Why Counts and Waiting Times Belong Together
A model of event counts can answer a waiting-time question, and a model of waiting times can answer a count question. For example, the probability that a fourth call arrives within an hour is the probability that at least four calls occur during that hour. Making this connection explicit is useful when a problem describes events in one way but asks about the other.
The connection requires a specific model: events occur according to a homogeneous Poisson process. Its rate is constant over time, and counts in disjoint time intervals are independent. If time is measured in units such as hours, the rate must be expressed per hour.
Write \(W_k\) for the time from time 0 until the \(k\)th event, where \(k\) is a positive integer. For instance, \(W_1\) is the waiting time to the first event and \(W_3\) is the waiting time to the third. The essential event is the equivalence between reaching an arrival threshold and counting events by a time threshold.
The first event time \(W_1\) has an exponential distribution with rate \(\lambda\): \(P(W_1>t)=e^{-\lambda t}\). More generally, the successive interarrival times in this model are independent exponential random variables with rate \(\lambda\). Thus \(W_k\) is the sum of the first \(k\) interarrival times. For exam calculations, the count equivalence above is often the most direct route.
Why the Count–Waiting-Time Formula Holds
The theorem follows from what the two events mean. By time \(t\), the \(k\)th event has occurred exactly when the count by time \(t\) is at least \(k\). Taking probabilities gives \(P(W_k\leq t)=P(N(t)\geq k)\). The complement is that fewer than \(k\) events have occurred, so the count is one of \(0,1,\ldots,k-1\).
This also justifies the first-arrival result: \(W_1>t\) means there are no events by time \(t\), so \(P(W_1>t)=P(N(t)=0)=e^{-\lambda t}\). The exponential distribution is therefore not a separate assumption when it is derived from this process; it follows from the Poisson count model.
The count and waiting-time events use a non-strict boundary in the theorem. In a Poisson process with continuous interarrival times, the probability that an event occurs at exactly a specified time is zero, so using \(W_k<t\) instead of \(W_k\leq t\) gives the same probability. Still, translating the event carefully prevents errors about whether the count should be “at least” or “at most.”
Worked Examples
Worked Example: A Third Event Within a Fixed Time
A hypothetical monitoring system records events according to a homogeneous Poisson process at a rate of \(0.5\) events per hour. What is the probability that the third event occurs within 4 hours?
Method. Convert the waiting-time event to a count event. The third event occurs within 4 hours exactly when there are at least three events in 4 hours. The common mistake to avoid is using the probability of exactly three events rather than at least three.
The Poisson mean over 4 hours is $$ \lambda t=(0.5)(4)=2. $$
Therefore, $$ \begin{aligned} P(W_3\leq 4) &=P(N(4)\geq 3)\\ &=1-P(N(4)=0)-P(N(4)=1)-P(N(4)=2)\\ &=1-e^{-2}\left(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}\right)\\ &=1-e^{-2}(1+2+2)\\ &=1-5e^{-2}\\ &=1-5(0.1353352832\ldots)\\ &=0.323323584\ldots. \end{aligned} $$
Rounded to four decimal places, the probability is 0.3233. As a check, this equals the upper tail \(P(N(4)\geq3)\), not the single probability \(P(N(4)=3)\).
Worked Example: No Events First, Then a Waiting-Time Threshold
A hypothetical repair desk receives requests according to a homogeneous Poisson process with rate \(0.6\) requests per day. Find the probability that there are no requests in the first 2 days and that the second request after day 2 arrives by day 5.
Method. Split the time into the disjoint intervals from day 0 to day 2 and from day 2 to day 5. The first condition is a zero count in the first interval. The second condition requires at least two requests in the next 3 days. Independent increments allow the probabilities to be multiplied. Do not treat the second condition as requiring exactly two requests.
The mean count in the first interval is $$ (0.6)(2)=1.2, $$ so the probability of no requests there is \(e^{-1.2}\). The mean count in the next 3 days is $$ (0.6)(3)=1.8. $$ The probability of at least two requests in that interval is $$ \begin{aligned} P(N(3)\geq2) &=1-P(N(3)=0)-P(N(3)=1)\\ &=1-e^{-1.8}(1+1.8)\\ &=1-2.8e^{-1.8}\\ &=1-2.8(0.1652988882\ldots)\\ &=0.537163116\ldots. \end{aligned} $$
By independence of the counts in the two intervals, $$ \begin{aligned} P(\text{no requests in first 2 days and at least 2 in next 3}) &=e^{-1.2}\left(1-2.8e^{-1.8}\right)\\ &=(0.3011942119\ldots)(0.537163116\ldots)\\ &=0.1618\text{ (to four decimal places).} \end{aligned} $$
The answer is 0.1618. Here “the second request after day 2 arrives by day 5” means at least two requests occur between days 2 and 5; later requests in that interval do not invalidate the condition.
Worked Example: Interpreting a Mean Interarrival Time
A hypothetical sensor records events according to a homogeneous Poisson process. The mean time between events is 12 minutes. Find the probability that the fourth event occurs within 30 minutes.
Method. Convert the mean interarrival time to a rate, then use the Poisson count over 30 minutes. The trap is to use 12 as the rate; 12 minutes is a mean waiting time, so the rate is its reciprocal.
The rate is $$ \lambda=\frac{1}{12}\text{ event per minute}, $$ and the mean count in 30 minutes is $$ \lambda t=\frac{1}{12}(30)=2.5. $$ The fourth event occurs by 30 minutes exactly when \(N(30)\geq4\). Hence $$ \begin{aligned} P(W_4\leq30) &=1-P(N(30)\leq3)\\ &=1-e^{-2.5}\left(1+2.5+\frac{2.5^2}{2!}+\frac{2.5^3}{3!}\right)\\ &=1-e^{-2.5}(1+2.5+3.125+2.6041666667\ldots)\\ &=1-e^{-2.5}(9.2291666667\ldots)\\ &=1-(0.0820849986\ldots)(9.2291666667\ldots)\\ &=0.242423866\ldots. \end{aligned} $$
Rounded to four decimal places, the probability is 0.2424. The reciprocal conversion is essential: the mean waiting time is 12 minutes, while the event rate is \(1/12\) per minute.
Rates, Intervals, and Conditional Waiting Questions
The product \(\lambda t\) must be dimensionless: it is an expected number of events. If a rate is given per day and a time is given in hours, convert one of them before calculating. For example, a rate of \(3\) events per day is \(3/24=0.125\) events per hour, not \(3\) events per hour.
A homogeneous Poisson process also makes future counts independent of past counts. Thus, after observing no events in an initial interval, the probability of no events in a later interval of length \(s\) is still \(e^{-\lambda s}\). It is not necessary to restart the model or change the rate. For a condition involving a specified count in the future interval, use that interval’s length to form its own Poisson mean.
Check that the model is a homogeneous Poisson process and express \(\lambda\) and all times in consistent units.
For a kth-arrival threshold, use \(W_k\leq t\) if at least \(k\) events must occur by time \(t\), or \(W_k>t\) if fewer than \(k\) events occur.
For a single interval, use a Poisson mean \(\lambda t\). For disjoint intervals, use their individual means and the independence of their counts.
Confirm that “by time \(t\)” became an upper tail of the count, and that the rate and time produce the intended expected count.
Common Exam Traps
A particularly common error is to confuse the event “the kth arrival occurs by \(t\)” with “exactly \(k\) arrivals occur by \(t\).” The correct translation is $$ P(W_k\leq t)=P(N(t)\geq k), $$ because the process may have more than \(k\) arrivals by time \(t\). For the complement, $$ P(W_k>t)=P(N(t)\leq k-1). $$
Another error is reversing the inequality. If the question asks whether the third event has not occurred by time \(t\), the count is at most two, not at least three. Finally, a stated mean interarrival time is not the rate: a mean of \(m\) time units corresponds to rate \(1/m\) per time unit under this model.
These translations are useful because they let you choose the simpler calculation. A waiting-time threshold can be evaluated using a finite Poisson sum, while an event-count question can sometimes be phrased as a waiting-time threshold. The model assumptions matter: if the event rate changes over time or counts in disjoint intervals are dependent, the homogeneous Poisson formulas used here need not apply. The methods in this tutorial are aligned with the SOA Exam P (Probability) syllabus.
Check Your Understanding
For each problem, use a homogeneous Poisson process and keep the time units consistent.
- A hypothetical system has an event rate of \(0.5\) per hour. What is the probability that the second event occurs within 3 hours? (Answer: 0.4422)
- A hypothetical alarm has a mean interarrival time of 4 days. What is the probability of no alarms during a 6-day period? (Answer: 0.2231)
- A hypothetical counter records events at a rate of 3 per day. What is the probability that the third event takes more than 1 day to occur? (Answer: 0.4232)
- A hypothetical process has rate \(0.4\) per hour. Given that no event occurs in the first 2 hours, what is the probability of at least one event in the next 1.5 hours? (Answer: 0.4512)
- A hypothetical process has rate \(0.2\) per hour. What is the probability of exactly one event in the first 2 hours and at least two further events by hour 5? (Answer: 0.0327)