Why Infinitesimal Conditions Exclude Multiple Arrivals

Why a Small-Interval Condition Matters

For a counting process, a short interval may contain zero arrivals, one arrival, or several arrivals. The condition $\mathbb{P}\{N(h)\ge2\}=o(h)$ says that the last possibility is negligible compared with an effect of order $h$: its probability divided by $h$ tends to zero as $h\downarrow0$. This is exactly the scale distinction needed when a first-order calculation keeps terms proportional to $h$ and discards smaller terms.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $(N(t))_{t\ge0}$ be a continuous-time counting process with state space $\{0,1,2,\ldots\}$. A counting process is nondecreasing, integer-valued, and has $N(0)=0$. For an interval $(s,s+h]$, its count is $N(s+h)-N(s)$. We use intervals of this form so that adjacent partition cells do not overlap.

Definition: A function $r(h)$ is $o(h)$ as $h\downarrow0$ if $r(h)/h\to0$. A local multiple-arrival condition is $\mathbb{P}\{N(h)\ge2\}=r(h)=o(h)$. If increments are stationary, the same probability $r(h)$ applies to every interval of length $h$. More generally, the partition argument below requires the uniform condition $\sup_{s\ge0}\mathbb{P}\{N(s+h)-N(s)\ge2\}=o(h)$.

The distinction between a condition at the origin and a uniform condition matters. A statement about $N(h)$ alone controls later intervals if increments are stationary, or if a suitable uniform bound is separately assumed. Without either, the probability of multiple arrivals could be small near time zero but large on some later interval.

Negligible in a First-Order Calculation

Suppose $g$ is a bounded function on the nonnegative integers, with $|g(k)|\le M$ for every $k$. Replacing a count by the indicator of whether the interval is occupied leaves the values at counts zero and one unchanged. Only counts of at least two can cause a difference. The following result makes that observation precise.

Proposition: If $\mathbb{P}\{N(h)\ge2\}=o(h)$, then for every bounded $g$, $$ \left|\mathbb{E}\big[g(N(h))\big]-\mathbb{E}\big[g(\mathbf{1}_{\{N(h)\ge1\}})\big]\right| \le 2M\,\mathbb{P}\{N(h)\ge2\}=o(h). $$

Proof. On the event $\{N(h)=0\}$, both arguments of $g$ equal zero. On $\{N(h)=1\}$, both equal one. On $\{N(h)\ge2\}$, the absolute difference of the two $g$-values is at most $2M$, because each has absolute value at most $M$. Therefore the absolute difference of expectations is at most the expectation of the absolute difference, which is at most $2M\mathbb{P}\{N(h)\ge2\}$. Dividing this bound by $h$ gives a quantity tending to zero, as required. $\square$

Thus, for bounded calculations, multiple arrivals can be omitted without changing the coefficient of the first-order term in $h$. This statement is about probabilities or bounded functions of the count. It does not, by itself, say that the expected number of arrivals occurring in multiple-arrival intervals is $o(h)$: a rare event can carry a very large count.

Worked Example: The Poisson Short-Interval Bound

Let $(N(t))_{t\ge0}$ be a Poisson process of rate $\lambda=3$, so $N(h)$ has a Poisson distribution with mean $3h$. For $h=\tfrac1{10}$, compute the exact probability of at least two arrivals and give a simpler upper bound.

Here the parameter of the Poisson count is $3h=\tfrac3{10}$. Its probabilities of zero and one are $e^{-3/10}$ and $\tfrac3{10}e^{-3/10}$, respectively. Hence

\mathbb{P}\{N(1/10)\ge2\} =1-e^{-3/10}-\tfrac3{10}e^{-3/10} =1-\tfrac{13}{10}e^{-3/10}.

For an upper bound, any integer $k\ge2$ satisfies $\mathbf{1}_{\{k\ge2\}}\le k(k-1)/2$. For a Poisson random variable with parameter $a$, direct summation gives $\mathbb{E}[N(N-1)]=a^2$: writing $p_k=e^{-a}a^k/k!$, the sum is $e^{-a}\sum_{k\ge2}a^k/(k-2)!=a^2$. Substituting $a=\tfrac3{10}$ therefore gives

\mathbb{P}\{N(1/10)\ge2\} \le \frac{\mathbb{E}[N(1/10)(N(1/10)-1)]}{2} =\frac{(3/10)^2}{2} =\frac9{200}.

Check: the parameter used in the bound is $3\cdot\tfrac1{10}=\tfrac3{10}$, and $(\tfrac3{10})^2/2=\tfrac9{200}$. More generally, the same computation gives $\mathbb{P}\{N(h)\ge2\}\le \lambda^2h^2/2$, which is $o(h)$ for fixed finite $\lambda$.

Accumulating the Bound over a Partition

A local probability of order $o(h)$ must be multiplied by the number of cells when considering an entire time interval. This does not usually make the resulting bound $o(h)$; rather, it makes it tend to zero as the mesh is refined.

Theorem: Fix $t>0$. Suppose that increments are stationary and $\mathbb{P}\{N(h)\ge2\}=r(h)=o(h)$. Divide $[0,t]$ into $m$ equal intervals of length $h=t/m$. Then the probability that at least one cell contains two or more arrivals is at most $m r(t/m)$, and this bound tends to zero as $m\to\infty$.

Proof. For $k=1,\ldots,m$, let $A_k=\{N(kt/m)-N((k-1)t/m)\ge2\}$. The event that some cell contains two or more arrivals is $\bigcup_{k=1}^m A_k$. By the union bound, its probability is at most $\sum_{k=1}^m\mathbb{P}(A_k)$. Stationary increments give $\mathbb{P}(A_k)=r(t/m)$ for every $k$, so the sum is $m r(t/m)$. Set $h=t/m$. Then $m r(t/m)=t\,r(h)/h$, which tends to zero by the definition of $r(h)=o(h)$. No independence between the cells is needed for this bound. $\square$

Equivalently, for sufficiently fine equal partitions, with probability approaching one every cell contains at most one arrival. If the uniform condition from the definition holds instead of stationary increments, the same proof works with the supremum over starting times in place of $r(h)$. The partition estimate controls the chance of any problematic cell; it does not claim that the probability for the whole interval is of order $o(h)$.

Worked Example: A Partition Bound for a Poisson Process

Let $(N(t))_{t\ge0}$ be a Poisson process of rate $2$, and divide the time interval $[0,1]$ into $m=8$ equal cells. Bound the probability that some cell contains at least two arrivals.

Each cell has length $h=\tfrac18$, so its Poisson parameter is $2h=\tfrac14$. In one cell the exact multiple-arrival probability is $1-e^{-1/4}(1+\tfrac14)=1-\tfrac54e^{-1/4}$. The factorial-moment bound gives a simpler upper bound:

\mathbb{P}\{\text{some cell has at least two arrivals}\} \le 8\,\frac{(1/4)^2}{2} =8\cdot\frac1{32} =\frac14.

The exact union-bound expression, which is also valid without independence, is $8(1-\tfrac54e^{-1/4})$. The simpler bound $\tfrac14$ follows from the displayed factorial-moment estimate applied separately to each cell, then summed. For general $m$, the cell length is $1/m$, the parameter is $2/m$, and the same calculation gives

\mathbb{P}\{\text{some cell has at least two arrivals}\} \le m\,\frac{(2/m)^2}{2} =\frac2m.

Check: at $m=8$, the general bound is $2/8=\tfrac14$, matching the direct bound. As $m\to\infty$, $2/m\to0$, as required by the partition theorem.

What the Probability Condition Does Not Control

The local condition bounds the chance of a multiple-arrival event, but it does not bound the size of the count on that event. The next example gives a concrete reason not to turn a probability estimate into an expectation estimate without additional assumptions.

Worked Example: A Rare Event with a Large Count

For each integer $n\ge2$, define a nonnegative integer-valued random variable $Y_n$ by $\mathbb{P}(Y_n=2)=1/n^2$, $\mathbb{P}(Y_n=n^2)=1/n^2$, and $\mathbb{P}(Y_n=0)=1-2/n^2$. These probabilities are nonnegative and sum to $1$, including at $n=2$. Regard $h=1/n$ as the interval length and compare the multiple-count probability with the expected excess above one.

Both positive values of $Y_n$ are at least two, so

\mathbb{P}(Y_n\ge2)=\frac1{n^2}+\frac1{n^2}=\frac2{n^2}, \qquad \frac{\mathbb{P}(Y_n\ge2)}{h} =\frac{2/n^2}{1/n} =\frac2n\longrightarrow0.

Thus the multiple-count probability is $o(h)$ along this sequence. But the expected excess count on the multiple-count event is

\mathbb{E}\big[(Y_n-1)\mathbf{1}_{\{Y_n\ge2\}}\big] =\frac{2-1}{n^2}+\frac{n^2-1}{n^2} =\frac1{n^2}+1-\frac1{n^2} =1.

Check: the excess is one when $Y_n=2$, and $n^2-1$ when $Y_n=n^2$; multiplying each by its probability gives the two terms shown. The expectation stays equal to $1$, even though the probability divided by $h$ tends to zero. Therefore, controlling expected multiple counts requires more than the probability condition alone.

Common Mistake and Practical Use

A common mistake is to infer that a local $o(h)$ probability yields an $o(h)$ bound over a fixed time interval. If $r(h)=o(h)$, the union bound over $t/h$ cells is $(t/h)r(h)=t\,r(h)/h=o(1)$, not necessarily $o(h)$. For instance, if $r(h)=h^{3/2}$, then $r(h)=o(h)$, while $(t/h)r(h)=t\sqrt{h}$, which tends to zero but is not $o(h)$. For a Poisson process the stronger estimate $r(h)\le\lambda^2h^2/2$ gives a partition bound of order $h$; that stronger conclusion comes from the stronger quadratic estimate, not merely from $o(h)$.

The useful lesson is to match the bound to the question. To neglect multiple arrivals in a bounded first-order probability calculation, $o(h)$ is enough. To show that a refined partition has no cell with multiple arrivals with high probability, apply the union bound and use stationarity or a uniform local estimate. To control expected counts contributed by rare multiple-arrival events, require an additional moment or tail bound.

Check Your Understanding

Use the local estimate and the partition bound carefully; distinguish probability statements from expectation statements.

If $r(h)=h^{3/2}$, what is the order of the union-bound expression over $t/h$ equal cells? Is it $o(h)$ or $o(1)$?
Why does the partition theorem not require independent increments?
For a Poisson process with rate $\lambda$, use $\mathbb{E}[N(h)(N(h)-1)]=(\lambda h)^2$ to bound $\mathbb{P}\{N(h)\ge2\}$.
Which condition lets a probability estimate for $N(h)$ apply equally to every cell in a partition?
What additional issue prevents $\mathbb{P}\{N(h)\ge2\}=o(h)$ from alone controlling the expected excess count?