From Poisson Increments to Exponential Gaps

From Counts to Waiting Times

A Poisson process describes arrivals by counting how many occur in each time interval. Waiting times describe the same events from another perspective: when does the first arrival occur, and how long is each subsequent gap? The first waiting time follows directly from the probability of zero arrivals. To prove that all successive gaps are independent, we also need to justify why the process after an arrival has the same probabilistic behavior as it does from time zero.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N_t)_{t\ge 0}\) be a Poisson process with rate \(\lambda>0\), adapted to its natural filtration \(\mathcal{F}_t=\sigma(N_s:0\le s\le t)\). Thus \(N_0=0\), the paths are nondecreasing, integer-valued and right-continuous, and increments over disjoint time intervals are independent. For \(0\le s<t\), \(N_t-N_s\) has a Poisson distribution with mean \(\lambda(t-s)\). In particular, \(N_t\) has a Poisson distribution with mean \(\lambda t\).

The assumption \(\lambda>0\) matters: it ensures that each fixed arrival time is finite almost surely. The zero-rate case has no arrivals and is not described by exponential gaps with a positive rate.

The First Waiting Time

The first arrival has occurred by time \(t\) exactly when \(N_t\ge1\). Thus its survival event is the event of no arrivals by time \(t\). Since \(N_t\) is Poisson with mean \(\lambda t\), this gives the full distribution of the first waiting time:

Consequently, \(\mathbb{P}(\tau_1\le t)=1-e^{-\lambda t}\). This establishes the exponential distribution of the first gap. It does not, by itself, establish independence of later gaps: a later gap begins at a random arrival time, rather than at a fixed time.

Why the Process Restarts at an Arrival

To extend the first-gap calculation, we will use a fact about Poisson increments at a random time. A random time \(\sigma\) is a stopping time for \((\mathcal{F}_t)\) if \(\{\sigma\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). The arrival time \(S_n\) is a stopping time because \(\{S_n\le t\}=\{N_t\ge n\}\). For a stopping time \(\sigma\), \(\mathcal{F}_\sigma\) denotes the information available up to \(\sigma\); an event \(A\) belongs to \(\mathcal{F}_\sigma\) if \(A\cap\{\sigma\le t\}\in\mathcal{F}_t\) for every \(t\).

Proof. First consider a stopping time \(\sigma\) taking values in the grid \(\{k\delta:k=0,1,2,\ldots\}\), where \(\delta>0\). Fix times \(0\le u_1<\cdots<u_r\) and a bounded measurable function \(f\) of the vector of future counts at those times. For any \(A\in\mathcal{F}_\sigma\), split according to the value of \(\sigma\). On \(\{\sigma=k\delta\}\), the event \(A\cap\{\sigma=k\delta\}\) belongs to \(\mathcal{F}_{k\delta}\). Deterministic independent increments imply that the vector \((N_{k\delta+u_j}-N_{k\delta})_{j=1}^r\) is independent of \(\mathcal{F}_{k\delta}\) and has the same distribution as \((N_{u_j})_{j=1}^r\). The independence from \(\mathcal{F}_{k\delta}\) follows by first checking events determined by finitely many past counts and then extending to their generated sigma-field. Summing over \(k\) therefore gives \(\mathbb{E}[\mathbf{1}_A f((N_{\sigma+u_j}-N_\sigma)_{j=1}^r)]=\mathbb{P}(A)\mathbb{E}[f((N_{u_j})_{j=1}^r)]\).

For a general almost surely finite stopping time \(\sigma\), set \(\sigma_m=2^{-m}\lceil 2^m\sigma\rceil\). Each \(\sigma_m\) is a grid-valued stopping time, \(\sigma_m\downarrow\sigma\), and \(\mathcal{F}_\sigma\subseteq\mathcal{F}_{\sigma_m}\). Apply the grid result to \(A\in\mathcal{F}_\sigma\) and \(\sigma_m\). Right-continuity of the paths gives \(N_{\sigma_m+u_j}\to N_{\sigma+u_j}\) and \(N_{\sigma_m}\to N_\sigma\) almost surely for every fixed \(u_j\). For bounded continuous \(f\), bounded convergence passes the grid identity to the limit. This proves independence and equality of the finite-dimensional distributions; the corresponding statement for the future process follows because its events are generated by its finite-dimensional observations. \(\square\)

We also check that the arrival times in this argument are finite and distinct. For \(n\ge1\), \(\{S_n>t\}=\{N_t<n\}\), so \(\mathbb{P}(S_n>t)=e^{-\lambda t}\sum_{r=0}^{n-1}(\lambda t)^r/r!\to0\) as \(t\to\infty\). Hence \(S_n<\infty\) almost surely. To exclude multiple arrivals at the same time, fix \(T>0\) and divide \([0,T]\) into \(m\) equal intervals of length \(h=T/m\). If a jump of size at least two occurs by time \(T\), one of the interval counts is at least two. For a Poisson random variable \(Z\) of mean \(\lambda h\), \(\mathbf{1}_{\{Z\ge2\}}\le Z(Z-1)/2\), and \(\mathbb{E}[Z(Z-1)]=(\lambda h)^2\). The probability of any such interval is therefore at most \(m(\lambda T/m)^2/2=\lambda^2T^2/(2m)\), which tends to zero. Thus all jumps are of size one almost surely, and \(N_{S_n}=n\).

Proof of the Exponential-Gap Theorem

Proof. The first gap \(\tau_1\) is exponential with rate \(\lambda\) by the zero-count calculation. Suppose \(n\ge1\). Since \(S_n\) is an almost surely finite stopping time, the lemma says that the process after \(S_n\), measured from its value \(N_{S_n}=n\), is independent of \(\mathcal{F}_{S_n}\) and has the same finite-dimensional distributions as a fresh Poisson process. The next gap satisfies \[ \{\tau_{n+1}>x\}=\{N_{S_n+x}-N_{S_n}=0\},\qquad x\ge0. \] The lemma makes the probability of this event \(e^{-\lambda x}\), and makes \(\tau_{n+1}\) independent of \(\mathcal{F}_{S_n}\). Every earlier gap \(\tau_1,\ldots,\tau_n\) is determined by the arrival history up to \(S_n\), so it is \(\mathcal{F}_{S_n}\)-measurable. Thus \(\tau_{n+1}\) is exponential with rate \(\lambda\) and independent of all earlier gaps. Applying this argument successively for \(n=1,2,\ldots\) proves that every finite collection of gaps is independent and that all have the same exponential distribution. \(\square\)

A Common Pitfall

A gap after an arrival is not the time from zero to the next arrival. For instance, at rate \(\lambda=1\), the incorrect calculation \(\mathbb{P}(\tau_1>1,\tau_2>1)=\mathbb{P}(N_1=0)=e^{-1}\) only checks whether the first arrival has not occurred by time \(1\); it says nothing about the second gap. The correct calculation uses the joint law of two independent exponential gaps:

The distinction is between a count in a fixed interval starting at zero and the waiting time measured from a random arrival time. Stationary independent increments, applied at that arrival time through the lemma, justify the second calculation.