Poisson Counts Given Their Sum

Why Condition on the Total?

Suppose several independent sources generate counts, and we know only their combined count. The total tells us how many events occurred, but not how they were divided among the sources. Conditioning on that total gives a probability model for the allocation. For independent Poisson counts, this allocation has a particularly useful form: it is multinomial, with category probabilities proportional to the Poisson rates.

This result connects two familiar ways to describe counts. The independent Poisson model specifies a rate for each source; once the total number of events is fixed, the relevant parameters are instead the sources’ shares of the combined rate. We will derive that relationship and use it to calculate specific conditional probabilities.

Setup and Conditional Law

Let $X_1,\ldots,X_k$ be independent random variables, where $k$ is a positive integer and $X_i\sim\operatorname{Poisson}(\lambda_i)$ with $\lambda_i\geq 0$. Define the total rate $\Lambda=\sum_{i=1}^k\lambda_i$ and total count $S=\sum_{i=1}^kX_i$. First assume $\Lambda>0$, and set $p_i=\lambda_i/\Lambda$. Thus each $p_i\geq0$ and $\sum_i p_i=1$.

Definition: An allocation of a total $m$ across $k$ categories is a vector $(x_1,\ldots,x_k)$ of nonnegative integers satisfying $\sum_{i=1}^k x_i=m$. Conditional on $S=m$, only these allocations are possible.

Theorem: For independent $X_i\sim\operatorname{Poisson}(\lambda_i)$, with $\Lambda>0$, and any nonnegative integer $m$, $$ P(X_1=x_1,\ldots,X_k=x_k\mid S=m) =\frac{m!}{x_1!\cdots x_k!}\prod_{i=1}^k p_i^{x_i}, \qquad \sum_{i=1}^k x_i=m. $$ If the counts do not sum to $m$, the conditional probability is zero. Here $p_i=\lambda_i/\Lambda$, and a factor with $p_i=0,x_i=0$ is interpreted as $1$. Equivalently, the conditional allocation has a multinomial distribution with size $m$ and probabilities $p_1,\ldots,p_k$.

The conditioning event has positive probability: since $\Lambda>0$, $P(S=m)>0$ for every nonnegative integer $m$. A category with $\lambda_i=0$ has $X_i=0$ with probability one, and therefore receives no events in any allocation of positive conditional probability.

Deriving the Conditional Multinomial Formula

Independence and the Poisson probability mass functions give, for nonnegative integers $x_1,\ldots,x_k$,

P(X_1=x_1,\ldots,X_k=x_k) =\prod_{i=1}^k e^{-\lambda_i}\frac{\lambda_i^{x_i}}{x_i!} =e^{-\Lambda}\frac{\prod_{i=1}^k\lambda_i^{x_i}}{\prod_{i=1}^k x_i!}.

To compute the probability of the conditioning event, sum over every allocation of $m$. The multinomial theorem states that for nonnegative numbers $a_1,\ldots,a_k$, $\sum_{x_1+\cdots+x_k=m}m!\prod_i a_i^{x_i}/\prod_i x_i!=(\sum_i a_i)^m$. Applying it with $a_i=\lambda_i$ gives

\begin{aligned} P(S=m) &=\sum_{x_1+\cdots+x_k=m} e^{-\Lambda}\frac{\prod_i\lambda_i^{x_i}}{\prod_i x_i!}\\ &=e^{-\Lambda}\frac{1}{m!} \sum_{x_1+\cdots+x_k=m} \frac{m!}{\prod_i x_i!}\prod_i\lambda_i^{x_i}\\ &=e^{-\Lambda}\frac{\Lambda^m}{m!}. \end{aligned}

In particular, this probability is positive when $\Lambda>0$. For an allocation summing to $m$, divide its joint probability by this total probability:

\begin{aligned} P(X_1=x_1,\ldots,X_k=x_k\mid S=m) &=\frac{e^{-\Lambda}\prod_i\lambda_i^{x_i}/\prod_i x_i!} {e^{-\Lambda}\Lambda^m/m!}\\ &=\frac{m!}{\prod_i x_i!} \prod_i\left(\frac{\lambda_i}{\Lambda}\right)^{x_i}\\ &=\frac{m!}{\prod_i x_i!}\prod_i p_i^{x_i}. \end{aligned}

The formula is zero for vectors that do not sum to $m$, because those vectors are incompatible with $S=m$. For allocations that do sum to $m$, the multinomial theorem with $\sum_i p_i=1$ shows that the displayed probabilities sum to one. This proves the conditional distribution claim. $\square$

Worked Allocations

Worked Example: Two Sources Given a Total of Four

Let $X_1\sim\operatorname{Poisson}(2)$ and $X_2\sim\operatorname{Poisson}(3)$ be independent. Find $P(X_1=1,X_2=3\mid X_1+X_2=4)$.

The combined rate is $\Lambda=2+3=5$, so the conditional category probabilities are $p_1=2/5$ and $p_2=3/5$. The specified allocation sums to four. Therefore,

\begin{aligned} P(X_1=1,X_2=3\mid X_1+X_2=4) &=\frac{4!}{1!3!}\left(\frac25\right)^1\left(\frac35\right)^3\\ &=4\cdot\frac25\cdot\frac{27}{125} =\frac{216}{625} =0.3456. \end{aligned}

The calculation depends on the total only through the fixed allocation size $4$; the rates enter through their proportions $2/5$ and $3/5$.

Worked Example: A Grouped Allocation Across Three Sources

Let $X_1,X_2,X_3$ be independent Poisson counts with rates $1,1,2$, respectively. Given $X_1+X_2+X_3=5$, find the probability that $X_3=2$ and $X_1+X_2=3$.

The combined rate is $4$, so the conditional probabilities are $p_1=1/4$, $p_2=1/4$, and $p_3=1/2$. The event consists of four allocations $(j,3-j,2)$, for $j=0,1,2,3$. Their probabilities are:

For $(0,3,2)$, the coefficient is $5!/(0!3!2!)=10$, giving $10(1/4)^3(1/2)^2=5/128$.
For $(1,2,2)$, the coefficient is $5!/(1!2!2!)=30$, giving $30(1/4)^3(1/2)^2=15/128$.
For $(2,1,2)$, the coefficient is $5!/(2!1!2!)=30$, giving $30(1/4)^3(1/2)^2=15/128$.
For $(3,0,2)$, the coefficient is $5!/(3!0!2!)=10$, giving $10(1/4)^3(1/2)^2=5/128$.

These four disjoint allocations are exactly the event in question. Adding their probabilities,

P(X_3=2\mid X_1+X_2+X_3=5) =\frac{5+15+15+5}{128} =\frac{40}{128} =\frac{5}{16}.

As a check, the common factor in the sum is $(1/4)^3(1/2)^2=1/256$, and the four multinomial coefficients sum to $10+30+30+10=80$. Thus the probability is $80/256=5/16$, the same result.

Grouped Counts and a Useful Consequence

When several categories are combined, their conditional probabilities add. More precisely, if $A$ is a subset of the categories and $Y=\sum_{i\in A}X_i$, then, conditional on $S=m$, $Y$ has a binomial distribution with size $m$ and success probability $q=\sum_{i\in A}p_i$. This follows directly by summing the multinomial allocation probabilities over all allocations whose counts in $A$ add to a fixed value $r$.

Result: For $0\leq r\leq m$, $$ P(Y=r\mid S=m)=\binom{m}{r}q^r(1-q)^{m-r}, \qquad q=\sum_{i\in A}p_i. $$ The probability is zero for $r<0$ or $r>m$.

To verify the summation, write $B$ for the categories not in $A$. For any allocation with total $r$ in $A$ and $m-r$ in $B$, its multinomial coefficient factors as $\binom{m}{r}$ times the within-group multinomial coefficients $r!/\prod_{i\in A}x_i!$ and $(m-r)!/\prod_{i\in B}x_i!$. Summing the factors involving $A$ over allocations totaling $r$ gives $(\sum_{i\in A}p_i)^r=q^r$ by the multinomial theorem. Summing those involving $B$ gives $(\sum_{i\in B}p_i)^{m-r}=(1-q)^{m-r}$. The remaining factor is $\binom{m}{r}$, proving the formula. $\square$

Worked Example: Probability of One Count From a Group

Let $X_1,X_2,X_3$ be independent Poisson counts with rates $2,3,1$. Given their total is $4$, find $P(X_1=1\mid X_1+X_2+X_3=4)$.

The total rate is $6$, and $p_1=2/6=1/3$. Apply the grouped-count result with $A=\{1\}$. The count $X_1$, conditional on the total, is binomial with size $4$ and probability $1/3$. Hence

\begin{aligned} P(X_1=1\mid X_1+X_2+X_3=4) &=\binom41\left(\frac13\right)\left(\frac23\right)^3\\ &=4\cdot\frac13\cdot\frac{8}{27} =\frac{32}{81}. \end{aligned}

The result includes every allocation with $X_1=1$: the remaining three events can be divided between categories 2 and 3 in several ways. Using the binomial marginal accounts for all of them without listing each allocation.

Zero Rates, Interpretation, and Common Pitfalls

If a rate $\lambda_i$ is zero while $\Lambda>0$, then $p_i=0$ and category $i$ receives no events with conditional probability one. The formula assigns probability zero to any allocation with $x_i>0$. If all rates are zero, then every $X_i=0$ with probability one: conditioning on $S=0$ gives the all-zero allocation with probability one, while conditioning on $S=m>0$ is undefined because the conditioning event has probability zero.

Conditioning changes the structure of the model. Before conditioning, the counts are independent; once their sum is fixed, an increase in one count leaves fewer events available for the others. Their conditional allocation is described by the multinomial law, not by independent Poisson probabilities.

Using rates instead of proportions. The conditional category probabilities are $\lambda_i/\Lambda$, not $\lambda_i$ themselves. They must sum to one.
Forgetting the allocation constraint. A vector with $\sum_i x_i\ne m$ has conditional probability zero given $S=m$.
Omitting the multinomial coefficient. The factor $m!/\prod_i x_i!$ counts the arrangements associated with the specified category counts.
Confusing an event with one allocation. If an event allows several allocations, calculate each disjoint allocation and add, or use the grouped-count result when applicable.
Making an arithmetic error in a sum. Check that the allocation probabilities for the event add to the stated value; for the full set of allocations, their sum must be one.

Key takeaway: Independent Poisson counts with rates $\lambda_1,\ldots,\lambda_k$, conditional on a total of $m$, allocate those $m$ events multinomially with probabilities proportional to the rates. Use the allocation formula for a specified vector and the binomial grouped-count result for a sum over categories.

Check Your Understanding

For each question, state the conditional probabilities or show the calculation that supports your answer.

Independent Poisson counts have rates $3$ and $5$. Given their sum is $4$, what is the probability of allocation $(2,2)$?
Three independent Poisson counts have rates $2,1,3$. Given their total is $5$, what are the multinomial probabilities for the three categories?
For rates $1,2,1$, condition on a total of $3$. Find the probability that the first count is $0$.
Why does a category with rate zero have count zero conditional on a positive total, provided the combined rate is positive?
When calculating the probability of an event that permits several allocations, what check can help detect an arithmetic error?

What You'll Learn