When Bernoulli Records Collapse to a Binomial Count

From Individual Outcomes to a Count

Suppose \(m\) labeled units each produce a binary outcome: success or failure. If every unit has the same success probability \(p\), and the outcomes are independent, the record is a collection of \(m\) independent Bernoulli observations. Sometimes the individual outcomes are recorded; sometimes only the total number of successes is available. The key question is whether replacing the full record by that total changes what the data say about \(p\).

Under the stated common-probability and independence assumptions, the count has a binomial distribution. This is more than a convenient description: its likelihood contains all the dependence on \(p\) present in the labeled records. The count does discard which units succeeded, so it does not preserve every detail of the observed data.

Deriving the Count Likelihood

For one particular labeled record \(\mathbf{y}=(y_1,\ldots,y_m)\), where each \(y_i\) is either 0 or 1, independence gives its probability. Write \(s=\sum_{i=1}^m y_i\). There are \(s\) factors of \(p\) and \(m-s\) factors of \(1-p\), so the likelihood of this particular record is

That expression is not yet the probability of observing the count \(s\). The event \(S=s\) includes every labeled record with exactly \(s\) successes. There are \(\binom{m}{s}\) such records, and each has the same probability under the common-\(p\) model. Adding their probabilities gives

Thus, if only the count is observed, its exact likelihood includes the binomial coefficient. For likelihood comparisons or estimation of \(p\) with \(m\) fixed, that coefficient is constant in \(p\), so the count likelihood and the full-record likelihood have the same \(p\)-dependent part.

What the Count Retains About the Parameter

A statistic is sufficient for a parameter when the observed statistic retains all the sample information about that parameter. Here this statement can be made precise in two complementary ways: the joint probability factors through \(S\), and, given \(S\), the arrangement of successes is distributed independently of \(p\).

Proof. For any record \(\mathbf{y}\), its joint probability is \(p^s(1-p)^{m-s}\), where \(s=\sum_i y_i\). This factors as a function of the record through \(s\), multiplied by a factor that does not otherwise depend on \(p\):

By the factorization theorem, \(S\) is sufficient for \(p\). For the conditional statement, take any record \(\mathbf{y}\) with \(s\) successes. Since \(0<p<1\), \(P_p(S=s)>0\), and

The conditional distribution of the arrangement is therefore parameter-free. Once the count is known, learning which particular units succeeded gives no additional information about \(p\) in this model. \(\square\)

Equality of Fisher Information

Sufficiency has a direct information consequence. For \(0<p<1\), the Fisher information in a differentiable likelihood is the expected square of its score, where the score is the derivative of the log-likelihood with respect to \(p\). The full record and its count have the same Fisher information for \(p\).

Derivation. For the full record, the log-likelihood is \(s\log p+(m-s)\log(1-p)\). Its score is

Because \(S\) is the sum of \(m\) independent Bernoulli variables, \(E[S]=mp\) and \(\operatorname{Var}(S)=mp(1-p)\). The score has mean zero, and its expected square is

For the count likelihood, the log-likelihood adds the term \(\log\binom{m}{s}\), which is constant in \(p\). Its score is consequently the same expression, \((s-mp)/[p(1-p)]\), and its expected square is the same. This proves the equality. The result concerns information about \(p\); it does not mean the count preserves the identities or ordering in the record. \(\square\)

What Aggregation Does and Does Not Preserve

Under the model’s assumptions, replacing \((Y_1,\ldots,Y_m)\) by \(S\) preserves the likelihood information about the single common probability \(p\). It also preserves any likelihood-based comparison about \(p\) that uses these observations and this model. But aggregation is not lossless in every sense: after retaining only \(S\), one cannot recover which units succeeded.

The premise matters. The derivation used independence and the same probability \(p\) for every outcome. If probabilities differ across units, a particular arrangement can matter: placing successes on units with different probabilities can change its likelihood even when the total count stays fixed. Whether a grouped binomial model is appropriate when probabilities may vary within a group requires separate consideration; the count derivation alone does not establish that premise.

Common Mistake

• Wrong reasoning: “The observed data contain four successes, so their likelihood is just \(p^4(1-p)^{m-4}\), whether I observed the individual records or only the count.”
• Correction: That expression is the likelihood of one particular labeled arrangement. If the observation is only the count, its exact probability sums over all arrangements, giving \(\binom{m}{4}p^4(1-p)^{m-4}\). The coefficient is irrelevant to inference about \(p\) when \(m\) is fixed, but it is part of the count’s probability.