Endpoint Conventions for Event Counts

Why Interval Endpoints Need Care

For a counting process, an interval such as \((a,b]\) includes arrivals after \(a\) and at or before \(b\). Changing it to \([a,b]\) can change the count on a particular sample path if an arrival occurs exactly at \(a\). For a Poisson process, however, a specified deterministic time has probability zero of being an arrival time. This makes the different endpoint conventions equivalent for count probabilities, even though the counts need not agree on every possible path.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N(t))_{t\ge0}\) be a Poisson process with rate \(\lambda\ge0\), with state space \(S=\{0,1,2,\ldots\}\) and continuous time index \(t\ge0\). Thus \(N(0)=0\), and for \(0\le a<b\), \(N(b)-N(a)\) has a Poisson distribution with mean \(\lambda(b-a)\). Write \(S_1,S_2,\ldots\) for the arrival times, so \(N(t)\) counts arrivals in \((0,t]\). For an interval or singleton \(I\), write \(N(I)=\#\{k:S_k\in I\}\). We use the standard Poisson-process convention that the process has finitely many arrivals in every bounded time interval.

The interval \((a,b]\) has a direct relation to the cumulative count: \(N((a,b])=N(b)-N(a)\). Other endpoint conventions add or remove arrivals at \(a\) or \(b\). The key fact is that those boundary arrivals have probability zero when the endpoints are fixed in advance. We first prove that fact, including the rate-zero case.

Proof. If \(\lambda=0\), then \(N(t)-N(0)\) has a Poisson distribution with mean \(0\), and hence equals \(0\) almost surely for each \(t\). In particular, no fixed time has an arrival. Now let \(\lambda>0\). At \(t=0\), \(N(0)=0\), so the claim holds. Fix \(t>0\), and choose any \(h\) with \(0<h<t\). If there is an arrival exactly at \(t\), there is at least one arrival in \((t-h,t]\). Therefore \[ 0\le \mathbb{P}(N(\{t\})>0) \le \mathbb{P}\bigl(N(t)-N(t-h)>0\bigr) =1-e^{-\lambda h}. \] The final equality follows from the Poisson distribution of the increment, whose mean is \(\lambda h\). Taking \(h\downarrow0\), the right-hand side tends to \(0\). The probability on the left does not depend on \(h\), so it must be \(0\). \(\square\)

Endpoint Invariance for Fixed Intervals

Proof. First suppose \(a<b\). Consider the event that there is no arrival at either endpoint, that is, \(N(\{a\})=N(\{b\})=0\). By the proposition, each endpoint has probability zero of containing an arrival. The probability that either endpoint contains an arrival is at most the sum of those two probabilities, so with probability one neither endpoint contains an arrival. On this event, including or excluding either endpoint cannot change the count. Thus all four interval counts agree almost surely. The count on \((a,b]\) is \(N(b)-N(a)\), which is Poisson with mean \(\lambda(b-a)\); the other counts have the same distribution because they agree with it almost surely. If \(a=b\), the intervals \((a,a)\), \((a,a]\), and \([a,a)\) are empty, so their counts are zero. The closed interval \([a,a]\) is \(\{a\}\), whose count is zero almost surely by the proposition. This also covers \(a=b=0\), since \(N(0)=0\). \(\square\)

The theorem asserts almost-sure equality, not equality on every sample path. It also relies on the endpoints being deterministic. These distinctions matter when writing count formulas: \(N(b)-N(a)\) gives the count on \((a,b]\) exactly, while its equality with a count using a different convention is generally only an almost-sure statement.

What the Equality Does—and Does Not—Say

When two counts agree almost surely, their distributions are the same. In particular, for every integer \(k\ge0\), the probability that either count equals \(k\) is identical. Expectations and other quantities determined by the count distribution also agree whenever they are defined. This is why formulas for Poisson counts can use the convenient convention \((a,b]\) without changing the probabilities for the other fixed-endpoint versions.

The theorem does not say that changing endpoints never changes a count. It says that for fixed endpoints the change occurs only on an event of probability zero. Nor does it automatically apply when an endpoint is random. For example, \(S_1\) is the first arrival time. The interval \([S_1,S_1]\) contains the first arrival, whereas \((S_1,S_1]\) is empty. These counts differ with probability one. The endpoint \(S_1\) is selected from the process itself; it is not a fixed deterministic time. The fixed-time proposition cannot be applied to it.

Common mistake. It is incorrect to replace the almost-sure identity for different endpoint conventions with a pathwise identity. For example, the exact pathwise formula for \(a>0\) is \[ N([a,b])=N(b)-N(a)+N(\{a\}). \] A mistaken calculation would drop the last term and claim \(N([a,b])=N(b)-N(a)\) for every path. On a path with an arrival at \(a=2\) and one more arrival in \((2,5]\), the closed-interval count is \(2\), but \(N(5)-N(2)=1\). The correction is to retain the endpoint term for a pathwise identity, then use \(\mathbb{P}(N(\{2\})>0)=0\) to conclude that the two counts agree almost surely under the Poisson model. A probability-zero exception is negligible for count probabilities, but it is not erased from the pathwise formula.

For a finite collection of deterministic endpoints, the same reasoning applies: the probability that any of them contains an arrival is at most the sum of their individual zero probabilities, and is therefore zero. This observation is useful when describing a fixed finite collection of interval counts. It does not assert independence between those counts; endpoint invariance concerns only the effect of including or excluding boundary times.