(N_t)_{t \geq 0}$ taking values in $\{0, 1, 2, \ldots\}$ is a Poisson process with rate (or intensity) $\lambda > 0$ if:

$N_0 = 0$ .
Independent increments: For any $0 \leq t_1 < t_2 < \cdots < t_n$ , the increments $N_{t_2} - N_{t_1}, N_{t_3} - N_{t_2}, \ldots, N_{t_n} - N_{t_{n-1}}$ are independent.
Stationary increments: For any $s, t \geq 0$ , the distribution of $N_{t+s} - N_s$ depends only on $t$ , not on $s$ .
Poisson distribution: For all $t \geq 0$ , $N_t \sim \text{Poisson}(\lambda t)$ , i.e.,

$\mathbb{P}(N_t = k) = e^{-\lambda t} \frac{(\lambda t)^k}{k!}, \quad k = 0, 1, 2, \ldots$

Intuitively, $N_t$ counts the number of "arrivals" (or "events") in the time interval $[0, t]$ . The rate $\lambda$ is the expected number of arrivals per unit time: $\mathbb{E}[N_t] = \lambda t$ .

RemarkEquivalent characterizations

The Poisson process can also be defined via:

Infinitesimal properties: For small $h > 0$ :
- $\mathbb{P}(N_h = 1) = \lambda h + o(h)$ ,
- $\mathbb{P}(N_h \geq 2) = o(h)$ ,
- $\mathbb{P}(N_h = 0) = 1 - \lambda h + o(h)$ .
Inter-arrival times: The times between successive events are i.i.d. exponential random variables with rate $\lambda$ .

These characterizations are all equivalent and provide different intuitions for the Poisson process.

Inter-arrival times

Theorem2.1Inter-arrival times are exponential

Let $T_1, T_2, \ldots$ be the times of the successive arrivals in a Poisson process with rate $\lambda$ , and let $\tau_n = T_n - T_{n-1}$ (with $T_0 = 0$ ) be the inter-arrival times. Then $\tau_1, \tau_2, \ldots$ are i.i.d. $\text{Exp}(\lambda)$ random variables.

Conversely, if $\tau_1, \tau_2, \ldots$ are i.i.d. $\text{Exp}(\lambda)$ , and we define $T_n = \sum_{k=1}^n \tau_k$ and $N_t = \max\{n : T_n \leq t\}$ , then $(N_t)$ is a Poisson process with rate $\lambda$ .

ExampleBus arrivals

Suppose buses arrive at a stop according to a Poisson process with rate $\lambda = 5$ per hour. Then:

The expected number of buses in 2 hours is $\mathbb{E}[N_2] = 5 \cdot 2 = 10$ .
The probability of exactly 3 buses arriving in 30 minutes is

$\mathbb{P}(N_{0.5} = 3) = e^{-2.5} \frac{(2.5)^3}{3!} \approx 0.2138.$

The time until the first bus arrives is $\text{Exp}(5)$ , with mean $1/5$ hour = 12 minutes.

Superposition and thinning

Theorem2.2Superposition

If $(N_t^{(1)})$ and $(N_t^{(2)})$ are independent Poisson processes with rates $\lambda_1$ and $\lambda_2$ , then their superposition

$N_t = N_t^{(1)} + N_t^{(2)}$

is a Poisson process with rate $\lambda_1 + \lambda_2$ .

This is the additive property of Poisson processes: merging two independent Poisson streams yields another Poisson stream.

Theorem2.3Thinning

If $(N_t)$ is a Poisson process with rate $\lambda$ , and each arrival is independently retained with probability $p$ (and discarded with probability $1-p$ ), then the retained arrivals form a Poisson process with rate $p\lambda$ .

Thinning is the converse of superposition: splitting a Poisson stream into independent substreams.

ExampleCustomer types in a queueing system

Customers arrive at a service center according to a Poisson process with rate $\lambda = 10$ per hour. Each customer is type A with probability $0.6$ and type B with probability $0.4$ , independently. Then:

Type A arrivals form a Poisson process with rate $10 \cdot 0.6 = 6$ per hour.
Type B arrivals form a Poisson process with rate $10 \cdot 0.4 = 4$ per hour.
The two streams are independent.

Conditional properties

Theorem2.4Uniform order statistics

Conditionally on $N_t = n$ , the arrival times $T_1, \ldots, T_n$ are distributed as the order statistics of $n$ i.i.d. $\text{Uniform}(0, t)$ random variables.

This beautiful result says that, given $n$ arrivals in $[0, t]$ , the arrival times are "uniformly scattered" throughout the interval.

ExampleConditional probability

Suppose $N_1 = 2$ (two arrivals in $[0,1]$ ). What is the probability that both arrivals occurred in the first half $[0, 0.5]$ ?

By the uniform order statistics property, each arrival independently falls in $[0, 0.5]$ with probability $0.5$ , so

$\mathbb{P}(T_1, T_2 \leq 0.5 \mid N_1 = 2) = (0.5)^2 = 0.25.$

Compound Poisson processes

Definition2.2Compound Poisson process

A compound Poisson process is a process of the form

$X_t = \sum_{i=1}^{N_t} Y_i,$

where $(N_t)$ is a Poisson process with rate $\lambda$ and $\{Y_i\}$ are i.i.d. random variables (independent of $N_t$ ), called the jump sizes.

If $Y_i$ represents the "size" or "value" of the $i$ -th arrival, then $X_t$ is the cumulative value up to time $t$ .

ExampleInsurance claims

An insurance company receives claims according to a Poisson process with rate $\lambda = 20$ per year. Each claim amount is i.i.d. with mean $\mu = 1000$ and variance $\sigma^2 = 500{,}000$ . The total claims up to time $t$ is

$X_t = \sum_{i=1}^{N_t} Y_i.$

Then:

$\mathbb{E}[X_t] = \mathbb{E}[N_t] \mathbb{E}[Y_1] = 20t \cdot 1000 = 20{,}000 t$ .
$\text{Var}(X_t) = \mathbb{E}[N_t] \mathbb{E}[Y_1^2] = 20t \cdot (1{,}000{,}000 + 500{,}000) = 30{,}000{,}000 t$ .

Waiting time paradox

ExampleWaiting time paradox (renewal paradox)

You arrive at a bus stop at a random time. The buses arrive according to a Poisson process with rate $\lambda$ . What is the expected time until the next bus?

Answer: $1/\lambda$ (the same as the mean inter-arrival time). This is a special feature of the memoryless property of the exponential distribution: the remaining time until the next arrival is always $\text{Exp}(\lambda)$ , regardless of when you arrive.

In contrast, for a non-Poisson renewal process (e.g., deterministic inter-arrival times), the expected waiting time for a random arrival is typically greater than half the inter-arrival time. This is the inspection paradox or waiting time paradox.

Summary

The Poisson process is characterized by:

Independent, stationary increments: Memoryless evolution.
Exponential inter-arrival times: $\tau_i \sim \text{Exp}(\lambda)$ , i.i.d.
Superposition and thinning: Merging and splitting preserve the Poisson property.
Uniform order statistics: Conditional arrivals are uniformly distributed.

These properties make the Poisson process tractable for analysis and ubiquitous in applications: queueing, insurance, telecommunications, reliability theory, and more.