Arrival Locations Given Counts in Subintervals

What Does Fixing the Counts Tell Us About the Locations?

A Poisson process specifies how many arrivals occur in a time interval, but it also has a conditional description of where those arrivals fall. If we know that a given interval contains exactly $n$ arrivals, their unordered locations behave like $n$ independent uniform samples from that interval. The actual arrival times are ordered, however, so their joint density is supported only on the region where the locations increase. With counts fixed in several disjoint intervals, the ordered locations from different intervals are independent.

We derive this result directly. In particular, the derivation will show why the density for ordered locations includes a factorial, why gaps between the specified intervals do not change the conditional laws, and how a zero count is handled.

Definition: 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 $N_0=0$. Its arrival times are $S_k=\inf\{t\ge0:N_t\ge k\}$, $k\ge1$. Consider a finite collection of pairwise disjoint intervals $(a_r,b_r]$, $r=1,\ldots,m$, with $0\le a_r<b_r<\infty$. Write $L_r=b_r-a_r$, and let $C_r=N_{b_r}-N_{a_r}$ be the count in interval $r$. Conditional on $C_r=n_r$ for all $r$, the arrival locations in interval $r$, listed in increasing order, are denoted $U_{r,1}<\cdots<U_{r,n_r}$. If $n_r=0$, this list is empty.

The conditioning event has positive probability for every choice of nonnegative integers $n_1,\ldots,n_m$: independent increments and the Poisson count formula give its probability as $\prod_{r=1}^m e^{-\lambda L_r}(\lambda L_r)^{n_r}/n_r!$, which is positive because $\lambda>0$ and $L_r>0$. Thus the conditional law below is well-defined. Endpoints have probability zero of containing an arrival, so using open or closed endpoints for a location event makes no difference.

The Conditional Density in One Interval

First consider an interval of length $L>0$, and shift its left endpoint to zero. The successive waiting times of a rate-$\lambda$ Poisson process are independent exponential random variables with density $x\mapsto\lambda e^{-\lambda x}$ for $x>0$. Consequently, for $n\ge1$, the joint density of the first $n$ arrival locations $0<u_1<\cdots<u_n$ together with the event that there are no further arrivals by time $L$ is $\lambda^n e^{-\lambda L}$. To see the last factor, the first $n$ waiting times contribute $\lambda^n e^{-\lambda u_n}$; the probability that the next waiting time exceeds $L-u_n$ is $e^{-\lambda(L-u_n)}$. Multiplying gives the stated expression.

Theorem (Conditional ordered locations): For a rate-$\lambda$ Poisson process on an interval of length $L>0$, conditional on exactly $n$ arrivals in the interval, the ordered locations have joint density $$ f(u_1,\ldots,u_n\mid N_L=n)=\frac{n!}{L^n}\mathbf{1}_{\{0<u_1<\cdots<u_n<L\}}, \qquad n\ge1. $$ Equivalently, before ordering, the $n$ locations are independent uniform samples on the interval. For $n=0$, the conditional list of locations is empty with probability one.

Proof. For $n\ge1$, the joint density of the ordered locations and the event $N_L=n$, on $0<u_1<\cdots<u_n<L$, is $\lambda^n e^{-\lambda L}$ by the waiting-time calculation above. Also, $N_L$ has the Poisson distribution, so $\mathbb{P}(N_L=n)=e^{-\lambda L}(\lambda L)^n/n!$. Dividing the joint density by this positive probability gives \[ \frac{\lambda^n e^{-\lambda L}}{e^{-\lambda L}(\lambda L)^n/n!}=\frac{n!}{L^n}. \] Outside the ordered region the density is zero because arrival times increase. For $n=0$, conditioning on no arrivals leaves no locations to specify. To verify the density is normalised for $n\ge1$, the ordered region has volume $L^n/n!$: the cube $(0,L)^n$, of volume $L^n$, is divided into $n!$ regions according to the possible strict orderings of its coordinates, and ties have volume zero. Thus the integral of the density is $(n!/L^n)(L^n/n!)=1$. $\square$

The factor $n!$ is essential: the density is for the ordered list, whose possible values occupy only one of the $n!$ order regions in the cube. In contrast, if $V_1,\ldots,V_n$ are the locations before sorting, their joint density is $1/L^n$ on the entire cube. Sorting those independent uniform samples produces the density $n!/L^n$ on the ordered region.

Several Disjoint Intervals

Theorem (Conditional locations in disjoint intervals): Fix nonnegative integers $n_1,\ldots,n_m$ and condition on $C_r=n_r$ for every $r$. The ordered location lists from distinct intervals are independent. In interval $r$, the translated locations $U_{r,1}-a_r,\ldots,U_{r,n_r}-a_r$ have density $n_r!/L_r^{n_r}$ on $0<u_1<\cdots<u_{n_r}<L_r$ when $n_r\ge1$. If $n_r=0$, that interval contributes an empty list. Equivalently, within each interval the unsorted locations are independent uniform samples over that interval, with the collections from different intervals independent.

Proof. Independent increments imply that the point patterns restricted to pairwise disjoint intervals are independent. In detail, events about a pattern that specify counts in finitely many subintervals are determined by increments of $N_t$ over disjoint intervals; independent increments factor their joint probabilities. Such count events determine the point patterns, so the restrictions to the specified intervals are independent. In particular, for any location events $B_r$ determined by the arrivals in interval $r$, independence gives \[ \mathbb{P}\left(\bigcap_{r=1}^m B_r,\ C_r=n_r\text{ for all }r\right) =\prod_{r=1}^m\mathbb{P}(B_r,\ C_r=n_r). \] The probability of the conditioning event is $\prod_{r=1}^m\mathbb{P}(C_r=n_r)>0$. Dividing by it yields \[ \mathbb{P}\left(\bigcap_{r=1}^m B_r\mid C_r=n_r\text{ for all }r\right) =\prod_{r=1}^m\mathbb{P}(B_r\mid C_r=n_r). \] This is conditional independence of the interval patterns. Applying the one-interval density result to each factor gives the asserted ordered density in interval $r$, shifted by $a_r$. If $n_r=0$, its conditional pattern is empty. $\square$

This factorisation also explains why the intervals need not fill a single time horizon. Arrivals in gaps between the specified intervals do not enter the conditional location lists. The independent increments property separates the patterns in the specified intervals from one another, whether the intervals touch or have gaps.

Worked Example: Two Locations in Each of Two Intervals

Take a Poisson process with rate $\lambda=\tfrac32$, and consider the intervals $(0,2]$ and $(3,5]$. Condition on counts $2$ and $1$, respectively. Let $U_1<U_2$ be the locations in the first interval, and let $V_1$ be the location in the second.

Both intervals have length $2$. The ordered density of $(U_1,U_2)$ is $2!/2^2=2/4=\tfrac12$ on $0<u_1<u_2<2$. Its support has area $2^2/2!=4/2=2$, so its integral is $\tfrac12\cdot2=1$. The density of $V_1$ is $1/2$ on $3<v_1<5$, whose integral is $\tfrac12(5-3)=1$. The two lists are conditionally independent.

For example, the event $U_1<1<U_2$ occupies a rectangle of area $1$ within the ordered region: $u_1$ ranges over $(0,1)$, and $u_2$ over $(1,2)$. Its conditional probability is therefore $\tfrac12\cdot1=\tfrac12$. The event $7/2<V_1<9/2$ has conditional probability $(9/2-7/2)/2=1/2$. By conditional independence, the probability that both events occur is $\tfrac12\cdot\tfrac12=\tfrac14$.

Check: The joint conditional density is $(\tfrac12)(\tfrac12)=\tfrac14$ on the product of the two location regions. The specified joint event has volume $1\cdot1=1$, so direct integration gives $\tfrac14$, agreeing with the factorised calculation.

Worked Example: A Gap and an Interval with No Arrivals

Let $\lambda=2$, and specify the intervals $(1,4]$, $(5,6]$, and $(7,9]$, with counts $2$, $0$, and $1$, respectively. Their lengths are $3$, $1$, and $2$. The probability of the conditioning event is \[ \left(e^{-6}\frac{6^2}{2!}\right)\left(e^{-2}\frac{2^0}{0!}\right)\left(e^{-4}\frac{4^1}{1!}\right) =(18e^{-6})(e^{-2})(4e^{-4})=72e^{-12}>0. \]

In the first interval, the translated ordered pair has density $2!/3^2=2/9$ on $0<u_1<u_2<3$. Both locations falling in $(1,2]$ corresponds, after translation, to $0<u_1<u_2<1$. This triangle has area $1^2/2=1/2$, so the conditional probability is $(2/9)(1/2)=1/9$. In the third interval, one location is uniform on an interval of length $2$, so the probability it falls in $(8,9]$, of length $1$, is $1/2$. The middle interval has no locations under the conditioning. Conditional independence gives probability $(1/9)(1/2)=1/18$ for the combined location event.

Check: The product of the ordered density $2/9$, the first event's area $1/2$, the one-location density $1/2$, and its event length $1$ is $(2/9)(1/2)(1/2)(1)=1/18$, as above. The empty list in the middle interval adds no factor other than probability $1$ under its specified zero count.

Worked Example: The Earliest of Three Arrivals

Suppose an interval has length $4$, and condition on exactly three arrivals. The ordered triple has density $3!/4^3=6/64=3/32$ on $0<u_1<u_2<u_3<4$. Find the conditional probability that the earliest arrival occurs by time $1$.

The total volume of the ordered region is $4^3/3!=64/6=32/3$. The event that the earliest arrival is later than $1$ is $1<u_1<u_2<u_3<4$. Translating all three coordinates by $1$ gives an ordered region of length $3$, with volume $3^3/3!=27/6=9/2$. Thus the event that the earliest arrival is at most $1$ has volume $32/3-9/2=64/6-27/6=37/6$. Multiplying by the density gives \[ \mathbb{P}(u_1\le1\mid N_4=3)=\frac{3}{32}\frac{37}{6}=\frac{37}{64}. \]

Check: Before ordering, all three locations are independent uniform samples on $(0,4)$. The probability they are all later than $1$ is $(3/4)^3=27/64$, so the probability that at least one is at most $1$ is $1-27/64=37/64$, the same answer.

Common Pitfall: Forgetting the Ordering Factor

A common error is to use density $1/L^n$ directly for the ordered locations. For two arrivals in an interval of length $2$, that would give density $1/2^2=1/4$ on $0<u_1<u_2<2$. But the ordered region has area $2^2/2!=2$, so this proposed density integrates to $(1/4)(2)=1/2$, not $1$. It cannot be a probability density. The correct density includes the factor $2!$: $2!/2^2=1/2$, and $(1/2)(2)=1$. The density $1/L^n$ belongs to the labelled, unsorted sample on the full cube; sorting restricts the locations to one order region and multiplies its density by $n!$.

The conclusion is precise: conditioning on counts fixes how many locations occur in each interval but does not favour any point within an interval over another. Within each interval the unsorted locations are uniform samples; sorting them gives ordered uniform locations. The independent-increments property ensures that conditioning on counts in other disjoint intervals does not alter this local conditional law.

Check Your Understanding

For each question, distinguish the ordered list from the unsorted locations and state the interval length used.

For three arrivals in an interval of length $2$, what is the joint density of the ordered locations, and what is the volume of its support?
Two disjoint intervals have lengths $3$ and $5$, with counts $1$ and $0$. Describe the conditional location law in each interval and their conditional dependence.
In an interval of length $6$, conditional on two arrivals, calculate the probability that both lie in a specified subinterval of length $2$.
Why does the density $1/L^n$ fail to normalise when used only on the ordered region?
Explain why a gap between two specified intervals does not change the conditional location distribution within either interval.

Arrival Locations Given Counts in Subintervals

What You'll Learn