A checklist for a proposed model
A formula for the distribution of \(N(t)\) at each fixed time does not by itself settle whether \((N(t))_{t\ge0}\) is a Poisson process. A proposed model must also have the right paths and the right dependence between counts on different time intervals. The useful approach is to check these features separately rather than infer them from a single marginal distribution.
Throughout, the candidate process is defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), has time index \(t\ge0\), and takes values in the nonnegative integers. A counting process has \(N(0)=0\), nondecreasing, right-continuous paths, and finitely many arrivals on every bounded time interval. For a simple counting process, every jump has size one. A homogeneous Poisson process of rate \(\lambda>0\) is a simple counting process whose increments on disjoint intervals are independent, whose increment law depends only on interval length, and for which \(N(t)-N(s)\) has the Poisson distribution with parameter \(\lambda(t-s)\) when \(0\le s<t\).
For a candidate with a proposed rate \(\lambda\), check the following in order: the initial value; the path conditions; independence of disjoint-interval counts; stationarity of increment distributions; and, for small \(h>0\), the probabilities of zero, one, or at least two arrivals. The local requirements are \( \mathbb{P}(N(t+h)-N(t)=1)=\lambda h+o(h)\) and \(\mathbb{P}(N(t+h)-N(t)\ge2)=o(h)\) as \(h\downarrow0\). When increments are stationary, these probabilities do not depend on \(t\). The first-order term specifies the rate; the second condition rules out a first-order chance of multiple arrivals.
What the short-interval check establishes
The next result explains how the local checks fit together with independent, stationary increments. It is the infinitesimal characterization in a form useful for auditing a candidate. The path and increment assumptions matter: short-interval probabilities alone do not establish independence.
Proof. Fix \(t\ge0\), and write \(p_n(t)=\mathbb{P}(N(t)=n)\) for integers \(n\ge0\). By stationarity of increments, the distribution of \(N(t+h)-N(t)\) is independent of \(t\). Let \(q_k(h)=\mathbb{P}(N(t+h)-N(t)=k)\). The local assumptions give \(q_1(h)=\lambda h+o(h)\), \(\sum_{k\ge2}q_k(h)=o(h)\), and therefore \(q_0(h)=1-\lambda h+o(h)\).
First establish continuity of each \(p_n\), before using any differential equation. For \(h>0\), if \(N(t+h)-N(t)=0\), then \(N(t+h)=N(t)\); hence the indicators of the events \(\{N(t+h)=n\}\) and \(\{N(t)=n\}\) can differ only if the increment is positive. Consequently, \(|p_n(t+h)-p_n(t)|\le\mathbb{P}(N(t+h)-N(t)>0)=\lambda h+o(h)\). For \(t>0\) and \(0<h<t\), the same argument on \((t-h,t]\) gives \(|p_n(t)-p_n(t-h)|\le\mathbb{P}(N(t)-N(t-h)>0)=\lambda h+o(h)\). At \(t=0\), right continuity follows from the first bound, and \(p_n(0)=\mathbf{1}_{\{n=0\}}\). Thus every \(p_n\) is continuous on \([0,\infty)\).
For \(n=0\), the event \(N(t+h)=0\) requires both \(N(t)=0\) and a zero increment; independence of increments yields \(p_0(t+h)=p_0(t)q_0(h)\). For \(n\ge1\), conditioning on the increment gives \(p_n(t+h)=\sum_{k=0}^{n}p_{n-k}(t)q_k(h)\). Terms with \(k>n\) cannot contribute because the process is nondecreasing. Subtract \(p_n(t)\), divide by \(h\), and let \(h\downarrow0\). For each fixed \(n\), the terms with \(2\le k\le n\) are \(o(h)\), since \(q_k(h)\le\sum_{j\ge2}q_j(h)=o(h)\). The resulting right derivatives are \(p_0'(t)=-\lambda p_0(t)\) and \(p_n'(t)=\lambda p_{n-1}(t)-\lambda p_n(t)\) for \(n\ge1\). Here the right-hand sides are continuous, by the continuity just established. The elementary calculus theorem for a continuous function with a continuous right derivative gives the corresponding integral equation, so these equations may be solved by integration.
The initial values are \(p_0(0)=1\) and \(p_n(0)=0\) for \(n\ge1\). The first equation gives \(p_0(t)=e^{-\lambda t}\). For \(n\ge1\), multiply the equation by \(e^{\lambda t}\); the product rule gives \(\frac{d}{dt}(e^{\lambda t}p_n(t))=\lambda e^{\lambda t}p_{n-1}(t)\). Inductively, if \(p_{n-1}(t)=e^{-\lambda t}(\lambda t)^{n-1}/(n-1)!\), integrating from \(0\) to \(t\) gives \(e^{\lambda t}p_n(t)=\int_0^t \lambda(\lambda u)^{n-1}/(n-1)!\,du=(\lambda t)^n/n!\). Therefore \(p_n(t)=e^{-\lambda t}(\lambda t)^n/n!\) for every \(n\ge0\). This is a probability distribution because its sum is \(e^{-\lambda t}e^{\lambda t}=1\). Stationarity gives the same law with \(t\) replaced by \(t-s\) for every increment \(N(t)-N(s)\). The assumed independent increments then give the required Poisson process properties. \(\square\)
Proof. The increment has the Poisson distribution with parameter \(\lambda h\). Its one-arrival probability is \(e^{-\lambda h}\lambda h\), and \(e^{-\lambda h}\to1\), so this equals \(\lambda h+o(h)\). Its probability of at least two arrivals is \(1-e^{-\lambda h}(1+\lambda h)\). The exponential expansion \(e^{-\lambda h}=1-\lambda h+O(h^2)\) gives \(e^{-\lambda h}(1+\lambda h)=1+O(h^2)\), so the multiple-arrival probability is \(O(h^2)=o(h)\). \(\square\)
Worked checks on candidate models
Worked Example: A Homogeneous Poisson Benchmark
Suppose a proposed process is specified to have simple counting paths, \(N(0)=0\), independent and stationary increments, and a Poisson increment with parameter \(2h\) on every interval of length \(h\). Check the rate and the local probabilities.
The increment law specifies the rate \(\lambda=2\). For an interval of length \(h\), the probabilities of zero and one arrivals are respectively \(e^{-2h}\) and \(2he^{-2h}\). For at least two arrivals the probability is \(1-e^{-2h}(1+2h)\). The expansions \(e^{-2h}=1-2h+O(h^2)\) and \(2he^{-2h}=2h+O(h^2)\) show that the zero-arrival probability is \(1-2h+O(h^2)\), the one-arrival probability is \(2h+O(h^2)\), and the multiple-arrival probability is \(O(h^2)\). Thus the infinitesimal checks agree with rate \(2\).
Check at \(h=\tfrac12\): the increment parameter is \(2\cdot\tfrac12=1\). The zero- and one-arrival probabilities are each \(e^{-1}\), and the probability of at least two is \(1-2e^{-1}\). These sum to \(e^{-1}+e^{-1}+1-2e^{-1}=1\). Together with the stated path and increment properties, the model passes the Poisson-process checks.
Worked Example: Stationary Increments Without Independence
Let \(U\) be uniform on \((0,1)\) on a probability space, and define \(N(t)\) to be the number of points of the random lattice \(\{U+k:k=0,1,2,\ldots\}\) in \((0,t]\). Check the paths, the increment properties, and the short-interval probabilities.
There are finitely many lattice points in every bounded interval; the paths start at zero, are nondecreasing and right-continuous, and every jump has size one. For any interval \((s,s+h]\), shifting the lattice by \(s\) changes its phase to \((U-s)\) modulo \(1\), which is again uniform on \((0,1)\). Hence the increment distribution depends on \(h\) but not on \(s\): increments are stationary.
If \(0<h<1\), the interval contains one lattice point with probability \(h\), and contains at least two with probability zero. Thus the one-arrival probability is exactly \(h\), and the multiple-arrival probability is zero; the local conditions hold with \(\lambda=1\).
Independence nevertheless fails. Let \(A=N(1/2)-N(0)\) and \(B=N(1)-N(1/2)\). Except at endpoints, which have probability zero, \(A=1\) when \(U<1/2\) and \(B=1\) when \(U>1/2\). Therefore \(A+B=1\) with probability one, while \(\mathbb{P}(A=1)=\mathbb{P}(B=1)=1/2\). If they were independent, \(\mathbb{P}(A=1,B=1)\) would be \(1/4\); in fact it is zero. This is a valid simple counting process with stationary increments and the stated local probabilities, but it is not Poisson because its increments are dependent.
Worked Example: A Time Change That Breaks Stationarity
Let \(M(u)\) be a rate-one Poisson process on a probability space and set \(N(t)=M(t^2)\) for \(t\ge0\). Check which candidate requirements hold and find where the proposed homogeneous model fails.
The paths start at zero and are nondecreasing, right-continuous, and locally finite. A time change preserves independence of increments over disjoint time intervals, because the corresponding intervals on the \(u\)-scale are disjoint. But an increment over \((s,s+h]\) has Poisson parameter \((s+h)^2-s^2=2sh+h^2\), which depends on \(s\). In particular, the interval \((0,1]\) has parameter \(1^2-0^2=1\), whereas \((1,2]\) has parameter \(2^2-1^2=3\). Their zero-arrival probabilities are \(e^{-1}\) and \(e^{-3}\), respectively, so the increments are not stationary.
For a short interval starting at \(s\), the one-arrival probability is \(e^{-(2sh+h^2)}(2sh+h^2)\). At a fixed \(s>0\), dividing by \(h\) and taking \(h\downarrow0\) gives \(2s\), not a rate constant independent of the starting time. At \(s=0\), the parameter is \(h^2\), so the one-arrival probability is \(e^{-h^2}h^2\), whose ratio to \(h\) tends to zero. The model has independent increments, but it fails stationarity and the constant-rate local check.
Worked Example: A Process With Double Jumps
Let \(M(t)\) be a rate-one Poisson process and define \(N(t)=2M(t)\). This candidate starts at zero and has independent stationary increments. Does it meet the path and short-interval requirements for a simple Poisson process?
Whenever \(M\) jumps, \(N\) jumps by two, so the unit-jump path requirement fails. Over an interval of length \(h\), the increment of \(N\) is always even. Its probability of exactly one arrival is therefore zero, not \(\lambda h+o(h)\) for any \(\lambda>0\). Its probability of at least two arrivals is the probability that \(M\) has at least one arrival, namely \(1-e^{-h}=h+O(h^2)\). This is not \(o(h)\), since \((1-e^{-h})/h\to1\). Both checks detect the double jumps.
Common mistakes and a final audit
Common mistake: checking only the mean count. For the random-lattice process above, \(\mathbb{E}[N(t)]=t\): for \(t=m+r\), where \(m\) is a nonnegative integer and \(0\le r<1\), the count is \(m+1\) with probability \(r\) and \(m\) with probability \(1-r\), so its expectation is \(r(m+1)+(1-r)m=m+r=t\). A tempting but incorrect conclusion is that the process is Poisson because its mean is linear in time and its local rate is \(1\). The correction is to test independence: for the two half-interval counts, \(\mathbb{P}(A=1,B=1)=0\), not \(\mathbb{P}(A=1)\mathbb{P}(B=1)=1/4\). Mean counts and local probabilities do not replace the independent-increments check.
When evaluating a candidate, keep the checks separate. A process may have independent but nonstationary increments, as the time-changed example shows. It may have stationary increments and the right first-order local probabilities while lacking independence, as the random-lattice example shows. It may have independent stationary increments yet fail the simple-path or multiple-arrival conditions, as the double-jump example shows.
- Verify \(N(0)=0\), right continuity, monotonicity, local finiteness, and unit jumps.
- Test independence on arbitrary disjoint intervals, not only adjacent intervals of one chosen length.
- Compare increment laws at different starting times while holding the interval length fixed.
- For short intervals, verify both the coefficient of \(h\) for one arrival and the \(o(h)\) bound for multiple arrivals.
- Use the infinitesimal characterization only when independent and stationary increments and the stated path assumptions are also in place.
Check Your Understanding
For each question, state which part of the candidate-model check is being tested.
- A counting process has \(N(0)=0\), simple paths, and independent stationary increments. Its one-arrival probability on a length-\(h\) interval is \(3h+o(h)\), and its multiple-arrival probability is \(o(h)\). What rate does the infinitesimal check identify?
- Why does stationary increments not imply independent increments? Use the two half-interval counts in the random-lattice example.
- For \(N(t)=M(t^2)\), compare the Poisson parameters of the increments on \((0,1]\) and \((1,2]\).
- For \(N(t)=2M(t)\), calculate the probability of at least two arrivals in an interval of length \(h\), and explain why it is not \(o(h)\).
- What continuity fact must be established about \(p_n(t)=\mathbb{P}(N(t)=n)\) before integrating the forward equations in the infinitesimal-check proof?