Dirichlet Spacings Given the Total Count

From Arrival Times to Gaps

Fix a time horizon $T>0$, and let $(N_t)_{t\ge0}$ be a homogeneous Poisson process of rate $\lambda>0$, on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, with state space the nonnegative integers and continuous time. Write $S_1<S_2<\cdots$ for its arrival times. Given that exactly $n$ arrivals occur in $[0,T]$, the $n$ arrival times divide the interval into $n+1$ gaps: the time from 0 to the first arrival, the times between successive arrivals, and the time from the last arrival to $T$. We derive the joint law of all those gaps and use it to answer questions about their lengths.

For $n\ge1$, define $D_0=S_1$, $D_i=S_{i+1}-S_i$ for $1\le i\le n-1$, and $D_n=T-S_n$, conditional on $N_T=n$. These gaps are positive with probability one and sum to $T$. Their normalized versions are $X_i=D_i/T$, so $\sum_{i=0}^n X_i=1$. Conditioning is meaningful because $\mathbb{P}(N_T=n)=e^{-\lambda T}(\lambda T)^n/n!>0$. When $n=0$, there is just one gap, $D_0=T$, and its normalized value is $X_0=1$.

Definition: For $n\ge1$, a vector $(X_0,\ldots,X_n)$ has the Dirichlet distribution with parameters $(1,\ldots,1)$ if, in coordinates $x_0,\ldots,x_{n-1}$ with $x_n=1-\sum_{i=0}^{n-1}x_i$, its density is $n!$ on the simplex $x_i>0$, $\sum_{i=0}^{n-1}x_i<1$, and zero elsewhere. The case $n=0$ is the point mass at $X_0=1$.

The Conditional Joint Law

Theorem: Given $N_T=n$, for $n\ge1$, the normalized spacings $(X_0,\ldots,X_n)$ have the Dirichlet distribution with parameters $(1,\ldots,1)$. Equivalently, the density of $(D_0,\ldots,D_{n-1})$ is $n!/T^n$ on $d_i>0$, $\sum_{i=0}^{n-1}d_i<T$, with $D_n=T-\sum_{i=0}^{n-1}d_i$. The conditional law does not depend on $\lambda$.

Proof. For ordered arrival times $0<s_1<\cdots<s_n<T$, the joint density of those arrivals together with no further arrival by $T$ is $\lambda^n e^{-\lambda T}$. Indeed, the first $n$ exponential waiting times contribute density $\lambda^n e^{-\lambda s_n}$, and the probability of no arrival from $s_n$ through $T$ is $e^{-\lambda(T-s_n)}$. Dividing by $\mathbb{P}(N_T=n)=e^{-\lambda T}(\lambda T)^n/n!$ gives conditional density $n!/T^n$ for $(S_1,\ldots,S_n)$ on the ordered region.

Now use $d_0=s_1$, $d_i=s_{i+1}-s_i$ for $1\le i\le n-1$. The inverse transformation is $s_k=\sum_{i=0}^{k-1}d_i$, for $1\le k\le n$. Its Jacobian determinant is 1: its derivative matrix is triangular with diagonal entries 1. Thus the density in the first $n$ gaps remains $n!/T^n$, on $d_i>0$, $\sum_{i=0}^{n-1}d_i<T$; the final gap is fixed by the total. Scaling $d_i=Tx_i$ multiplies volume by $T^n$, so the density in the first $n$ normalized gaps is $n!$ on the unit simplex. This is the stated Dirichlet law, and the cancellation of $\lambda$ shows that the conditional law is independent of the rate. $\square$

The density is constant over the simplex, but the gaps are not independent: their sum is fixed. The following consequence gives exact probabilities for simultaneous lower bounds. The inequalities are strict; for $n\ge1$, replacing them by non-strict inequalities does not change the probabilities because the boundary has probability zero.

Spacing probability: Let $a_0,\ldots,a_n\ge0$. For $n\ge1$, $$ \mathbb{P}(X_i>a_i\text{ for all }0\le i\le n\mid N_T=n) = \begin{cases} \left(1-\sum_{i=0}^n a_i\right)^n,&\sum_{i=0}^n a_i<1,\\ 0,&\sum_{i=0}^n a_i\ge1. \end{cases} $$

Proof. If $\sum_i a_i\ge1$, the strict inequalities would force $\sum_iX_i>1$, which is impossible. If $\sum_i a_i<1$, translate each coordinate by writing $X_i=a_i+Y_i$. The residual coordinates satisfy $Y_i>0$ and sum to $c=1-\sum_i a_i$. In the first $n$ coordinates, this simplex has volume $c^n/n!$: scaling the unit simplex by $c$ multiplies its $n$-dimensional volume $1/n!$ by $c^n$. The original simplex has volume $1/n!$. Since the density is constant, the probability is the volume ratio $c^n$. $\square$

Worked Spacing Calculations

Worked Example: Two Endpoint Gaps Must Exceed Thresholds

Take a Poisson process of rate $\lambda=2$ on $[0,3]$, conditional on $N_3=2$. Find the probability that the first gap exceeds $1$ and the final gap exceeds $3/4$.

The conditioning event has probability $e^{-6}6^2/2!=18e^{-6}>0$. There are three normalized gaps. The thresholds for $X_0$ and $X_2$ are $1/3$ and $1/4$, since $1/3=1/3$ and $(3/4)/3=1/4$; the middle gap has threshold zero. Their sum is $1/3+1/4=7/12<1$. The spacing probability formula gives

\mathbb{P}(D_0>1,\ D_2>3/4\mid N_3=2) = \left(1-\frac13-\frac14\right)^2 = \left(\frac5{12}\right)^2 = \frac{25}{144}.

Check: In the two-dimensional simplex, the original area is $1/2$. Subtracting the thresholds leaves a simplex with total side scale $5/12$, hence area $(5/12)^2/2=25/288$. The ratio of areas is $(25/288)/(1/2)=25/144$, as above.

Worked Example: One Gap Exceeds Half the Interval

On $[0,2]$, take a rate-one Poisson process conditional on $N_2=3$. Find the probability that a specified gap, say $D_1$, exceeds $1$.

The conditioning event has probability $e^{-2}2^3/3!=\tfrac43e^{-2}>0$. There are four normalized gaps, and $D_1>1$ is $X_1>1/2$. Set the threshold for $X_1$ to $1/2$ and the other three thresholds to zero. Their sum is $1/2<1$, so

\mathbb{P}(D_1>1\mid N_2=3) = \left(1-\frac12\right)^3 = \frac18.

Check: The three remaining normalized gaps must sum to less than $1/2$. In the three-dimensional simplex, scaling that remaining simplex by $1/2$ multiplies its volume by $(1/2)^3=1/8$, matching the calculation. Any one of the four gaps has the same probability by the symmetry of the Dirichlet density.

Means, Variances, and Covariances

Simplex integrals yield all integer moments. For nonnegative integers $r_0,\ldots,r_n$, put $R=\sum_{i=0}^n r_i$. The integral of $\prod_i x_i^{r_i}$ over the unit simplex is $\prod_i r_i!/(n+R)!$. To obtain it, integrate one coordinate at a time and use $\int_0^1 u^a(1-u)^b\,du=a!b!/(a+b+1)!$ for nonnegative integers $a,b$; this identity follows by repeated integration by parts. Multiplying the simplex integral by its density $n!$ gives the moment formula below.

Theorem (Integer moments): For $n\ge1$ and nonnegative integers $r_0,\ldots,r_n$, $$ \mathbb{E}\left[\prod_{i=0}^n X_i^{r_i}\mid N_T=n\right] = \frac{n!\prod_{i=0}^n r_i!}{(n+R)!}. $$ Consequently, for distinct $i,j$, $$ \mathbb{E}[X_i\mid N_T=n]=\frac1{n+1},\qquad \operatorname{Var}(X_i\mid N_T=n)=\frac{n}{(n+1)^2(n+2)},\qquad \operatorname{Cov}(X_i,X_j\mid N_T=n)=-\frac1{(n+1)^2(n+2)}. $$

Proof. The constant density is $n!$, and integrating the monomial over the simplex gives $\prod_i r_i!/(n+R)!$, so the displayed integer-moment formula follows. Taking $r_i=1$ and all other exponents zero gives $\mathbb{E}[X_i\mid N_T=n]=n!/(n+1)!=1/(n+1)$. Taking $r_i=2$ gives $\mathbb{E}[X_i^2\mid N_T=n]=2n!/(n+2)!=2/((n+1)(n+2))$. Subtracting the squared mean gives variance $2/((n+1)(n+2))-1/(n+1)^2=n/((n+1)^2(n+2))$. For $i\ne j$, taking $r_i=r_j=1$ gives $\mathbb{E}[X_iX_j\mid N_T=n]=1/((n+1)(n+2))$. Subtracting $1/(n+1)^2$ gives covariance $-1/((n+1)^2(n+2))$. $\square$

Since $D_i=TX_i$, each spacing has mean $T/(n+1)$, variance $T^2n/((n+1)^2(n+2))$, and distinct spacings have covariance $-T^2/((n+1)^2(n+2))$, all conditional on $N_T=n$.

Worked Example: Moments of Three Gaps

Suppose $N_6=2$. There are three physical gaps, each with mean $6/(2+1)=2$. For a normalized gap, the first moment is $2!/(3!)=1/3$, and the second moment is $2\cdot2!/(4!)=1/6$. Thus its variance is $1/6-(1/3)^2=1/18$. Multiplying by $6^2=36$, each physical gap has variance $36/18=2$.

For distinct normalized gaps, the product moment is $2!/(4!)=1/12$, so their covariance is $1/12-(1/3)(1/3)=-1/36$. The physical covariance is $6^2(-1/36)=-1$.

Check: The three means sum to $2+2+2=6$, the fixed total interval length. Since the sum of the gaps is constant, its variance must be zero; the computed values give $3(2)+6(-1)=0$, where there are three variance terms and six ordered cross-covariance terms. This confirms the signs and scaling.

A Common Mistake: Treating the Gaps as Independent

Unconditioned Poisson interarrival times are independent exponential random variables, but the gaps on a fixed interval, conditional on the total count, are not independent: they must sum to $T$. For example, in the first worked example, the individual conditional probabilities are $\mathbb{P}(X_0>1/3\mid N_3=2)=(1-1/3)^2=4/9$ and $\mathbb{P}(X_2>1/4\mid N_3=2)=(1-1/4)^2=9/16$. Multiplying them as if the gaps were independent gives $(4/9)(9/16)=1/4$, not the correct joint probability $25/144$. The joint event must be computed using the shared simplex constraint.

For $n\ge1$, the probability that one specified normalized gap exceeds $x$ is $(1-x)^n$ for $0\le x<1$, and is zero for $x\ge1$; the strict event is empty at $x=1$. When $n=0$, there is no random division into multiple gaps: the only normalized gap equals 1 with probability one. Keeping this degenerate case separate avoids applying the positive-dimensional density formula where it does not apply.

Check Your Understanding

Use the conditional Dirichlet law and verify each calculation from the simplex constraints.

Given $N_T=1$, what is the probability that the first gap exceeds $T/4$?
Given $N_T=2$, find the probability that all three gaps exceed $T/6$.
For $n=3$, compute the conditional mean and variance of a normalized gap.
Explain why two distinct physical gaps have negative covariance when $N_T=n\ge1$.
What is the normalized spacing vector when $N_T=0$, and why is it not covered by the positive-dimensional density formula?