The First Arrival Time

From Zero Arrivals to the First Arrival

For a Poisson process, the first arrival is a random time, while the number of arrivals by a fixed time is a random count. These two descriptions are linked by a simple event identity: by time \(t\), the first arrival has not occurred exactly when the process has counted zero arrivals. This connection converts the known distribution of a Poisson count into the full distribution of the first arrival time.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a Poisson process of rate \(\lambda>0\), with state space \(\mathbb{N}_0=\{0,1,2,\ldots\}\), continuous time \(t\ge0\), and \(N_0=0\). In particular, \(N_t\) has a Poisson distribution with mean \(\lambda t\), so \(\mathbb{P}(N_t=k)=e^{-\lambda t}(\lambda t)^k/k!\) for \(k\in\mathbb{N}_0\). Its sample paths count arrivals, are nondecreasing, and include an arrival at time \(t\) in \(N_t\).

The endpoint convention matters. Since \(N_t\) includes arrivals occurring at \(t\), the event that the first arrival occurs exactly at \(t\) belongs to \(\{N_t\ge1\}\), not to \(\{N_t=0\}\). Consequently, the event identity uses \(T_1>t\), rather than \(T_1\ge t\).

Proof. Fix \(t\ge0\). If \(N_t=0\), the nondecreasing count has remained zero through time \(t\), so there has been no arrival by \(t\), and the first arrival time satisfies \(T_1>t\) (including the possibility \(T_1=\infty\)). Conversely, if \(T_1>t\), there has been no arrival at or before \(t\); because \(N_t\) counts all arrivals through \(t\), this means \(N_t=0\). The events are therefore equal. Taking probabilities and using the Poisson distribution at time \(t\) gives \(\mathbb{P}(T_1>t)=\mathbb{P}(N_t=0)=e^{-\lambda t}(\lambda t)^0/0!=e^{-\lambda t}\). The complementary event is \(\{T_1\le t\}\), so its probability is \(1-e^{-\lambda t}\). \(\square\)

At \(t=0\), the formulas give \(\mathbb{P}(T_1>0)=1\) and \(\mathbb{P}(T_1\le0)=0\). For \(t<0\), \(T_1\ge0\) implies \(\mathbb{P}(T_1\le t)=0\). Taken together, these facts specify the cumulative distribution function on the whole real line.

Distribution, Density, and Quantiles

The survival function records the probability of waiting longer than a given time; the cumulative distribution function records the probability of having arrived by that time. For \(t\ge0\), the two add to one: \(e^{-\lambda t}+(1-e^{-\lambda t})=1\). Differentiating the cumulative distribution function on \(t>0\) gives the density. There is no atom at zero because \(\mathbb{P}(T_1\le0)=0\), and for \(\lambda>0\) the survival probability tends to zero as \(t\) tends to infinity.

Proof. On \(t>0\), differentiating \(F_{T_1}(t)=1-e^{-\lambda t}\) gives \(f_{T_1}(t)=\lambda e^{-\lambda t}\). There is no point mass at \(0\), since \(\mathbb{P}(T_1=0)=\mathbb{P}(T_1\le0)=0\). The density integrates to one: \(\int_0^\infty \lambda e^{-\lambda t}\,dt=[-e^{-\lambda t}]_0^\infty=0-(-1)=1\), where the limit \(e^{-\lambda t}\to0\) uses \(\lambda>0\). Thus the density accounts for the entire distribution. \(\square\)

For a probability level \(p\) with \(0<p<1\), the \(p\)-quantile \(q_p\) is the time satisfying \(F_{T_1}(q_p)=p\). Solving the distribution formula gives \(e^{-\lambda q_p}=1-p\), and hence \(q_p=-\ln(1-p)/\lambda\). The restrictions on \(p\) ensure that \(1-p\) is strictly between zero and one, so this quantile is finite and positive.

Probabilities over an interval can be found by subtracting cumulative probabilities. For \(0\le a<b\), this gives \(\mathbb{P}(a<T_1\le b)=F_{T_1}(b)-F_{T_1}(a)=e^{-\lambda a}-e^{-\lambda b}\). The strict lower endpoint and inclusive upper endpoint are deliberate: this event excludes arrival by \(a\) and includes arrival by \(b\). Since the distribution has no point masses at finite times, changing an endpoint from strict to inclusive does not change the probability, but the event identity should still be written with the correct endpoint convention.

Assumptions and a Common Pitfall

The positive-rate assumption is necessary for a proper finite first-arrival distribution. If \(\lambda=0\), then \(N_t=0\) at every finite \(t\) with probability one. Thus \(T_1=\infty\) with probability one: the expression \(1-e^{-\lambda t}\) is zero for every finite \(t\), and the first arrival has no probability density integrating to one over finite times. The formulas above are for \(\lambda>0\).

A common mistake is to use the probability of exactly one arrival as the probability that the first arrival has occurred. For example, when \(\lambda=2\) and \(t=1\), the incorrect calculation would be \(\mathbb{P}(N_1=1)=2e^{-2}\). But the event \(T_1\le1\) includes one arrival, two arrivals, and every larger count, so its probability is \(\mathbb{P}(N_1\ge1)=1-\mathbb{P}(N_1=0)=1-e^{-2}\). These are not the same: the first-arrival event is a positive count, not an exactly-one count.

The first-arrival distribution follows from a single-time zero-count probability. This does not, by itself, assert anything about the dependence between waiting times after successive arrivals. The useful conclusion here is specific: for each fixed \(t\), asking whether the first arrival has happened is equivalent to asking whether the count at \(t\) is positive. The event identity and the Poisson count formula then determine the survival function, cumulative distribution function, and density.