Part 7 - Poisson Processes I

Introduction

In this part we will introduce Poisson processes and their properties.

Poisson Processes

Intuition (Poisson Processes)

In the beginning, we defined a stochastic process as a collection $\{X_t, t \in \mathcal{I}\}$ of random variable with a common state space $\mathcal{S}$ . Initially, the state space was discrete (i.e., finite or countable), last time we extended this to continuous state spaces.

In this part, we will extend even further and make the set $\mathcal{I}$ is a non-countable set, for example, all positive real numbers, or all subsets of $\mathbb{R}^2$ .

Intuition (Poisson Process)

A random variable with valuyes $0, 1, 2, \ldots$ with a Poisson distribution can be used to model the count of events happening independently, e.g., within some time interval.

We know that if $X \sim \mathrm{Poisson}(\lambda)$ , then,

\pi(x) = e^{-\lambda} \frac{\lambda^x}{x!},

and,

\mathbb{E}[X] = \lambda, \quad \mathrm{Var}(X) = \lambda.

A Poisson process models not only the count for a specific time interval, but also the exact time of every event.

Definition 1 (Counting Process)

A counting process $\{N_t\}_{t \in \mathcal{I}}$ is a stochastic process where $\mathcal{I} = \mathbb{R}_0^+$ , where the state space is the non-negative integers, and where $0 \leq s \leq t$ implies $N_s \leq N_t$ .

Generally, when $s < t, N_t - N_s$ counts the number of “events” (in the loose meaning “things happening”) in $(s, t]$ .

A realization of $N_t$ is then a function of $t$ that is a right-continuous step function.

Definition 2 (Poisson Process I)

A Poisson process $\{N_t\}_{t \geq 0}$ with parameter $\lambda > 0$ is a counting process, fulfilling,

$N_0 = 0$ .
$N_t \sim \mathrm{Poisson}(\lambda t)$ for all $t > 0$ .
Stationary Increments: $N_{t + s} - N_s$ has the same distribution as $N_t$ for all $s > 0, t > 0$ .
Independent Increments: $N_t - N_s$ and $N_r - N_q$ are independent, when $0 \leq q < r \leq s < t$ .

Note

It is not obvious that such a process necessarily exists. Further, $\mathbb{E}[N_t] = \lambda t$ , thus what one is counting occurs with a rate of $\lambda$ items per time unit.

Example 1 (Poisson Process Example I)

Messages are sent to a computer server at a rate of 7 messages per minute. Assuming a Poisson process $\{N_t\}_{t \geq 0}$ , after 3 minutes 22 messages have arrived.

What is the probability of at least 18 messages arriving in the next 2 minutes?

Solution

We first note that,

\begin{align*} P(N_5 - N_3 \geq 18 \mid N_3 = 22) & = P(N_5 - N_3 \geq 18 \mid N_3 - N_0 = 22) \newline & = P(N_5 - N_3 \geq 18) \newline & = P(N_2 \geq 18) \newline & = 1 - P(N_2 \leq 17) \newline \end{align*}

Since $N_2 \sim \mathrm{Poisson}(\lambda t = 7 \cdot 2 = 14)$ , we have,

1 - \sum_{k = 0}^{17} e^{-14} \frac{14^k}{k!} = 0.172.

Equivalently, in R, 1 - ppois(17, lambda = 14).

Example 2 (Poisson Process Example II)

Customers arrive at a rate of 6 customers per hour to a restaurant. Assuming a Poisson process $\{N_t\}_{t \geq 0}$ , what is the probability that exactly 10 customers arrive after 2 hours and a total of 22 customers have arrived after 4 hours?

Solution

We first note that,

\begin{align*} P(N_2 = 10, N_4 = 22) & = P(N_2 - N_0 = 10, N_4 - N_2 = 12) \newline & = P(N_2 - N_0 = 10) P(N_4 - N_2 = 12) \newline & = P(N_2 = 10) P(N_2 = 12) \newline \end{align*}

Since $N_2 \sim \mathrm{Poisson}(\lambda t = 6 \cdot 2 = 12)$ , we have,

\left(e^{-12} \frac{12^{10}}{10!}\right) \left(e^{-12} \frac{12^{12}}{12!}\right) = 0.01199

Equivalently, in R, (dpois(10, lambda = 12)) * (dpois(12, lambda = 12)).

Exponential Distribution and Poisson Processes

Recall (Exponential Distribution)

A random variable $X$ with non-negative real values as possible values has an exponential distribution with parameter $\lambda$ if the density is,

\pi(x) \coloneqq \lambda e^{-\lambda x}.

The cumulative probability distribution is,

F(x) \coloneqq 1 - e^{-\lambda x}.

The expectation is and variance are,

\mathbb{E}[X] = \frac{1}{\lambda}, \quad \mathrm{Var}(X) = \frac{1}{\lambda^2}.

Note (Memoryless of the Exponential Distribution)

A random variable $X$ is called memoryless if,

P(X > s + t \mid X > s) = P(X > t)

for all $s > 0, t > 0$ .

The exponential distribution is memoryless, and is the only memoryless continuous random variable. Thus, we need to take this into consideration when modeling waiting times.

Equivalent Definitions of Poisson Processes

Definition 3 (Poisson Process II)

Let $X_1, X_2, \ldots$ be a sequence of i.i.d. exponential random variables with parameter $\lambda$ . Let $N_0 = 0$ and, for $t > 0$ ,

N_t \coloneqq \max\{n : X_1 + \ldots + X_n \leq t\}.

Then, $\{N_t\}_{t \geq 0}$ is a Poisson process with parameter $\lambda$ .

Proof

Firstly, if we start with a Poisson process (definition I), and let $X_1, X_2, \ldots$ be the inter-arrival times, i.e., the time between event 1 and event 2, event 2 and event 3, etc., then these are i.i.d. exponential random variables with parameter $\lambda$ . Thus, $N_t$ will be defined as above.

Conversely, if we construct $N_t$ as above, all properties of definition I can be verified.

Let $\{M_t\}_{t \geq 0}$ be defined as $M_t \coloneqq N_{t + s} - N_s$ for some $s > 0$ .

It is a counting process by definition.
$N_0 = 0$ by definition.
Stationary increments follow from the memoryless property of the exponential distribution: $M_k = N_{s + t + k} - N_{s + t} = \max\{n : X_{N_{s + t} + 1} + \ldots + X_{N_{s + t} + n} \leq k\}$ , which has the same distribution as $N_k$ .
Independent increments follow from the independence of the $X_i$ ’s.
$M_k = M_{k + s} - N_s = \max\{n : X_{N_s + 1} + \ldots + X_{N_s + n} \leq k\}$ , which has a $\mathrm{Poisson}(\lambda k)$ distribution.

We call $S_n \coloneqq X_1 + \ldots + X_n$ the arrival times of the process. Let $M = \min\{X_1, \ldots, X_n\}$ , where, independently for each $i$ , $X_i \sim \mathrm{Exponential}(\lambda_i)$ , then,

M \sim \mathrm{Exponential}(\lambda_1 + \ldots + \lambda_n), \quad P(M = X_k) = \frac{\lambda_k}{\lambda_1 + \ldots + \lambda_n}.

Proof

Since the $X_i$ ’s are independent,

\begin{align*} P(X_1 > t) P(X_2 > t) \ldots P(X_n > t) & = e^{-\lambda_1 t} e^{-\lambda_2 t} \ldots e^{-\lambda_n t} \newline & = e^{-(\lambda_1 + \ldots + \lambda_n) t} \newline \end{align*}

Thus, $M \sim \mathrm{Exponential}(\lambda_1 + \ldots + \lambda_n)$ .

Further,

\begin{align*} P(M = X_1) & = P(X_2 \geq X_1, \ldots, X_n \geq X_1) \newline & = \mathbb{E}[\mathbb{I}[X_2 \geq X_1, \ldots, X_n \geq X_1]] \newline & = \mathbb{E}[\mathbb{E}[\mathbb{I}[X_2 \geq X_1, \ldots, X_n \geq X_1] \mid X_1]] \newline & = \mathbb{E}[P(X_2 \geq X_1, \ldots, X_n \geq X_1 \mid X_1)] \newline & = \mathbb{E}[e^{-\lambda_2 X_1} \ldots e^{-\lambda_n X_1} \mid X_1] \newline & = \mathbb{E}[e^{-(\lambda_2 + \ldots + \lambda_n) X_1}] \newline & = \int_{0}^{\infty} e^{-(\lambda_2 + \ldots + \lambda_n) x} \lambda_1 e^{-\lambda_1 x} \ dx \newline & = \lambda_1 \int_{0}^{\infty} e^{-(\lambda_1 + \lambda_2 + \ldots + \lambda_n) x} \ dx \newline & = \frac{\lambda_1}{\lambda_1 + \lambda_2 + \ldots + \lambda_n} \underbrace{\left[-\frac{1}{\lambda_1 + \lambda_2 + \ldots + \lambda_n} e^{-(\lambda_1 + \lambda_2 + \ldots + \lambda_n) x}\right]_{0}^{\infty}}_{= 1} \newline & = \frac{\lambda_1}{\lambda_1 + \lambda_2 + \ldots + \lambda_n} \ _\blacksquare \end{align*}

Definition 4 (Poisson Process III)

A Poisson process $\{N_t\}_{t \geq 0}$ with parameter $\lambda > 0$ is a counting process, fulfilling,

$N_0 = 0$ .
The process has stationary and independent increments.

\begin{align*} P(N_h = 0) & = 1 - \lambda h + o(h) \newline P(N_h = 1) & = \lambda h + o(h) \newline P(N_h > 1) & = o(h) \newline \end{align*}

for all $h > 0$ , where $o(h)$ is a function such that $\lim_{h \to 0} \frac{o(h)}{h} = 0$ .