Pooling Categories in a Multinomial Experiment

Why Pool Categories?

A multinomial experiment records how many of $n$ trials fall in each of several mutually exclusive categories. Sometimes the question of interest does not distinguish every original category. A report might combine “mild” and “moderate” responses as “non-severe,” or combine two related product types into one broader class. Pooling is useful because it preserves a simple probability model: the combined counts are still multinomial, and the probability of a pooled category is the sum of the probabilities of the original categories it contains.

The key requirement is that the original experiment has the multinomial structure: trials are independent, each trial uses the same category probabilities, and exactly one category occurs on each trial. Pooling changes how outcomes are labelled; it does not change the trials themselves.

Definition and Pooling Theorem

Definition: Suppose $n$ independent trials each fall in exactly one of $k$ categories, with category probabilities $p_1,\ldots,p_k$, where $p_j\geq 0$ and $\sum_{j=1}^k p_j=1$. Let $X_j$ be the count in category $j$. Then $X=(X_1,\ldots,X_k)$ has a multinomial distribution with parameters $n$ and $(p_1,\ldots,p_k)$, written $X\sim\operatorname{Multinomial}(n;p_1,\ldots,p_k)$, if $$ P(X_1=x_1,\ldots,X_k=x_k) = \frac{n!}{x_1!\cdots x_k!}\prod_{j=1}^k p_j^{x_j} $$ for nonnegative integers $x_1,\ldots,x_k$ that sum to $n$, and the probability is zero otherwise. In this formula, a factor $0^0$ is interpreted as $1$.

To describe pooling, divide the original category indices into $m$ nonempty, disjoint groups $G_1,\ldots,G_m$ whose union is $\{1,\ldots,k\}$. Define the pooled count $Y_g$ to be the sum of the original counts in group $G_g$. Its probability $q_g$ is the sum of the original category probabilities in that group.

Theorem: If $X\sim\operatorname{Multinomial}(n;p_1,\ldots,p_k)$ and $G_1,\ldots,G_m$ form a partition of the original categories, then $$ Y_g=\sum_{j\in G_g}X_j,\qquad q_g=\sum_{j\in G_g}p_j $$ satisfy $Y=(Y_1,\ldots,Y_m)\sim\operatorname{Multinomial}(n;q_1,\ldots,q_m)$.

Proof by Recoding Each Trial

Let $C_i$ denote the original category on trial $i$, for $i=1,\ldots,n$. The $C_i$ are independent and have the same category probabilities $p_1,\ldots,p_k$. Replace each original label $j$ by the group label $g$ for which $j\in G_g$. Call the resulting label $D_i$. For any group $g$, $$ P(D_i=g)=P(C_i\in G_g)=\sum_{j\in G_g}p_j=q_g. $$

Because each $D_i$ is a function of $C_i$, and the original trial outcomes are independent, the recoded outcomes $D_1,\ldots,D_n$ are also independent. They have the same probabilities $q_1,\ldots,q_m$ on every trial. Their counts are exactly $Y_1,\ldots,Y_m$. Thus the recoded trials meet the definition of a multinomial experiment with $m$ categories. Applying the multinomial probability formula gives, for nonnegative $y_1,\ldots,y_m$ that sum to $n$, $$ P(Y_1=y_1,\ldots,Y_m=y_m) = \frac{n!}{y_1!\cdots y_m!}\prod_{g=1}^m q_g^{y_g}. $$ This is the required multinomial distribution. If some $q_g=0$, that group’s count is zero with probability one; the same formula applies with the stated $0^0$ convention. $\square$

Why the Added Probabilities Give the Same Formula

The trial-recoding proof explains the model directly. The same conclusion can be checked by summing the original multinomial probabilities over every allocation of each pooled count among its component categories. This algebra makes explicit why adding the probabilities is sufficient.

Fix feasible pooled counts $y_1,\ldots,y_m$, with $\sum_g y_g=n$. Within each group $G_g$, the original counts $x_j$ must sum to $y_g$. Therefore the probability of the pooled count vector is $$ P(Y=y) = \sum_{\substack{\sum_{j\in G_g}x_j=y_g\\g=1,\ldots,m}} \frac{n!}{\prod_{j=1}^k x_j!}\prod_{j=1}^k p_j^{x_j}. $$

The sum separates across groups after taking out $n!$: $$ P(Y=y) = n!\prod_{g=1}^m \left( \sum_{\substack{x_j\geq 0,\ j\in G_g\\\sum_{j\in G_g}x_j=y_g}} \prod_{j\in G_g}\frac{p_j^{x_j}}{x_j!} \right). $$ By the multinomial theorem, the sum in parentheses for group $g$ equals $$ \frac{\left(\sum_{j\in G_g}p_j\right)^{y_g}}{y_g!} = \frac{q_g^{y_g}}{y_g!}. $$ Substitution gives $$ P(Y=y)=\frac{n!}{\prod_{g=1}^m y_g!}\prod_{g=1}^m q_g^{y_g}, $$ which is the multinomial probability formula for the pooled categories. If $q_g=0$, every $p_j$ in that group is zero; the only possible value is $y_g=0$, and the formula remains valid under the $0^0$ convention. $\square$

Worked Examples

Worked Example: Combining Two Response Categories

On each of six independent trials, an outcome is classified as A, B, or C, with probabilities $0.20$, $0.30$, and $0.50$. Combine A and B into a single category called “selected.” Find the distribution of the selected count $R$, and calculate $P(R=2)$.

The probability of “selected” on one trial is $$ P(A\text{ or }B)=0.20+0.30=0.50. $$ The other pooled category is C, with probability $0.50$. The pooling theorem gives a two-category multinomial model with six trials and probabilities $0.50,0.50$. Thus $R\sim\operatorname{Binomial}(6,0.50)$, and $$ P(R=2)=\binom{6}{2}(0.50)^2(0.50)^4 =15(0.50)^6 =\frac{15}{64} =0.234375. $$

For a direct check, the original counts for A and B can add to 2 in three ways: $(0,2)$, $(1,1)$, and $(2,0)$; the C count must be 4. The original multinomial probabilities sum to $$ \frac{6!}{0!2!4!}(0.30)^2(0.50)^4 + \frac{6!}{1!1!4!}(0.20)(0.30)(0.50)^4 + \frac{6!}{2!0!4!}(0.20)^2(0.50)^4 $$ $$ =15(0.09)(0.0625)+30(0.06)(0.0625)+15(0.04)(0.0625) =0.084375+0.1125+0.0375 =0.234375. $$ The direct sum agrees with the pooled calculation.

Worked Example: Recoding Four Labels into Three Groups

In eight independent trials, the original outcome probabilities for A, B, C, and D are $0.10$, $0.20$, $0.25$, and $0.45$. A is grouped with C, while B and D remain separate. Find the probability that the pooled counts are 3 for the A-or-C group, 2 for B, and 3 for D.

The new probabilities are $$ q_1=P(A\text{ or }C)=0.10+0.25=0.35,\qquad q_2=0.20,\qquad q_3=0.45. $$ They sum to $0.35+0.20+0.45=1$, as required. The pooled count vector $(3,2,3)$ is feasible because its entries sum to $8$. Its probability is $$ \frac{8!}{3!2!3!}(0.35)^3(0.20)^2(0.45)^3. $$ The multinomial coefficient is $$ \frac{8!}{3!2!3!}=\frac{40320}{6\cdot2\cdot6}=560. $$ The probability factors are $(0.35)^3=0.042875$, $(0.20)^2=0.04$, and $(0.45)^3=0.091125$. Therefore $$ 560(0.042875)(0.04)(0.091125) =560(0.000156279375) =0.08751645. $$ The result concerns the combined A-or-C count; it does not require specifying how many of those three outcomes were A and how many were C.

Worked Example: Pooling Several Categories into a Broad Class

A four-category outcome has probabilities $0.12$, $0.18$, $0.30$, and $0.40$ on each of four independent trials. Combine the first three categories into one broad class, leaving the fourth category as the other group. Calculate the probability that exactly two trials fall in each group.

The broad class has probability $$ q_1=0.12+0.18+0.30=0.60, $$ and the other group has probability $q_2=0.40$. The pooled count vector $(2,2)$ has probability $$ \frac{4!}{2!2!}(0.60)^2(0.40)^2 =6(0.36)(0.16) =0.3456. $$ To see the combination of original categories inside the broad class, conditional details aside, the total single-trial probability of landing in any of its three labels is $0.60$; the two broad-class trials can occupy any of the $\binom{4}{2}=6$ pairs of trial positions. Each such position pattern has probability $(0.60)^2(0.40)^2=0.0576$. Summing the six equal probabilities gives $6(0.0576)=0.3456$, the same result.

Common Mistake and Practical Importance

A common mistake is to add the probabilities of only some original categories but then keep the remaining probabilities unchanged in a way that makes the new list fail to sum to one. Pooling must define a complete recoding: every original category belongs to exactly one new group. Since the groups partition the original categories, $$ \sum_{g=1}^m q_g = \sum_{g=1}^m\sum_{j\in G_g}p_j = \sum_{j=1}^k p_j =1. $$

Pooling also does not repair a misspecified experiment. If trials have different category probabilities, or if their outcomes are dependent, the original counts need not have a multinomial distribution. Relabelling outcomes alone does not establish the independence and identical probability conditions. When those conditions do hold, pooling is valuable because it lets a question be answered at the level of detail actually needed, while retaining an exact multinomial model.

Pooling can be viewed as replacing a detailed label by a coarser one. Every pooled count is the sum of the counts in its group, each new probability is the corresponding sum of probabilities, and the total number of trials remains $n$. No further adjustment to the multinomial coefficient or to the sample size is needed.

Check Your Understanding

Use the pooling theorem and show the probability calculations where appropriate.

A multinomial experiment has probabilities $0.15, 0.25, 0.60$. What is the probability of the group formed by pooling the first two categories?
For $n=5$, pool categories with probabilities $0.20$ and $0.30$, leaving a category of probability $0.50$. Find the probability that the pooled counts are $(1,4)$.
Explain why the groups used to recode categories must be disjoint and cover every original category.
If a pooled group has probability zero, what pooled count can it have with positive probability?
Does pooling categories make dependent trials into a multinomial experiment? State the conditions that still matter.