Multinomial Sampling and Fixed Totals

Why the Total Is Fixed in a One-Way Goodness-of-Fit Test

A one-way goodness-of-fit test compares observed category counts with counts expected under a specified probability distribution. In the usual design, a fixed number $n$ of observations is classified into one of $k$ categories. The counts must add to $n$: every observed outcome belongs to exactly one category. This is why the model is often described as conditioning on a fixed total.

A fixed total is not, by itself, enough to make the counts multinomial. The probability model also depends on how the observations were generated. Independent categorical outcomes with the same category probabilities give a multinomial model. A sample drawn without replacement from a finite population generally does not, because the draws are dependent. Recognizing that distinction helps determine whether the usual one-way goodness-of-fit framework matches the study design.

Definition: Suppose $n$ independent trials each produce exactly one outcome in one of $k$ mutually exclusive and exhaustive categories. If the probability of category $i$ is the same on every trial and equals $p_i$, with $\sum_{i=1}^{k}p_i=1$, the count vector $(Y_1,\ldots,Y_k)$ has a multinomial distribution with size $n$ and category probabilities $p_1,\ldots,p_k$. Its probability at nonnegative counts $y_1,\ldots,y_k$ satisfying $\sum_i y_i=n$ is $$ P(Y_1=y_1,\ldots,Y_k=y_k) =\frac{n!}{y_1!\cdots y_k!}\prod_{i=1}^{k}p_i^{y_i}. $$

The factor $n!/(y_1!\cdots y_k!)$ counts how many different orderings of the $n$ trial outcomes produce those category counts. The product gives the probability of any one such ordering when the trials are independent and have the same category probabilities. Thus the multinomial formula combines the probability of an ordering with the number of orderings that produce the count vector.

Independent Outcomes and Dependent Counts

The individual trial outcomes can be independent even though the category counts are not. Once the total $n$ is fixed, increasing the count in one category leaves fewer outcomes available for the others. In particular, $Y_1+\cdots+Y_k=n$ always holds, so the counts cannot vary freely. This dependence among counts is part of the multinomial model; it does not contradict independence of the underlying trials.

For a multinomial count $Y_i$, the expected count is $E(Y_i)=np_i$. For two different categories $i$ and $j$, the covariance is $\operatorname{Cov}(Y_i,Y_j)=-np_ip_j$. The negative sign reflects the fixed-total constraint: a trial assigned to category $i$ cannot also be assigned to category $j$.

Worked Example: Counts From Independent Categorical Trials

Suppose, hypothetically, that a researcher classifies each of 10 independently selected specimens into one of three categories. For every specimen, the probabilities are $p_1=0.50$, $p_2=0.30$, and $p_3=0.20$. Find the probability of observing counts $(5,3,2)$, and find the expected count in each category.

The multinomial assumptions are that there are exactly 10 trials, each specimen receives one category, the category probabilities are the same for every specimen, and the specimens’ outcomes are independent. Under these assumptions, $$ P(Y_1=5,Y_2=3,Y_3=2) =\frac{10!}{5!3!2!}(0.50)^5(0.30)^3(0.20)^2. $$

The number of orderings is $10!/(5!3!2!)=2520$. Also, $(0.50)^5=0.03125$, $(0.30)^3=0.027$, and $(0.20)^2=0.04$. Therefore, $$ P(Y_1=5,Y_2=3,Y_3=2) =2520(0.03125)(0.027)(0.04)=0.08505. $$

The expected counts are $10(0.50)=5$, $10(0.30)=3$, and $10(0.20)=2$. These expected values happen to equal the observed counts. That coincidence does not mean the counts were guaranteed, or that the probability of this count vector is 1; the computed probability is 0.08505.

In a one-way goodness-of-fit test, the null hypothesis specifies the category probabilities, such as $H_0:p_i=p_{i,0}$ for each category $i$. Under the multinomial model, the expected count for category $i$ is $np_{i,0}$. The Pearson goodness-of-fit statistic compares the observed counts with those expected counts; its chi-square calibration also requires the conditions discussed in the earlier material on chi-square calibration and sparse expected counts. The multinomial model describes how the counts arise, while those additional conditions concern whether the chi-square approximation is suitable.

Conditioning on a Random Total

Sometimes counts are first generated by a process in which the total number of events is random. Conditioning on the observed total can still lead to a multinomial model. One important example is a set of independent Poisson counts, one for each category. If $X_i$ has a Poisson distribution with mean $\lambda_i$, and the $X_i$ are independent, then the total $T=\sum_i X_i$ has a Poisson distribution with mean $\Lambda=\sum_i\lambda_i$. Given $T=n$, the category counts have a multinomial distribution with probabilities $p_i=\lambda_i/\Lambda$.

The following derivation explains the role of the condition. For counts $y_1,\ldots,y_k$ summing to $n$, the independent Poisson model gives their joint probability. Dividing by the probability of the observed total produces the conditional probability:

\begin{aligned} P(X_1=y_1,\ldots,X_k=y_k\mid T=n) &=\frac{\prod_{i=1}^{k}\left(e^{-\lambda_i}\lambda_i^{y_i}/y_i!\right)} {e^{-\Lambda}\Lambda^n/n!}\\ &=\frac{n!}{y_1!\cdots y_k!} \prod_{i=1}^{k}\left(\frac{\lambda_i}{\Lambda}\right)^{y_i}. \end{aligned}

In the simplification, $\prod_i e^{-\lambda_i}=e^{-\Lambda}$, which cancels the exponential term in the denominator, and $\sum_i y_i=n$. The result has the multinomial form, with probabilities $\lambda_i/\Lambda$. Conditioning removes information about the overall event rate $\Lambda$; the resulting comparison concerns the relative allocation of the total among categories.

Worked Example: Conditioning Independent Poisson Counts

Imagine two categories of events with independent Poisson counts, having means 12 and 8 per observation period. Suppose the total count is observed to be 20. Find the conditional probability that 15 events are in the first category and 5 are in the second.

Here $\Lambda=12+8=20$, so the conditional category probabilities are $p_1=12/20=0.60$ and $p_2=8/20=0.40$. Given $T=20$, the count vector is multinomial. Therefore, $$ P(X_1=15,X_2=5\mid T=20) =\frac{20!}{15!5!}(0.60)^{15}(0.40)^5. $$

The coefficient is $20!/(15!5!)=15{,}504$. Substituting and evaluating gives $$ 15{,}504(0.60)^{15}(0.40)^5\approx 0.074647, $$ or about 0.07465 to five decimal places. This is a probability conditional on the total being 20; it does not describe the probability of observing 20 events in the first place.

Why Sampling Without Replacement Is Different

A sample of a fixed size taken without replacement from a finite population has a fixed total, but its draws are generally dependent. After an item is selected, the population composition changes. Consequently, the category probability on a later draw can depend on which categories were selected earlier. The resulting count model is multivariate hypergeometric rather than multinomial.

Definition: Suppose a finite population has $N$ objects, of which $M_i$ belong to category $i$, where $\sum_i M_i=N$. If $n$ objects are sampled without replacement, the probability of counts $y_1,\ldots,y_k$, with $\sum_i y_i=n$, is $$ P(Y_1=y_1,\ldots,Y_k=y_k) =\frac{\prod_{i=1}^{k}\binom{M_i}{y_i}}{\binom{N}{n}}. $$

Worked Example: A Small Sample Without Replacement

A hypothetical box contains two red tokens and two blue tokens. Two tokens are drawn without replacement. Find the probability of drawing one token of each color, and compare it with the probability under two independent draws with replacement.

Without replacement, there are $\binom{4}{2}=6$ equally likely pairs of tokens. To get one red and one blue, choose one of the two red tokens and one of the two blue tokens, giving $\binom{2}{1}\binom{2}{1}=4$ favorable pairs. Thus $$ P(Y_{\text{red}}=1,Y_{\text{blue}}=1) =\frac{\binom{2}{1}\binom{2}{1}}{\binom{4}{2}} =\frac{4}{6}=\frac{2}{3}. $$

With replacement, each draw has red and blue probabilities of $1/2$, and the draws are independent. The multinomial probability of one of each color would be $$ \frac{2!}{1!1!}(0.50)^1(0.50)^1=0.50. $$

The probabilities differ: $2/3$ without replacement versus $0.50$ with replacement. Both designs have a total of two draws, but only the independent, same-probability design supports the multinomial calculation exactly. The finite-population calculation accounts for the changing composition of the box.

For a large finite population sampled without replacement, the dependence between draws may be small enough that a multinomial approximation is useful, but this is an approximation, not an exact consequence of having a fixed sample size. Whether the approximation is adequate depends on the sampling fraction and the intended accuracy; there is no universal cutoff that makes every such sample multinomial. If the sampling design or population structure matters materially, use a model that represents it rather than treating the draws as independent.

Assessing Whether the Multinomial Model Fits the Design

Before applying a one-way goodness-of-fit test, translate the data collection into a model check. Ask whether each observation is classified once into exactly one category; whether the total number of observations is fixed or whether inference is explicitly conditional on the total; whether outcomes are independent; and whether the same category probabilities apply to every observation. The usual multinomial model requires all of these features.

Identify the observational units and categories.
Each unit must contribute exactly one outcome to one of the categories being analyzed.

Check how the total is set.
A planned sample size gives a fixed $n$. For a random total, decide whether the analysis is conditional on the observed total and whether that conditioning is meaningful for the question.

Check independence and common probabilities.
Repeated independent trials with the same probability vector support the multinomial model. A without-replacement sample or trials with changing probabilities generally do not support it exactly.

Then assess the test’s approximation conditions.
Even with an appropriate multinomial sampling model, verify the expected-count conditions for the chi-square approximation, or choose an exact or simulation-based method when needed.

Why the Distinction Matters

The multinomial model provides the probability distribution for the full vector of category counts under a specified set of probabilities. A one-way goodness-of-fit test uses that structure to assess whether the observed allocation is compatible with the null probabilities, subject to the conditions for the chosen reference distribution. When sampling without replacement, expected counts based on the same category proportions may still be informative, but the multinomial probability model is not exact. Treating it as exact can misstate the variability of the counts and therefore affect inferential conclusions.

Fixed total does not mean independent counts. The counts must sum to $n$, so they are constrained even when individual outcomes are independent.
Fixed total does not guarantee a multinomial model. Sampling without replacement and changing probabilities across trials can violate the model assumptions.
Conditioning changes the question. With independent Poisson category counts, conditioning on the total tests how events are divided among categories, not whether the total event rate is unusual.

A common mistake is to reason, “The sample size was fixed, so the counts must be multinomial.” The correction is to check the sampling mechanism: a fixed total is necessary for the usual fixed-$n$ formulation, but the observations must also be independent categorical trials with common probabilities. A without-replacement sample from a finite population instead has a multivariate hypergeometric distribution, except when a multinomial approximation is justified.

Key takeaway: A one-way goodness-of-fit analysis typically conditions on $n$, the number of classified observations. The multinomial model is appropriate when those observations are independent categorical outcomes with a common probability vector—or when a justified conditioning argument yields that model. A fixed total alone does not establish multinomial sampling.

Check Your Understanding

Use the sampling design, not just the fact that the counts add to a total, to decide whether a multinomial model is justified.

A researcher independently samples 40 parts from a process, classifying each as one of three defect types or no defect. What assumptions would support a multinomial model for the four counts?
Why can category counts be dependent even when the individual trial outcomes are independent?
Independent Poisson counts have means $\lambda_1,\lambda_2,\lambda_3$. Given their sum $T=n$, what are the multinomial category probabilities?
A researcher draws 15 cards from a standard deck without replacement and counts suits. Why is the exact count model not multinomial?
What information about the process is set aside when independent Poisson category counts are analyzed conditional on their observed total?