Counting Processes and Their Sample Paths

What a Counting Process Records

A counting process records how many events have occurred by each time. Examples include the number of arrivals at a service desk by time $t$, the number of failures observed by time $t$, or the number of births in a population up to time $t$. The basic object is not a particular probability distribution: it is a process whose possible sample paths behave like cumulative counts. In this tutorial, we establish the path conditions and the precise connection between jumps and event counts on intervals.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. A process $(N_t)_{t\ge0}$ has time index $t\ge0$ and state space $\{0,1,2,\ldots\}$. The conditions below are required to hold for almost every $\omega\in\Omega$; when discussing a single path, write $N_t$ for $N_t(\omega)$. Right-continuity means $N_t=\lim_{u\downarrow t}N_u$, and a left limit at $t>0$ means that $\lim_{u\uparrow t}N_u$ exists.

Definition: A counting process is a process $(N_t)_{t\ge0}$ with finite, nonnegative integer-valued states, $N_0=0$, nondecreasing sample paths, and right-continuous sample paths. A path is nondecreasing if $N_s\le N_t$ whenever $0\le s\le t$. Its left limits at positive times exist automatically, because each path is nondecreasing and finite on every bounded time interval.

The convention $N_0=0$ says that no events have been counted at time zero. It is part of the usual definition, not an automatic consequence of the other path conditions. If a record begins with $c>0$ events already present, one can instead study a cumulative process initialized at $c$; it is useful to distinguish that process from a counting process initialized at zero.

For $0\le s\le t$, the number of events in the interval $(s,t]$ is $N_t-N_s$. The left endpoint is excluded and the right endpoint is included. This convention agrees with right-continuous paths: an event occurring at time $u$ changes the count at $u$, so it is included in $N_t-N_s$ exactly when $s<u\le t$. In particular, the count on $(s,s]$ is $N_s-N_s=0$.

Jumps and Event Times

For $t>0$, define the jump size by $\Delta N_t=N_t-N_{t-}$, where $N_{t-}=\lim_{u\uparrow t}N_u$. Monotonicity gives $\Delta N_t\ge0$. Since both the path values and its left limit are integers, every positive jump has integer size at least one. A jump of size two records two events at the same time if events are counted with multiplicity. A counting process whose jumps are all either zero or one is called a simple counting process; simplicity is an additional condition, not part of the definition above.

For each positive integer $k$, define the $k$th event time by $T_k=\inf\{t\ge0:N_t\ge k\}$, taking the infimum of the empty set to be $\infty$. If $N_t$ jumps from $2$ to $5$ at one time, then $T_3=T_4=T_5$: three events are counted at that time. Thus the sequence of event times need not be strictly increasing. If every jump has size one, finite event times are strictly increasing.

Theorem (Path representation and interval counts): For any counting-process path and any finite $t\ge0$, there are finitely many positive jump times in $(0,t]$, and $$ N_t=\sum_{0<u\le t}\Delta N_u. $$ More generally, for $0\le s\le t$, $$ N_t-N_s=\sum_{s<u\le t}\Delta N_u. $$ The sums include every positive jump time in the indicated interval. Equivalently, $N_t-N_s$ is the number, counted with multiplicity, of event times $T_k$ satisfying $s<T_k\le t$.

Proof. Fix a path and a finite $t$. For each positive integer $k$, let $T_k=\inf\{u\ge0:N_u\ge k\}$. If $T_k<\infty$, then $N_{T_k}\ge k$: by the definition of infimum there are times $v_m$ with $T_k\le v_m<T_k+1/m$ and $N_{v_m}\ge k$; right-continuity at $T_k$ implies $N_{T_k}\ge k$. Also $N_u<k$ for every $u<T_k$. Since $N_0=0$, $T_k>0$ whenever it is finite. Thus each attained level $k$ is reached at a positive jump time. Conversely, at a positive jump time $u$, the levels strictly above $N_{u-}$ and at most $N_u$ are reached at $u$. There are exactly $N_u-N_{u-}=\Delta N_u$ such levels. Therefore the number of levels $k$ with $T_k\le t$ is $N_t$, and grouping those levels by their event times gives $N_t=\sum_{0<u\le t}\Delta N_u$. The sum is finite because it counts exactly $N_t$ levels. Subtracting the corresponding identity at $s$ gives the interval formula. Finally, $T_k\le t$ exactly for the levels $1\le k\le N_t$; hence the levels with $s<T_k\le t$ are exactly those counted by $N_t-N_s$. $\square$

This representation explains two complementary ways to describe a path: its values $N_t$, or the times and sizes of all its jumps. It also shows why counting events at the right endpoint matters. A jump at $u$ contributes to $N_t-N_s$ if and only if $u$ belongs to $(s,t]$.

Worked Paths

Worked Example: Two Events at Distinct Times

On a probability space consisting of one outcome, define $N_t=\mathbf{1}_{\{t\ge1\}}+\mathbf{1}_{\{t\ge3\}}$ for $t\ge0$. Check the definition and find the count in $(1,3]$.

For $0\le t<1$, $N_t=0$; for $1\le t<3$, $N_t=1$; and for $t\ge3$, $N_t=2$. These values are finite nonnegative integers, $N_0=0$, and they never decrease as $t$ increases. Each indicator is right-continuous, so the path is right-continuous. The positive jumps occur at times $1$ and $3$, each with size $1$. Thus this is a simple counting process.

At the endpoints, $N_1=1$ and $N_3=2$. Therefore the count in $(1,3]$ is $N_3-N_1=2-1=1$. The event at time $1$ is excluded and the event at time $3$ is included. The jump sum gives the same result: the only positive jump time in $(1,3]$ is $3$, and its jump size is $1$.

Check: On $(0,3]$, both jumps are included, so $N_3-N_0=2-0=2$, equal to $1+1=2$.

Worked Example: A Simultaneous Batch of Events

Define $N_t=2\mathbf{1}_{\{t\ge1\}}+\mathbf{1}_{\{t\ge2\}}$. Determine whether it is a counting process, list its jumps and event times, and find the count in $(1,2]$.

The path equals $0$ for $0\le t<1$, $2$ for $1\le t<2$, and $3$ for $t\ge2$. It starts at zero, has finite nonnegative integer values, is nondecreasing, and is right-continuous. It is therefore a counting process, but not a simple one: its jump at time $1$ has size $2$, and its jump at time $2$ has size $1$.

The event times, with multiplicity, are $T_1=1$, $T_2=1$, and $T_3=2$. Since $N_1=2$ and $N_2=3$, the interval count is $N_2-N_1=3-2=1$. This agrees with the event-time count: exactly one of $T_1,T_2,T_3$, namely $T_3$, satisfies $1<T_k\le2$.

Check: On $(0,2]$, the count is $N_2-N_0=3-0=3$, and the jump sizes sum to $2+1=3$.

Worked Example: A Path That Decreases

Consider $N_t=\mathbf{1}_{\{1\le t<2\}}$. It is integer-valued and $N_0=0$, but it is not a counting process: $N_1=1$ and $N_2=0$, so $N_1\le N_2$ fails. The path counts one event at time $1$, then loses that count at time $2$, which is incompatible with a cumulative event count. The interval difference $N_2-N_1=0-1=-1$ is negative, confirming that it cannot represent the number of events in $(1,2]$.

Check: The decrease occurs because the indicator is $1$ at $t=1$, but $0$ at $t=2$; the formula requires a cumulative count to retain past events.

Worked Example: A Monotone Path That Is Not Right-Continuous

Let $N_t=\mathbf{1}_{\{t>1\}}$. This path is nonnegative, integer-valued, nondecreasing, and has $N_0=0$, but it is not right-continuous at $t=1$. In fact, $N_1=0$, whereas $N_u=1$ for every $u>1$, so $\lim_{u\downarrow1}N_u=1\ne N_1$. Under the stated convention, it is not a counting process.

The strict inequality in the indicator places the increase immediately after time $1$, rather than at time $1$. Replacing it with $N_t=\mathbf{1}_{\{t\ge1\}}$ makes the path right-continuous and counts the event at time $1$. Its interval convention is then consistent: the event belongs to $(s,t]$ exactly when $s<1\le t$.

Initial Values, Finiteness, and Common Pitfalls

The zero initial value and the right-continuity convention settle boundary questions that otherwise cause ambiguity. If $N_0=0$, no event at time zero is included in the usual count. If a model explicitly includes events at time zero, its initial value must record them or the convention must be stated separately. At positive times, right-continuity assigns a jump to its time of occurrence; consequently, the count over $(s,t]$ uses $N_t-N_s$, not $N_t-N_{s-}$.

There cannot be infinitely many positive jumps on a bounded interval for a counting-process path. The path representation proves this directly: by time $t$, the sum of positive integer jump sizes is $N_t$, which is finite. In particular, the number of distinct positive jump times in $(0,t]$ is at most $N_t$. This finite-jump property is a consequence of the path conditions, not an extra assumption. It does not imply any particular probabilistic law for the times or sizes of the jumps.

Common pitfall: summing only some of the jumps. Take $N_t=\mathbf{1}_{\{t\ge1\}}+\mathbf{1}_{\{t\ge2\}}$. On $(0,2]$, the total count is $N_2-N_0=2-0=2$. If one lists only the jump at $1$, the sum of listed jump sizes is $1$, which is not the interval count. The jump-sum identity requires the list to contain all positive jump times in the interval. Here those times are $1$ and $2$, and the correct sum is $\Delta N_1+\Delta N_2=1+1=2$. More generally, whenever a jump sum is used, specify the interval and include every positive jump time in it.

Another common error is to require every counting process to have unit jumps. Unit jumps describe simple counting processes, but a valid counting process can count simultaneous events through a larger jump. Conversely, integer-valuedness alone is insufficient: a path that decreases violates cumulative counting, and a monotone path that fails the chosen right-continuity convention also fails the definition used here.

Check Your Understanding

For each question, use the stated endpoint convention $(s,t]$ and check path conditions before interpreting differences as event counts.

For $N_t=\mathbf{1}_{\{t\ge2\}}+\mathbf{1}_{\{t\ge4\}}$, list the positive jump times and compute the number of events in $(2,4]$.
Does $N_t=3\mathbf{1}_{\{t\ge1\}}$ define a counting process? Is it simple? State its jump size.
For a counting process with $N_5=7$ and $N_2=3$, how many events occurred in $(2,5]$? Explain why the answer cannot be negative.
Why does a path with $N_0=1$ fail the definition given here, even if it is nondecreasing, integer-valued, and right-continuous?
For a jump at time $u$, which of the intervals $(s,t]$ include it: $s=u<t$, $s<u=t$, or $s<t<u$?