The Infinitesimal Definition

Describing Events on a Short Time Scale

A counting process records how many events have occurred by each time. Its behavior over a very short interval can be summarized by three possibilities: no event, exactly one event, or two or more events. For a homogeneous Poisson process, the probability of exactly one event is approximately the rate times the interval length, while the probability of multiple events is negligible compared with that length.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N_t)_{t\ge0}$ be a counting process with state space $\mathbb{N}_0=\{0,1,2,\ldots\}$, continuous time $t\ge0$, and $N_0=0$. Its paths are nondecreasing, right-continuous, integer-valued, and finite on bounded time intervals. The natural filtration is $\mathcal{F}_t^N=\sigma(N_s:0\le s\le t)$. For $s\ge0$ and $h>0$, the count in the interval $(s,s+h]$ is $N_{s+h}-N_s$.

Definition: A homogeneous Poisson process with rate $\lambda>0$ has independent increments and stationary increments, and its short-interval probabilities satisfy, as $h\downarrow0$, $$ \mathbb{P}(N_{s+h}-N_s=1)=\lambda h+o(h),\qquad \mathbb{P}(N_{s+h}-N_s\ge2)=o(h). $$ Here $o(h)$ denotes a quantity $r(h)$ such that $r(h)/h\to0$. Independent increments means that counts on disjoint time intervals are independent; stationary increments means that the distribution of an interval count depends only on the interval length. Thus the displayed probabilities do not depend on $s$.

The definition separates two kinds of information. The short-interval probabilities specify the local event rate and rule out multiple events to first order. Independent and stationary increments specify how counts across different intervals fit together. The short-interval conditions are asymptotic: they do not say that a finite-interval probability is exactly $\lambda h$, or that multiple events are impossible.

Consequences of the Short-Interval Conditions

Proposition: If the short-interval conditions hold, then $$ \mathbb{P}(N_{s+h}-N_s=0)=1-\lambda h+o(h) \quad\text{and}\quad \mathbb{P}(N_{s+h}-N_s\ge1)=\lambda h+o(h). $$

Proof. Write $Y=N_{s+h}-N_s$. Since $Y$ takes values in $\mathbb{N}_0$, the events $\{Y=0\}$, $\{Y=1\}$, and $\{Y\ge2\}$ are disjoint and have union $\Omega$. Therefore $\mathbb{P}(Y=0)=1-\mathbb{P}(Y=1)-\mathbb{P}(Y\ge2)$. Substituting the two defining conditions gives $1-(\lambda h+o(h))-o(h)=1-\lambda h+o(h)$, since a sum of two quantities that are $o(h)$ is also $o(h)$. Also, $\{Y\ge1\}=\{Y=1\}\cup\{Y\ge2\}$, so its probability is $(\lambda h+o(h))+o(h)=\lambda h+o(h)$. $\square$

The second expression gives a useful interpretation of the rate. For each $s$, the probability of at least one event in an interval of length $h$, divided by $h$, tends to $\lambda$. This is a local probability statement; it is not the expected count in a finite interval, nor does it determine how counts in separate intervals depend on one another.

Proposition (Uniqueness of the infinitesimal rate): A process cannot satisfy the single-event condition with two different rates.

Proof. Suppose the same process satisfies $\mathbb{P}(N_{s+h}-N_s=1)=\lambda h+o(h)$ and $\mathbb{P}(N_{s+h}-N_s=1)=\gamma h+o(h)$. Divide each equality by $h>0$ and let $h\downarrow0$. The first gives a limit of $\lambda$, and the second gives a limit of $\gamma$, for the same function of $h$. A limit, when it exists, is unique, so $\lambda=\gamma$. $\square$

Worked Short-Interval Calculations

Worked Example: Checking a Poisson Count at Rate Four

For a Poisson process with rate $\lambda=4$, verify the infinitesimal conditions for an interval of length $h$. A count over that interval has a Poisson distribution with parameter $4h$, so the probabilities of zero and one event are $e^{-4h}$ and $4h e^{-4h}$.

Using the expansion $e^{-4h}=1-4h+8h^2+O(h^3)$, the probability of no events is $1-4h+8h^2+O(h^3)=1-4h+o(h)$. For exactly one event, multiply that expansion by $4h$: $4h e^{-4h}=4h-16h^2+O(h^3)=4h+o(h)$.

For two or more events, subtract the zero- and one-event probabilities from one: \[ \mathbb{P}(N_{s+h}-N_s\ge2) =1-e^{-4h}(1+4h). \] Multiplying $(1-4h+8h^2+O(h^3))(1+4h)$ gives $1-8h^2+O(h^3)$. Hence the multiple-event probability is $8h^2+O(h^3)=o(h)$. Check: dividing the three leading expressions by $h$, the single-event probability tends to $4$, the multiple-event probability tends to $0$, and the no-event probability tends to $0$ after subtracting $1$ and dividing by $h$, with slope $-4$.

Worked Example: The Local Conditions Do Not Ensure Independent Increments

Consider a process that first selects a random rate $Z$, with $\mathbb{P}(Z=1)=\mathbb{P}(Z=3)=\tfrac12$, and then, conditional on $Z$, runs a Poisson process with rate $Z$. The unconditional process has stationary increments, but we check whether its local probabilities have the rate term $\lambda=2$.

Conditional on $Z=z$, the probability of exactly one event in an interval of length $h$ is $zh e^{-zh}$. Averaging over the two possible rates gives \[ \mathbb{P}(N_{s+h}-N_s=1) =\tfrac12 h e^{-h}+\tfrac12(3h e^{-3h}). \] Using $e^{-h}=1-h+O(h^2)$ and $e^{-3h}=1-3h+O(h^2)$, this is $\tfrac12(h-h^2+O(h^3))+\tfrac12(3h-9h^2+O(h^3)) =2h-5h^2+O(h^3)=2h+o(h)$.

For conditional rate $z$, the probability of at least two events is $1-e^{-zh}(1+zh)=\tfrac12z^2h^2+O(h^3)$. Averaging the cases $z=1$ and $z=3$ gives $\tfrac12(\tfrac12h^2+O(h^3))+\tfrac12(\tfrac92h^2+O(h^3)) =\tfrac52h^2+O(h^3)=o(h)$. Thus the local conditions hold with rate $2$.

Nevertheless, increments on $(0,1]$ and $(1,2]$ are not independent. Let $A$ and $B$ be the events that the count is zero on those respective intervals. Conditional on $Z$, each has probability $e^{-Z}$, and the two are conditionally independent. Therefore $\mathbb{P}(A\cap B)=\tfrac12(e^{-2}+e^{-6})$, while $\mathbb{P}(A)\mathbb{P}(B)=((e^{-1}+e^{-3})/2)^2$. The difference is $(e^{-1}-e^{-3})^2/4>0$, so the two events are dependent. This process satisfies the local conditions but is not a homogeneous Poisson process, because independent increments are also required.

Worked Example: Testing a Candidate Local Probability Table

For sufficiently small $h>0$, suppose a proposed interval-count distribution assigns probabilities \[ p_0=1-2h-h^2,\qquad p_1=2h,\qquad p_2=h^2, \] and zero probability to counts above two. Check its local conditions. These probabilities are nonnegative when $0<h\le\sqrt{2}-1$, and their sum is $(1-2h-h^2)+2h+h^2=1$.

The single-event probability is $p_1=2h$, so it has rate term $2h$ exactly. The probability of multiple events is $p_2=h^2$; its ratio to $h$ is $h$, which tends to zero. Also $p_0=1-2h-h^2=1-2h+o(h)$, since $(-h^2)/h=-h\to0$. Thus the table satisfies the local probability conditions with $\lambda=2$. This check concerns the distribution in one short interval only: by itself, it does not establish that these distributions can be assembled into a process with independent stationary increments.

Common Mistake: Treating the Error as Zero

A frequent error is to replace “the probability of two or more events is $o(h)$” with “the probability of two or more events is zero.” The distinction matters. In the rate-four example, that probability is $8h^2+O(h^3)$: it is generally positive for $h>0$, but divided by $h$ it tends to zero. The condition makes multiple events negligible at first order; it does not forbid them on a short interval.

Another check is to test the ratio directly. If a proposed model has $\mathbb{P}(N_{s+h}-N_s\ge2)=h$, then the ratio of this probability to $h$ is $1$ for every $h>0$, not a quantity tending to zero. It therefore fails the multiple-event condition, even if its probability of exactly one event is $\lambda h+o(h)$. In contrast, a multiple-event probability $h^2$ has ratio $h\to0$, as in the candidate table above.

The infinitesimal definition is useful because it isolates what happens at the smallest time scale: a single event occurs with first-order probability $\lambda h$, while multiple events contribute no first-order term. To identify a homogeneous Poisson process, however, local behavior must be combined with independent and stationary increments. A short-interval calculation is a necessary test, not a substitute for those structural assumptions.

Check Your Understanding

Use the definition and verify each limit by dividing the relevant probability by $h$.

If the single-event probability is $5h+o(h)$ and the multiple-event probability is $3h^2$, what are the rate and the first-order no-event probability?
For a proposed multiple-event probability $4h$, does the negligible-multiple-events condition hold? Show the ratio that decides.
Why does a mixture of Poisson processes with a randomly selected rate fail to be a homogeneous Poisson process, even when its short-interval rate term is the mean of the selected rate?
In the rate-four calculation, what is the coefficient of $h^2$ in the probability of exactly one event?
What additional increment properties, beyond the short-interval probabilities, are part of the homogeneous Poisson process definition?

What You'll Learn