Quasi-Binomial Estimation and Inference

Why a Binomial Mean Model May Need More Variation

For grouped binary data, an observation often consists of $Y_i$ successes among $m_i$ trials. A binomial model with success probability $p_i$ gives the count variance $m_i p_i(1-p_i)$. In practice, observations in a group may be more variable than that model allows. A quasi-binomial model keeps the same mean structure but multiplies the variance by a dispersion parameter. It therefore changes the estimating equations’ uncertainty calculations without asserting a complete probability distribution for the counts.

This distinction matters in inference. The fitted mean can remain the same as under a binomial model, while standard errors become larger when the estimated dispersion exceeds one. The calculations below use independent grouped observations, fixed trial counts, and a correctly specified mean model. Inference is approximate and relies on having enough independent groups for the estimating-equation approximation to be useful.

Definition: For independent grouped counts $Y_1,\ldots,Y_G$, with fixed trial counts $m_i$, a quasi-binomial mean and variance specification is $$ E[Y_i]=\mu_i=m_i p_i,\qquad \operatorname{Var}(Y_i)=\phi m_i p_i(1-p_i), $$ where $0<p_i<1$ and $\phi>0$ is a common dispersion parameter. Equivalently, for the observed proportion $\bar Y_i=Y_i/m_i$, $$ E[\bar Y_i]=p_i,\qquad \operatorname{Var}(\bar Y_i)=\frac{\phi p_i(1-p_i)}{m_i}. $$

The term “quasi-binomial” describes these moment assumptions and the resulting estimating procedure; it does not, by itself, specify that $Y_i$ has a binomial distribution. When $\phi=1$, the variance agrees with the binomial variance. Values above one represent extra-binomial variation relative to that benchmark. A value below one represents less variation.

Deriving the Estimating Equation

Let $g$ be a differentiable link function, with $g(p_i)=\eta_i=x_i^\mathsf{T}\beta$. Here $x_i$ is a known $q$-vector of covariates and $\beta$ is the $q$-vector of regression coefficients. Write $p_i=g^{-1}(\eta_i)$, and let $p_i'=d p_i/d\eta_i$. The quasi-score estimating function uses the derivative of the mean and the specified variance:

U(\beta)=\sum_{i=1}^G \frac{\partial\mu_i}{\partial\beta} \frac{Y_i-\mu_i}{\operatorname{Var}(Y_i)}.

The expression is a vector equation, with one component for each coefficient. The dispersion factor is included in the variance, so it scales the equation but does not alter its root when $\phi$ is a common positive constant.

Result: Under the quasi-binomial variance specification and link model above, the estimating equation is $$ U(\beta)=\frac{1}{\phi}\sum_{i=1}^G x_i\frac{p_i'}{p_i(1-p_i)}(Y_i-m_i p_i)=0. $$ For the logit link, $p_i'=p_i(1-p_i)$, so this reduces to $$ \sum_{i=1}^G x_i(Y_i-m_i p_i)=0. $$

To verify the general expression, differentiate $\mu_i=m_i p_i$ with respect to $\beta$. The chain rule gives $\partial\mu_i/\partial\beta=m_i p_i'x_i$. Substituting this derivative and $\operatorname{Var}(Y_i)=\phi m_i p_i(1-p_i)$ into the defining quasi-score cancels $m_i$:

\begin{aligned} U(\beta) &=\sum_{i=1}^G \frac{m_i p_i'x_i}{\phi m_i p_i(1-p_i)} (Y_i-m_i p_i)\\ &=\frac{1}{\phi}\sum_{i=1}^G x_i\frac{p_i'}{p_i(1-p_i)}(Y_i-m_i p_i). \end{aligned}

For the logit link, $\eta_i=\log\{p_i/(1-p_i)\}$, and differentiation gives $p_i'=p_i(1-p_i)$. The fraction in the general equation is then one. Since multiplication by $1/\phi$ does not change the solutions of an equation set equal to zero, the logit quasi-score root is the same as the binomial log-likelihood score root. This equality of coefficient estimates does not imply equality of their standard errors.

Worked Example: A Common Proportion With a Logit Link

Suppose two independent groups have $m_1=20$ and $m_2=30$ trials, with $Y_1=14$ and $Y_2=21$ successes. Fit a common proportion $p$, so the model has only an intercept and uses the logit link. Suppose a working dispersion value of $\phi=1.5$ is supplied.

For an intercept-only logit model, the estimating equation is $\sum_i(Y_i-m_i p)=0$. Thus $35-(20+30)p=0$, and the fitted proportion is $\hat p=35/50=0.70$. The fitted log odds are $\hat\beta=\log(0.70/0.30)=0.8473$.

The information quantity without the dispersion multiplier is $H=\sum_i m_i\hat p(1-\hat p)=50(0.70)(0.30)=10.5$. Therefore the model-based variance estimate for the intercept is $\hat\phi/H=1.5/10.5=0.142857$, and its standard error is $\sqrt{0.142857}=0.3780$. On the proportion scale, the delta-method standard error is $\sqrt{\hat\phi\hat p(1-\hat p)/50}=\sqrt{0.0063}=0.0794$.

A direct 95% Wald interval on the proportion scale is $0.70\mathbin{\pm}1.96(0.0794)=[0.544,0.856]$, approximately. It stays inside the allowable probability range in this example. In other cases a direct Wald interval can extend below zero or above one, which is a reason to form an interval on the link scale and transform its endpoints.

Standard Errors and the Role of Dispersion

For the logit link, define $H(\beta)=\sum_i m_i p_i(1-p_i)x_i x_i^\mathsf{T}$. The derivative of the logit estimating function with respect to $\beta$ is $-H(\beta)$. The variability of the unscaled estimating function $\sum_i x_i(Y_i-m_i p_i)$ is $\phi H(\beta)$: independence makes the cross-covariances zero, and each residual has variance $\phi m_i p_i(1-p_i)$. These two facts give the large-sample covariance of the root.

Result: If the independent observations satisfy the stated mean and variance model, the mean model is correctly specified, $H(\beta)$ is nonsingular in the limit, and the usual regularity conditions for estimating-equation asymptotics hold, then the model-based covariance estimate for a logit coefficient estimator is $$ \widehat{\operatorname{Cov}}(\hat\beta) =\hat\phi\left\{\sum_{i=1}^G m_i\hat p_i(1-\hat p_i)x_i x_i^\mathsf{T}\right\}^{-1}. $$

For completeness, let $S(\beta)=\sum_i x_i(Y_i-m_i p_i)$, and write $H$ for its negative derivative near the true coefficient. A first-order expansion of $S(\hat\beta)=0$ about the true coefficient $\beta$ gives $\hat\beta-\beta$ approximately $H^{-1}S(\beta)$. Independence and the variance model give $\operatorname{Var}\{S(\beta)\}=\phi H$. Consequently, $\operatorname{Cov}(\hat\beta)$ is approximately $H^{-1}(\phi H)H^{-1}=\phi H^{-1}$. Replacing unknown quantities by their fitted values yields the displayed estimate. This derivation explains why the dispersion multiplies the covariance, not the standard error: standard errors are multiplied by $\sqrt{\hat\phi}$.

A common moment estimate of dispersion is the Pearson statistic divided by residual degrees of freedom: $\hat\phi=X_P^2/(G-r)$, where $X_P^2=\sum_i (Y_i-m_i\hat p_i)^2/\{m_i\hat p_i(1-\hat p_i)\}$ and $r$ is the rank of the fitted design matrix. This is a working estimate, not a claim that the ratio is exactly unbiased in every finite sample. It requires positive residual degrees of freedom. In particular, a saturated model with $G=r$ cannot estimate dispersion from this ratio.

Worked Example: A Log-Odds Contrast and Its Interval

Consider two groups fitted by a saturated two-group logistic mean model. Group A has 20 successes among 30 trials, and group B has 10 successes among 30 trials. Let $\beta_0$ be the log odds in group B and $\beta_1$ the difference in log odds, A minus B. For this calculation, use a dispersion value $\phi=1.4412$, supplied from residual information beyond this saturated two-group table.

The fitted probabilities are $\hat p_A=20/30=2/3$ and $\hat p_B=10/30=1/3$. Their fitted log odds are $\log((2/3)/(1/3))=\log 2=0.6931$ and $\log((1/3)/(2/3))=-0.6931$. Hence $\hat\beta_1=0.6931-(-0.6931)=1.3863$, and the estimated odds ratio is $\exp(1.3863)=4$.

The information contributions for the two group-specific logits are $H_A=30(2/3)(1/3)=6.6667$ and $H_B=30(1/3)(2/3)=6.6667$. The variance of the difference is $\hat\phi(1/H_A+1/H_B)=1.4412(0.15+0.15)=0.43236$, so its standard error is $\sqrt{0.43236}=0.6575$. A 95% Wald interval for the log odds ratio is $1.3863\mathbin{\pm}1.96(0.6575)=[0.0976,2.6750]$, approximately. Exponentiating the endpoints gives an odds-ratio interval of approximately $[1.10,14.52]$.

The example illustrates both the role and the limits of a dispersion value: multiplying the binomial-based covariance by $\phi$ widens the interval, but the coefficient estimate remains the same. Since the table is saturated, the table alone has no residual degrees of freedom from which to estimate $\phi$.

Wald Intervals on the Link and Response Scales

For a scalar coefficient $\beta_j$, a large-sample Wald interval is $\hat\beta_j\mathbin{\pm}z_{1-\alpha/2}\operatorname{SE}(\hat\beta_j)$, where $z_{1-\alpha/2}$ is a standard normal quantile. The endpoints are first formed on the coefficient scale. If the coefficient is a log odds or another transformed parameter, applying the inverse transformation to each endpoint gives an interval on the corresponding scale. For a fitted probability, use its gradient with respect to $\beta$ and the estimated coefficient covariance to obtain a delta-method standard error, or form an interval for the linear predictor and transform its endpoints through the inverse link.

Worked Example: A Fitted Probability Interval From a Logit Model

Suppose a two-group logistic model uses group 1 as the reference category. In that group, 12 of 40 trials are successes, so the fitted probability is $0.30$. Use $\phi=1.2$, and suppose the goal is an approximate interval for the group 1 probability.

The fitted reference logit is $\hat\beta_0=\log(0.30/0.70)=-0.8473$. Its information contribution is $H_1=40(0.30)(0.70)=8.4$; because this group’s logit is $\beta_0$, its estimated variance is $1.2/8.4=0.142857$, giving standard error $0.3780$. The 95% interval for the logit is $-0.8473\mathbin{\pm}1.96(0.3780)=[-1.5881,-0.1065]$, approximately.

The inverse-logit function is $\operatorname{expit}(t)=1/(1+\exp(-t))$. Applying it to the two endpoints gives $\operatorname{expit}(-1.5881)\approx0.1696$ and $\operatorname{expit}(-0.1065)\approx0.4734$. Thus the transformed interval is approximately $[0.170,0.473]$. It remains within $[0,1]$ because it was constructed on the logit scale.

Common Pitfalls and Interpretation

The principal practical consequence of quasi-binomial estimation is straightforward: the mean model specifies fitted proportions, while the variance specification determines how uncertain those estimates are. A dispersion estimate larger than one inflates standard errors by its square root, all else equal. It does not change the fitted values for a common dispersion under the logit link.

Calling the model a binomial likelihood. Quasi-binomial estimation uses a mean and variance specification. Those moments alone do not determine the count distribution or justify likelihood-based calculations that require a full probability law.
Multiplying the standard error by $\hat\phi$. The covariance is multiplied by $\hat\phi$, so standard errors are multiplied by $\sqrt{\hat\phi}$.
Assuming the dispersion changes the logit coefficient estimates. A common dispersion scales the logit estimating equation but does not change its root. Its role here is in estimated uncertainty.
Estimating dispersion from a saturated table. With no residual degrees of freedom, the Pearson ratio described above is undefined. A separate source of information or a less saturated mean model is needed.
Interpreting an approximate Wald interval as exact. The normal approximation can be poor with few independent groups or probabilities near zero or one. Transforming a link-scale interval respects the probability range but does not remove the approximation behind the interval.

Key takeaway: For grouped quasi-binomial data, model the mean through a link, use $\phi m_i p_i(1-p_i)$ as the count variance, and carry the dispersion into the coefficient covariance. Under the logit link, the fitted coefficients solve the familiar grouped logistic score equation, but their standard errors and confidence intervals reflect the extra-binomial variance.

Check Your Understanding

Work through these questions using the definitions and equations above.

For a group with $m=80$, fitted proportion $p=0.25$, and dispersion $\phi=1.4$, calculate the model variance of the count and of the observed proportion.
Explain why a common dispersion parameter does not change the root of the logit quasi-score equation.
For a logit model, what matrix quantity determines the coefficient covariance apart from the multiplier $\phi$?
Why can a saturated grouped model fail to provide a Pearson-based estimate of dispersion?
If an estimated dispersion changes from 1 to 2.25 while the fitted mean model stays fixed, by what factor do the model-based standard errors change?