Poisson Marginals Do Not Ensure a Poisson Process

Why One-Time Distributions Are Not Enough

A counting process records how many events have occurred by each time. Knowing the distribution of the count at every single time may seem to describe the process completely. It does not: those distributions say nothing by themselves about how counts at different times are coupled. In particular, they do not imply that counts accumulated over disjoint time intervals are independent.

For a rate-\(\lambda\) Poisson process, the count at time \(t\) has a Poisson distribution with mean \(\lambda t\). But that is only one part of the definition. A Poisson process also has independent increments: counts over disjoint time intervals are independent. We will construct a process with rate \(1\) Poisson marginals at every time, and then calculate events that show its increments are dependent.

The example will use the probability space \(((0,1),\mathcal{B}(0,1),\mathbb{P})\), where \(\mathbb{P}\) is uniform probability, and the state space \(S=\{0,1,2,\ldots\}\). The time index is continuous, \(t\ge0\). Write \(F_n\) for the distribution function of a Gamma random variable with shape \(n\) and rate \(1\), and let \(U\) be the coordinate on \((0,1)\), so \(U\) is uniform. Define the quantile \(F_n^{-1}(u)=\inf\{x:F_n(x)\ge u\}\), and use the same \(U\) for every \(n\).

We first check that this really defines a counting process and establish its one-time distributions. A Gamma random variable with shape \(n\) and rate \(1\) has the same distribution as a sum of \(n\) independent rate-\(1\) exponential random variables. Adding one more positive exponential variable shows that the shape-\((n+1)\) Gamma distribution is stochastically larger than the shape-\(n\) distribution. Consequently \(F_{n+1}(x)\le F_n(x)\) for every \(x\), and their quantiles satisfy \(F_{n+1}^{-1}(u)\ge F_n^{-1}(u)\). Therefore \(T_{n+1}\ge T_n\).

For any fixed finite \(t\), the probabilities \(F_n(t)=\mathbb{P}(\operatorname{Pois}(t)\ge n)\) tend to zero as \(n\) tends to infinity. For each \(U\in(0,1)\), eventually \(F_n(t)<U\), so \(T_n>t\). Hence there are only finitely many \(T_n\) in every bounded time interval. Also \(T_n>0\), so \(N_0=0\). The paths are nondecreasing and integer-valued, as required.

Proof of the marginal claim. For each \(n\ge1\), the definition of the quantile gives \(\mathbb{P}(T_n\le t)=F_n(t)\). Since the arrival times are ordered, \(N_t\ge n\) if and only if \(T_n\le t\). Thus \(\mathbb{P}(N_t\ge n)=F_n(t)=\mathbb{P}(\operatorname{Pois}(t)\ge n)\). These equalities of tail probabilities for all \(n\) determine the distribution on the nonnegative integers: for \(k\ge0\), the probability of exactly \(k\) is the tail probability at \(k\) minus the tail probability at \(k+1\), with the tail probability at \(0\) equal to \(1\). Therefore \(N_t\sim\operatorname{Pois}(t)\). \(\square\)

Two Disjoint Increments Are Dependent

Let \(I_1=N_1\) be the count in \([0,1]\), and \(I_2=N_2-N_1\) the count in \((1,2]\). We will show that \(I_1\) and \(I_2\) are not independent. The key is to express arrival-time events as intervals for the shared \(U\).

For \(t\ge0\), set \(a_n(t)=F_n(t)=\mathbb{P}(\operatorname{Pois}(t)\ge n)\). The Gamma distribution functions are continuous and strictly increasing for positive arguments, so \(T_n\in(1,2]\) is equivalent, up to endpoints of probability zero, to \(a_n(1)<U\le a_n(2)\). For the first three values of \(n\), direct Poisson-tail calculations give

The first interval \((a_1(1),a_1(2)]\) is the event that \(T_1\in(1,2]\). The next two intervals, for \(n=2\) and \(n=3\), overlap: \(a_3(2)\ge a_2(1)\) is equivalent to \(e\ge5/2\). Also \(a_2(2)<a_1(1)\) is equivalent to \(e<3\), so this joined pair of intervals is disjoint from the interval for \(n=1\). For completeness, \(5/2<e<3\): the series \(e=\sum_{k\ge0}1/k!\) gives \(e>1+1+\tfrac12+\tfrac16>5/2\), while \(k!\ge2^{k-1}\) for \(k\ge2\), with strict inequality for some \(k\), gives \(e<1+1+\sum_{k\ge2}2^{-(k-1)}=3\).

Proof. The marginal assertion was proved by matching the tail probabilities \(\mathbb{P}(N_t\ge n)\) to those of \(\operatorname{Pois}(t)\). The calculation above proves that \(I_1=N_1\) and \(I_2=N_2-N_1\) are dependent. Independent increments are part of the definition of a Poisson process. Therefore the constructed process is not a Poisson process, despite having Poisson marginals at every fixed time. \(\square\)

What the Example Does—and Does Not—Show

The construction preserves every distribution \(N_t\), but it does not preserve the joint laws of counts at different times. A one-time marginal describes the random variable \(N_t\) for one fixed \(t\); it does not determine how \(N_s\) and \(N_t\) are related when \(s<t\). Independent increments are a separate requirement. Here the arrival times are ordered, so the paths are valid counting paths, yet the shared uniform variable links different parts of the path.

Common mistake. A tempting but invalid calculation is to see \(N_1\sim\operatorname{Pois}(1)\) and \(N_2\sim\operatorname{Pois}(2)\), then conclude that \(N_2-N_1\sim\operatorname{Pois}(1)\) and is independent of \(N_1\). The distribution of a difference cannot be found by subtracting the marginal distributions, and independence does not follow from either marginal. In this example, the explicit calculation gives \(\mathbb{P}(I_2\ge1)>1-e^{-1}\), while \(\mathbb{P}(I_2\ge1\mid I_1=0)=1-e^{-1}\). The unequal probabilities directly expose the error.

When checking whether a process is Poisson, verify the increment conditions rather than only checking the distribution at selected times. In particular, Poisson one-time marginals are necessary for a Poisson process, but they are not sufficient.