Multinomial Support and Probability Constraints

Counts Must Satisfy a Total

In a multinomial experiment, each of $n$ trials places exactly one observation into one of $k$ categories. The count in category $i$ is $X_i$. Because every trial contributes to exactly one category, the counts cannot vary independently: their sum must equal the number of trials. This simple constraint determines the basic feasible count space before any probabilities are calculated.

The probability vector has a parallel constraint. Each category probability must be nonnegative, and the probabilities across all categories must sum to one. These conditions are not merely convenient checks: they express that the categories are mutually exclusive and collectively exhaustive outcomes of one trial.

Definition: For a fixed nonnegative integer $n$ and a positive integer $k$, the feasible count space is $$ \mathcal{X}_{n,k} = \left\{(x_1,\ldots,x_k): x_i\in\{0,1,2,\ldots\}\text{ for every }i,\quad \sum_{i=1}^k x_i=n\right\}. $$ A valid multinomial parameter vector is $\boldsymbol{p}=(p_1,\ldots,p_k)$ with $p_i\geq 0$ for every $i$ and $\sum_{i=1}^k p_i=1$.

The nonnegative-integer requirement matters: a count cannot be negative or fractional. The sum constraint matters too: a vector with total $n-1$ leaves one trial unaccounted for, while a vector with total $n+1$ assigns more observations than there were trials. The symbol $k$ denotes the number of categories; it does not have to equal the number of observed categories in a particular data set.

Feasible Space and Actual Support

It is useful to distinguish two meanings that are sometimes both called “support.” The feasible count space $\mathcal{X}_{n,k}$ depends only on $n$ and $k$. The positive-probability support also depends on the parameter vector: it consists of the feasible vectors that actually have positive probability. If every $p_i$ is positive, every feasible vector has positive probability. If one or more category probabilities are zero, any vector assigning a positive count to such a category has probability zero.

Theorem: Suppose $\boldsymbol{X}=(X_1,\ldots,X_k)$ has a multinomial distribution with valid parameter vector $\boldsymbol{p}$. Its positive-probability support is $$ \left\{\boldsymbol{x}\in\mathcal{X}_{n,k}: x_i=0\text{ whenever }p_i=0\right\}. $$ In particular, if all $p_i>0$, the positive-probability support is the entire feasible count space.

To verify the statement, use the multinomial probability formula $$ P(\boldsymbol{X}=\boldsymbol{x}) = \frac{n!}{x_1!\cdots x_k!}\prod_{i=1}^k p_i^{x_i}, \qquad \boldsymbol{x}\in\mathcal{X}_{n,k}. $$ The multinomial coefficient is positive for every feasible vector. A factor $p_i^{x_i}$ is zero precisely when $p_i=0$ and $x_i>0$; when $x_i=0$, the factor contributes one, including at $p_i=0$. Thus the displayed probability is positive exactly for the feasible vectors described in the theorem. $\square$

For $n=0$, the feasible space contains only $(0,\ldots,0)$. Under any valid multinomial parameter vector, this vector has probability one, so it is also the positive-probability support. Boundary probabilities do not create an exception to the support characterization: every count is zero.

Worked Example: Checking a Count Vector and Calculating Its Probability

A sample has $n=5$ observations classified into three categories, with probabilities $p_1=0.2$, $p_2=0.5$, and $p_3=0.3$. Check whether $(2,1,2)$ is feasible, then find its probability.

First check the parameters: each is nonnegative, and $0.2+0.5+0.3=1$, so they form a valid probability vector. Next check the counts: $2,1,2$ are nonnegative integers, and $2+1+2=5$, so the vector is feasible. All three probabilities are positive, so this feasible vector is also in the positive-probability support.

Using the multinomial formula, $$ P(\boldsymbol{X}=(2,1,2)) = \frac{5!}{2!1!2!}(0.2)^2(0.5)^1(0.3)^2. $$

The coefficient is $5!/(2!1!2!)=120/4=30$. The product of probability factors is $0.04\times0.5\times0.09=0.0018$. Therefore, $$ P(\boldsymbol{X}=(2,1,2))=30(0.0018)=0.054. $$ The result is between zero and one, as a probability must be.

Parameter Constraints and Normalization

The validity conditions on a multinomial parameter vector are $p_i\geq0$ for every category and $\sum_i p_i=1$. Together these imply $p_i\leq1$ for each $i$: since all the other entries are nonnegative, no one entry can exceed the total of one. Thus probabilities need not be checked against one separately once nonnegativity and the sum constraint have been checked.

The multinomial formula also explains why these conditions normalize the probabilities over all feasible count vectors. More generally, suppose $q_1,\ldots,q_k$ are any nonnegative numbers, not yet assumed to sum to one, and form the same formula using the $q_i$. The multinomial theorem gives the total of these values over the feasible space.

Theorem: Let $q_i\geq0$ for every $i$, and define $$ m(\boldsymbol{x})=\frac{n!}{x_1!\cdots x_k!}\prod_{i=1}^k q_i^{x_i}, \qquad \boldsymbol{x}\in\mathcal{X}_{n,k}. $$ If $n\geq1$, then $\sum_{\boldsymbol{x}\in\mathcal{X}_{n,k}}m(\boldsymbol{x})=1$ if and only if $\sum_{i=1}^k q_i=1$. If $n=0$, the sum of these values is one for every nonnegative vector $(q_1,\ldots,q_k)$.

Proof. Write $s=\sum_{i=1}^k q_i$. By the multinomial theorem, for $n\geq1$, $$ \sum_{\boldsymbol{x}\in\mathcal{X}_{n,k}} \frac{n!}{x_1!\cdots x_k!}\prod_{i=1}^k q_i^{x_i} = \left(\sum_{i=1}^k q_i\right)^n=s^n. $$ Since $s\geq0$, $s^n=1$ holds exactly when $s=1$. This proves both directions for $n\geq1$. For $n=0$, the only count vector is $(0,\ldots,0)$; its coefficient and product are both one, so the total is one regardless of the $q_i$. $\square$

The zero-trial case requires careful interpretation. The formula’s values sum to one for any nonnegative entries when $n=0$, but that fact does not change the definition of a multinomial parameter vector: a valid parameter vector must still sum to one. With no trials, the distribution of the count vector is the point mass at $(0,\ldots,0)$, whatever valid category probabilities govern the trial scheme.

Worked Example: Rejecting an Invalid Parameter Vector

A proposed model has $n=4$ trials and category entries $q=(0.2,0.5,0.4)$. Check the parameters and the proposed count vector $(1,1,2)$.

The entries are nonnegative, but $0.2+0.5+0.4=1.1$, not one. Therefore $q$ is not a valid multinomial probability vector. The count vector itself has nonnegative integer entries and total $1+1+2=4$, so it is feasible as a count vector; that does not make the proposed probability model valid.

The distinction can also be seen by summing the proposed formula over all feasible vectors. Here $n=4\geq1$, so the total is $(1.1)^4=1.4641$, not one. The formula using these entries cannot be a probability mass function. A count vector can satisfy the support constraint while the parameter vector fails its separate constraint.

Boundary Probabilities

A category may have probability zero or one; both are allowed boundary values. If $p_j=0$, that category is never observed under the model, so every positive-probability count vector must have $x_j=0$. If $p_j=1$, the sum-to-one condition forces all other probabilities to be zero, and all $n$ observations must fall in category $j$. These are valid, though degenerate, multinomial distributions.

Worked Example: Support at a Boundary Parameter

Let $n=4$, $k=3$, and $\boldsymbol{p}=(1,0,0)$. The entries are nonnegative and sum to one, so the parameter vector is valid. Consider the feasible vector $(2,1,1)$: its entries are nonnegative integers and sum to four, but it is not in the positive-probability support because it assigns positive counts to categories with probability zero.

The only positive-probability vector is $(4,0,0)$. Its probability is $$ \frac{4!}{4!0!0!}(1)^4(0)^0(0)^0=1, $$ where the usual convention is that a zero-probability factor raised to the zero count contributes one. Every other feasible vector puts a positive count in category 2 or 3 and therefore has probability zero.

Worked Example: The Zero-Trial Case

Take $n=0$, $k=2$, and first consider the valid probability vector $(0.3,0.7)$. The only feasible count vector is $(0,0)$, and its probability is one. Now consider nonnegative entries $(2,0)$. The expression over the one feasible vector still gives $1$, because both counts are zero and the coefficient is $0!/(0!0!)=1$. However, $(2,0)$ is not a valid multinomial parameter vector: its entries sum to two.

This example separates two questions. The count formula at zero trials has total one for any nonnegative entries, but the definition of valid category probabilities continues to require a sum of one. Do not use the exceptional normalization at $n=0$ to relax the parameter definition.

A Reliable Validity Check

When assessing a proposed multinomial specification, check counts and parameters separately. This avoids a frequent error: a feasible count vector does not certify that the proposed model is valid, and valid probabilities do not make an infeasible count vector possible.

Check the experiment size and category count.
Confirm that $n$ is a nonnegative integer and $k$ is a positive integer.

Check the parameter vector.
Every $p_i$ must be nonnegative, and their sum must be one, including when $n=0$.

Check the count vector.
Every $x_i$ must be a nonnegative integer, and the counts must sum to $n$.

Check whether its probability is positive.
If any category with $p_i=0$ has $x_i>0$, the vector is feasible but has probability zero; otherwise its probability is positive.

A common mistake is to say that every nonnegative integer vector is a possible multinomial outcome. The correction is that such a vector is only a candidate: its entries must sum to $n$. A second mistake is to treat the full feasible space as the positive-probability support even when some $p_i=0$. The correction is to apply the zero-probability restriction after checking feasibility. Finally, the normalization calculation must be handled with its stated condition: for $n\geq1$, the nonnegative entries sum to one exactly when the displayed masses sum to one; for $n=0$, those masses sum to one for any nonnegative entries, even though valid multinomial parameters are still required to sum to one.

Key takeaway: Feasible multinomial counts are nonnegative integers totaling $n$. Valid category probabilities are nonnegative and total one. The actual positive-probability support removes feasible vectors that assign positive counts to zero-probability categories; at zero trials the only count vector is the all-zero vector.

Check Your Understanding

For each question, distinguish the count constraint from the parameter constraint.

For $n=6$ and $k=3$, is $(2,0,4)$ feasible? Is $(2,0,5)$ feasible? Explain.
Are $(0.25,0.25,0.5)$ and $(0.25,0.25,0.6)$ valid probability vectors? State the check that decides each case.
If $n=3$ and $\boldsymbol{p}=(0.4,0,0.6)$, which condition must a feasible vector satisfy to have positive probability?
For $n\geq1$, what is the total of the multinomial-formula values when nonnegative entries $q_i$ sum to $1.2$?
When $n=0$, do nonnegative entries summing to two form a valid multinomial parameter vector, even though the formula’s values sum to one? Explain the distinction.