Conditional Count Allocations Across Time Intervals

Fixing the Total Changes How Arrivals Are Allocated

For a Poisson process, counts in disjoint time intervals are independent before conditioning. But once the total number of arrivals over a fixed horizon is known, the counts in the separate intervals cannot vary independently: they must add up to that total. The conditional distribution describes exactly how the fixed total is allocated. Its probabilities depend on the lengths of the intervals, not on the process rate.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space carrying a homogeneous Poisson process $(N(t))_{t\geq 0}$ of rate $\lambda>0$. Thus $N(0)=0$, the process has independent increments, and for $0\leq s\leq t$, $N(t)-N(s)$ has a Poisson distribution with mean $\lambda(t-s)$. Fix $T>0$ and a partition $0=t_0<t_1<\cdots<t_m=T$. Define the interval lengths $\Delta_i=t_i-t_{i-1}>0$ and counts $X_i=N(t_i)-N(t_{i-1})$, for $i=1,\ldots,m$. Then $\sum_{i=1}^m X_i=N(T)$.

Theorem (Conditional count allocation): For any integer $n\geq 0$ and nonnegative integers $x_1,\ldots,x_m$ with $\sum_{i=1}^m x_i=n$, $$ \mathbb{P}\bigl(X_1=x_1,\ldots,X_m=x_m\mid N(T)=n\bigr) =\frac{n!}{x_1!\cdots x_m!}\prod_{i=1}^m\left(\frac{\Delta_i}{T}\right)^{x_i}. $$ If the counts do not sum to $n$, this conditional probability is zero. Thus, given $N(T)=n$, the vector of interval counts has a multinomial distribution with $n$ trials and cell probabilities $p_i=\Delta_i/T$.

The conditioning is well-defined: since $\lambda T>0$, $\mathbb{P}(N(T)=n)=e^{-\lambda T}(\lambda T)^n/n!>0$ for every $n\geq0$. The probabilities $p_i$ are positive and sum to one because the interval lengths are positive and $\sum_i\Delta_i=T$.

Proof. If $\sum_i x_i=n$, independent increments give $\mathbb{P}(X_1=x_1,\ldots,X_m=x_m)=\prod_i e^{-\lambda\Delta_i}(\lambda\Delta_i)^{x_i}/x_i!$. The event that all these counts equal $x_i$ implies $N(T)=n$. Dividing this joint probability by $\mathbb{P}(N(T)=n)=e^{-\lambda T}(\lambda T)^n/n!$, and using $\sum_i\Delta_i=T$ and $\sum_i x_i=n$, gives $$ \frac{\prod_i e^{-\lambda\Delta_i}(\lambda\Delta_i)^{x_i}/x_i!} {e^{-\lambda T}(\lambda T)^n/n!} =\frac{n!}{\prod_i x_i!}\prod_i\left(\frac{\Delta_i}{T}\right)^{x_i}. $$ If $\sum_i x_i\ne n$, the two events cannot occur together, so the conditional probability is zero. Finally, the displayed probabilities sum to one: by the multinomial expansion, their sum over all nonnegative $x_i$ summing to $n$ is $(\sum_i\Delta_i/T)^n=1^n=1$. This also covers $n=0$: the only possible vector is all zeros, and its probability is one. $\square$

The formula separates two roles. The total count $n$ determines how many arrivals must be allocated. The interval proportions determine the allocation probabilities. The rate $\lambda$ disappears because the same rate appears in both the joint probability and the probability of the conditioning event.

Calculating Conditional Allocation Probabilities

Worked Example: A Specified Allocation Across Three Intervals

Suppose the process has rate $\lambda=2$ on $[0,3]$, partitioned at $1$ and $3/2$. Given $N(3)=4$, find the probability that the three interval counts are $1,1,2$, in that order.

The interval lengths are $1$, $1/2$, and $3/2$, which sum to $3$. Their proportions are therefore $1/3$, $1/6$, and $1/2$. The requested counts sum to $4$, so the conditional allocation formula applies:

\begin{aligned} \mathbb{P}(X_1=1,X_2=1,X_3=2\mid N(3)=4) &=\frac{4!}{1!1!2!} \left(\frac13\right)\left(\frac16\right)\left(\frac12\right)^2\\ &=12\cdot\frac{1}{72} =\frac16. \end{aligned}

Check: The interval proportions sum to $1/3+1/6+1/2=1$, and the allocation probability is $\frac16$. The factor $12$ counts the distinct orders in which four arrivals can be assigned with these three cell counts: $4!/(1!1!2!)=12$.

Often the question concerns one interval rather than the full allocation. Adding the multinomial probabilities over all allocations with a specified count in one cell gives a binomial distribution. More generally, for any subset $J\subseteq\{1,\ldots,m\}$, the combined count $\sum_{i\in J}X_i$, conditional on $N(T)=n$, is binomial with $n$ trials and success probability $\sum_{i\in J}\Delta_i/T$. This follows by grouping the cells in $J$ into one category and the remaining cells into another.

Result (Conditional marginal and dependence): Write $p_i=\Delta_i/T$. Given $N(T)=n$, $$ \mathbb{E}[X_i\mid N(T)=n]=np_i,\qquad \operatorname{Var}(X_i\mid N(T)=n)=np_i(1-p_i), $$ and for $i\ne j$, $$ \operatorname{Cov}(X_i,X_j\mid N(T)=n)=-np_ip_j. $$

Proof. For fixed $n$, the conditional allocation formula gives the probability generating function $\mathbb{E}[\prod_i z_i^{X_i}\mid N(T)=n]=(\sum_i p_i z_i)^n$, by expanding the $n$th power. Differentiating once with respect to $z_i$ and setting every $z_k=1$ gives $\mathbb{E}[X_i\mid N(T)=n]=np_i$. Differentiating twice with respect to $z_i$ gives $\mathbb{E}[X_i(X_i-1)\mid N(T)=n]=n(n-1)p_i^2$. Hence $\mathbb{E}[X_i^2\mid N(T)=n]=n(n-1)p_i^2+np_i$; subtracting $n^2p_i^2$ yields $np_i(1-p_i)$. For $i\ne j$, differentiating once in $z_i$ and once in $z_j$ gives $\mathbb{E}[X_iX_j\mid N(T)=n]=n(n-1)p_ip_j$. Subtracting the product of the conditional means, $n^2p_ip_j$, gives $-np_ip_j$. These formulas also hold for $n=0$, when every count is zero. $\square$

Worked Example: A Conditional Marginal Count

Partition $[0,2]$ into intervals of lengths $1/2$ and $3/2$. Given $N(2)=5$, find the probability that the first interval contains exactly two arrivals.

The first interval's proportion is $(1/2)/2=1/4$, and the other interval's proportion is $(3/2)/2=3/4$. Summing over the possible counts in the second interval, or using the binomial marginal, gives

\begin{aligned} \mathbb{P}(X_1=2\mid N(2)=5) &=\binom52\left(\frac14\right)^2\left(\frac34\right)^3\\ &=10\cdot\frac1{16}\cdot\frac{27}{64} =\frac{270}{1024} =\frac{135}{512}. \end{aligned}

Check: If $X_1=2$, then $X_2=3$, so the two-cell allocation formula gives $\frac{5!}{2!3!}(1/4)^2(3/4)^3=\frac{135}{512}$, the same result.

Worked Example: Conditional Means and Covariance

Partition $[0,6]$ into intervals of lengths $1,2,3$. Given that $N(6)=5$, find the conditional mean of each count and the conditional covariance between the first and second counts.

The proportions are $p_1=1/6$, $p_2=2/6=1/3$, and $p_3=3/6=1/2$. They sum to $1/6+1/3+1/2=1$. The conditional means are $\mathbb{E}[X_1\mid N(6)=5]=5(1/6)=5/6$, $\mathbb{E}[X_2\mid N(6)=5]=5(1/3)=5/3$, and $\mathbb{E}[X_3\mid N(6)=5]=5(1/2)=5/2$. The first two counts have covariance

\operatorname{Cov}(X_1,X_2\mid N(6)=5) =-5\left(\frac16\right)\left(\frac13\right) =-\frac5{18}.

Check: The three conditional means sum to $5/6+5/3+5/2=(5+10+15)/6=5$, as they must because the counts sum to five. The negative covariance reflects that, with the total fixed, a larger count in one interval leaves fewer arrivals available for another.

Why Conditioning Creates Dependence

Before conditioning, the counts $X_1,\ldots,X_m$ are independent Poisson random variables. After conditioning on their sum, the negative covariances show that they are generally not independent. For example, if $m\geq2$, $n>0$, and all interval proportions are positive, then $\operatorname{Cov}(X_i,X_j\mid N(T)=n)=-np_ip_j<0$ for distinct intervals. The fixed-sum constraint is the source of this dependence.

Common mistake. It is incorrect to use the unconditional distribution of an interval count after learning the total. Consider two intervals of length $1$ in $[0,2]$, with $\lambda=1$, and condition on $N(2)=2$. The unconditional probability that the first interval has one arrival is $e^{-1}$, but the conditional probability is $\binom21(1/2)(1/2)=1/2$. Directly, the joint probability of one arrival in each interval is $e^{-2}$, while $\mathbb{P}(N(2)=2)=2e^{-2}$, so the conditional probability is $e^{-2}/(2e^{-2})=1/2$. The unconditional value $e^{-1}$ answers a different question: it does not incorporate the information that the total is exactly two.

If $T=0$ or $\lambda=0$, the total count is zero with probability one, so conditioning on $N(T)=n$ is meaningful only for $n=0$; the positive-horizon, positive-rate theorem above deliberately excludes those degenerate cases. Likewise, partitions in the theorem have strictly positive interval lengths. If a proposed partition includes a zero-length cell, that cell's count is necessarily zero and it can be removed before applying the formula.

Check Your Understanding

Use the conditional allocation formula and verify each probability or moment from its stated inputs.

For a partition of $[0,4]$ into intervals of lengths $1$ and $3$, find the conditional probability that the counts are $2$ and $1$, given $N(4)=3$.
For the same partition, find the conditional probability that the first interval contains no arrivals, given $N(4)=3$.
For three intervals with proportions $1/2,1/3,1/6$, calculate the conditional mean of the third count when the total is $6$.
Explain why two distinct interval counts have nonpositive conditional covariance when the total count is fixed.
What is the conditional count vector when $n=0$, and why does the multinomial formula assign it probability one?

Conditional Count Allocations Across Time Intervals

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