Waiting-Time Probability Calculations

From Arrival Counts to Waiting-Time Questions

A Poisson process can describe either how many events have occurred by a given time or how long to wait for an event. These are two views of the same process. For example, “no arrivals during the next two hours” concerns a count, while “the third arrival occurs within two hours” concerns an arrival time. The first question uses exponential survival; the second uses a Gamma distribution, also called an Erlang distribution when its shape parameter is an integer.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a Poisson process with state space \(\{0,1,2,\ldots\}\), initial value \(N_0=0\), and rate \(\lambda\). For \(\lambda>0\), its successive interarrival times \(\tau_1,\tau_2,\ldots\) are independent exponential random variables with rate \(\lambda\), as established in the earlier material on Poisson interarrivals. The \(k\)th arrival time is \(S_k=\tau_1+\cdots+\tau_k\), equivalently \(S_k=\inf\{t\ge0:N_t\ge k\}\). We use the convention that an arrival by deadline \(t\) means \(S_k\le t\), or, equivalently, \(N_t\ge k\).

The rate has units of arrivals per unit time, so the product \(\lambda t\) is dimensionless. When using an exponential formula, check that the rate and time use compatible units. For a zero-rate process, \(N_t=0\) for every \(t\ge0\) almost surely: there are no arrivals, and \(S_k=\infty\) for every \(k\ge1\). The formulas involving exponential gaps and Gamma densities below assume \(\lambda>0\).

Event-Free Probabilities

For a positive-rate process, the probability of no arrivals during an interval of length \(t\) is the probability that its first interarrival time exceeds \(t\). It is also the probability that the Poisson count at time \(t\) is zero.

The strict inequality in \(S_1>t\) is the complement of the event \(S_1\le t\). For \(\lambda>0\), the arrival time has a continuous distribution, so \(\mathbb{P}(S_1=t)=0\); replacing \(>\) by \(\ge\) in this probability does not change its value. The count convention still matters when describing events: “by time \(t\)” includes an arrival at \(t\), while “after time \(t\)” does not.

By stationary increments, the number of arrivals in an interval of length \(h\ge0\) has the same distribution as \(N_h\). Thus the probability of no arrivals in an interval of length \(h\) is \(e^{-\lambda h}\). This calculation concerns only the interval’s length; the location of the interval does not change the answer.

The Arrival Time of the \(k\)th Event

Waiting for the \(k\)th arrival means adding \(k\) successive exponential gaps. Independence of those gaps gives a density for their sum. The result is useful whenever the question asks whether several arrivals have occurred by a deadline, or asks for the distribution of a particular arrival time.

Proof. For \(k=1\), \(S_1=\tau_1\), whose density is \(\lambda e^{-\lambda t}\); this is the stated density because \(t^{0}/0!=1\). Suppose the density formula holds for some integer \(k\ge1\). Since \(\tau_{k+1}\) is independent of \(S_k\), the density of \(S_{k+1}=S_k+\tau_{k+1}\), for \(t>0\), is their convolution:

This proves the density formula by induction. To obtain the distribution function, define \(H(t)=1-e^{-\lambda t}\sum_{r=0}^{k-1}(\lambda t)^r/r!\) for \(t\ge0\). At \(t=0\), \(H(0)=1-1=0\). Differentiating the finite sum, the terms from differentiating the powers cancel the terms from differentiating the exponential except for the last power, giving \(H'(t)=\lambda^k t^{k-1}e^{-\lambda t}/(k-1)!\) for \(t>0\). This is the density just proved. The integral of that density from \(0\) to \(t\) and \(H(t)\) therefore have the same derivative and the same value at \(0\), so they are equal. Finally, by the definition of \(S_k\), the event \(S_k\le t\) is exactly the event \(N_t\ge k\). This completes the proof. \(\square\)

The finite sum in the distribution function is also the probability of fewer than \(k\) arrivals, with a complement taken. In fact, the Poisson count formula gives \(\mathbb{P}(N_t=r)=e^{-\lambda t}(\lambda t)^r/r!\). Summing over \(r=0,\ldots,k-1\) and subtracting from one gives the same expression. This provides a second way to compute arrival-by-deadline probabilities.

Choosing the Right Event

Many errors in waiting-time problems come from using the probability of exactly \(k\) arrivals when the question asks for the \(k\)th arrival by a deadline. The distinction is captured by the event statements below.

Common mistake. Suppose the question asks for the probability that the third arrival occurs by time \(t\). A tempting but incorrect calculation is to use the probability of exactly three arrivals, \(e^{-\lambda t}(\lambda t)^3/3!\). That event excludes outcomes with four or more arrivals, even though the third arrival has already occurred in all of them. The correct event is \(N_t\ge3\), so its probability is \(1-\mathbb{P}(N_t=0)-\mathbb{P}(N_t=1)-\mathbb{P}(N_t=2)\). Substituting the Poisson count probabilities gives \(1-e^{-\lambda t}\{1+\lambda t+(\lambda t)^2/2!\}\), exactly the \(S_3\) deadline formula.

For a zero-rate process, the event “no arrivals by \(t\)” has probability \(1\), and “at least \(k\) arrivals by \(t\)” has probability \(0\) for every \(k\ge1\). Do not substitute \(\lambda=0\) into a Gamma density as if a positive-rate arrival-time distribution still existed: there are no finite arrival times in that case.