Uniform Order Statistics Given a Fixed Poisson Count

Conditioning on the Number of Arrivals

Knowing how many arrivals occur by time \(T\) does not determine when they occur. For a homogeneous Poisson process, however, conditioning on that count gives a particularly simple distribution for their locations: they are distributed like independent uniform points on the observation interval, placed in increasing order. This fact turns questions about arrival times into calculations over a bounded region of Euclidean space.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N_t)_{t\ge0}\) be a homogeneous Poisson process of rate \(\lambda>0\), with state space \(S=\{0,1,2,\ldots\}\) and continuous time \(t\ge0\). Thus \(N_0=0\), increments over disjoint intervals are independent, and \(N_t-N_s\) has the Poisson distribution with mean \(\lambda(t-s)\) when \(0\le s<t\). For \(k\ge1\), define \(S_k=\inf\{t\ge0:N_t\ge k\}\), the time of the \(k\)th arrival. Fix \(T>0\). We condition on \(N_T=n\), an event of probability \(e^{-\lambda T}(\lambda T)^n/n!>0\), so the conditional law is well-defined.

The ordering in the density is essential: \(S_1,\ldots,S_n\) are already labeled by arrival order, so their possible values lie in the simplex \(0<s_1<\cdots<s_n<T\), not in the whole cube \((0,T)^n\). The factor \(n!\) accounts for the \(n!\) possible orderings of \(n\) distinct uniform values.

Deriving the Conditional Density

For a Poisson process, the successive interarrival times \(\tau_1=S_1\) and \(\tau_k=S_k-S_{k-1}\) for \(k\ge2\) are independent exponential random variables with rate \(\lambda\). This standard consequence of the independent-increment and no-arrival properties can also be seen from \(\mathbb{P}(\tau_1>u)=\mathbb{P}(N_u=0)=e^{-\lambda u}\); the same no-arrival calculation after an arrival gives the same waiting-time law, independently of the past. We use this interarrival representation to derive the theorem.

Proof. Assume first \(n\ge1\). The event \(N_T=n\) is the event \(S_n\le T<S_{n+1}\), up to probability-zero equality at \(T\). Write \(u_k=S_k-S_{k-1}\) with \(S_0=0\). For \(u_1,\ldots,u_n>0\) with \(u_1+\cdots+u_n<T\), the joint density of the first \(n\) interarrival times is \(\lambda^n e^{-\lambda(u_1+\cdots+u_n)}\). Independently, the probability that the next interarrival exceeds \(T-(u_1+\cdots+u_n)\) is \(e^{-\lambda(T-u_1-\cdots-u_n)}\). Multiplying gives the joint density of the first \(n\) gaps together with the event of no further arrival by \(T\): $$ \lambda^n e^{-\lambda(u_1+\cdots+u_n)} e^{-\lambda(T-u_1-\cdots-u_n)} =\lambda^n e^{-\lambda T}. $$ The transformation from gaps \((u_1,\ldots,u_n)\) to arrival times \((s_1,\ldots,s_n)\), where \(s_k=u_1+\cdots+u_k\), has Jacobian determinant \(1\): its matrix is triangular with diagonal entries \(1\). Its range is precisely \(0<s_1<\cdots<s_n<T\). Consequently, the joint density of \((S_1,\ldots,S_n)\) together with \(N_T=n\) is \(\lambda^n e^{-\lambda T}\) on that simplex and \(0\) elsewhere. The Poisson count law gives \(\mathbb{P}(N_T=n)=e^{-\lambda T}(\lambda T)^n/n!\). Dividing the joint density by this positive probability yields \(n!/T^n\) on the simplex, as claimed.

To check the equivalence with ordered uniforms, let \(U_1,\ldots,U_n\) be independent Uniform\((0,T)\) random variables. Their joint density is \(1/T^n\) on the cube \((0,T)^n\). Ties have probability zero, and the cube, except for those ties, divides into \(n!\) disjoint regions according to the ordering of the coordinates. On the region \(0<u_{(1)}<\cdots<u_{(n)}<T\), each of the \(n!\) permutations contributes density \(1/T^n\) to the ordered vector \((u_{(1)},\ldots,u_{(n)})\). Its density is therefore \(n!/T^n\), the same as the conditional density just derived. Equality of these densities proves equality of the distributions. \(\square\)

This conclusion follows directly from the formula: for fixed \(T\) and \(n\), neither \(\lambda\) nor any expression involving it appears in \(n!/T^n\). The restriction to a fixed observation interval matters. The theorem describes the locations conditional on the count within that interval; it does not assert that an unconditioned process has a fixed number of arrivals.

Using the Density to Compute Probabilities

For any event described by inequalities on the ordered arrival times, its conditional probability is the integral of \(n!/T^n\) over the part of the simplex satisfying those inequalities. The calculations below show how to set the region and how to check the resulting answer against the equivalent uniform-sample description.

Normalization and a Common Mistake

The density must integrate to one over the ordered simplex. That simplex occupies \(1/n!\) of the cube \((0,T)^n\): away from ties, each of the \(n!\) orderings occupies the same volume, and together they partition the cube. Its volume is therefore \(T^n/n!\). Multiplying by the density gives \((n!/T^n)(T^n/n!)=1\). This also confirms that the factorial in the theorem is necessary.

Common mistake: omitting the factorial. A tempting but incorrect calculation is to divide the joint density \(\lambda^n e^{-\lambda T}\) by \(\mathbb{P}(N_T=n)\) and report \(1/T^n\) on the ordered simplex. For \(n=2\) and \(T=3\), that would give density \(1/9\). The simplex has area \(3^2/2!=9/2\), so this proposed density integrates to \((1/9)(9/2)=1/2\), not \(1\). The correct division includes \(n!\): the density is \(2!/3^2=2/9\), and \((2/9)(9/2)=1\). The missing factor arises because the conditional count probability contains \((\lambda T)^n/n!\).

The theorem is useful whenever a probability concerns the locations of a fixed number of Poisson arrivals within a fixed time window. It replaces a conditional Poisson-process calculation with a uniform-sample calculation while preserving the arrival ordering. Keep the two parts of the statement distinct: the original uniform values are independent before sorting, but the ordered values are not independent of one another.