37.05.11 · probability / 05-markov-chains

The Poisson Process: Equivalent Characterizations

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Anchor (Master): Norris 1997 *Markov Chains* (Cambridge) §2.4; Kingman 1993 *Poisson Processes* (Oxford) §2-§5 (superposition, thinning, the Poisson point process, order statistics); Daley-Vere-Jones 2003 *An Introduction to the Theory of Point Processes* 2e Vol. I Ch. 2

Intuition Beginner

A Poisson process is the simplest model of random events arriving in time: calls reaching a switchboard, clicks of a Geiger counter, customers entering a shop. You scatter points along a time line so events happen at a steady average rate but with no schedule. The single number that controls everything is the rate, written $λ$ : the average number of events per unit time, so over an hour you expect about $λ$ events, and the bigger $λ$ is, the more crowded the line of points becomes.

What makes the model so clean is two assumptions you can picture separately. First, the average rate never changes with time, so a window of a given length is statistically the same wherever you slide it. Second, events in non-overlapping stretches of time have nothing to do with each other: knowing how many calls arrived this morning tells you nothing about how many arrive this afternoon. These two ideas, a steady rate and independent disjoint windows, pin the model down almost completely.

Remarkably, three different-looking descriptions all pin down the same process. You can say how many events fall in each window (the count in a window of length $t$ follows the Poisson recipe with average $λ t$ ). You can describe the process instant by instant (in a tiny slice of length $h$ , one event appears with chance about $λh$ , and two-at-once essentially never happen). Or you can describe the gaps between consecutive events (each gap is the memoryless exponential clock, and the gaps are independent and identically distributed). Each description sounds like a separate rule, yet each forces the other two.

Two more operations keep you inside the family. Merge two independent Poisson streams and you get a Poisson stream whose rate is the sum (superposition). Flip a coin at each event and keep only the heads, and the kept events again form a Poisson stream, now with a thinned rate (thinning).

The one-sentence takeaway: a Poisson process drops random points on a time line at average rate $λ$ with independent disjoint windows, and the same process can be described by Poisson counts, by a tiny-interval rate, or by exponential gaps — three names for one thing, stable under merging and coin-flip thinning.

Visual Beginner

Picture a horizontal time line with events marked as ticks, and read off the three ways to describe it.

Top: the count picture — disjoint windows hold independent Poisson counts with mean $λ t$ . Bottom-left: the gap picture — the spaces between ticks are independent exponential waits. Right: the instant picture — a tiny slice holds at most one event with chance about $λh$ . The merging and coin-flip panels show that superposition adds rates and thinning scales the rate by the keep-probability.

Worked example Beginner

Suppose a help desk receives calls as a Poisson process at rate $λ = 3$ calls per hour. We compute a few quantities by hand.

Step 1. The expected number of calls in one hour is $λ \times 1 = 3$ , and in a half hour it is $λ \times 0.5 = 1.5$ . The count in any window of length $t$ averages $λ t$ .

Step 2. The chance of exactly $0$ calls in one hour uses the Poisson recipe: the chance of $k$ events when the average is $m = λ t = 3$ is $e^{- m} m^{k} / k!$ . For $k = 0$ this is $e^{- 3} \times 1 = e^{- 3} \approx 0.050$ . So about a $5%$ chance of a silent hour.

Step 3. The chance of exactly $1$ call in one hour is $e^{- 3} \times 3^{1} /1! = 3 e^{- 3} \approx 0.149$ .

Step 4. The average gap between consecutive calls is $1/ λ = 1/3$ of an hour, that is $20$ minutes. The gaps are exponential waits with rate $3$ .

Step 5. Now thin the stream: suppose each call is an emergency with probability $p = 0.2$ , independently. The emergencies form their own Poisson process at the thinned rate $p λ = 0.2 \times 3 = 0.6$ emergencies per hour, so on average one emergency every $1/0.6 \approx 1.67$ hours.

What this tells us: the rate $λ$ alone fixed the average count, the Poisson recipe gave the exact chances of $0$ and $1$ calls, the same rate gave the average $20$ -minute gap, and a coin flip at each event produced a smaller Poisson process. Every number came from the one parameter $λ$ and the two simple assumptions.

Check your understanding Beginner

Exercise (easy, multiple choice).

In a Poisson process, what does the independence of disjoint windows mean?

A. The number of events in non-overlapping stretches of time are unrelated to each other B. Every window contains the same number of events C. Events can only happen at whole-number times D. The first event always comes after exactly one time unit

Hint

Think about whether knowing this morning's count changes what you expect this afternoon.

Answer

A. Counts in non-overlapping stretches are unrelated. Independent increments means the number of events in disjoint time windows carry no information about one another. Feedback-correct: this is one of the two defining assumptions, alongside a steady average rate. Feedback-wrong: B, C, and D describe behaviors a Poisson process does not have — counts vary randomly, events land at any real time, and the first event is a random exponential wait.

Formal definition Intermediate+

Throughout, events occur in continuous time $t \in [0, \infty)$ on a probability space $(Ω, F, P)$ . The counting process is $N = (N_{t})_{t \geq 0}$ , where $N_{t}$ records the number of events in $(0, t]$ ; it is right-continuous, non-decreasing, integer-valued, with $N_{0} = 0$ , and increases by unit jumps. The arrival times are $T_{n} = in f {t : N_{t} \geq n}$ for $n \geq 1$ (with $T_{0} = 0$ ), and the inter-arrival times (gaps) are $S_{n} = T_{n} - T_{n - 1}$ for $n \geq 1$ . The counting process and the arrival sequence are inverse to one another: $N_{t} = \sum_{n \geq 1} 1 {T_{n} \leq t}$ and ${N_{t} \geq n} = {T_{n} \leq t}$ . The recollections of 26.02.01 — the Poisson law $Poisson (m)$ with $P (K = k) = e^{- m} m^{k} / k!$ and the exponential law $Exp (λ)$ with tail $e^{- λ t}$ — are used freely, as is the continuous-time Markov apparatus of 37.05.08.

Fix a rate $λ \in (0, \infty)$ . The three definitions below are stated separately; the Key theorem proves they describe the same law.

Definition A (independent stationary increments with Poisson marginals). $N$ has independent increments if for $0 \leq t_{0} < t_{1} < \dots < t_{m}$ the increments $N_{t_{1}} - N_{t_{0}}, \dots, N_{t_{m}} - N_{t_{m - 1}}$ are independent, and *stationary increments* if the law of $N_{t + s} - N_{t}$ depends only on $s$ . The process is a *Poisson process of rate $λ$ * under Definition A if it has independent stationary increments and $N_{t + s} - N_{t} \sim Poisson (λ s)$ for all $t \geq 0$ , $s > 0$ .

Definition B (infinitesimal / Markov definition). $N$ is a Poisson process of rate $λ$ under Definition B if it has independent stationary increments, $N_{0} = 0$ , and as $h ↓ 0$ , uniformly in $t$ , $P (N_{t + h} - N_{t} = 1) = λh + o (h), P (N_{t + h} - N_{t} \geq 2) = o (h), P (N_{t + h} - N_{t} = 0) = 1 - λh + o (h) .$ Equivalently, $N$ is the continuous-time Markov chain on $I = {0, 1, 2, \dots}$ whose $Q$ -matrix 37.05.08 has $q_{n, n + 1} = λ$ , $q_{nn} = - λ$ , and $q_{nj} = 0$ otherwise: a pure birth chain with every birth rate equal to $λ$ .

Definition C (i.i.d. exponential inter-arrival times). $N$ is a Poisson process of rate $λ$ under Definition C if the inter-arrival times $(S_{n})_{n \geq 1}$ are independent and identically $Exp (λ)$ -distributed, and $N_{t} = sup {n : T_{n} \leq t}$ with $T_{n} = S_{1} + \dots + S_{n}$ . Equivalently, the arrival times are the partial sums of an i.i.d. $Exp (λ)$ sequence, so $T_{n} \sim Gamma (n, λ)$ with density $λ^{n} t^{n - 1} e^{- λ t} / (n - 1)!$ on $t > 0$ .

Counterexamples to common slips Intermediate+

Stationary increments alone do not give Poisson. A process can have stationary independent increments yet have $N_{t + s} - N_{t}$ Bernoulli- or compound-distributed; the Poisson marginal is an additional input in Definition A. What Definition B shows is that the infinitesimal one-at-a-time condition forces the marginal to be Poisson, so the marginal is not free once the rare-jump assumption is imposed.
Memorylessness is what couples the three views. The exponential gaps of Definition C are the only continuous inter-arrival law compatible with the lack of memory that independent increments demand; replacing $Exp (λ)$ by any other gap law (a renewal process) keeps the i.i.d.-gap structure but destroys both the independent-increments property and the Poisson marginals.
The order-statistics property is conditional, not unconditional. Given $N_{t} = n$ , the $n$ arrival times are distributed as the order statistics of $n$ i.i.d. $Uniform [0, t]$ points — but unconditionally the arrival times are spread out by the Gamma laws, not uniform. The uniformity appears only after conditioning on the count.
Thinning needs independent marking. Keeping each event with probability $p$ yields a rate- $p λ$ Poisson process only when the marks are independent of each other and of the process. Marking that depends on the inter-arrival times (e.g. keep an event iff its gap exceeds a threshold) produces a non-Poisson kept stream.

Key theorem with proof Intermediate+

Theorem (equivalence of the three definitions). For a counting process $N$ with $N_{0} = 0$ and a fixed rate $λ \in (0, \infty)$ , Definitions A, B, and C are equivalent: each implies the other two, and all three determine the same law on path space ^{[Norris 1997 §2.4]}.

Proof. We prove C $\Rightarrow$ A $\Rightarrow$ B $\Rightarrow$ C.

Step 1 (C $\Rightarrow$ A: Poisson marginal from exponential gaps). Assume the gaps $(S_{n})$ are i.i.d. $Exp (λ)$ , so $T_{n} = S_{1} + \dots + S_{n} \sim Gamma (n, λ)$ . Using ${N_{t} = n} = {T_{n} \leq t < T_{n + 1}}$ and conditioning on $T_{n} = u$ with $S_{n + 1} \sim Exp (λ)$ independent, $P (N_{t} = n) = \int_{0}^{t} \frac{λ ^{n} u ^{n - 1} e ^{- λ u}}{( n - 1 )!} P (S_{n + 1} > t - u) d u = \int_{0}^{t} \frac{λ ^{n} u ^{n - 1} e ^{- λ u}}{( n - 1 )!} e^{- λ (t - u)} d u = e^{- λ t} \frac{λ ^{n}}{( n - 1 )!} \int_{0}^{t} u^{n - 1} d u,$ and $\int_{0}^{t} u^{n - 1} d u = t^{n} / n$ , so $P (N_{t} = n) = e^{- λ t} (λ t)^{n} / n!$ , i.e. $N_{t} \sim Poisson (λ t)$ . For the increments and their independence, the memoryless property of $Exp (λ)$ is decisive: at any fixed time $t$ , conditional on the past, the residual gap $T_{N_{t} + 1} - t$ is again $Exp (λ)$ and independent of $(N_{s})_{s \leq t}$ , because $P (S > t^{'} + r ∣ S > t^{'}) = e^{- λ r}$ erases the elapsed wait. Hence the post- $t$ process is a fresh copy of the same exponential-gap construction, independent of the past and with the same law shifted by $t$ ; this is precisely stationary independent increments, and the marginal computation gives $N_{t + s} - N_{t} \sim Poisson (λ s)$ . Thus C $\Rightarrow$ A.

Step 2 (A $\Rightarrow$ B: the infinitesimal rates). Assume Definition A. Stationary independent increments are inherited directly. From $N_{h} \sim Poisson (λh)$ , $P (N_{h} = 0) = e^{- λh} = 1 - λh + o (h), P (N_{h} = 1) = λh e^{- λh} = λh + o (h), P (N_{h} \geq 2) = 1 - e^{- λh} (1 + λh) = o (h),$ the last by Taylor expansion $e^{- λh} (1 + λh) = 1 - \frac{1}{2} (λh)^{2} + o (h^{2})$ . Stationarity makes these estimates uniform in $t$ . These are exactly the infinitesimal conditions of Definition B, and they identify the generator entries $q_{n, n + 1} = λ$ , $q_{nn} = - λ$ : a pure birth chain with constant rate $λ$ in the sense of 37.05.08. Thus A $\Rightarrow$ B.

Step 3 (B $\Rightarrow$ C: exponential gaps from the Markov structure). Assume Definition B. Read $N$ as the continuous-time Markov chain with $Q$ -matrix $q_{n, n + 1} = λ$ , $q_{nn} = - λ$ 37.05.08. Its holding time in each state $n$ is $Exp (q_{n})$ with $q_{n} = λ$ , and the jump chain is deterministic, $n \mapsto n + 1$ . By the jump-chain/holding-time construction of 37.05.08, the times spent in successive states $0, 1, 2, \dots$ are independent $Exp (λ)$ variables, and these times are exactly the inter-arrival gaps $S_{n}$ . Hence $(S_{n})$ are i.i.d. $Exp (λ)$ and $T_{n} = S_{1} + \dots + S_{n}$ , which is Definition C. Since each birth rate equals the constant $λ$ and $\sum_{n} 1/ λ = \infty$ , the chain is non-explosive 37.05.08, so $N_{t} < \infty$ for all $t$ almost surely and the construction is consistent on all of $[0, \infty)$ . Thus B $\Rightarrow$ C, closing the cycle. $□$

Bridge. This equivalence builds toward the entire theory of point processes and appears again in the order-statistics property, superposition, and thinning, each of which is most transparent in whichever of the three pictures suits it. The foundational reason the three definitions coincide is that the memoryless exponential is the unique continuous inter-arrival law the independent-increments structure permits: this is exactly the same uniqueness that fixed the holding-time law in the continuous-time Markov chain of 37.05.08, so the Poisson process is the constant-rate pure birth chain viewed three ways. The count picture (Definition A) generalises to the spatial Poisson point process where time is replaced by a measure space; the infinitesimal picture (Definition B) is dual to the forward equation $p_{n}^{'} (t) = λ p_{n - 1} (t) - λ p_{n} (t)$ that the Poisson semigroup satisfies; the gap picture (Definition C) is the renewal-theoretic skeleton. Putting these together, the three definitions are one law, and the central insight is that the rate $λ$ is the single invariant tying the Poisson count, the infinitesimal rate, and the exponential gap into a single object.

Exercises Intermediate+

Exercise 4 (medium, symbolic).

Let $N$ and $M$ be independent Poisson processes of rates $λ$ and $μ$ . Show that the superposition $N + M$ (the process counting events of either) is a Poisson process of rate $λ + μ$ .

Hint

Sums of independent Poisson variables are Poisson; independent increments are inherited from independence.

Answer

The superposition $L_{t} = N_{t} + M_{t}$ has $L_{0} = 0$ and, over disjoint time windows, increments that are sums of the corresponding independent increments of $N$ and $M$ ; independence of those across windows and across the two processes gives $L$ independent stationary increments. For a window of length $s$ , $N_{t + s} - N_{t} \sim Poisson (λ s)$ and $M_{t + s} - M_{t} \sim Poisson (μ s)$ are independent, and the sum of independent $Poisson (a)$ and $Poisson (b)$ is $Poisson (a + b)$ (convolve, or multiply probability generating functions $e^{a (z - 1)} e^{b (z - 1)} = e^{(a + b) (z - 1)}$ ). Hence $L_{t + s} - L_{t} \sim Poisson ((λ + μ) s)$ , so $L$ satisfies Definition A with rate $λ + μ$ .

Exercise 5 (medium, symbolic).

Let $N$ be a Poisson process of rate $λ$ , and mark each event independently as type 1 with probability $p$ and type 2 with probability $1 - p$ . Show the type-1 events form a Poisson process of rate $p λ$ .

Hint

Condition on $N_{t} = n$ ; given the count, the number of type-1 events is Binomial $(n, p)$ . Mix a Binomial over a Poisson count.

Answer

Let $N_{t}^{(1)}$ count type-1 events in $(0, t]$ . Given $N_{t} = n$ , the marks are i.i.d., so $N_{t}^{(1)} ∣ {N_{t} = n} \sim Binomial (n, p)$ . Mixing over $N_{t} \sim Poisson (λ t)$ , $P (N_{t}^{(1)} = k) = n \geq k \sum e^{- λ t} \frac{( λ t ) ^{n}}{n !} (k n) p^{k} (1 - p)^{n - k} = e^{- λ t} \frac{( p λ t ) ^{k}}{k !} n \geq k \sum \frac{(( 1 - p ) λ t ) ^{n - k}}{( n - k )!} = e^{- p λ t} \frac{( p λ t ) ^{k}}{k !},$ so $N_{t}^{(1)} \sim Poisson (p λ t)$ . Independence of the marks across disjoint windows transfers $N$ 's independent increments to $N^{(1)}$ , giving Definition A with rate $p λ$ . The same computation shows the type-2 stream is $Poisson ((1 - p) λ)$ and, because the joint generating function factors, the two thinned streams are independent.

Exercise 6 (medium, multiple choice).

Which statement about the order-statistics property of a rate- $λ$ Poisson process is correct?

A. The arrival times $T_{1}, \dots, T_{n}$ are unconditionally i.i.d. $Uniform [0, t]$ B. Given $N_{t} = n$ , the arrival times $(T_{1}, \dots, T_{n})$ are distributed as the order statistics of $n$ i.i.d. $Uniform [0, t]$ variates C. Given $N_{t} = n$ , the arrival times are i.i.d. $Exp (λ)$ D. The property holds only when $λ = 1$

Hint

The uniformity is conditional on the count and respects the ordering $T_{1} < \dots < T_{n}$ .

Answer

B. Conditional on $N_{t} = n$ , the unordered arrival epochs are i.i.d. $Uniform [0, t]$ , so the ordered times $(T_{1}, \dots, T_{n})$ are their order statistics. Option A drops the conditioning and the ordering. Option C confuses the gaps (exponential) with the arrival times. Option D is false: the rate cancels in the conditional law, which is why the conditional distribution is free of $λ$ .

Exercise 7 (hard, symbolic).

Prove the order-statistics property for $n = 2$ : given $N_{t} = 2$ , the joint density of $(T_{1}, T_{2})$ is $2/ t^{2}$ on ${0 < u_{1} < u_{2} < t}$ , the order-statistics density of two i.i.d. $Uniform [0, t]$ points.

Hint

Use the gap representation: $T_{1} = S_{1}$ , $T_{2} = S_{1} + S_{2}$ , and ${N_{t} = 2} = {T_{2} \leq t < T_{3}}$ . Write the joint density of $(S_{1}, S_{2}, S_{3})$ and change variables.

Answer

The gaps $S_{1}, S_{2}, S_{3}$ are i.i.d. $Exp (λ)$ with joint density $λ^{3} e^{- λ (s_{1} + s_{2} + s_{3})}$ on the positive octant. Set $u_{1} = s_{1}$ , $u_{2} = s_{1} + s_{2}$ (so $T_{1} = u_{1}$ , $T_{2} = u_{2}$ , with Jacobian $1$ ), and require ${N_{t} = 2}$ , i.e. $u_{2} \leq t$ and $T_{3} = u_{2} + s_{3} > t$ . For $0 < u_{1} < u_{2} < t$ the density of $(T_{1}, T_{2}, S_{3})$ is $λ^{3} e^{- λ u_{2}} e^{- λ s_{3}}$ ; integrating $s_{3}$ over $(t - u_{2}, \infty)$ gives $\int_{t - u_{2}}^{\infty} λ e^{- λ s_{3}} d s_{3} = e^{- λ (t - u_{2})}$ , so the joint density of $(T_{1}, T_{2})$ on ${N_{t} = 2}$ is $λ^{2} e^{- λ u_{2}} e^{- λ (t - u_{2})} = λ^{2} e^{- λ t}$ . Dividing by $P (N_{t} = 2) = e^{- λ t} (λ t)^{2} /2!$ yields the conditional density $λ^{2} e^{- λ t} / (e^{- λ t} λ^{2} t^{2} /2) = 2/ t^{2}$ on ${0 < u_{1} < u_{2} < t}$ . This is exactly the order-statistics density of two i.i.d. $Uniform [0, t]$ points (the constant $n! / t^{n} = 2/ t^{2}$ ), and the rate $λ$ has cancelled.

Exercise 8 (hard, symbolic).

Let $N$ be a rate- $λ$ Poisson process and attach to its $n$ -th event an i.i.d. jump $Y_{n}$ with $E [Y_{1}] = ν$ and $Var (Y_{1}) = σ^{2}$ , independent of $N$ . The compound Poisson process is $X_{t} = \sum_{n = 1}^{N_{t}} Y_{n}$ . Compute $E [X_{t}]$ and $Var (X_{t})$ .

Hint

Condition on $N_{t}$ and use Wald-type identities, or differentiate the characteristic function $E [e^{i θ X_{t}}] = exp (λ t (E [e^{i θ Y_{1}}] - 1))$ .

Answer

Condition on $N_{t} = n$ : $E [X_{t} ∣ N_{t} = n] = n ν$ and $Var (X_{t} ∣ N_{t} = n) = n σ^{2}$ . By the tower rule, $E [X_{t}] = E [N_{t}] ν = λ t ν$ . For the variance, the law of total variance gives $Var (X_{t}) = E [Var (X_{t} ∣ N_{t})] + Var (E [X_{t} ∣ N_{t}]) = E [N_{t}] σ^{2} + ν^{2} Var (N_{t}) = λ t σ^{2} + ν^{2} λ t = λ t (σ^{2} + ν^{2}) = λ t E [Y_{1}^{2}],$ using $E [N_{t}] = Var (N_{t}) = λ t$ for the Poisson count. Equivalently, the cumulant generating function $lo g E [e^{θ X_{t}}] = λ t (E [e^{θ Y_{1}}] - 1)$ has first two derivatives at $0$ equal to $λ t E [Y_{1}] = λ t ν$ and $λ t E [Y_{1}^{2}]$ , the mean and variance.

Advanced results Master

Beyond the equivalence, the theory organizes around the conditioning (order-statistics) property, the closure operations of superposition and thinning, the compound process built by attaching marks, and the passage from the time line to a general measure space, where the Poisson process becomes the Poisson point process whose count picture is the only one of the three definitions that survives.

Theorem 1 (order-statistics / conditioning property). Let $N$ be a rate- $λ$ Poisson process. Conditional on $N_{t} = n$ , the vector of arrival times $(T_{1}, \dots, T_{n})$ has the distribution of the order statistics of $n$ i.i.d. $Uniform [0, t]$ random variables; equivalently, its conditional density is $n! / t^{n}$ on the simplex ${0 < u_{1} < \dots < u_{n} < t}$ . The conditional law is free of $λ$ . This is the precise sense in which a Poisson process is "completely random": once the number of points in a window is fixed, their positions are as uniform and exchangeable as possible.

Theorem 2 (superposition). If $N^{(1)}, \dots, N^{(k)}$ are independent Poisson processes of rates $λ_{1}, \dots, λ_{k}$ , their superposition $N = \sum_{i} N^{(i)}$ is a Poisson process of rate $λ_{1} + \dots + λ_{k}$ . The construction is associative and commutative, so the Poisson processes of all rates form a one-parameter convolution semigroup under merging, the rate being the additive label. Each contributing point can be tagged by which stream produced it, and given a point of $N$ the tag is $λ_{i} / \sum_{j} λ_{j}$ , independent across points — the inverse operation of thinning.

Theorem 3 (thinning / colouring). Let $N$ be a rate- $λ$ Poisson process and mark each point independently with one of $k$ colours, colour $c$ with probability $p_{c}$ ( $\sum_{c} p_{c} = 1$ ). Then the colour- $c$ points form a Poisson process of rate $p_{c} λ$ , and the $k$ coloured processes are mutually independent. The independence is the surprising half: although the coloured counts in a fixed window sum to the total count, decoupling them by independent marking removes all dependence, because conditional on $N_{t} = n$ the colour assignment is multinomial and the Poisson mixing diagonalises the joint generating function into a product.

Theorem 4 (compound Poisson process). Attach to the $n$ -th point an i.i.d. mark $Y_{n} \sim F$ , independent of $N$ , and set $X_{t} = \sum_{n = 1}^{N_{t}} Y_{n}$ . Then $(X_{t})$ has stationary independent increments with characteristic function $E [e^{i θ X_{t}}] = exp (λ t (F (θ) - 1))$ , where $F (θ) = E [e^{i θ Y_{1}}]$ . Its mean is $λ t E [Y_{1}]$ and variance $λ t E [Y_{1}^{2}]$ . The compound Poisson laws are exactly the infinitely divisible laws with finite Lévy measure $λ F$ , and they are the pure-jump building blocks of the Lévy–Khintchine representation: every Lévy process is a Brownian motion with drift plus a (possibly infinite-activity) limit of compound Poisson processes.

Theorem 5 (spatial Poisson point process). Let $(S, A, μ)$ be a $σ$ -finite measure space. A Poisson point process with mean measure $μ$ is a random integer-valued measure $Π$ such that (i) for disjoint $A_{1}, \dots, A_{m} \in A$ the counts $Π (A_{1}), \dots, Π (A_{m})$ are independent (complete independence), and (ii) $Π (A) \sim Poisson (μ (A))$ for every $A$ with $μ (A) < \infty$ . Existence holds for every $σ$ -finite $μ$ , and when $S = [0, \infty)$ with $μ = λ Leb$ this recovers the rate- $λ$ Poisson process of Definitions A–C. On a general space only the count picture (Definition A) generalises: there is no linear order, so neither the infinitesimal-rate description (Definition B) nor the inter-arrival-gap description (Definition C) has an unconditional analogue, though the order-statistics property persists as the statement that, given $Π (A) = n$ , the $n$ points are i.i.d. with law $μ (\cdot \cap A) / μ (A)$ .

Synthesis. The foundational reason the subject coheres is that one rate $λ$ — promoted to a mean measure $μ$ in the spatial setting — encodes the entire law, and putting these together, the three definitions are exactly the same information presented as a count law, an infinitesimal rate, and an exponential gap law, with the memoryless exponential the unique gap the independent-increments structure allows. The order-statistics property is the central insight that conditioning on the count strips out the rate and leaves uniform, exchangeable positions: this is exactly what makes superposition and thinning transparent, since merging adds rates and colouring multiplies the rate by the colour probability, and these two operations are dual — superposition is the inverse of thinning, the colour tag of a merged point being the rate share. The compound Poisson process generalises the count by attaching i.i.d. marks, and this is dual to the Lévy–Khintchine description in which compound Poisson laws are the finite-activity pure-jump pieces of every infinitely divisible law. The bridge to the general theory is the spatial Poisson point process, where the count definition is the foundational reason the construction survives the loss of a time order: complete independence plus Poisson marginals defines a random measure on any $σ$ -finite space, and the rate- $λ$ process on the half-line is its one-dimensional shadow with $μ = λ Leb$ .

Full proof set Master

Proposition 1 (order-statistics property, general $n$ ). For a rate- $λ$ Poisson process, conditional on $N_{t} = n$ the joint density of $(T_{1}, \dots, T_{n})$ is $n! / t^{n}$ on ${0 < u_{1} < \dots < u_{n} < t}$ , the order-statistics law of $n$ i.i.d. $Uniform [0, t]$ points.

Proof. The gaps $S_{1}, \dots, S_{n + 1}$ are i.i.d. $Exp (λ)$ with joint density $λ^{n + 1} e^{- λ \sum_{i = 1}^{n + 1} s_{i}}$ . The map $(s_{1}, \dots, s_{n + 1}) \mapsto (u_{1}, \dots, u_{n}, s_{n + 1})$ with $u_{k} = s_{1} + \dots + s_{k}$ has unit Jacobian, so $(T_{1}, \dots, T_{n}, S_{n + 1})$ has density $λ^{n + 1} e^{- λ (u_{n} + s_{n + 1})}$ on ${0 < u_{1} < \dots < u_{n}, s_{n + 1} > 0}$ . The event ${N_{t} = n}$ is ${u_{n} \leq t < u_{n} + s_{n + 1}}$ . Integrating out $s_{n + 1}$ over $(t - u_{n}, \infty)$ gives $\int_{t - u_{n}}^{\infty} λ e^{- λ s_{n + 1}} d s_{n + 1} = e^{- λ (t - u_{n})}$ , so the density of $(T_{1}, \dots, T_{n})$ restricted to ${N_{t} = n}$ is $λ^{n} e^{- λ u_{n}} e^{- λ (t - u_{n})} = λ^{n} e^{- λ t}$ on the simplex ${0 < u_{1} < \dots < u_{n} < t}$ , constant in the $u_{k}$ . Dividing by $P (N_{t} = n) = e^{- λ t} (λ t)^{n} / n!$ yields the conditional density $λ^{n} e^{- λ t} / (e^{- λ t} (λ t)^{n} / n!) = n! / t^{n}$ , independent of $λ$ . This is the order-statistics density of $n$ i.i.d. $Uniform [0, t]$ variables. $□$

Proposition 2 (sum of independent Poisson laws). If $K_{1} \sim Poisson (a)$ and $K_{2} \sim Poisson (b)$ are independent, then $K_{1} + K_{2} \sim Poisson (a + b)$ .

Proof. The probability generating function of $Poisson (m)$ is $G (z) = E [z^{K}] = \sum_{k} e^{- m} m^{k} z^{k} / k! = e^{m (z - 1)}$ . By independence $E [z^{K_{1} + K_{2}}] = e^{a (z - 1)} e^{b (z - 1)} = e^{(a + b) (z - 1)}$ , the generating function of $Poisson (a + b)$ . Generating functions determine the law of a non-negative integer variable, so $K_{1} + K_{2} \sim Poisson (a + b)$ . $□$

Proposition 3 (thinning produces independent Poisson streams). Let $N$ be rate- $λ$ Poisson and colour each point red with probability $p$ , blue with probability $1 - p$ , independently. Then $(N_{t}^{R}, N_{t}^{B})$ are independent with $N_{t}^{R} \sim Poisson (p λ t)$ , $N_{t}^{B} \sim Poisson ((1 - p) λ t)$ .

Proof. Condition on $N_{t} = n$ : the colours are i.i.d., so $(N_{t}^{R}, N_{t}^{B}) ∣ {N_{t} = n}$ is $Multinomial (n; p, 1 - p)$ , i.e. $P (N_{t}^{R} = j, N_{t}^{B} = k ∣ N_{t} = j + k) = (j j + k) p^{j} (1 - p)^{k}$ . Then $P (N_{t}^{R} = j, N_{t}^{B} = k) = e^{- λ t} \frac{( λ t ) ^{j + k}}{( j + k )!} (j j + k) p^{j} (1 - p)^{k} = (e^{- p λ t} \frac{( p λ t ) ^{j}}{j !}) (e^{- (1 - p) λ t} \frac{(( 1 - p ) λ t ) ^{k}}{k !}),$ using $e^{- λ t} = e^{- p λ t} e^{- (1 - p) λ t}$ and $(λ t)^{j + k} / (j + k)! \cdot (j j + k) = (λ t)^{j + k} / (j! k!)$ . The joint mass factors into the product of the two marginals, so $N_{t}^{R}$ and $N_{t}^{B}$ are independent with the stated Poisson laws. Independent increments across disjoint windows transfer from $N$ because the colouring is independent across points. $□$

Proposition 4 (compound Poisson characteristic function and moments). For $X_{t} = \sum_{n = 1}^{N_{t}} Y_{n}$ with $(Y_{n})$ i.i.d. of law $F$ independent of the rate- $λ$ process $N$ , $E [e^{i θ X_{t}}] = exp (λ t (F (θ) - 1))$ , $E [X_{t}] = λ t E [Y_{1}]$ , and $Var (X_{t}) = λ t E [Y_{1}^{2}]$ .

Proof. Condition on $N_{t} = n$ and use independence of the $Y_{n}$ : $E [e^{i θ X_{t}} ∣ N_{t} = n] = F (θ)^{n}$ . Averaging over $N_{t} \sim Poisson (λ t)$ , $E [e^{i θ X_{t}}] = n \geq 0 \sum e^{- λ t} \frac{( λ t ) ^{n}}{n !} F (θ)^{n} = e^{- λ t} e^{λ t F (θ)} = exp (λ t (F (θ) - 1)) .$ Write $ψ (θ) = λ t (F (θ) - 1)$ for the cumulant exponent. Then $ψ^{'} (0) = λ t F^{'} (0) = λ t i E [Y_{1}]$ gives $E [X_{t}] = λ t E [Y_{1}]$ , and $ψ^{''} (0) = λ t F^{''} (0) = - λ t E [Y_{1}^{2}]$ gives the second cumulant $Var (X_{t}) = λ t E [Y_{1}^{2}]$ , since the variance is the second cumulant. $□$

Proposition 5 (existence of the spatial Poisson point process). For any $σ$ -finite measure $μ$ on $(S, A)$ there exists a point process $Π$ with complete independence and $Π (A) \sim Poisson (μ (A))$ for $μ (A) < \infty$ .

Proof. First suppose $μ (S) = m < \infty$ . Draw $K \sim Poisson (m)$ and, given $K = n$ , place points $ξ_{1}, \dots, ξ_{n}$ i.i.d. with law $μ / m$ ; set $Π = \sum_{j = 1}^{K} δ_{ξ_{j}}$ . For disjoint $A_{1}, \dots, A_{k}$ with union $A_{0}$ of complement-aware total, the conditional counts are multinomial, and the Poisson mixing factorises exactly as in Proposition 3 (the $k$ -colour version): $(Π (A_{1}), \dots)$ are independent with $Π (A_{i}) \sim Poisson (μ (A_{i}))$ . For $σ$ -finite $μ$ , partition $S = ⨆_{n} S_{n}$ with $μ (S_{n}) < \infty$ , build independent finite Poisson processes $Π_{n}$ on each piece, and superpose: $Π = \sum_{n} Π_{n}$ . Complete independence and the Poisson marginals are preserved under the disjoint superposition by Proposition 2 applied countably (the counts over any $A$ split as an independent sum $\sum_{n} Π_{n} (A)$ of Poisson variables with summable means $\sum_{n} μ (A \cap S_{n}) = μ (A)$ ). $□$

Connections Master

The continuous-time Markov chain framework 37.05.08 is the parent structure: the Poisson process is the constant-rate pure birth chain with $Q$ -matrix $q_{n, n + 1} = λ$ , $q_{nn} = - λ$ , so its exponential holding times are the inter-arrival gaps, its deterministic jump chain is the march $n \mapsto n + 1$ , and its non-explosion ( $\sum_{n} 1/ λ = \infty$ ) is what makes $N_{t}$ finite for all $t$ ; the equivalence B $\Leftrightarrow$ C is exactly the jump-chain/holding-time construction specialised to constant rate.
The elementary probability rules and named distributions 26.02.01 supply the Poisson and exponential laws, the Gamma arrival-time law as a sum of exponentials, the Binomial-Poisson thinning identity, and the generating-function and conditioning machinery (multiplication rule, total variance) that every proof here runs on; the Poisson process is the canonical lift of those static distributions into a process indexed by continuous time.
The order-statistics property points forward to the general theory of exchangeability and to the spatial point process: given the count, the points are exchangeable and uniform, which is the discrete shadow of the statement that a Poisson point process conditioned on its count is an i.i.d. sample from the normalised mean measure, connecting this unit to the measure-theoretic point-process formalism 37.01.01.

Historical & philosophical context Master

The Poisson distribution appears in Siméon Denis Poisson's 1837 treatise on the probability of judgments ^{[Poisson 1837]}, derived as the limit of the binomial when the number of trials grows and the success probability shrinks with their product held fixed. Poisson studied the distribution, not the process: the dynamical object — random points scattered in time with independent increments — was assembled later, as the rare-event limit of the binomial was recognised as the marginal of a process whose disjoint windows are independent.

The process itself crystallised in early-twentieth-century applied work. Ladislaus Bortkiewicz's 1898 Das Gesetz der kleinen Zahlen popularised the Poisson law for rare events, Agner Krarup Erlang's 1909 telephone-traffic studies introduced the exponential inter-arrival and Poisson-arrival model that founded queueing theory, and Ernest Rutherford and Hans Geiger's 1910 radioactive-decay counts gave the canonical physical realisation, with the exponential gaps between clicks. The unification of the count, infinitesimal, and gap descriptions into the single object treated here, and the closure properties of superposition and thinning, are the modern synthesis of Norris ^{[Norris 1997]} and Kingman ^{[Kingman 1993]}.

The spatial generalisation is due in its modern form to John Frank Charles Kingman, whose 1993 monograph ^{[Kingman 1993]} organised the Poisson point process on an abstract measure space around complete independence and the Poisson marginal, with the mean measure $μ$ replacing the rate; the compound Poisson process and its role as the finite-activity piece of the Lévy–Khintchine decomposition trace to the work on infinitely divisible laws and were brought into statistical modelling by, among others, Lewis's 1964 branching-Poisson failure model ^{[Lewis 1964]}.

Bibliography Master

@book{Norris1997,
  author    = {Norris, James R.},
  title     = {Markov Chains},
  series    = {Cambridge Series in Statistical and Probabilistic Mathematics},
  publisher = {Cambridge University Press},
  year      = {1997}
}

@book{Kingman1993,
  author    = {Kingman, John Frank Charles},
  title     = {Poisson Processes},
  series    = {Oxford Studies in Probability},
  publisher = {Oxford University Press},
  year      = {1993}
}

@book{Poisson1837,
  author    = {Poisson, Sim\'eon Denis},
  title     = {Recherches sur la probabilit\'e des jugements en mati\`ere criminelle et en mati\`ere civile},
  publisher = {Bachelier},
  address   = {Paris},
  year      = {1837}
}

@book{DaleyVereJones2003,
  author    = {Daley, Daryl J. and Vere-Jones, David},
  title     = {An Introduction to the Theory of Point Processes, Volume I: Elementary Theory and Methods},
  edition   = {2},
  publisher = {Springer},
  year      = {2003}
}

@article{Lewis1964,
  author  = {Lewis, Peter A. W.},
  title   = {A branching {P}oisson process model for the analysis of computer failure patterns},
  journal = {Journal of the Royal Statistical Society, Series B},
  volume  = {26},
  year    = {1964},
  pages   = {398--456}
}

Prerequisites

37.05.08
26.02.01

Tier anchors

beginner: Norris 1997 *Markov Chains* (Cambridge) §2.4; informal picture of random points dropped on a time line where disjoint stretches are independent and the count in any window is Poisson
intermediate: Norris 1997 *Markov Chains* (Cambridge) §2.4 (the three definitions and their equivalence); Kingman 1993 *Poisson Processes* (Oxford) §2-§3
master: Norris 1997 *Markov Chains* (Cambridge) §2.4; Kingman 1993 *Poisson Processes* (Oxford) §2-§5 (superposition, thinning, the Poisson point process, order statistics); Daley-Vere-Jones 2003 *An Introduction to the Theory of Point Processes* 2e Vol. I Ch. 2

References

Norris — Markov Chains · Cambridge University Press 1997, §2.4 (the Poisson process; three equivalent definitions; superposition)
Kingman — Poisson Processes · Oxford University Press 1993, §2-§5 (the Poisson process, superposition, thinning, Poisson point processes, the conditioning property)
Poisson — Recherches sur la probabilite des jugements en matiere criminelle et en matiere civile · Bachelier, Paris 1837
Daley-Vere-Jones — An Introduction to the Theory of Point Processes, 2e Vol. I · Springer 2003, Ch. 2 (the Poisson process and its characterizations)
Lewis — A branching Poisson process model for the analysis of computer failure patterns · Journal of the Royal Statistical Society B 26 (1964), 398-456

Estimated time

beginner: 18m
intermediate: 55m
master: 90m