Event Probabilities from Ordered Uniform Arrival Times

From Arrival-Time Constraints to Probabilities

When a Poisson process has a fixed number of arrivals by a deadline, questions about the arrival times become geometric: they ask what fraction of a region of ordered points satisfies a particular constraint. This perspective handles questions such as whether the third arrival occurs before a specified time, or whether every pair of consecutive arrivals is separated by at least a given interval.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) supporting a Poisson process \((N_t)_{t\ge0}\) with state space \(S=\{0,1,2,\ldots\}\) and rate \(\lambda>0\). Fix a horizon \(T>0\). Write \(S_i\) for the time of the \(i\)th arrival. We condition on \(N_T=n\), where \(n\ge1\); this event has positive probability \(e^{-\lambda T}(\lambda T)^n/n!>0\). Given \(N_T=n\), the ordered arrival times have the same distribution as the order statistics of \(n\) independent uniform random variables on \([0,T]\).

In particular, the joint density of \((S_1,\ldots,S_n)\), conditional on \(N_T=n\), is \(n!/T^n\) on the ordered region \(0<s_1<\cdots<s_n<T\), and zero elsewhere. Indeed, \(n\) independent uniform points have density \(1/T^n\) on the cube \([0,T]^n\); except on the probability-zero set of ties, the cube is divided into \(n!\) regions corresponding to the possible orderings. This density lets us calculate event probabilities by counting volume.

Proof. Conditional on \(N_T=n\), let \(C_a\) be the number of arrivals in \([0,a]\). The uniform-point representation makes \(C_a\) a binomial random variable with \(n\) trials and success probability \(a/T\): for each point, the probability of lying in \([0,a]\) is \(a/T\), independently of the other points. The \(k\)th ordered arrival satisfies \(S_k\le a\) exactly when \(C_a\ge k\). Summing the binomial probabilities for \(r=k,\ldots,n\) gives the formula. At \(a=0\) or \(a=T\), it also gives the correct probabilities, respectively \(0\) and \(1\). \(\square\)

Minimum Gaps and a Shifted Ordered Region

A constraint on gaps between successive arrivals is not a collection of independent conditions: the ordered times share the same finite horizon. Instead, translate the ordered points so that the minimum gap is absorbed into their locations. This reduces the constrained region to an ordinary ordered simplex on a shorter interval.

Proof. The conditional density of the ordered arrivals is \(n!/T^n\). On the gap-constrained region, set \(y_i=s_i-(i-1)\delta\) for \(1\le i\le n\). Then \(s_1\ge0\) gives \(y_1\ge0\), and \(s_{i+1}-s_i\ge\delta\) gives \(y_{i+1}\ge y_i\). The endpoint condition \(s_n\le T\) gives \(y_n\le T-(n-1)\delta=H\). Conversely, any \(0\le y_1\le\cdots\le y_n\le H\) maps back to a point satisfying the original constraints. The translation has Jacobian \(1\). If \(H>0\), the translated region has volume \(H^n/n!\): the cube \([0,H]^n\) is divided into \(n!\) equal ordered regions, apart from ties, which have zero volume. Multiplication by the density \(n!/T^n\) proves the gap probability. If \(H<0\), the endpoint and gap constraints cannot both hold. If \(H=0\) and \(n>1\), all the ordered times would have to occupy a single boundary configuration, a set of zero volume, so its probability is zero.

For the additional deadline, \(s_k\le a\) becomes \(y_k\le a-(k-1)\delta=b\). In the translated region, the \(y_i\) have the same distribution as the order statistics of \(n\) independent uniform points on \([0,H]\). Thus \(y_k\le b\) exactly when at least \(k\) of those points lie in \([0,b]\), an event with the binomial probability displayed in the theorem. Multiplying by the probability \((H/T)^n\) of entering the gap-constrained region gives the stated joint probability. If \(b<0\), no translated point can satisfy the deadline; if \(b\ge H\), every point in the translated region satisfies it. \(\square\)

Worked Calculations

Why the Conditioning Matters

These calculations are conditional on a fixed count. Their answers do not depend on the Poisson rate \(\lambda\), because, after conditioning on \(N_T=n\), the arrival locations are uniform order statistics on \([0,T]\). This does not mean that the unconditional probability of the same event is independent of \(\lambda\): to obtain an unconditional probability, one must also average over the possible values of \(N_T\), using their Poisson probabilities.

For a deadline, it is often simplest to count how many points lie before that deadline. For minimum gaps, use the shift \(y_i=s_i-(i-1)\delta\), then apply the deadline calculation on the shortened interval. These methods handle weak inequalities as written. Because the conditional joint distribution has a density, replacing a boundary equality such as \(S_k=a\) or \(S_{i+1}-S_i=\delta\) by a strict inequality does not change the probability.

Common mistake. In the three-arrival example on \([0,10]\), a tempting but incorrect calculation treats the two consecutive gaps as independent uniform random variables on \([0,10]\), giving \((8/10)^2=16/25\). The gaps are not independent uniform variables: they are constrained jointly by the ordered locations and the fixed endpoint \(10\). The correct calculation uses the translated ordered region, giving \((6/10)^3=27/125\). The discrepancy is substantial: \(16/25=80/125\), not \(27/125\).