Why Trial Counts Are Not Generic Weights

Why the Meaning of a Weight Matters

A group with 3 successes in 10 trials is not a single Bernoulli observation equal to 1, nor is its observed success proportion a Bernoulli outcome. The count comes from 10 trials, and that sampling structure determines its likelihood and variance. Yet a weighted Bernoulli likelihood can produce the same score equations as the grouped binomial likelihood. The key is to match both the weight and the response correctly.

This distinction matters because matching score equations means the parameter estimates may agree, but it does not turn every weighted response into a binomial count or make the two sampling interpretations interchangeable. The calculation below isolates exactly what agrees and what does not.

Two Likelihood Contributions

Suppose groups \(i=1,\ldots,m\) are independent. In group \(i\), \(n_i\) is a positive integer number of trials and \(y_i\) is the number of successes, where \(y_i\in\{0,\ldots,n_i\}\). All trials in that group have success probability \(p_i\). Let \(p_i=h(\eta_i)\), where \(h\) is a differentiable inverse link, \(\eta_i=x_i^{\mathsf T}\beta+o_i\), \(x_i\) is a fixed design vector, and \(o_i\) is a fixed offset. We consider parameter values for which \(0<p_i<1\).

Under the grouped binomial model, \(Y_i\sim\operatorname{Binomial}(n_i,p_i)\), and the log-likelihood contribution is

Alternatively, record the observed proportion \(r_i=y_i/n_i\) and form a Bernoulli log-likelihood kernel weighted by \(n_i\):

This weighted expression is a likelihood kernel in \(p_i\), not a claim that the fractional value \(r_i\) is itself a Bernoulli observation. Substituting \(r_i=y_i/n_i\) gives \(n_i r_i=y_i\) and \(n_i(1-r_i)=n_i-y_i\). Thus the two expressions contain the same terms that depend on \(\beta\); the binomial expression also contains a combinatorial term that does not depend on \(\beta\).

When the Score Equations Coincide

For the binomial model, differentiate the parameter-dependent part of the contribution with respect to \(\eta_i\). Since \(p_i=h(\eta_i)\), the chain rule gives

The offset is fixed, so \(\partial\eta_i/\partial\beta=x_i\). The full score contribution is therefore the last expression multiplied by \(x_i\). For the weighted kernel, the derivative with respect to \(\eta_i\) is

When \(r_i=y_i/n_i\), this is exactly the binomial derivative. The equality holds for any differentiable inverse link on the stated interior parameter range; it is not limited to the logit link.

Proof. For each \(i\), substitution of \(r_i=y_i/n_i\) into the weighted contribution gives \(\ell_i^{\mathrm{wt}}=y_i\log p_i+(n_i-y_i)\log(1-p_i)\). Hence \(\ell_i^{\mathrm{bin}}-\ell_i^{\mathrm{wt}}=\log{n_i\choose y_i}\). Summing over groups gives the stated constant difference. Differentiating that constant with respect to \(\beta\) gives zero, so the score vectors agree. Equivalently, the chain-rule derivatives above agree group by group, and multiplication by \(x_i\) and summation preserve equality. \(\square\)

Consequently, where the maxima exist, the two expressions have the same maximizing parameter values. Their observed derivatives of every order also agree wherever those derivatives exist, since their difference is constant in \(\beta\). This statement concerns the likelihood as a function of the model parameters; the data probability model still has to be specified correctly.

Why the Binomial Constant Is There

The binomial coefficient has a sampling interpretation. Suppose there are \(n\) independent Bernoulli trials, each with success probability \(p\), and the recorded data are the ordered trial outcomes. Any particular ordered sequence with \(y\) successes has probability \(p^y(1-p)^{n-y}\). If instead the data record only the count \(y\), there are \({n\choose y}\) different ordered sequences that produce that count. Adding their probabilities gives

This explains why the probability of a count includes the combinatorial factor, while the likelihood for one particular ordered sequence does not. The factor is constant with respect to \(p\), so it does not change the score or likelihood ratios for this model. But it is part of the probability assigned to the recorded count.

Proof. There are \({n\choose y}\) ways to choose which \(y\) of the \(n\) trial positions are successes. By independence, each such sequence has probability \(p^y(1-p)^{n-y}\). These sequences are mutually exclusive and exhaust the event that the count is \(y\). Summing their equal probabilities gives \({n\choose y}p^y(1-p)^{n-y}\), as claimed. \(\square\)

The same counting argument also clarifies the interpretation of the weighted kernel: after aggregation, its parameter-dependent part matches the likelihood of the count, but the kernel alone does not include the factor that makes it the binomial probability mass function.

When a Trial Count Is Not the Right Weight

A common mistake is to take a single binary observation and multiply its Bernoulli log-likelihood by the number of people in its group, without first replacing the binary response by the group’s success proportion. That calculation treats every member of the group as if their outcome were the one observed binary value. It is not the grouped binomial likelihood.

A generic weight \(w_i\) and an observed binary response \(Y_i\in\{0,1\}\) contribute \(w_i\{Y_i\log p_i+(1-Y_i)\log(1-p_i)\}\). That is a different expression from a grouped binomial contribution unless the response and weight encode the matching proportion and trial count. In particular, assigning a group’s trial count to a single binary response does not reconstruct the group’s successes and failures.

The distinction is also reflected in variability: under the binomial sampling model, \(\operatorname{Var}(Y_i)=n_i p_i(1-p_i)\), so the proportion has variance \(\operatorname{Var}(Y_i/n_i)=p_i(1-p_i)/n_i\). A weighted fractional-response kernel does not, by itself, assert that a fractional observation arose from that binomial sampling process. Use the binomial model when the data are success counts out of known numbers of trials; use a weighted Bernoulli expression only when its response and weight have the intended meaning.