Exponential Interarrival Times

From Arrivals to Gaps

A counting process records how many events have occurred by each time, while an interarrival time records how long the process waits between consecutive events. For a homogeneous Poisson process, these gaps are exponentially distributed: a larger rate means shorter typical waits, and the probability of a gap falling in a chosen time range can be computed directly from the exponential distribution.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a homogeneous Poisson process with state space \(S=\{0,1,2,\ldots\}\), time index \(t\ge0\), and rate \(\lambda>0\). Thus \(N_0=0\), its increments over disjoint time intervals are independent, and \(N_{t+h}-N_t\) has a Poisson distribution with mean \(\lambda h\) for \(t,h\ge0\). In particular, \(\mathbb{P}(N_h=0)=e^{-\lambda h}\). The rate \(\lambda\) is measured in events per unit time.

For a random variable \(\tau\sim\operatorname{Exp}(\lambda)\), the distribution function is \(F(t)=\mathbb{P}(\tau\le t)=1-e^{-\lambda t}\) for \(t\ge0\); for \(t<0\), \(F(t)=0\). Its density is \(f(t)=\lambda e^{-\lambda t}\) for \(t\ge0\), and \(f(t)=0\) for \(t<0\). The survival probability \(\mathbb{P}(\tau>t)=e^{-\lambda t}\) is often the most direct route to a waiting-time probability.

These formulas distinguish the rate from the mean. The rate \(\lambda\) has units of inverse time, while \(\tau\) has units of time. For example, if time is measured in hours and \(\lambda=3\) events per hour, then the exponent for a wait of \(t\) hours is \(-3t\), not \(-3\) unless \(t=1\).

Calculating Probabilities for Waiting-Time Ranges

For \(0\le a<b\), subtract the probability of waiting no more than \(a\) from the probability of waiting no more than \(b\). Because the exponential distribution has a density, the probability of an exact endpoint is zero, so strict or non-strict endpoint choices give the same answer.

The same formula can be read as a difference of survival probabilities: a wait longer than \(a\), but not longer than \(b\). It is important to use the same time units for \(a\), \(b\), and \(1/\lambda\). If \(a=0\), the formula reduces to \(1-e^{-\lambda b}\); if \(b\) increases without bound, it tends to \(e^{-\lambda a}\), the probability of waiting more than \(a\).

Mean and Variance of a Gap

Proof. For any nonnegative random variable \(Y\), the identities \(Y=\int_0^\infty \mathbf{1}_{\{t<Y\}}\,dt\) and \(Y^2=\int_0^\infty 2t\mathbf{1}_{\{t<Y\}}\,dt\) hold pointwise. Taking expectations and interchanging expectation and integration for nonnegative functions gives \(\mathbb{E}[Y]=\int_0^\infty\mathbb{P}(Y>t)\,dt\) and \(\mathbb{E}[Y^2]=\int_0^\infty2t\mathbb{P}(Y>t)\,dt\). For \(\tau\), the survival probability is \(e^{-\lambda t}\). Therefore \(\mathbb{E}[\tau]=\int_0^\infty e^{-\lambda t}\,dt=1/\lambda\). Also \(\mathbb{E}[\tau^2]=\int_0^\infty2te^{-\lambda t}\,dt=2/\lambda^2\); the last integral follows by integration by parts, with the boundary term \(te^{-\lambda t}\) tending to zero at infinity. Hence \(\operatorname{Var}(\tau)=\mathbb{E}[\tau^2]-(\mathbb{E}[\tau])^2=2/\lambda^2-1/\lambda^2=1/\lambda^2\). \(\square\)

The mean \(1/\lambda\) gives a typical time between events, not a fixed or guaranteed waiting time. The variance also depends on the rate: increasing \(\lambda\) decreases both the mean and the variance. These formulas give a useful dimensional check: \(1/\lambda\) has units of time, and \(1/\lambda^2\) has units of time squared.

Several Successive Gaps

The interarrival characterization also specifies independence: \(\tau_1,\tau_2,\ldots\) are independent and identically distributed as \(\operatorname{Exp}(\lambda)\) for a homogeneous Poisson process. Independence matters when a question concerns more than one gap. For two specified ranges, the probability that the first gap lies in one range and the second lies in another is the product of the two individual probabilities. This multiplication requires independent gaps; matching marginal exponential distributions alone would not be enough.

Common Pitfalls and Interpretation

A common error is to mix time units when calculating \(\lambda t\). For instance, at rate \(4\) per hour, writing the probability of no arrival in 15 minutes as \(e^{-4(15)}\) treats 15 as hours. The correction is to convert 15 minutes to \(\tfrac14\) hour first: the probability of waiting longer than 15 minutes is \(e^{-4(\tfrac14)}=e^{-1}\). The exponent must be dimensionless.

Another distinction is between the time to the first arrival and a generic interarrival time. The first arrival time is \(\tau_1=S_1\); later gaps are \(\tau_k=S_k-S_{k-1}\). Under the homogeneous Poisson assumptions, every one has the same exponential distribution and the gaps are independent. That conclusion need not hold for a process with a changing rate or dependent increments. In such models, the exponential formulas above cannot be applied automatically to every gap.

The exponential model is useful because it turns waiting-time questions into simple survival and interval calculations. For a single gap, check the rate and time units, decide whether the event is a lower-tail, upper-tail, or range event, and then substitute into \(1-e^{-\lambda t}\) or \(e^{-\lambda t}\). For multiple gaps, first check whether independence is part of the model before multiplying probabilities.