Poisson Counts in Disjoint Time Intervals

Why the Interval Matters

A count of events is meaningful only when its observation window is specified. For example, a service desk might receive an average of three calls per hour, but the expected number of calls in ten minutes differs from the expected number in an hour. A model for counts over time must account for both the event rate and the length of the interval.

A constant-rate Poisson process does more than describe the count in one window. It also specifies how counts behave across separate windows: counts in disjoint intervals are independent, and the distribution of each count depends on that interval's length. These properties make it possible to calculate probabilities involving several time periods without treating them as one undifferentiated total.

The Constant-Rate Poisson Process

Let \(N(t)\) denote the number of events observed from time 0 through time \(t\), where \(t\) is measured in a specified time unit. A constant-rate Poisson process has rate \(\lambda\), measured in events per time unit. Its assumptions are that the rate remains constant over the period being modeled, counts in disjoint time intervals are independent, and the distribution of an interval's count depends on its length rather than on its location in time.

Definition: A constant-rate Poisson process with rate \(\lambda>0\) has \(N(0)=0\), independent increments, and stationary increments. In particular, for any interval of length \(t\geq 0\), the number of events in that interval has a Poisson distribution with mean \(\lambda t\). Counts in any collection of pairwise disjoint intervals are mutually independent.

In this model, if \(N(I)\) is the count in an interval \(I\) of length \(t\), then \(N(I)\sim\operatorname{Poisson}(\lambda t)\), and for any nonnegative integer \(k\),

P(N(I)=k)=\frac{e^{-\lambda t}(\lambda t)^k}{k!}.

The product \(\lambda t\) is the expected count for that interval. Its units cancel: for example, a rate in calls per hour multiplied by a duration in hours gives a mean measured in calls. If a rate is stated per day but the interval is measured in hours, convert one of them before calculating.

For intervals that touch at an endpoint, it is conventional to assign the endpoint to only one interval, such as by using intervals that include their left endpoint but not their right. In a continuous-time Poisson model, the probability of an event at any particular fixed time is zero, so this convention does not affect the probabilities.

Worked Example: A Quiet Window Followed by Two Calls

Suppose a hypothetical service desk receives calls according to a constant-rate Poisson process at 3 calls per hour. Assume the rate is stable through the 30-minute period and that calls in disjoint time windows satisfy the process assumptions. What is the probability of no calls in the first 10 minutes and exactly two calls in the following 20 minutes?

The windows are disjoint, so their counts are independent. Their lengths in hours are \(10/60=1/6\) and \(20/60=1/3\). The corresponding Poisson means are

\mu_1=3\left(\frac{1}{6}\right)=0.5,\qquad \mu_2=3\left(\frac{1}{3}\right)=1.

The first window has zero calls with probability \(e^{-0.5}\). The second has exactly two calls with probability \(e^{-1}(1)^2/2!=e^{-1}/2\). Independence lets us multiply:

P(N_1=0,\ N_2=2) =\left(e^{-0.5}\right)\left(\frac{e^{-1}}{2}\right) =\frac{e^{-1.5}}{2} \approx 0.1116.

Under the stated model, the probability of no calls in the first 10 minutes and exactly two calls in the next 20 minutes is about 0.112, or 11.2%.

Independent Increments: What Is and Is Not Independent

An increment is the count added over a time interval. If \(a<b<c<d\), then the counts over \([a,b)\) and \([c,d)\) are independent because the intervals are disjoint. The same holds for adjacent intervals such as \([a,b)\) and \([b,c)\), using a consistent endpoint convention. Their means may differ if their lengths differ, but their counts are still independent.

By contrast, counts in overlapping intervals are generally not independent. For example, the count from 9:00 to 10:00 and the count from 9:30 to 10:30 both include events from 9:30 to 10:00. An event in that shared half-hour contributes to both counts, linking them. The independent-increments rule applies to disjoint intervals, not merely to intervals that have different start times.

For several disjoint windows, the model assigns a Poisson distribution to each count using that window's own length, and the joint probability of specified counts is the product of their individual probabilities. The total count across those windows is also Poisson, with mean equal to the sum of the interval means. The count in a longer interval is not a separate independent copy of the count in its component intervals: it is their sum.

Worked Example: Comparing Two Interval Lengths

Suppose a hypothetical monitoring system registers events at a constant rate of 4 per hour. Assume the rate and event-generating conditions remain stable. Compare the probabilities of exactly two events in a 15-minute window and in a 45-minute window.

For 15 minutes, \(t=15/60=0.25\) hour, so the mean is \(\mu=4(0.25)=1\). For 45 minutes, \(t=45/60=0.75\) hour, so the mean is \(\mu=4(0.75)=3\). Using the Poisson probability formula,

P(N=2\text{ in 15 min}) =\frac{e^{-1}(1)^2}{2!} =\frac{e^{-1}}{2} \approx 0.1839,

P(N=2\text{ in 45 min}) =\frac{e^{-3}(3)^2}{2!} =\frac{9e^{-3}}{2} \approx 0.2240.

The probability of exactly two events is about 18.4% in the 15-minute window and 22.4% in the 45-minute window. A longer interval does not make every particular count probability larger: the probability of exactly two first rises in this comparison, but it eventually falls as the mean becomes much larger than two.

A different question, whether there are at least two events, gives a probability that increases in this comparison. For a Poisson count with mean \(\mu\), the only outcomes below two are zero and one, so

P(N\geq 2)=1-P(N=0)-P(N=1) =1-e^{-\mu}(1+\mu).

For the 15-minute window this is \(1-e^{-1}(2)\approx0.2642\). For the 45-minute window it is \(1-e^{-3}(4)\approx0.8009\). These are probabilities for two separate hypothetical windows, not probabilities that both windows contain at least two events.

Using Interval Length to Compare Probabilities

The interval length affects the mean through \(\mu=\lambda t\). With \(\lambda\) fixed, doubling the interval length doubles the mean. This changes the entire count distribution, not just its average. A longer interval makes high counts more plausible, but the probability of any one exact count can move up or down depending on the starting mean and the chosen count.

One probability that has a simple, consistent relationship with interval length is the chance of at least one event. Since the only way to have fewer than one event is to have zero, the Poisson formula gives

P(N(I)\geq 1)=1-P(N(I)=0)=1-e^{-\lambda t}.

For a fixed positive rate, this probability increases with \(t\). To justify the comparison, if \(t_2>t_1\), then \(-\lambda t_2<-\lambda t_1\). The exponential function is increasing, so \(e^{-\lambda t_2}<e^{-\lambda t_1}\), and therefore \(1-e^{-\lambda t_2}>1-e^{-\lambda t_1}\).

This result is useful when the practical question is whether anything happens during a window. It does not imply that every exact-count probability increases with window length. The appropriate probability must be calculated for the event in the question.

Worked Example: Counts Across Two Adjacent Periods

Suppose a hypothetical sensor records events at a constant rate of 6 per day. Assume its operating conditions and rate remain stable over the five hours of interest. Find the probability of exactly one event in the first two hours and at least two events in the following three hours.

The windows are adjacent and disjoint. Convert their durations to days: \(2/24=1/12\) day and \(3/24=1/8\) day. Their Poisson means are

\mu_1=6\left(\frac{1}{12}\right)=0.5,\qquad \mu_2=6\left(\frac{1}{8}\right)=0.75.

The probability of exactly one event in the first period is

P(N_1=1)=\frac{e^{-0.5}(0.5)^1}{1!} =0.5e^{-0.5}\approx0.3033.

For the second period, use the complement of zero or one event:

P(N_2\geq2) =1-\left[e^{-0.75}+\frac{e^{-0.75}(0.75)^1}{1!}\right] =1-e^{-0.75}(1.75) \approx0.1734.

The counts are independent because the periods are disjoint, so the probability of both requested conditions is

P(N_1=1,\ N_2\geq2) =(0.3033)(0.1734) \approx0.05257.

Thus, the probability of exactly one event in the first two hours and at least two in the next three hours is about 5.26%.

A Short Justification for the Joint Calculation

The multiplication used in the examples follows directly from the definition of independent increments. For disjoint intervals \(I_1,\ldots,I_m\), let their lengths be \(t_1,\ldots,t_m\), and let their counts be \(N_1,\ldots,N_m\). Each \(N_i\) has a Poisson distribution with mean \(\lambda t_i\); mutual independence means that the probability of any specified list of counts factors:

P(N_1=k_1,\ldots,N_m=k_m) =\prod_{i=1}^{m}P(N_i=k_i) =\prod_{i=1}^{m}\frac{e^{-\lambda t_i}(\lambda t_i)^{k_i}}{k_i!}.

If a condition is a range of counts rather than one specified count, such as \(N_2\geq2\), first calculate the probability of that condition for its interval. Independence then allows that probability to be multiplied by the probability for the other interval's condition. This reasoning requires disjoint intervals and the stated constant-rate process model.

Common Mistake: Treating Every Pair of Windows as Independent

A common error is to say, “The two intervals start at different times, so their counts are independent.” Different start times do not guarantee disjointness. If the windows overlap, some events contribute to both counts, and multiplying their separate probabilities is not justified by independent increments.

The correction is to check the intervals themselves before applying independence. If they are disjoint, calculate each mean from its own duration and rate, then multiply the relevant probabilities. If they overlap, the shared portion must be accounted for; do not use the disjoint-interval multiplication rule as though the counts were independent.

A second pitfall is to use the rate as the Poisson mean without adjusting for duration. A rate of 4 events per hour gives a mean of 1 event in 15 minutes, not 4. Always calculate \(\lambda t\) with compatible time units before using the Poisson formula.

Key takeaway: In a constant-rate Poisson process, an interval of length \(t\) has count distribution \(\operatorname{Poisson}(\lambda t)\), and counts in disjoint intervals are independent. Use each interval's own length to find its mean, and check disjointness before multiplying probabilities.

Check Your Understanding

Use the constant-rate Poisson process properties to answer these questions.

A process has rate 8 events per hour. What is the Poisson mean for a 15-minute interval? What is the probability of zero events in that interval?
Two adjacent windows have lengths 10 minutes and 20 minutes. Are their counts independent under a constant-rate Poisson process? Explain what property supports your answer.
Why can counts from 9:00–10:00 and 9:30–10:30 not be treated as independent increments?
For a fixed positive rate, explain why the probability of at least one event increases when the observation interval gets longer.
Can the probability of exactly three events be assumed to increase whenever the interval gets longer? Explain why or why not.