Multinomial Source Counts Given the Total

Why Fixing the Total Changes the Count Model

Suppose events arrive from several independent sources, and an observer records both the combined total and the source of each event. Before conditioning, counts from different sources are independent Poisson random variables. Once the combined total is fixed, however, the source counts must add to that total. Their conditional distribution is therefore not a collection of independent Poisson counts: it is a multinomial allocation.

We derive this allocation law directly from the independent Poisson probabilities. The derivation also shows how the source rates determine the allocation probabilities, and why a source with rate zero receives no events. Exact examples then illustrate point probabilities and probabilities defined by several possible allocations.

Definition: On a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $N_r(t)$, $1\le r\le m$, be independent Poisson processes with rates $\lambda_r\ge0$. For a fixed interval of length $h>0$, let $C_r$ be the number of events from source $r$ in that interval, and let $N=\sum_{r=1}^m C_r$ be the total count. Write $\Lambda=\sum_{r=1}^m\lambda_r$. When $\Lambda>0$, define $p_r=\lambda_r/\Lambda$. The vector $(p_1,\ldots,p_m)$ is a probability vector, since each $p_r\ge0$ and $\sum_r p_r=1$.

The length $h$ affects the distribution of the total count, but it will cancel from the conditional source allocation. For $\Lambda>0$, every event $\{N=n\}$, for integer $n\ge0$, has positive probability, so conditioning on it is well-defined. If $\Lambda=0$, all source counts are zero with probability one; only conditioning on $N=0$ is possible.

The Conditional Multinomial Law

Theorem: Suppose $\Lambda>0$. For every integer $n\ge0$ and nonnegative integers $n_1,\ldots,n_m$ with $\sum_{r=1}^m n_r=n$, $$ \mathbb{P}(C_1=n_1,\ldots,C_m=n_m\mid N=n) =\frac{n!}{n_1!\cdots n_m!}\prod_{r=1}^m p_r^{n_r}. $$ If any $p_r=0$ and $n_r>0$, this probability is zero. A factor with $p_r=n_r=0$ is interpreted as $1$. If the counts do not sum to $n$, the conditional probability is zero. When $\Lambda=0$, $N=0$ and every $C_r=0$ with probability one.

Proof. Fix $n\ge0$ and a vector of nonnegative integers $(n_1,\ldots,n_m)$ summing to $n$. Independence and the Poisson count law give $$ \mathbb{P}(C_1=n_1,\ldots,C_m=n_m) =e^{-\Lambda h}\prod_{r=1}^m\frac{(\lambda_r h)^{n_r}}{n_r!}. $$ To find the probability of the conditioning event, sum this expression over all nonnegative vectors summing to $n$. Expanding the $n$th power by the multinomial theorem gives $$ \mathbb{P}(N=n) =e^{-\Lambda h}\frac{h^n}{n!} \sum_{\sum_r n_r=n}\frac{n!}{n_1!\cdots n_m!}\prod_{r=1}^m\lambda_r^{n_r} =e^{-\Lambda h}\frac{(\Lambda h)^n}{n!}. $$ The multinomial expansion here follows by multiplying $n$ copies of $\lambda_1+\cdots+\lambda_m$: a term with $n_r$ choices of $\lambda_r$ occurs $n!/(n_1!\cdots n_m!)$ times. Since $\Lambda>0$ and $h>0$, the denominator $\mathbb{P}(N=n)$ is positive. Dividing the joint probability by it and using $\sum_r n_r=n$ yields $$ \frac{e^{-\Lambda h}\prod_r(\lambda_r h)^{n_r}/n_r!} {e^{-\Lambda h}(\Lambda h)^n/n!} =\frac{n!}{n_1!\cdots n_m!}\prod_r\left(\frac{\lambda_r}{\Lambda}\right)^{n_r}. $$ If a rate is zero, its count is zero with probability one, so allocations assigning it a positive count have probability zero; the displayed formula agrees with this convention. If $\Lambda=0$, every $\lambda_r=0$, so every source count is zero almost surely. This proves all cases. $\square$

The coefficient $n!/(n_1!\cdots n_m!)$ counts the different source-label sequences with those source totals. The product gives the probability of any one such sequence when each of the $n$ labels has probabilities $p_1,\ldots,p_m$. This is the same categorical allocation described by the independent source labels conditional on the total. In particular, the conditional law depends on the relative rates, not on the interval length.

Worked Example: One Allocation from Three Equal-Rate Sources

Three independent sources each have rate $1$ per unit time. In an interval of length $2$, find the probability that, given four total events, source 1 contributes two events, source 2 contributes one, and source 3 contributes one.

The total rate is $\Lambda=1+1+1=3$, so $p_1=p_2=p_3=1/3$. The proposed counts sum to $2+1+1=4$, as required. The conditional probability is $$ \frac{4!}{2!1!1!}\left(\frac13\right)^2 \left(\frac13\right)^1\left(\frac13\right)^1 =12\cdot\frac1{81}=\frac4{27}. $$ Here the coefficient is $4!/(2!1!1!)=24/2=12$, and the product of the four source probabilities is $1/81$.

Check by conditioning the independent Poisson counts. Each source count has mean $1\cdot2=2$. The joint probability of counts $(2,1,1)$ is $$ \left(e^{-2}\frac{2^2}{2!}\right) \left(e^{-2}\frac{2^1}{1!}\right) \left(e^{-2}\frac{2^1}{1!}\right) =e^{-6}(2)(2)(2)=8e^{-6}. $$ The total count has mean $3\cdot2=6$, so $$ \mathbb{P}(N=4)=e^{-6}\frac{6^4}{4!} =e^{-6}\frac{1296}{24}=54e^{-6}. $$ The ratio is $8e^{-6}/(54e^{-6})=4/27$, agreeing with the multinomial calculation.

Worked Example: Unequal Source Probabilities

Three independent sources have rates $1,2,3$ per unit time. In an interval of length $1$, calculate the probability that, conditional on three total events, the source counts are $(0,1,2)$.

The total rate is $1+2+3=6$, giving source probabilities $p_1=1/6$, $p_2=2/6=1/3$, and $p_3=3/6=1/2$. The counts sum to $0+1+2=3$. Thus $$ \mathbb{P}((C_1,C_2,C_3)=(0,1,2)\mid N=3) =\frac{3!}{0!1!2!}\left(\frac16\right)^0 \left(\frac13\right)^1\left(\frac12\right)^2 =3\cdot1\cdot\frac13\cdot\frac14=\frac14. $$

Check using the independent counts. Their Poisson means over this interval are $1,2,3$, and the joint probability of $(0,1,2)$ is $$ (e^{-1})(2e^{-2})\left(e^{-3}\frac{3^2}{2!}\right) =e^{-6}\cdot2\cdot\frac92=9e^{-6}. $$ The total count has mean $6$, so its probability of being $3$ is $$ e^{-6}\frac{6^3}{3!}=e^{-6}\frac{216}{6}=36e^{-6}. $$ Dividing gives $9e^{-6}/(36e^{-6})=1/4$, as before.

Conditional Means, Variances, and Dependence

The multinomial allocation also gives useful summaries without listing every possible allocation. Conditional on $N=n$, each of the $n$ source labels has probability $p_r$ of being $r$. Represent the count from source $r$ as the sum of $n$ indicators, one for each label. For distinct sources, a single label cannot belong to both, which creates negative covariance.

Theorem: Under the conditional multinomial law with $\Lambda>0$, for $N=n$, $$ \mathbb{E}[C_r\mid N=n]=np_r,\qquad \operatorname{Var}(C_r\mid N=n)=np_r(1-p_r), $$ and for $r\ne s$, $$ \operatorname{Cov}(C_r,C_s\mid N=n)=-np_rp_s. $$ These formulas include $n=0$ and zero-probability sources.

Proof. Given $N=n$, let $I_{kr}$ be $1$ if the $k$th source label is $r$, and $0$ otherwise, for $1\le k\le n$. Then $C_r=\sum_{k=1}^n I_{kr}$. Each indicator has conditional mean $p_r$ and variance $p_r(1-p_r)$, since it is Bernoulli with success probability $p_r$. The labels are independent conditional on the total, so summing their means and variances gives $\mathbb{E}[C_r\mid N=n]=np_r$ and $\operatorname{Var}(C_r\mid N=n)=np_r(1-p_r)$. For $r\ne s$, $I_{kr}I_{ks}=0$, because a single label cannot be both sources. Hence $$ \operatorname{Cov}(I_{kr},I_{ks}\mid N=n) =0-p_rp_s=-p_rp_s. $$ Indicators from different labels are conditionally independent and have covariance zero. Summing the covariances over the $n$ matching labels therefore gives $\operatorname{Cov}(C_r,C_s\mid N=n)=-np_rp_s$. If $n=0$, all sums are empty and all counts and moments are zero. If $p_r=0$, the corresponding indicators and count are zero almost surely, and the formulas still hold. $\square$

Worked Example: A Probability with Multiple Allocations

Three equal-rate sources are observed, conditional on a total of five events. Find the probability that source 1 contributes exactly two events and sources 2 and 3 each contribute at least one.

Each source has probability $1/3$ for a label. Once $C_1=2$, the remaining counts sum to three. The conditions $C_2\ge1$ and $C_3\ge1$ allow exactly two allocations: $(C_1,C_2,C_3)=(2,1,2)$ and $(2,2,1)$. Each has multinomial coefficient $$ \frac{5!}{2!1!2!}=\frac{120}{4}=30, $$ and each has probability $30(1/3)^5=30/243=10/81$. The two allocations are disjoint, so their probabilities add: $$ \mathbb{P}(C_1=2,\ C_2\ge1,\ C_3\ge1\mid N=5) =\frac{10}{81}+\frac{10}{81}=\frac{20}{81}. $$

Check by grouping sources 2 and 3. The probability that each of the five labels is source 1 is $1/3$, while the probability of a label from sources 2 or 3 is $2/3$. The probability that exactly two labels are source 1 is therefore $$ \frac{5!}{2!3!}\left(\frac13\right)^2\left(\frac23\right)^3 =10\cdot\frac19\cdot\frac8{27}=\frac{80}{243}. $$ Given that the other three labels are from sources 2 or 3, each is equally likely to be either source. The probability that all three have the same source is $2(1/2)^3=1/4$, so the probability that both sources appear is $1-1/4=3/4$. Multiplying gives $(80/243)(3/4)=240/972=20/81$, confirming the result.

As a separate moment check in this example, the conditional mean of $C_1$ is $5(1/3)=5/3$, and its conditional variance is $5(1/3)(2/3)=10/9$. For the first allocation, the probability $10/81$ is computed above; summing over all possible allocations, rather than only the constrained ones, gives the full multinomial distribution.

A Common Conditioning Mistake

A frequent error is to treat the source counts as independent after conditioning on their total. For example, with three equal-rate sources and $N=4$, one might incorrectly multiply the conditional marginal probabilities for $C_1=2$ and $C_2=1$. The marginal distribution of $C_1$ is binomial with parameters $4$ and $1/3$, so $$ \mathbb{P}(C_1=2\mid N=4) =\frac{4!}{2!2!}\left(\frac13\right)^2\left(\frac23\right)^2 =6\cdot\frac19\cdot\frac49=\frac8{27}. $$ Similarly, $$ \mathbb{P}(C_2=1\mid N=4) =4\cdot\frac13\left(\frac23\right)^3=\frac{32}{81}. $$ Multiplying these marginal probabilities would give $256/2187$, but it is not the joint probability: the counts must also satisfy $C_3=1$ when $C_1=2$ and $C_2=1$. The correct conditional joint probability is $4/27$, as calculated earlier. The negative covariance formula explains why independence is not expected: for distinct sources, $\operatorname{Cov}(C_r,C_s\mid N=n)=-np_rp_s$, which is strictly negative when $n>0$ and both source probabilities are positive.

Conditioning on the total is useful whenever the combined number of events is known or fixed by design, but the source allocation is still uncertain. The key procedure is to use the relative rates as categorical probabilities, enumerate every allocation satisfying the event of interest, and sum their multinomial probabilities. Always check that the allocations sum to the conditioned total and that the source probabilities sum to one.

Check Your Understanding

For each question, state the relevant conditioning event and show enough arithmetic to verify the result.

Four equal-rate sources are observed given a total of three events. What is the conditional probability of counts $(1,1,1,0)$?
Two sources have rates $2$ and $1$. Given a total of four events, what is the probability that the first source contributes three?
For three sources with probabilities $1/2,1/3,1/6$, calculate the conditional probability of counts $(1,2,0)$ given a total of three.
Explain why the counts from two distinct sources have nonpositive conditional covariance when the total is fixed.
If all source rates are zero, which total-count event has positive probability, and what are the source counts given that event?

Multinomial Source Counts Given the Total

What You'll Learn