Poisson Processes

A continuous-time counting process with independent increments and Poisson-distributed counts.

1. The idea

A Poisson process ${N (t)}_{t \geq 0}$ with rate $λ$ counts events over time such that $N (t) \sim Poisson (λ t)$ for every $t$ , $N (0) = 0$ , and the counts in disjoint time intervals are independent.

Equivalently, the process can be built from iid Exponential $(λ)$ inter-arrival times: the $n$ -th event happens at $T_{n} = \sum_{i = 1}^{n} X_{i}$ where each $X_{i} \sim Exp (λ)$ .

The independent-increments property means the number of events in $[s, s + t]$ has the same Poisson $(λ t)$ distribution regardless of what happened before time $s$ — a continuous-time analogue of memorylessness.

2. Why it matters

Order flow, trade arrivals, and default events are commonly modeled as Poisson processes, and independent increments let you decompose a long time horizon into independent chunks for easy calculation.

3. Core mathematics

Definition

N (t) \sim Poisson (λ t), N (0) = 0

Independent increments

N (t_{2}) - N (t_{1}) ⊥ N (t_{4}) - N (t_{3}) for disjoint intervals

Stationary increments

N (t + s) - N (t) \sim Poisson (λ s)

Inter-arrival times

X_{i} \sim ii d Exp (λ), T_{n} = i = 1 \sum n X_{i}

$n$ -th arrival time

T_{n} \sim Gamma (n, λ)

4. How to recognize it

Think of this when…

"Events occur at a constant average rate $λ$ over time" — a continuous-time counting setup
Questions asking for the count in one window and a separate, non-overlapping window
Requests to find the distribution of the time of the $n$ -th event
Combined with thinning/superposition or exponential-race language

5. Worked example

Worked example

Trades arrive as a Poisson process with rate $λ = 5$ per minute. Find the probability that exactly $3$ trades occur in $[0, 1]$ and exactly $2$ occur in $[1, 2]$ (disjoint minute-long windows).

1
By stationary increments, $N (1) - N (0) \sim Poisson (5)$ and $N (2) - N (1) \sim Poisson (5)$ .
2
By independent increments, these two counts are independent, so multiply their probabilities.
$P (N (1) = 3) = \frac{e ^{- 5} 5 ^{3}}{3 !}, P (N (2) - N (1) = 2) = \frac{e ^{- 5} 5 ^{2}}{2 !}$
3
Combine using independence.
$P = \frac{e ^{- 5} 5 ^{3}}{6} \times \frac{e ^{- 5} 5 ^{2}}{2} = \frac{e ^{- 10} \cdot 125 \cdot 25}{12} \approx 0.0158 \times 0.0842$
4
Multiply the two Poisson probabilities numerically (each computed separately: $P (N (1) = 3) \approx 0.1404$ , $P (2nd = 2) \approx 0.0842$ ).
$P \approx 0.1404 \times 0.0842 \approx 0.0118$
Answer: $P \approx 0.0118$ .

6. Common mistakes

Treating overlapping intervals as independent when they share time
Forgetting to scale $λ$ by the interval length before applying the Poisson PMF
Confusing the Poisson process's inter-arrival times (Exponential) with the count itself (Poisson)
Assuming the process has memory of past arrivals when computing future-interval probabilities

7. Quick check

For a Poisson process with rate $λ = 3$ /min, what is the distribution of the time of the 2nd arrival?

Which property allows counts in $[0, 5]$ and $[5, 10]$ to be treated as independent Poissons?

8. Practice

9 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Poisson Thinning Exponential Race Memorylessness

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