Poisson Process

Definition

Let $\{N(t),t\ge 0\}$ be a counting process.

If

1. $N(0)=0$
2. The process has independent increments
3. $P(N(t+h)-N(t)=1)=\lambda h+o(h)$
4. $P(N(t+h)-N(t)\ge 2)=o(h)$

Then $\{N(t),t\ge 0\}$ is a Poisson Process of rate $\lambda >0$

Interpretation

A Poisson process models events occurring randomly in continuous time (unlike Poisson Distribution which is discrete) at a constant average rate $\lambda$.

About 3rd and 4th condition

The MAIN point of conditions 3 and 4 is to “concern” ourselves only with modeling events occurring one at a time, not two or more at a time (the probability of multiple simultaneous events is vanishingly small)

Conditions 3 and 4 say: in a tiny interval $h$, the chance of exactly one event is proportional to $h$, and the chance of two or more is negligible.

Equivalent Definition

The counting process $\{N(t),t\ge 0\}$ is a Poisson process of rate $\lambda >0$ if:

1. $N(0)=0$
2. The process has independent increments
3. $N(t+s)-N(s)\sim \text{Po}(\lambda t)$ for all $s,t\ge 0$

Axioms

For infinitesimal $h$:

• $P(\text{0 events in }[t,t+h])=1-\lambda h+o(h)$
• $P(\text{1 event in }[t,t+h])=\lambda h+o(h)$
• $P(\ge 2\text{ events in }[t,t+h])=o(h)$

Where $f(h)=o(h)$ means ${\lim }_{h\to 0}\frac{f(h)}{h}=0$.

Properties

1. $N(t)\sim \text{Po}(\lambda t)$ with $E[N(t)]=\lambda t$
2. Time of first event $T_1\sim \text{Exp}(\lambda )$
3. Inter-arrival times $T_n\stackrel{\text{i.i.d.}}{\sim }\text{Exp}(\lambda )$
4. Waiting time $W_n\sim \text{Gamma}(n,\lambda )$

Concrete Example

Scenario: Customers arrive at a store according to a Poisson process with rate $\lambda =5$ per hour.

Axiom Interpretation

For a small time interval $h=0.01$ hour (36 seconds):

Axiom	Formula	Concrete Value
Exactly 1 arrival	$\lambda h+o(h)$	$5\times 0.01+o(0.01)\approx 0.05$
$\ge 2$ arrivals	$o(h)$	Negligible (e.g., $h^2=0.0001$)

Why $o(h)$ matters: The exact probability of one arrival isn’t precisely $\lambda h$, but $\lambda h+o(h)$. The $o(h)$ term captures “noise that vanishes faster than linearly.” When computing rates:

$\frac{\lambda h+o(h)}{h}=\lambda +\frac{o(h)}{h}\stackrel{h\to 0}{\to }\lambda$

Numerical Computations

• Expected arrivals in 2 hours: $E[N(2)]=\lambda \cdot 2=10$
• Probability of exactly 3 arrivals in 1 hour:

$P(N(1)=3)=\frac{e^{-\lambda }{\lambda }^3}{3!}=\frac{e^{-5}\cdot 5^3}{6}=\frac{e^{-5}\cdot 125}{6}\approx 0.140$

• Probability of at least 1 arrival in 30 minutes ($t=0.5$):

$P(N(0.5)\ge 1)=1-P(N(0.5)=0)=1-e^{-\lambda \cdot 0.5}=1-e^{-2.5}\approx 0.918$

Intuition Check

The rate $\lambda =5$ means we expect 5 customers per hour on average. But:

• In a tiny 36-second window, the chance of exactly one customer is ~5%
• In that same window, the chance of two or more is $o(h)$ — vanishingly small

This is why Poisson processes model “rare events in continuous time” — events occur one at a time, well-separated.

Related

• Counting Process
• Inter-arrival Times
• Waiting Times
• Thinning
• Nonhomogeneous Poisson Process
• Compound Poisson Process