The Poisson Process

How many customers arrive at a shop by time tt? How many radioactive decays occur in one second? How many earthquakes strike a region in a decade? These questions share a common mathematical skeleton. Events occur one at a time, at random moments. They are scattered "uniformly at random" along the time axis, in a precise statistical sense. The Poisson process is the canonical model for such phenomena. It is arguably the simplest interesting continuous-time stochastic process one can write down. The idea matters because it is a hinge connecting discrete counting to continuous randomness. On one side sits the Exponential distribution, governing the waiting time between events. On the other side sits the full counting process N(t)N(t). Its increments over disjoint intervals behave independently and depend only on interval length. This rigidity — independent, stationary increments plus a mild regularity condition ruling out simultaneous events — forces the distribution of N(t)N(t) to be Poisson. It also forces interarrival times to be i.i.d. exponential. The process is simultaneously a limit of many independent rare events (a Poisson approximation to Binomial). It is also an exact continuous-time construction from exponential clocks. The big picture is that the Poisson process is the "null model" of pure randomness in time. There is no memory, no clustering, no periodicity — just a constant instantaneous rate λ\lambda of event occurrence. Every deviation from this null model (clustering, self-excitation, time-varying rates) is measured relative to it. This mirrors how the normal distribution serves as a reference point for symmetric noise. It is also the entry point to continuous-time Markov chains. After a scaling limit, it leads to Introduction to Brownian Motion. This article gives the precise definition of a Poisson process via three equivalent characterizations. It proves the fundamental equivalence theorem linking counting increments to exponential interarrival times. It works through two asymmetric numerical examples. Finally, it shows exactly which hypothesis fails when the model breaks — for instance under time-inhomogeneous or clustered arrivals.

Key takeaways
  • A Poisson process counts random events occurring at rate λ>0\lambda>0 per unit time, with no memory of the past.
  • Three equivalent definitions exist: axiomatic (independent, stationary Poisson increments), constructive (via i.i.d. exponential gaps), and infinitesimal (via a differential rate condition).
  • The number of events in an interval of length tt is Poisson(λt)\text{Poisson}(\lambda t); increments over disjoint intervals are independent.
  • The gaps between consecutive events are i.i.d. Exponential(λ)\text{Exponential}(\lambda) random variables — a direct consequence of memorylessness.
  • Dropping stationarity gives an inhomogeneous Poisson process; dropping independence can produce clustering (Cox/doubly-stochastic processes).
  • Key idea: independence + stationarity + no simultaneous arrivals is rigid enough to force the Poisson distribution and exponential gaps exactly.

Where this lives today

In queueing theory, Poisson arrivals combined with exponential service times give the M/M/1 queue, whose steady-state behavior is solvable exactly via birth-death Markov chain methods. In reliability engineering, Poisson processes model the arrival of shocks or failures to a system, with the interarrival gaps governing time-to-failure distributions. In actuarial science, the Cramér–Lundberg ruin model uses a compound Poisson process for claim arrivals to an insurer. In spatial statistics and stochastic geometry, the Poisson point process generalizes the temporal version to Rd\mathbb{R}^d, modeling locations of stars, trees, or wireless network nodes. In physics, radioactive decay counts and photon arrivals in quantum optics are modeled as Poisson processes, connecting to shot noise in electronics. In finance, jump-diffusion models (Merton's model) add a compound Poisson jump component to Brownian motion to capture sudden price shocks — this is precisely the bridge toward Introduction to Brownian Motion, where the Poisson process's discrete jumps give way to a continuous-path limit under different scaling. In computer science and cryptography, Poisson processes model packet arrivals in network traffic analysis and random polling/sampling schemes; the memoryless property underlies the analysis of hash collision timing and randomized load balancing.

Ask the tutor: How does the compensated Poisson process N(t)−λtN(t)-\lambda t behave under rescaling, and how does that limit connect to Brownian motion?

DefinitionThe Poisson Process

A counting process is a stochastic process {N(t):t≥0}\{N(t): t \ge 0\} with N(0)=0N(0)=0, taking values in {0,1,2,… }\{0,1,2,\dots\}, right-continuous, non-decreasing, and increasing only by jumps of size 11 (a.s.), interpreted as the number of events in [0,t][0,t]. A counting process {N(t):t≥0}\{N(t): t\ge 0\} is a Poisson process with rate λ>0\lambda > 0 if: 1. (Independent increments) For any 0≤t1<t2<⋯<tn0 \le t_1 < t_2 < \cdots < t_n, the random variables

N(t2)−N(t1), N(t3)−N(t2), …, N(tn)−N(tn−1)N(t_2)-N(t_1),\ N(t_3)-N(t_2),\ \dots,\ N(t_n)-N(t_{n-1})

are mutually independent. 2. (Stationary increments) For any s,t≥0s,t \ge 0, the distribution of N(t+s)−N(s)N(t+s)-N(s) depends only on tt, not on ss. 3. (Local behavior / no simultaneous jumps) As h↓0h \downarrow 0,

P(N(h)=1)=λh+o(h),P(N(h)≥2)=o(h).P(N(h) = 1) = \lambda h + o(h), \qquad P(N(h) \ge 2) = o(h).

An equivalent constructive definition: let X1,X2,…X_1, X_2, \dots be i.i.d. Exponential(λ)\text{Exponential}(\lambda) random variables, set Tn=X1+⋯+XnT_n = X_1 + \cdots + X_n (the arrival times, T0=0T_0=0), and define

N(t)=max⁡{n≥0:Tn≤t}.N(t) = \max\{n \ge 0 : T_n \le t\}.

Intuition

Picture the timeline as a road on which "hits" land one at a time. The defining feature is forgetfulness: knowing that no event has occurred up to time ss tells you nothing about how much longer you must wait — the process resets. This is exactly the memoryless property of the exponential distribution, lifted from a single waiting time to an entire process of waiting times. Condition 3 above says that in an infinitesimally small window hh, the chance of exactly one event is proportional to hh (rate λ\lambda), and the chance of two or more events in that same tiny window is negligible even compared to hh itself — events don't pile up on top of each other. Combined with independence across disjoint windows, chopping [0,t][0,t] into nn tiny pieces of length t/nt/n turns N(t)N(t) into (approximately) a sum of nn independent Bernoulli(λt/n)(\lambda t/n) trials — a Binomial(n,λt/n)(n, \lambda t/n) variable. Letting n→∞n \to \infty produces exactly the Poisson limit, which is why the count N(t)N(t) ends up Poisson distributed with mean λt\lambda t.

Figure
Sample path of a Poisson process with rate λ = 1.4, showing jump times T1,…,T5T_1,\dots,T_5 and exponential gaps XiX_i
t 0 T₁ T₂ T₃ T₄ T₅ X₁ X₂ X₃ 0 1 2 3 4 5 N(t): right-continuous step function, jumps of size 1

Central Theorem

TheoremEquivalence of the Axiomatic and Constructive Poisson Process

Let λ>0\lambda > 0. The following two constructions of a counting process {N(t):t≥0}\{N(t): t \ge 0\} are equivalent, meaning each implies the other (in distribution):

(A) NN has independent increments, stationary increments, and satisfies the local conditions

P(N(h)=1)=λh+o(h),P(N(h)≥2)=o(h)(h↓0).P(N(h)=1) = \lambda h + o(h), \qquad P(N(h) \ge 2) = o(h) \quad (h \downarrow 0).

(B) N(t)=max⁡{n:Tn≤t}N(t) = \max\{n : T_n \le t\} where Tn=X1+⋯+XnT_n = X_1 + \cdots + X_n and X1,X2,…X_1, X_2, \dots are i.i.d. Exponential(λ)\text{Exponential}(\lambda). Moreover, under either (A) or (B):

N(t)∼Poisson(λt)for every t≥0,N(t) \sim \text{Poisson}(\lambda t) \quad \text{for every } t \ge 0,

and increments over disjoint intervals (s,t](s,t] satisfy N(t)−N(s)∼Poisson(λ(t−s))N(t)-N(s) \sim \text{Poisson}(\lambda(t-s)).

Proof

We prove (B) ⇒\Rightarrow the distributional claim, then (B) ⇒\Rightarrow (A), and finally (A) ⇒\Rightarrow (B). The technique is a mix of direct computation (via convolution of Gamma/Exponential densities) and a Poisson-limit argument via Binomial approximation.

Step 1: Under (B), N(t)∼Poisson(λt)N(t) \sim \text{Poisson}(\lambda t). Since the XiX_i are i.i.d. Exponential(λ)\text{Exponential}(\lambda), Tn=X1+⋯+XnT_n = X_1+\cdots+X_n has the Gamma(n,λ)\text{Gamma}(n,\lambda) distribution, with density

fTn(s)=λnsn−1e−λs(n−1)!,s>0.f_{T_n}(s) = \frac{\lambda^n s^{n-1} e^{-\lambda s}}{(n-1)!}, \quad s > 0.

This is proved by induction: T1=X1T_1 = X_1 has density λe−λs\lambda e^{-\lambda s}, matching the formula at n=1n=1. If TnT_n has the claimed density, then since Tn+1=Tn+Xn+1T_{n+1} = T_n + X_{n+1} with Xn+1X_{n+1} independent of TnT_n, the convolution formula gives

fTn+1(s)=∫0sfTn(u) λe−λ(s−u) du=∫0sλnun−1e−λu(n−1)!λe−λ(s−u)du=λn+1e−λs(n−1)!∫0sun−1 du=λn+1sne−λsn!,f_{T_{n+1}}(s) = \int_0^s f_{T_n}(u)\, \lambda e^{-\lambda(s-u)}\, du = \int_0^s \frac{\lambda^n u^{n-1} e^{-\lambda u}}{(n-1)!} \lambda e^{-\lambda(s-u)} du = \frac{\lambda^{n+1} e^{-\lambda s}}{(n-1)!}\int_0^s u^{n-1}\,du = \frac{\lambda^{n+1} s^n e^{-\lambda s}}{n!},

completing the induction. Now observe the key identity linking counts and arrival times:

{N(t)≥n}={Tn≤t},\{N(t) \ge n\} = \{T_n \le t\},

because N(t)≥nN(t)\ge n exactly when the nn-th event has already occurred by time tt. Hence

P(N(t)=n)=P(N(t)≥n)−P(N(t)≥n+1)=P(Tn≤t)−P(Tn+1≤t).P(N(t) = n) = P(N(t)\ge n) - P(N(t) \ge n+1) = P(T_n \le t) - P(T_{n+1}\le t).

Using P(Tn≤t)=∫0tfTn(s) dsP(T_n\le t) = \int_0^t f_{T_n}(s)\,ds, integrate by parts on P(Tn+1≤t)P(T_{n+1}\le t) with u=sn/n!u = s^n/n!, dv=λn+1e−λsdsdv = \lambda^{n+1}e^{-\lambda s}ds:

P(Tn+1≤t)=∫0tλn+1sne−λsn! ds=[−λnsne−λsn!]0t+∫0tλnsn−1e−λs(n−1)! ds=−(λt)ne−λtn!+P(Tn≤t).P(T_{n+1}\le t) = \int_0^t \frac{\lambda^{n+1}s^n e^{-\lambda s}}{n!}\,ds = \left[-\frac{\lambda^n s^n e^{-\lambda s}}{n!}\right]_0^t + \int_0^t \frac{\lambda^n s^{n-1}e^{-\lambda s}}{(n-1)!}\,ds = -\frac{(\lambda t)^n e^{-\lambda t}}{n!} + P(T_n \le t).

Therefore

P(N(t)=n)=P(Tn≤t)−P(Tn+1≤t)=(λt)ne−λtn!,n=0,1,2,…P(N(t)=n) = P(T_n\le t) - P(T_{n+1}\le t) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}, \qquad n = 0,1,2,\dots

This is exactly the Poisson(λt)\text{Poisson}(\lambda t) probability mass function. (For n=0n=0: P(N(t)=0)=P(T1>t)=e−λtP(N(t)=0) = P(T_1 > t) = e^{-\lambda t}, matching the formula.)

Step 2: (B) ⇒\Rightarrow (A). Stationary and independent increments. Fix s>0s>0. Given N(s)=kN(s) = k, the process restarted at time ss counts arrivals via the residual interarrival times. By the memoryless property of the exponential distribution, the residual time until the next arrival after ss, given no arrival in (Tk,s](T_k, s] so far, is again Exponential(λ)\text{Exponential}(\lambda) and independent of everything before time ss. Precisely, condition on N(s)=kN(s)=k and on T1,…,TkT_1,\dots,T_k; the residual R=Tk+1−sR = T_{k+1}-s satisfies, for r>0r>0,

P(R>r∣Tk+1>s,T1,…,Tk)=P(Xk+1>s−Tk+r∣Xk+1>s−Tk)=e−λrP(R > r \mid T_{k+1} > s, T_1,\dots,T_k) = P(X_{k+1} > s - T_k + r \mid X_{k+1} > s-T_k) = e^{-\lambda r}

by memorylessness of Xk+1∼Exponential(λ)X_{k+1}\sim\text{Exponential}(\lambda). Since this holds regardless of TkT_k, and all subsequent gaps Xk+2,Xk+3,…X_{k+2}, X_{k+3},\dots are independent of the past and unaffected by conditioning, the process {N(s+t)−N(s):t≥0}\{N(s+t)-N(s): t\ge 0\} is itself a Poisson process constructed from the fresh i.i.d. exponential sequence (R,Xk+2,Xk+3,… )(R, X_{k+2}, X_{k+3}, \dots), independent of {N(u):u≤s}\{N(u): u\le s\}. This is true for every kk, hence unconditionally: {N(s+t)−N(s):t≥0}\{N(s+t)-N(s):t\ge0\} is independent of {N(u):u≤s}\{N(u):u\le s\} and has the same distribution as {N(t):t≥0}\{N(t):t\ge 0\} itself (by Step 1 applied to the fresh process). This gives independent, stationary increments and, applied inductively, independence across arbitrarily many disjoint increments. Local behavior. By Step 1, N(h)∼Poisson(λh)N(h)\sim\text{Poisson}(\lambda h), so

P(N(h)=1)=λh e−λh=λh(1−λh+O(h2))=λh+o(h),P(N(h)=1) = \lambda h\, e^{-\lambda h} = \lambda h(1 - \lambda h + O(h^2)) = \lambda h + o(h),
P(N(h)≥2)=1−e−λh−λhe−λh=1−(1−λh+λ2h22+O(h3))−λh(1−λh+O(h2))=O(h2)=o(h).P(N(h)\ge 2) = 1 - e^{-\lambda h} - \lambda h e^{-\lambda h} = 1 - (1-\lambda h + \tfrac{\lambda^2h^2}{2}+O(h^3)) - \lambda h(1-\lambda h + O(h^2)) = O(h^2) = o(h).

Step 3: (A) ⇒\Rightarrow (B). Assume (A). We first show N(t)∼Poisson(λt)N(t) \sim \text{Poisson}(\lambda t) directly from the axioms, via a Binomial approximation limit, then recover the exponential interarrival structure. Poisson distribution of N(t)N(t). Fix t>0t>0 and partition [0,t][0,t] into nn equal subintervals of length t/nt/n. Let Ij(n)I_j^{(n)} be the indicator that at least one event falls in the jj-th subinterval, j=1,…,nj=1,\dots,n. By stationarity, P(Ij(n)=1)=P(N(t/n)≥1)=λ(t/n)+o(1/n)P(I_j^{(n)}=1) = P(N(t/n)\ge 1) = \lambda(t/n) + o(1/n) as n→∞n\to\infty. By independent increments, I1(n),…,In(n)I_1^{(n)},\dots,I_n^{(n)} are independent. Let Sn=∑j=1nIj(n)S_n = \sum_{j=1}^n I_j^{(n)}, a Binomial(n,pn)\text{Binomial}(n, p_n) variable with pn=λt/n+o(1/n)p_n = \lambda t/n + o(1/n), so npn→λtnp_n \to \lambda t. By the classical Poisson limit theorem for triangular arrays (a Binomial with npn→μnp_n\to\mu converges in distribution to Poisson(μ)\text{Poisson}(\mu) — proved by direct computation of the factorial-moment or characteristic function limit), Sn⇒Poisson(λt)S_n \Rightarrow \text{Poisson}(\lambda t). It remains to show Sn→N(t)S_n \to N(t) in probability, so that N(t)N(t) has the same limiting distribution. We have Sn≤N(t)S_n \le N(t) always (each event contributes at least one indicator), and

P(Sn<N(t))≤P(some subinterval contains≥2 events)≤n⋅P(N(t/n)≥2)=n⋅o(t/n)→0P(S_n < N(t)) \le P(\text{some subinterval contains} \ge 2 \text{ events}) \le n \cdot P(N(t/n)\ge 2) = n\cdot o(t/n) \to 0

as n→∞n\to\infty, using the local condition P(N(h)≥2)=o(h)P(N(h)\ge 2)=o(h) with h=t/nh=t/n. Hence Sn→N(t)S_n \to N(t) in probability, so N(t)N(t) has the same distribution as the limit of SnS_n, namely Poisson(λt)\text{Poisson}(\lambda t). The identical argument on any interval (s,t](s,t], using stationarity, gives N(t)−N(s)∼Poisson(λ(t−s))N(t)-N(s)\sim\text{Poisson}(\lambda(t-s)). Interarrival times are exponential. Let T1=inf⁡{t:N(t)≥1}T_1 = \inf\{t : N(t) \ge 1\}. Then

P(T1>t)=P(N(t)=0)=e−λt,P(T_1 > t) = P(N(t) = 0) = e^{-\lambda t},

so T1∼Exponential(λ)T_1 \sim \text{Exponential}(\lambda). Given T1=sT_1 = s, the process N′(t):=N(s+t)−N(s)N'(t) := N(s+t) - N(s) is, by independent and stationary increments, itself distributed as a fresh copy of NN, independent of {N(u):u≤s}\{N(u): u \le s\} — in particular independent of T1T_1. Hence the second interarrival gap X2:=T2−T1X_2 := T_2 - T_1 is the first-arrival time of an independent copy of the process, so X2∼Exponential(λ)X_2 \sim \text{Exponential}(\lambda) and is independent of X1=T1X_1 = T_1. Iterating, X1,X2,…X_1, X_2, \ldots are i.i.d. Exponential(λ)\text{Exponential}(\lambda), and Tn=X1+⋯+XnT_n = X_1+\cdots+X_n, recovering exactly construction (B). This closes the logical loop: (A) ⇔\Leftrightarrow (B), and both give N(t)∼Poisson(λt)N(t)\sim\text{Poisson}(\lambda t) with independent Poisson increments over disjoint intervals. ■\blacksquare

∎
CorollaryInterarrival and waiting time distributions

Under either definition, the interarrival times X1,X2,…X_1, X_2, \dots are i.i.d. Exponential(λ)\text{Exponential}(\lambda), and the nn-th arrival time TnT_n has the Gamma(n,λ)\text{Gamma}(n,\lambda) distribution with density λnsn−1e−λs/(n−1)!\lambda^n s^{n-1}e^{-\lambda s}/(n-1)! for s>0s>0.

Worked examples

Example 1 (Merging and thinning with asymmetric rates). A help desk receives billing calls as a Poisson process at rate λ1=2.3\lambda_1 = 2.3 per hour and technical-support calls, independently, as a Poisson process at rate λ2=1.1\lambda_2 = 1.1 per hour. Find (a) the distribution of the total call process, (b) the probability the first call received is a billing call, and (c) the probability that exactly 3 calls arrive in a 45-minute window. Solution. (a) Let N1(t),N2(t)N_1(t), N_2(t) be the two independent Poisson processes. The merged process N(t)=N1(t)+N2(t)N(t) = N_1(t)+N_2(t) has independent increments (sums of independent-increment processes over the same partition remain independent across disjoint intervals, since increments of N1N_1 and N2N_2 over any fixed sub-collection of disjoint intervals are jointly independent) and stationary increments inherited from each. Its local behavior: P(N(h)=1)=P(N1(h)=1,N2(h)=0)+P(N1(h)=0,N2(h)=1)=(λ1h+o(h))(1+o(1))+(λ2h+o(h))(1+o(1))=(λ1+λ2)h+o(h)P(N(h)=1) = P(N_1(h)=1,N_2(h)=0)+P(N_1(h)=0,N_2(h)=1) = (\lambda_1 h+o(h))(1+o(1)) + (\lambda_2h+o(h))(1+o(1)) = (\lambda_1+\lambda_2)h+o(h), and P(N(h)≥2)=o(h)P(N(h)\ge2)=o(h) by a union bound over the four ways to get ≥2\ge2 combined events, each o(h)o(h). So by the theorem, N(t)N(t) is Poisson process with rate λ=λ1+λ2=3.4\lambda = \lambda_1+\lambda_2 = 3.4 per hour. (b) The time to the first billing call is X(1)∼Exponential(2.3)X^{(1)}\sim\text{Exponential}(2.3), to the first tech call X(2)∼Exponential(1.1)X^{(2)}\sim\text{Exponential}(1.1), independent. The first call overall is billing iff X(1)<X(2)X^{(1)} < X^{(2)}. For independent exponentials,

P(X(1)<X(2))=∫0∞λ1e−λ1xP(X(2)>x) dx=∫0∞λ1e−λ1xe−λ2x dx=λ1λ1+λ2.P(X^{(1)} < X^{(2)}) = \int_0^\infty \lambda_1 e^{-\lambda_1 x} P(X^{(2)}>x)\,dx = \int_0^\infty \lambda_1 e^{-\lambda_1 x}e^{-\lambda_2 x}\,dx = \frac{\lambda_1}{\lambda_1+\lambda_2}.

Numerically, P=2.3/3.4=0.6765P = 2.3/3.4 = 0.6765 (to four decimal places). (c) With merged rate λ=3.4\lambda=3.4/hour and window length t=0.75t = 0.75 hours, λt=2.55\lambda t = 2.55. Then

P(N(0.75)=3)=(2.55)3e−2.553!=16.581375×e−2.556.P(N(0.75)=3) = \frac{(2.55)^3 e^{-2.55}}{3!} = \frac{16.581375 \times e^{-2.55}}{6}.

Since e−2.55≈0.078082e^{-2.55} \approx 0.078082, we get P≈16.581375×0.0780826≈1.294866≈0.2158P \approx \dfrac{16.581375\times 0.078082}{6} \approx \dfrac{1.29486}{6} \approx 0.2158.

Example 2 (Conditional uniformity of arrival times). Meteor sightings during a night follow a Poisson process at rate λ=4\lambda = 4 per hour. Given that exactly 22 meteors were seen in the interval [0,3][0,3] (hours), find the joint density of their arrival times (T1,T2)(T_1, T_2) and use it to compute P(T1<1)P(T_1 < 1). Solution. A standard but non-obvious fact: given N(t)=nN(t)=n, the nn arrival times are distributed as the order statistics of nn i.i.d. Uniform(0,t)(0,t) variables, and this holds regardless of λ\lambda. We verify it here for n=2n=2, t=3t=3 directly rather than quoting it as folklore. The joint density of (T1,T2)(T_1,T_2) on {0<t1<t2}\{0<t_1<t_2\} can be computed from the interarrival representation T1=X1,T2=X1+X2T_1=X_1, T_2=X_1+X_2 with X1,X2X_1,X_2 i.i.d. Exponential(λ\lambda); the Jacobian of (x1,x2)↦(t1,t2)=(x1,x1+x2)(x_1,x_2)\mapsto(t_1,t_2)=(x_1,x_1+x_2) is 11, so

fT1,T2(t1,t2)=λe−λt1⋅λe−λ(t2−t1)=λ2e−λt2,0<t1<t2.f_{T_1,T_2}(t_1,t_2) = \lambda e^{-\lambda t_1}\cdot \lambda e^{-\lambda(t_2-t_1)} = \lambda^2 e^{-\lambda t_2}, \qquad 0<t_1<t_2.

The event {N(3)=2}\{N(3)=2\} equals {T2≤3<T3}\{T_2 \le 3 < T_3\}; conditioning requires the joint density of (T1,T2,T3)(T_1,T_2,T_3), but we can shortcut using P(N(3)=2)P(N(3)=2) from the Poisson mass function and the relation

P(T1∈dt1,T2∈dt2,N(3)=2)=P(T1∈dt1,T2∈dt2)⋅P(X3>3−t2)=λ2e−λt2 dt1 dt2⋅e−λ(3−t2),P(T_1\in dt_1, T_2\in dt_2, N(3)=2) = P(T_1\in dt_1, T_2 \in dt_2)\cdot P(X_3 > 3-t_2) = \lambda^2 e^{-\lambda t_2}\, dt_1\,dt_2\cdot e^{-\lambda(3-t_2)},

for 0<t1<t2<30<t_1<t_2<3, since given T2=t2≤3T_2=t_2\le3, the event N(3)=2N(3)=2 means the third gap X3X_3 exceeds 3−t23-t_2, independent of (T1,T2)(T_1,T_2). This simplifies to

fT1,T2,N(3)=2(t1,t2)=λ2e−3λ,0<t1<t2<3,f_{T_1,T_2,N(3)=2}(t_1,t_2) = \lambda^2 e^{-3\lambda}, \qquad 0<t_1<t_2<3,

a constant, independent of t1,t2t_1,t_2! Dividing by P(N(3)=2)=(3λ)2e−3λ2!=9λ2e−3λ2P(N(3)=2) = \frac{(3\lambda)^2 e^{-3\lambda}}{2!} = \frac{9\lambda^2 e^{-3\lambda}}{2} gives the conditional density

fT1,T2∣N(3)=2(t1,t2)=λ2e−3λ9λ2e−3λ2=29,0<t1<t2<3.f_{T_1,T_2\mid N(3)=2}(t_1,t_2) = \frac{\lambda^2 e^{-3\lambda}}{\frac{9\lambda^2 e^{-3\lambda}}{2}} = \frac{2}{9}, \qquad 0 < t_1 < t_2 < 3.

This is exactly the density of the order statistics of 2 i.i.d. Uniform(0,3)(0,3) variables (whose joint density is 2/32=2/92/3^2 = 2/9 on the same triangle), confirming the claim, and notably independent of λ=4\lambda=4. To find P(T1<1∣N(3)=2)P(T_1<1 \mid N(3)=2), integrate:

P(T1<1∣N(3)=2)=∫01∫t1329 dt2 dt1=∫0129(3−t1) dt1=29[3t1−t122]01=29(3−12)=29⋅52=59.P(T_1<1\mid N(3)=2) = \int_0^1\int_{t_1}^3 \frac{2}{9}\,dt_2\,dt_1 = \int_0^1 \frac{2}{9}(3-t_1)\,dt_1 = \frac{2}{9}\left[3t_1 - \frac{t_1^2}{2}\right]_0^1 = \frac{2}{9}\left(3-\tfrac12\right) = \frac{2}{9}\cdot\frac{5}{2} = \frac{5}{9}.

So P(T1<1∣N(3)=2)=5/9≈0.5556P(T_1<1\mid N(3)=2) = 5/9 \approx 0.5556, matching what one would get treating T1T_1 as min⁡\min of 2 i.i.d. Uniform(0,3): P(min⁡<1)=1−(2/3)2=1−4/9=5/9P(\min < 1) = 1-(2/3)^2 = 1-4/9=5/9. ✓\checkmark

CounterexampleDropping stationarity: a time-inhomogeneous arrival process

Consider a call center where the arrival rate genuinely varies over the day: rate λ(t)=2+sin⁡(t)\lambda(t) = 2 + \sin(t) calls per hour (always positive, so well defined), with independent increments and the local condition P(N(t,t+h]=1)=λ(t)h+o(h)P(N(t,t+h]=1) = \lambda(t)h+o(h), P(N(t,t+h]≥2)=o(h)P(N(t,t+h]\ge2)=o(h) preserved, but stationarity dropped. This is a legitimate, commonly used model (the inhomogeneous Poisson process), yet the central theorem's conclusion fails exactly as stated. Concretely, N(t+s)−N(s)N(t+s)-N(s) is not independent of ss in distribution: one can show

N(t+s)−N(s)∼Poisson ⁣(∫ss+tλ(u) du),N(t+s) - N(s) \sim \text{Poisson}\!\left(\int_s^{s+t}\lambda(u)\,du\right),

so, e.g., taking t=1t=1, the mean number of calls in [0,1][0,1] is ∫01(2+sin⁡u) du=2+(1−cos⁡1)≈2.4597\int_0^1(2+\sin u)\,du = 2+(1-\cos 1)\approx 2.4597, while the mean number of calls in [π/2,π/2+1][\pi/2, \pi/2+1] is ∫π/2π/2+1(2+sin⁡u) du=2+(cos⁡(π/2)−cos⁡(π/2+1))=2+sin⁡(1)≈2.8415\int_{\pi/2}^{\pi/2+1}(2+\sin u)\,du = 2+(\cos(\pi/2)-\cos(\pi/2+1)) = 2+\sin(1)\approx 2.8415. These differ, so the increment distribution genuinely depends on the starting time ss — stationarity fails. Consequently the interarrival times are not i.i.d. exponential: the gap starting near a rate peak is stochastically shorter than the gap starting near a rate trough. The equivalence (A)⇔\Leftrightarrow(B) in the theorem breaks precisely at the step using stationarity to argue the restarted process has the same law as the original — that step is false here, since restarting at s=π/2s=\pi/2 gives a process with a different (time-shifted) rate function than restarting at s=0s=0. This shows stationarity is not a redundant technical nicety: without it, N(t)N(t) is still Poisson-distributed at each fixed tt (with time-varying mean ∫0tλ(u)du\int_0^t\lambda(u)du), but the clean i.i.d.-exponential-gap structure, and the "process is a fresh copy after any stopping point" self-similarity, are both lost.

Scope, edge cases & common traps

SituationWhat holdsWhat fails / caution
λ\lambda constant, standard Poisson processFull theorem: Poisson counts, i.i.d. exponential gaps—
Time-varying λ(t)\lambda(t) (inhomogeneous)N(t)∼Poisson(∫0tλ)N(t)\sim\text{Poisson}(\int_0^t\lambda)Stationarity fails; gaps not i.i.d.
Merged independent Poisson processesSum is Poisson, rate addsMust check independence of the two component processes
Thinned process (each event kept w.p. pp)Still Poisson, rate λp\lambda pKept/discarded sub-processes are independent — a separate, provable fact
Doubly-stochastic (Cox) process, random Λ\LambdaN(t)N(t) marginally may look Poisson-likeIndependent increments generally fails — clustering induced by shared randomness
N(t)N(t) observed only at integer timesIncrements still PoissonBut interarrival times, not just counts, need full continuous path to recover

The mathematician behind it

Origins · 1 minSiméon Denis Poisson and how this was discovered
SD
Siméon Denis Poisson
1781–1840 · French mathematics and mathematical physics, École Polytechnique

Poisson worked across mathematical physics, celestial mechanics, electricity, and probability. His 1837 treatise Recherches sur la probabilité des jugements introduced the limiting law now bearing his name, originally as an approximation to the binomial distribution for rare events across many trials — not yet as a continuous-time process. The stochastic process built from his distribution, with independent stationary increments, was developed much later by others (notably A. K. Erlang in queueing contexts, and the rigorous continuous-time probability theory of the 1930s), but the name has stuck to the entire framework: distribution, process, and limit theorem alike.

History
- 1837 — Poisson derives the Poisson distribution as a limit of the binomial, in a legal-probability context.
- 1909 — A. K. Erlang applies Poisson-type arrival models to telephone exchange traffic, effectively introducing the continuous-time counting process.
- 1930s — Kolmogorov and Khinchin place continuous-time stochastic processes, including the Poisson process, on rigorous measure-theoretic footing.
- 1940s–1950s — Feller's treatment cements the equivalence between axiomatic (independent/stationary increments) and constructive (exponential interarrival) definitions.
- Present — the Poisson process is foundational in queueing theory, reliability, spatial statistics, and as the driving noise for jump-diffusion models in finance and physics.
Why must interarrival times be exponential rather than some other distribution?

The exponential distribution is the unique continuous distribution with the memoryless property: P(X>s+t∣X>s)=P(X>t)P(X>s+t\mid X>s)=P(X>t). A Poisson process is defined by having no memory of elapsed time — the future evolution after any point depends only on the fact that "no event has happened yet," not on how long we've waited. If gaps followed, say, a Gamma or Weibull distribution instead, the process would "remember" how long since the last event (aging or anti-aging behavior), violating the stationary/independent increments axioms. The equivalence theorem shows this memorylessness is not an add-on assumption — it is forced by independent + stationary increments plus the local no-simultaneous-arrivals condition.

Is the Poisson process a Markov chain?

Yes, in continuous time. {N(t)}\{N(t)\} is a continuous-time Markov chain on state space {0,1,2,… }\{0,1,2,\dots\} with the property that, given the current state nn, the process jumps to n+1n+1 after an Exponential(λ)\text{Exponential}(\lambda) holding time, independent of how it got to nn. This is the simplest possible continuous-time Markov chain: a "pure birth process" with constant birth rate λ\lambda at every state. The transition matrix framework used for discrete-time chains generalizes here to a transition rate (generator) matrix, with P(N(t+h)=n+1∣N(t)=n)=λh+o(h)P(N(t+h)=n+1\mid N(t)=n) = \lambda h + o(h) playing the role of the discrete transition probability.

What happens if λ\lambda is itself random?

This gives a Cox process (doubly stochastic Poisson process): conditional on a random intensity Λ\Lambda (or intensity function Λ(t)\Lambda(t)), NN is Poisson with that rate. Marginally (averaging over Λ\Lambda), the process typically displays overdispersion relative to a true Poisson process — the variance of N(t)N(t) exceeds its mean, unlike the exact equality Var(N(t))=E[N(t)]=λt\text{Var}(N(t))=\text{E}[N(t)]=\lambda t that holds for a genuine Poisson process. Increments are also no longer independent, since they share the same unknown Λ\Lambda. Cox processes model clustered phenomena such as earthquake aftershocks or bursty network traffic, where the "no memory, no clustering" null model of the true Poisson process is too rigid.

How does thinning work, and why does it preserve the Poisson property?

If each event of a Poisson(λ\lambda) process is independently kept with probability pp and discarded with probability 1−p1-p, the kept events form a Poisson(λp\lambda p) process, and the discarded events independently form a Poisson(λ(1−p)\lambda(1-p)) process, with the two sub-processes mutually independent. This follows because, on any small interval of length hh, P(one kept event)=p⋅(λh+o(h))=λp h+o(h)P(\text{one kept event}) = p\cdot(\lambda h+o(h)) = \lambda p\, h+o(h), and events of the two colors in the same or different intervals inherit independence from the original process's independent increments plus the independent coin flips. Thinning is the exact converse operation to merging (superposition), and both preserve the Poisson family — a rare closure property.

Why is the o(h) condition on double jumps stated as o(h)o(h) rather than exactly 00?

Requiring P(N(h)≥2)=0P(N(h)\ge2)=0 exactly for all small hh would be too strong and in fact impossible for a process with continuous-time jumps of any positive rate — over any interval, however short, there is always some tiny but positive chance of two events landing extremely close together if we insist on approximating with a continuous model. The o(h)o(h) condition captures the right idea: the probability of a "collision" vanishes faster than the probability of a single event as the window shrinks, so simultaneous events have probability zero in the limit, without requiring the impossible exact statement at every finite hh.

Does the Poisson process have independent increments over overlapping intervals?

No — independence is only asserted (and only true) for disjoint intervals. If intervals overlap, the increments trivially share information: for instance, N(2)−N(0)N(2)-N(0) and N(3)−N(1)N(3)-N(1) both depend on N(1)N(1) and N(2)N(2), so they are correlated (in fact N(2)−N(1)N(2)-N(1) contributes to both). The independent-increments axiom is stated precisely for a partition into non-overlapping pieces t1<t2<⋯<tnt_1<t_2<\cdots<t_n, and this is essential — misapplying it to overlapping windows is a common source of error in exercises.

How does the Poisson process relate to Brownian motion?

Both are the two canonical building blocks of continuous-time stochastic processes with independent, stationary increments (Lévy processes): the Poisson process has increments that are discrete jumps (Poisson-distributed), while Brownian motion has increments that are continuous, Gaussian-distributed. In fact, a suitably centered and rescaled sum of many independent Poisson processes — or the compensated process N(t)−λtN(t) - \lambda t, rescaled by 1/n1/\sqrt{n} and sped up by nn — converges to a Brownian motion by a functional central limit theorem, foreshadowing Introduction to Brownian Motion. Where the Poisson process models rare discrete shocks, Brownian motion models the aggregate diffusive limit of infinitely many infinitesimally small ones.

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