Within-Group Homogeneity and Valid Aggregation

Why Within-Group Homogeneity Matters

A grouped binomial record replaces several individual binary outcomes with a trial total and a success count. This is a useful reduction only when the count has the distribution the binomial model assumes. In particular, individuals grouped together must have a common success probability after conditioning on the information that defines the group, and their outcomes must be conditionally independent.

The condition is about probabilities, not observed outcomes. A group can contain both successes and failures and still be homogeneous if each trial has the same probability of success. Conversely, a group can have a single recorded covariate label yet be heterogeneous if individuals differ in unrecorded or ignored predictors that change their conditional probabilities.

The conditional qualification matters in regression. For example, a study might define a group by treatment and age category. The common-probability condition is assessed after conditioning on the group’s specified predictors. If individuals in the same group have different relevant predictor values, a shared group probability may not be justified.

What Common Probability Gives You

Suppose the group has \(n\) conditionally independent outcomes, each with success probability \(p\). For any particular pattern with \(k\) successes and \(n-k\) failures, independence gives probability \(p^k(1-p)^{n-k}\). The count \(S\) equals \(k\) for exactly \(\binom{n}{k}\) such patterns. Adding their probabilities gives

This pattern-counting argument explains why the binomial formula depends on both assumptions. Independence gives the probability of a pattern as a product, and homogeneity makes that product the same for every pattern with \(k\) successes. If success probabilities differ, patterns with the same number of successes need not have the same probability.

Heterogeneous Probabilities Change the Count Distribution

When independent Bernoulli outcomes have different probabilities \(p_1,\ldots,p_n\), their sum is generally called a Poisson-binomial count. Its probability-generating function is the product of the individual generating functions:

This product generally does not equal the generating function of a binomial count with one common probability. Replacing the \(p_i\) values with their average can preserve the expected count, but it does not generally preserve the distribution.

Proof. If all \(p_i=q\), the pattern-counting argument above gives the binomial probability formula, proving one direction. For the converse, suppose \(S\sim\operatorname{Binomial}(n,q)\). Write \(\bar p=n^{-1}\sum_{i=1}^{n}p_i\). Linearity of expectation gives \(E[S]=\sum_i p_i=n\bar p\); the assumed binomial distribution gives \(E[S]=nq\). Hence \(\bar p=q\).

Independence gives \(\operatorname{Var}(S)=\sum_i p_i(1-p_i)\). Using \(\sum_i p_i=nq\), rewrite this variance as

The assumed binomial distribution has variance \(nq(1-q)\). Equating the two variance expressions yields \(\sum_i(p_i-q)^2=0\). Each term is nonnegative, so every term is zero and \(p_i=q\) for all \(i\). This proves the converse. \(\square\)

The theorem concerns independent trials with fixed conditional probabilities. It gives a precise reason that averaging heterogeneous probabilities is not a general remedy: the average can match the mean, but unless all probabilities equal that average, the count has a different variance and is not binomial.

Independence Is a Separate Requirement

A common probability does not by itself guarantee a binomial count. For example, let two outcomes be perfectly dependent: \(Y_1=Y_2=Z\), where \(Z\) is Bernoulli with probability \(0.5\). Both outcomes have the same success probability, but the count is 0 with probability \(0.5\) and 2 with probability \(0.5\); it can never equal 1. A \(\operatorname{Binomial}(2,0.5)\) count instead assigns probabilities \(0.25,0.5,0.25\) to counts 0, 1, and 2. Thus matching marginal probabilities without conditional independence is insufficient.

In applications, dependence may arise from shared unmeasured conditions, repeated observations on the same person, or other clustering. This tutorial’s binomial conclusions apply when outcomes are independent conditional on the specified group information. If that assumption is not credible, the ordinary binomial model needs more than a common probability to justify it.

Common Mistake: Matching the Mean Is Not Enough

A tempting argument is: “The individual probabilities average to \(q\), so the total count is \(\operatorname{Binomial}(n,q)\).” The first part can correctly identify the mean, since \(\sum_i p_i=nq\); the conclusion does not follow. The four-probability example had mean 2, just like \(\operatorname{Binomial}(4,0.5)\), but its variance was \(0.8\) rather than 1. The correction is to check whether the probabilities are genuinely common within the group and whether conditional independence is justified. Only then does the usual binomial count model follow.