Multinomial Sampling Of Genotype Counts

Why Model Genotype Counts?

A population model assigns probabilities to possible genotypes. A sample does not generally contain exactly those proportions: the observed counts fluctuate from sample to sample. A probability model for those counts makes it possible to describe the expected amount of fluctuation and how the counts move together.

The model in this tutorial applies to genotype categories with specified probabilities. The probabilities could come from a Hardy–Weinberg model, an inbreeding model, assortative mating, or another population description. The multinomial sampling model does not itself require Hardy–Weinberg equilibrium. It describes how a sample of individuals is drawn, given the genotype probabilities.

The Multinomial Model for Genotype Counts

Suppose there are \(k\) mutually exclusive and exhaustive genotype categories. For example, at a locus with two alleles, the categories could be \(AA\), \(Aa\), and \(aa\). Let \(\pi_j\) be the population probability of category \(j\), for \(j=1,\ldots,k\). These probabilities satisfy \(0\leq \pi_j\leq 1\) and \(\sum_{j=1}^k \pi_j=1\).

Take a sample of \(n\) individuals. The multinomial model assumes that each sampled individual belongs to exactly one of the categories, that each individual has the same category probabilities \((\pi_1,\ldots,\pi_k)\), and that individuals’ category assignments are independent. These assumptions describe independent sampling from a population distribution, as in sampling with replacement or from a population sufficiently large for the independent-sampling approximation. Sampling without replacement from a small, fixed population need not have this model.

For any possible count vector \((n_1,\ldots,n_k)\) summing to \(n\), its probability is

The factorial factor counts the different ways the \(n\) sampled individuals can be arranged into categories with those counts. The product gives the probability of any one such arrangement under independent sampling. A category with probability zero has count zero with certainty; in the probability formula, the usual convention is that a factor \(0^0\) equals 1.

Expected Counts and Variances

The expected count in category \(j\) is the sample size times that category’s probability. Its variance is \(n\pi_j(1-\pi_j)\). The standard deviation of the count, if needed, is the square root of that variance and is measured in individuals.

An expected count need not be an integer. It is a long-run average over repeated samples of size \(n\), not a prediction that one particular sample must have that count. The variance describes the spread of the count over those repeated samples.

Covariances Between Genotype Counts

Counts in different genotype categories share a fixed total. If more individuals fall in one category, fewer are available for the others. For two distinct categories \(i\) and \(j\), the covariance is

When both probabilities are positive and \(n>0\), this covariance is negative. It means that samples with a count above its mean in one category tend, in the covariance sense, to have a count below its mean in the other. Covariance is not a statement that every individual sample follows this pattern.

Why the Moment Formulas Hold

The formulas follow by representing each count as a sum of indicators. For sampled individual \(r\), define \(I_{rj}=1\) if that individual is in category \(j\), and \(I_{rj}=0\) otherwise. Then \(N_j=\sum_{r=1}^n I_{rj}\). Each \(I_{rj}\) is a Bernoulli random variable with probability \(\pi_j\) of being 1, so its expectation is \(\pi_j\) and its variance is \(\pi_j(1-\pi_j)\).

Linearity of expectation gives \(E[N_j]=\sum_{r=1}^n E[I_{rj}]=n\pi_j\). Since the indicators for a fixed category are independent across sampled individuals, the variance of their sum is the sum of their variances, giving \(\operatorname{Var}(N_j)=n\pi_j(1-\pi_j)\).

For distinct categories \(i\) and \(j\), one individual cannot belong to both. Thus \(I_{ri}I_{rj}=0\), and

Indicators belonging to different sampled individuals are independent, so their covariances are zero. Summing the within-individual covariances over the \(n\) individuals therefore gives \(\operatorname{Cov}(N_i,N_j)=-n\pi_i\pi_j\). This argument also shows why the formula gives zero covariance if either probability is zero.

A Useful Check: The Total Count Is Fixed

Because \(N_1+\cdots+N_k=n\) in every sample, the sum of the count variances and twice the sum of the distinct pairwise covariances must equal zero. For the three-category example above, the variance sum is \(30.0+25.2+19.2=74.4\). The sum of the three covariances is \(-18.0-12.0-7.2=-37.2\), so

This agrees with the variance of the fixed total, which is zero. More generally, the count covariance formulas satisfy this constraint because \(\sum_j\pi_j=1\). It is a useful arithmetic check, but it does not replace checking each probability or calculation.

Common Mistakes and Practical Interpretation

• Treating the genotype probabilities as sample proportions. The \(\pi_j\) values are the population probabilities specified by the model. The observed proportions \(N_j/n\) can differ from them because of sampling variation.
• Assuming the multinomial model requires Hardy–Weinberg equilibrium. It does not. Any valid set of genotype probabilities can be used, provided the sampling assumptions hold.
• Calling all distinct counts dependent without checking for zero probabilities. Positive probabilities for both categories and \(n>0\) give negative covariance and dependence. A zero-probability category has a constant zero count, which is independent of the others.
• Interpreting a negative covariance as a rule for every sample. The covariance summarizes how two counts vary jointly over repeated samples; it does not require every sample with one count above expectation to have the other below expectation.
• Forgetting the fixed total. The counts cannot vary independently when their sum is fixed and the category probabilities are positive. Their negative covariances are part of the model, not a calculation defect.