Three Ways to Describe the Same Arrival Process
A Poisson process can be described through the counts it produces, the gaps between its arrivals, or its behavior on very short time intervals. These descriptions are equivalent when their assumptions are stated carefully. In particular, a rule for short-interval probabilities alone does not guarantee independent increments: that property must be included in the infinitesimal characterization.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((N(t))_{t\ge0}\) be a counting process: \(N(0)=0\), its paths are nondecreasing, integer-valued and right-continuous, and its jumps count arrivals. Assume throughout the equivalence theorem that the rate \(\lambda\) is fixed and satisfies \(\lambda>0\). The arrival times are \(S_k=\inf\{t:N(t)\ge k\}\), and the interarrival times are \(\tau_1=S_1\) and \(\tau_k=S_k-S_{k-1}\) for \(k\ge2\). The process is nonexplosive if it has only finitely many arrivals in every bounded time interval.
The independent-increments assumption in the second definition is part of the characterization, not a consequence of the displayed short-interval probabilities. In each direction below, the process is assumed to satisfy the counting-process conditions just stated. The equivalence concerns positive rate; the zero-rate degenerate case is separate.
The Equivalence and Its Proof
Proof. We prove the three implications in turn.
(i) implies (iii). If \(Y=N(s+h)-N(s)\), then \(Y\) is Poisson with mean \(\lambda h\), so \(\mathbb{P}(Y=0)=e^{-\lambda h}\), \(\mathbb{P}(Y=1)=\lambda h e^{-\lambda h}\), and \(\mathbb{P}(Y\ge2)=1-e^{-\lambda h}(1+\lambda h)\). As \(h\downarrow0\), \(e^{-\lambda h}=1-\lambda h+O(h^2)\). Thus the one-arrival probability is \(\lambda h+O(h^2)\), and the probability of at least two is \(O(h^2)\). These expressions depend only on \(h\), so the error bounds are uniform in \(s\). Independent increments are assumed in (i).
(iii) implies (i). Fix a starting time \(s\), and let \(Y(u)=N(s+u)-N(s)\) for \(u\ge0\). Independent increments imply that the distribution of \(Y(u)\) evolves from \(Y(0)=0\) by adding an increment over each interval of length \(h\). Write \(p_n(u)=\mathbb{P}(Y(u)=n)\), with \(p_{-1}(u)=0\). By the infinitesimal assumptions, the probabilities of zero and one arrival in a length-\(h\) interval are \(1-\lambda h+o(h)\) and \(\lambda h+o(h)\); the probability of two or more is \(o(h)\). Consequently, for each \(n\ge0\), $$ p_n(u+h)-p_n(u) =\lambda h\,p_{n-1}(u)-\lambda h\,p_n(u)+o(h). $$ The uniform error condition justifies this difference quotient on bounded time intervals. These are the forward equations, with initial values \(p_0(0)=1\) and \(p_n(0)=0\) for \(n\ge1\). For \(n=0\), the equation is \(p_0'(u)=-\lambda p_0(u)\), whose solution is \(p_0(u)=e^{-\lambda u}\). For \(n\ge1\), solving \(p_n'(u)+\lambda p_n(u)=\lambda p_{n-1}(u)\) successively gives \(p_n(u)=e^{-\lambda u}(\lambda u)^n/n!\). Indeed, substitution gives the differential equation, and \(p_n(0)=0\); uniqueness follows by multiplying the equation for the difference of any two solutions by \(e^{\lambda u}\), which makes that difference constant and then zero. Thus \(Y(u)\) is Poisson with mean \(\lambda u\). Since \(s\) was arbitrary, every increment has the required law. Independent increments were assumed in (iii), so (i) follows.
(i) implies (ii). The probability of no arrivals by time \(t\) is \(\mathbb{P}(S_1>t)=\mathbb{P}(N(t)=0)=e^{-\lambda t}\), so the first gap is exponential. To establish independence of all gaps, consider ordered arrival times \(0<s_1<\cdots<s_k<T\). The joint density of the first \(k\) arrival times together with no further arrival by \(T\) is $$ \lambda^k e^{-\lambda T}. $$ To obtain it, place one arrival in each of \(k\) disjoint small intervals around \(s_1,\ldots,s_k\), and no arrivals in the rest of \([0,T]\). Independent increments give, to first order, a factor \(\lambda\,ds_j\) for each small interval and a factor \(e^{-\lambda(T-\sum_j ds_j)}\) for the remaining time. Dividing by the product of the small interval lengths and taking their lengths to zero gives the stated density; the probability of two or more arrivals in any one such interval is of smaller order. Equivalently, the density of the first \(k\) ordered arrival times, without requiring no later arrivals by \(T\), is \(\lambda^k e^{-\lambda s_k}\), since the probability of no further arrivals from \(s_k\) to \(T\) is \(e^{-\lambda(T-s_k)}\), and this factor cancels the corresponding survival restriction.
Now change variables to the gaps \(x_1=s_1\), \(x_j=s_j-s_{j-1}\) for \(2\le j\le k\). The transformation has Jacobian \(1\), and \(s_k=x_1+\cdots+x_k\). The density becomes $$ \lambda^k e^{-\lambda(x_1+\cdots+x_k)} =\prod_{j=1}^k \lambda e^{-\lambda x_j},\qquad x_j>0. $$ This is the product of \(k\) exponential densities. Since it holds for every \(k\), the interarrival times are iid exponential with rate \(\lambda\).
(ii) implies (i). For iid exponential gaps, the joint density of the first \(k\) ordered arrival times and the event of no further arrival by \(T\) is also \(\lambda^k e^{-\lambda T}\): multiply the \(k\) gap densities by the probability \(e^{-\lambda(T-s_k)}\) that the next gap exceeds \(T-s_k\). For a partition \(0=t_0<t_1<\cdots<t_r=T\), prescribe \(m_j\) arrivals in interval \((t_{j-1},t_j]\). Integrating this density over the ordered arrival locations in those intervals gives $$ \mathbb{P}\big(N(t_j)-N(t_{j-1})=m_j,\ 1\le j\le r\big) =e^{-\lambda T}\prod_{j=1}^r \frac{(\lambda(t_j-t_{j-1}))^{m_j}}{m_j!}. $$ For \(m_j=0\), the corresponding factor is \(1\); for positive \(m_j\), the factor \((t_j-t_{j-1})^{m_j}/m_j!\) is the volume of the ordered region for the \(m_j\) arrival locations in that interval. Because \(T=\sum_j(t_j-t_{j-1})\), the right-hand side factors as the product of Poisson probabilities with means \(\lambda(t_j-t_{j-1})\). Hence the increments on the partition intervals are independent and have the required distributions. Every finite collection of disjoint intervals can be included in a partition, proving (i). Finally, the process is nonexplosive: for iid exponential gaps, the independent events \(\{\tau_k\ge1/\lambda\}\) each have probability \(e^{-1}>0\); the probability that none occurs after any fixed index is zero, since \((1-e^{-1})^r\to0\). Infinitely many gaps are therefore at least \(1/\lambda\), so their partial sums diverge. This completes the proof. \(\square\)
Using Each Characterization
Worked Example: Poisson Counts Give Exponential Gaps
Suppose a process has stationary independent increments with rate \(\lambda=2\). Find the probability of no arrivals by time \(3/2\), and the joint density of its first two gaps.
The count \(N(3/2)\) is Poisson with mean \(\lambda t=2(3/2)=3\). Therefore \(\mathbb{P}(\tau_1>3/2)=\mathbb{P}(N(3/2)=0)=e^{-3}\). By the theorem, the first two gaps have joint density $$ f_{\tau_1,\tau_2}(x_1,x_2) =(2e^{-2x_1})(2e^{-2x_2}) =4e^{-2(x_1+x_2)},\qquad x_1,x_2>0. $$ For example, the density at \(x_1=1/2,\ x_2=1\) is \(4e^{-2(1/2+1)}=4e^{-3}\). Its factors are the rate-2 exponential densities at those two arguments, so this calculation checks both the rate and the independence factorization.
Worked Example: Exponential Gaps Give Independent Counts
Let the gaps be iid exponential with rate \(1\). Find the probability of exactly one arrival in \((0,1/2]\) and exactly two arrivals in \((1/2,3/2]\).
The interval lengths are \(1/2\) and \(3/2-1/2=1\), so the respective Poisson means are \(1(1/2)=1/2\) and \(1(1)=1\). The joint probability is $$ \left(e^{-1/2}\frac{(1/2)^1}{1!}\right) \left(e^{-1}\frac{1^2}{2!}\right) =\frac{e^{-3/2}}{4}. $$ The factors multiply because the theorem gives independent counts on disjoint intervals. As a check, the first factor is the Poisson probability for one event with mean \(1/2\), and the second is the probability for two events with mean \(1\); their product has the combined exponential factor \(e^{-(1/2+1)}=e^{-3/2}\).
Worked Example: Infinitesimal Conditions Determine Count Probabilities
Assume independent increments and homogeneous infinitesimal behavior with rate \(\lambda=1\). Find the probabilities of zero, one, or two arrivals by time \(t=1\).
The forward equations give \(p_0(t)=e^{-t}\), \(p_1(t)=te^{-t}\), and \(p_2(t)=t^2e^{-t}/2\). Substituting \(t=1\) gives $$ \mathbb{P}(N(1)=0)=e^{-1},\qquad \mathbb{P}(N(1)=1)=e^{-1},\qquad \mathbb{P}(N(1)=2)=\frac{e^{-1}}{2}. $$ The sum of these three probabilities is \(e^{-1}(1+1+1/2)=5/(2e)\), which is less than \(1\); the remaining probability is for three or more arrivals. This agrees with the full Poisson distribution, whose probabilities sum to \(e^{-1}\sum_{n\ge0}1/n!=1\).
Why the Assumptions Matter
Common mistake: omitting independent increments from the infinitesimal definition. Here is a concrete counterexample. First choose a random rate \(\Theta\), equal to \(1\) or \(3\) with probability \(1/2\) each. Conditional on \(\Theta\), let \(N(t)\) be a Poisson process of rate \(\Theta\). For a short interval of length \(h\), $$ \mathbb{P}(\Delta N=1) =\tfrac12(he^{-h})+\tfrac12(3he^{-3h}) =2h+O(h^2), $$ and \(\mathbb{P}(\Delta N\ge2)=\tfrac12[1-e^{-h}(1+h)]+\tfrac12[1-e^{-3h}(1+3h)]=O(h^2)\). Thus the local probabilities have rate \(\lambda=2\). But counts on disjoint intervals are not independent: for lengths \(a,b>0\), their covariance is \(ab\operatorname{Var}(\Theta)=ab\), not zero. The dependence arises because both counts share the same random rate. So local probabilities without independent increments do not establish the equivalence.
The three descriptions are useful for different tasks. Increment laws give direct probabilities for observation windows; exponential gaps are natural for modeling waiting times; infinitesimal conditions connect local event behavior to the full count distribution through the forward equations. The equivalence is valid only with the assumptions attached to each description. In particular, the process starts at zero, the rate is constant and positive, and independent increments are explicit wherever they are needed.
Check Your Understanding
For each question, identify the characterization and the assumptions being used.
- If a process has stationary independent increments and rate \(\lambda\), what is \(\mathbb{P}(N(t)=0)\), and what distribution does this imply for \(S_1\)?
- Under the homogeneous infinitesimal assumptions, write the forward equation for \(p_2(t)=\mathbb{P}(N(t)=2)\).
- If iid exponential gaps have rate \(3\), what is the probability of exactly one arrival in an interval of length \(2/3\)?
- Why is the probability of at least two arrivals in a short interval required to be \(o(h)\), rather than merely tending to zero?
- In the random-rate example, what feature makes counts on disjoint intervals dependent despite the correct first-order local rate?