Binomial Marginals of Multinomial Counts

Why a Multinomial Count Can Be Binomial

A multinomial experiment records how many of $n$ trials fall into each of several categories. Although the full count vector has multiple components, a question about just one category asks only whether each trial landed there or somewhere else. That distinction is binary on every trial. The count in that category therefore has a binomial distribution.

The same reasoning applies when the question combines several categories: on each trial, record whether its outcome belongs to the selected set. The resulting count is again a binomial count. The key is to use the probability of the selected category or set as the binomial success probability.

Definition: Suppose $n$ independent trials each have the same $k$ possible categorical outcomes, with probabilities $p_1,\ldots,p_k$, where $p_i\geq 0$ and $\sum_{i=1}^{k}p_i=1$. Let $X_i$ be the number of trials in category $i$. Then $(X_1,\ldots,X_k)$ has a multinomial distribution with sample size $n$ and category probabilities $p_1,\ldots,p_k$. Its support consists of nonnegative integer vectors satisfying $\sum_{i=1}^{k}x_i=n$, and its probability mass function is $$ P(X_1=x_1,\ldots,X_k=x_k) = \frac{n!}{x_1!\cdots x_k!}\prod_{i=1}^{k}p_i^{x_i}. $$

The identical-trial assumption matters: each trial must have the same category probabilities, and the trial outcomes must be independent. A count formed from trials with different category probabilities need not have a binomial distribution.

The Marginal Distribution of One Cell

To study a single cell $i$, ignore the distinctions among all the other categories and classify each trial as either “in category $i$” or “not in category $i$.” The first outcome has probability $p_i$, and the second has probability $1-p_i$. Across the $n$ independent trials, the count $X_i$ is the number of times the first outcome occurs.

Theorem: Under the multinomial model, for each category $i$, $$ X_i\sim \operatorname{Bin}(n,p_i). $$ Thus the binomial sample size is the total number of trials $n$, and the binomial success probability is that cell’s category probability $p_i$.

Here is a derivation directly from the multinomial probability mass function. Fix an integer $r$ with $0\leq r\leq n$. To find $P(X_i=r)$, sum the joint mass over all possible counts in the other $k-1$ categories, whose total must be $n-r$. By the multinomial theorem, $$ \begin{aligned} P(X_i=r) &=\sum_{\substack{x_j\geq 0,\ j\neq i\\\sum_{j\neq i}x_j=n-r}} \frac{n!}{r!\prod_{j\neq i}x_j!}\,p_i^r\prod_{j\neq i}p_j^{x_j}\\ &=\frac{n!}{r!}p_i^r \frac{(1-p_i)^{n-r}}{(n-r)!}\\ &=\binom{n}{r}p_i^r(1-p_i)^{n-r}. \end{aligned} $$ The second equality follows by expanding $(\sum_{j\neq i}p_j)^{n-r}$, since $\sum_{j\neq i}p_j=1-p_i$. For $r$ outside $\{0,\ldots,n\}$, the probability is zero. This is exactly the binomial mass function.

Worked Example: A Single Cell Count

Suppose $n=12$ independent trials have four possible outcomes with probabilities $0.20,\ 0.30,\ 0.10,\ 0.40$. Let $X_2$ count the trials in category 2. Identify its distribution and calculate $P(X_2=4)$.

The category probabilities sum to $0.20+0.30+0.10+0.40=1$, as required. The probability of category 2 on each trial is $p_2=0.30$. Therefore $$ X_2\sim \operatorname{Bin}(12,0.30). $$

Using the binomial mass function, $$ \begin{aligned} P(X_2=4) &=\binom{12}{4}(0.30)^4(0.70)^8\\ &=495(0.0081)(0.05764801)\\ &=0.231139696095. \end{aligned} $$ Thus the probability is approximately $0.2311$. The other three category counts do not need to be specified to calculate this marginal probability.

The Count in a Selected Set of Categories

Now let $A$ be a nonempty subset of the category labels $\{1,\ldots,k\}$, and define the pooled count $$ Y_A=\sum_{i\in A}X_i. $$ This counts all trials whose outcomes belong to $A$. Write $$ q_A=\sum_{i\in A}p_i $$ for the probability that one trial lands in the selected set. Each trial either belongs to $A$, with probability $q_A$, or belongs to its complement, with probability $1-q_A$.

Theorem: For any subset $A$ of the category labels, $$ Y_A=\sum_{i\in A}X_i\sim \operatorname{Bin}(n,q_A), \qquad q_A=\sum_{i\in A}p_i. $$ If $q_A=0$, then $Y_A=0$ with probability one. If $q_A=1$, then $Y_A=n$ with probability one.

To prove the result, denote the category on trial $j$ by $C_j$, for $j=1,\ldots,n$. Define $I_j=1$ when $C_j\in A$, and $I_j=0$ otherwise. Since the trials are independent, the indicators $I_1,\ldots,I_n$ are independent. Each satisfies $P(I_j=1)=q_A$, so each is Bernoulli with success probability $q_A$. Counting trials in $A$ gives $$ Y_A=\sum_{j=1}^{n}I_j. $$ A sum of $n$ independent Bernoulli variables with the same success probability $q_A$ has the binomial distribution $\operatorname{Bin}(n,q_A)$. This proves the claim when $0<q_A<1$. When $q_A=0$, every $I_j=0$ with probability one; when $q_A=1$, every $I_j=1$ with probability one. These give the stated boundary cases. $\square$

The same conclusion can be checked by summing the multinomial mass. For $0<q_A<1$, fix $r$ and sum over all count vectors for which the categories in $A$ total $r$ and the remaining categories total $n-r$. Applying the multinomial theorem separately within the two groups gives $$ \begin{aligned} P(Y_A=r) &=\frac{n!}{r!(n-r)!} \left(\sum_{i\in A}p_i\right)^r \left(\sum_{i\notin A}p_i\right)^{n-r}\\ &=\binom{n}{r}q_A^r(1-q_A)^{n-r}. \end{aligned} $$ For an empty complement, the indicator argument above covers the case $q_A=1$ without needing to interpret a boundary expression involving a zero raised to the zeroth power.

Worked Example: Combining Two Categories

In $n=10$ independent categorical trials, suppose the four category probabilities are $0.12,\ 0.18,\ 0.25,\ 0.45$. Let $Y$ be the total count in the first two categories. Find its distribution and calculate $P(Y=3)$.

The probability of landing in either of the first two categories on one trial is $$ q=0.12+0.18=0.30. $$ Therefore the pooled count is $$ Y\sim\operatorname{Bin}(10,0.30). $$ For exactly three trials in the selected set, $$ \begin{aligned} P(Y=3) &=\binom{10}{3}(0.30)^3(0.70)^7\\ &=120(0.027)(0.0823543)\\ &=0.266827932. \end{aligned} $$ The remaining $7$ trials can fall in either of the other two categories; their combined probability on each trial is $0.25+0.45=0.70$.

Identifying the Parameters and Checking the Model

For a cell count, take $p$ to be the probability of that one category. For a pooled count, add the probabilities of all categories included in the count. In either case, the number of trials remains $n$; pooling changes the success probability, not the number of trials.

Worked Example: A Nonconsecutive Set and Its Complement

Suppose $n=8$ trials have four category probabilities $0.05,\ 0.15,\ 0.25,\ 0.55$. Let $Y$ count outcomes in categories 1 and 3, and calculate $P(Y=2)$. Also identify the count outside those categories.

The selected categories need not be consecutive. Their combined probability is $$ q=0.05+0.25=0.30, $$ so $$ Y\sim\operatorname{Bin}(8,0.30). $$ The probability of exactly two selected outcomes is $$ \begin{aligned} P(Y=2) &=\binom{8}{2}(0.30)^2(0.70)^6\\ &=28(0.09)(0.117649)\\ &=0.29647548. \end{aligned} $$ The count outside categories 1 and 3 is $8-Y$, and its success probability per trial is $0.15+0.55=0.70$. It has distribution $\operatorname{Bin}(8,0.70)$; in fact, its value is exactly $8-Y$.

This reduction is useful whenever the question concerns only whether an outcome belongs to a particular category or group. It replaces a many-category description with a two-outcome count while preserving the relevant probability. It does not claim that all the separate cell counts are independent: they are constrained to sum to $n$. Each marginal can be binomial even though the full vector remains multinomial.

Common Mistake: Averaging or Multiplying Category Probabilities

A frequent error is to use the average probability of the selected categories, or to multiply their probabilities. Neither operation gives the chance that one trial belongs to at least one of several distinct categories. The categories are mutually exclusive outcomes of one trial, so their probabilities add. For example, if a set contains categories with probabilities $0.12$ and $0.18$, the probability that a trial lands in the set is $0.12+0.18=0.30$, not their average $0.15$ or their product $0.0216$. Use $n$ as the number of trials and this sum as the binomial success probability.

Key takeaway: A single multinomial cell count is $\operatorname{Bin}(n,p_i)$. The total in a selected set $A$ is $\operatorname{Bin}(n,\sum_{i\in A}p_i)$. In both cases, first identify the event counted on one trial; its probability is the binomial parameter.

Check Your Understanding

For each question, identify the binomial parameters before calculating any probability.

There are $15$ independent trials with category probabilities $0.10,\ 0.25,\ 0.65$. What is the distribution of the count in category 2?
For those same trials, what is the distribution of the total count in categories 1 and 2?
In $9$ trials with four categories of probabilities $0.20,\ 0.30,\ 0.10,\ 0.40$, define $Y$ as the count in categories 2 and 4. Identify $Y$'s distribution.
What happens to a pooled count if the selected categories have total probability zero? What if their total probability is one?
Why does a binomial marginal for each individual cell not imply that the multinomial cell counts are mutually independent?