Common Errors in Quasi-Binomial and Quasi-Poisson Analysis

Introduction: Errors About What Dispersion Changes

Quasi-binomial and quasi-Poisson models allow the variance to exceed the ordinary binomial or Poisson variance while retaining a specified mean model. This flexibility is useful, but it is easy to draw the wrong conclusion from it. A dispersion estimate changes the uncertainty calculations; it does not, by itself, multiply the fitted regression coefficients or change the fitted means.

Several related mistakes can undermine an otherwise sensible analysis: treating a grouped binomial proportion as if it came from one trial, ignoring exposure or weights, using a likelihood-ratio test when no full likelihood has been specified, or describing a larger standard error as a change in the estimated mean. The examples below show how to detect these errors and what calculation to use instead.

Keep the Mean Model, Variance, and Inference Distinct

Definition: In a quasi-GLM, the mean is specified as \(E[Y_i]=\mu_i\), with a link relation such as \(g(\mu_i)=x_i^\mathsf{T}\beta+o_i\), where \(x_i\) is the covariate vector and \(o_i\) is an optional offset. A variance function specifies \(\operatorname{Var}(Y_i)=\phi V(\mu_i)\), possibly including known prior weights or binomial trial totals. The dispersion parameter \(\phi\) scales the variance; the quasi-model does not, by that specification alone, give a full probability distribution or likelihood.

For Poisson counts, the quasi-Poisson variance function is \(\operatorname{Var}(Y_i)=\phi\mu_i\). For grouped binomial counts \(Y_i\) out of \(m_i\) trials, the quasi-binomial variance is \(\operatorname{Var}(Y_i)=\phi m_i p_i(1-p_i)\), where \(p_i=E[Y_i]/m_i\). Equivalently, for the observed proportion \(R_i=Y_i/m_i\), its variance is \(\phi p_i(1-p_i)/m_i\). These expressions show why the number of trials is not a cosmetic detail.

The central calculation is the quasi-score equation. For independent observations with a common scalar \(\phi\), it has the form

U(\beta)=\frac{1}{\phi}X^\mathsf{T}D(\beta)(y-\mu)=0,

where \(D(\beta)\) contains the mean- and variance-function weights. Multiplication of the whole equation by the nonzero constant \(\phi\) does not change its roots. Thus, when the mean structure, observations, weights, and offsets are the same, the dispersion factor does not change the coefficient estimate. It does change the estimated covariance under the usual model-based quasi-GLM calculation.

Result: If a quasi-score equation is \(U(\beta)=\phi^{-1}U_0(\beta)=0\) with one positive scalar \(\phi\), its solutions are exactly the solutions of \(U_0(\beta)=0\). The fitted linear predictors and means therefore remain the same when only \(\phi\) changes. This statement assumes the same data, mean model, and other weights; changing those can change the fitted coefficients.

The reason is direct: \(\phi>0\), so \(U(\beta)=0\) if and only if \(\phi U(\beta)=U_0(\beta)=0\). This is not a claim that every adjustment described as “overdispersion” leaves every fit unchanged. It applies to a common scalar multiplier in the quasi-score, not to changing the response, design, offsets, link, or observation weights.

Worked Example: Dispersion Does Not Multiply the Coefficient

Suppose four independent counts are fitted with a Poisson log-link intercept only, and the observed counts are \(2,4,6,8\). The quasi-Poisson estimating equation for the intercept \(\beta_0\) is

\sum_{i=1}^4 (y_i-\mu_i)=0,\qquad \mu_i=e^{\beta_0}.

The equation gives \(2+4+6+8-4e^{\beta_0}=0\), so \(4e^{\beta_0}=20\), \(\hat\mu=5\), and \(\hat\beta_0=\log(5)\approx1.609\). With dispersion \(\phi=2\), the score is divided by 2, but its zero remains at the same \(\hat\beta_0\). Doubling the coefficient to \(3.218\) would instead imply a mean of \(e^{3.218}=25\), which does not solve the score equation.

Using the usual model-based covariance for this intercept-only Poisson fit, \(\widehat{\operatorname{Var}}(\hat\beta_0)=\hat\phi/\sum_i\hat\mu_i=2/20=0.1\), and the standard error is \(\sqrt{0.1}\approx0.316\). With \(\phi=1\), it would be \(\sqrt{1/20}\approx0.224\). The coefficient and fitted means are unchanged; the estimated uncertainty is larger.

Why Standard Errors Change

For a GLM mean model, write \(W\) for the usual iterative-weight matrix, constructed from the variance function, link, and any specified prior weights. The model-based quasi-GLM covariance, with known \(\phi\), is \(\phi(X^\mathsf{T}WX)^{-1}\), evaluated at the fitted mean. When dispersion is estimated, \(\phi\) in this expression is replaced by its estimate. This is a model-based calculation: it relies on the stated variance structure and is not automatically a robust covariance estimate for arbitrary dependence or misspecification.

Result: Under the quasi-GLM covariance formula, if the fitted means and \(W\) are held fixed, multiplying \(\phi\) by \(c>0\) multiplies the coefficient covariance matrix by \(c\) and each coefficient standard error by \(\sqrt{c}\). It does not multiply the coefficient estimates by \(c\).

To verify this, let \(A=X^\mathsf{T}WX\), assumed nonsingular. The covariance at dispersion \(\phi\) is \(\phi A^{-1}\); at dispersion \(c\phi\), it is \(c\phi A^{-1}=c(\phi A^{-1})\). A standard error is the square root of a diagonal covariance entry, so it is multiplied by \(\sqrt{c}\). The result concerns uncertainty for the same fitted mean model, not changes caused by refitting a different model.

Worked Example: A Standard Error Is Not a Fitted Mean

For a fitted log-link model, suppose a particular covariate has estimate \(\hat\beta=0.40\) and model-based standard error \(0.10\) when \(\hat\phi=1\). If the estimated dispersion becomes 4 while the fitted mean and weight matrix stay fixed, the covariance is multiplied by 4 and the standard error by \(\sqrt{4}=2\). The new standard error is \(0.20\); the coefficient estimate remains \(0.40\).

For a one-unit increase in the covariate, the estimated mean ratio is \(\exp(0.40)\approx1.492\) both before and after this dispersion change. The uncertainty around that estimate is greater under \(\hat\phi=4\), but the point estimate of the ratio has not become \(\exp(4\times0.40)\approx4.953\). That latter calculation incorrectly treats a variance adjustment as a coefficient adjustment.

Binomial Trial Totals Must Stay With the Response

A grouped binomial observation records a success count \(Y_i\) and a trial total \(m_i\). Its estimated success proportion is \(Y_i/m_i\), but this proportion alone does not retain how much information was observed. A proportion based on 2 trials is generally less precise than the same proportion based on 200 trials. In a grouped binomial GLM, the totals must therefore be represented correctly, whether the software takes success and failure counts or a proportion with the corresponding trial weight.

The distinction follows from the score. For a logit mean \(p_i\), the grouped-binomial score contribution for the linear predictor is proportional to \(Y_i-m_i p_i\). Replacing \(Y_i\) with \(Y_i/m_i\) while dropping \(m_i\) changes the estimating equation: it treats groups with different trial totals as if they carried equal information.

Worked Example: Unequal Trial Totals Change the Intercept

Consider two groups in a binomial model with an intercept only. Group 1 has \(Y_1=2\) successes in \(m_1=10\) trials, and group 2 has \(Y_2=50\) successes in \(m_2=100\) trials. The correct fitted probability solves

$$ (Y_1-m_1p)+(Y_2-m_2p)=0. $$

Substitution gives \(2-10p+50-100p=0\), hence \(110p=52\) and \(\hat p=52/110\approx0.473\). Equivalently, the fitted probability is the total successes divided by total trials.

If the two observed proportions, \(0.2\) and \(0.5\), are incorrectly treated as two equally weighted observations, the intercept-only estimate is their unweighted mean, \((0.2+0.5)/2=0.35\). The difference, \(0.473\) versus \(0.35\), is not a dispersion adjustment: it comes from omitting the trial totals and thereby fitting a different estimating equation.

Do Not Use a Likelihood-Ratio Test for a Quasi-Model

A likelihood-ratio test compares maximized likelihoods from specified probability models. A quasi-likelihood specifies a mean–variance relationship and a quasi-score, but it need not specify a probability mass function or density whose likelihood can be maximized. Consequently, a deviance difference from quasi-GLMs is not automatically a likelihood-ratio statistic, and the usual chi-square reference for a likelihood-ratio test is not justified just by having computed that difference.

For nested quasi-GLMs, a scaled deviance comparison may instead be used as an approximate \(F\) test under the relevant model and sampling assumptions, with dispersion estimated as appropriate. This is an approximate quasi-GLM inference procedure, not an exact likelihood-ratio test. Its use does not establish that a full probability distribution has been specified.

Worked Example: A Deviance Difference Is Not Automatically a Likelihood Ratio

Suppose a reduced quasi-Poisson model has deviance \(D_R=30\), and a nested full model has deviance \(D_F=20\) with 10 residual degrees of freedom. Suppose the deviance-based dispersion estimate for the full model is \(\hat\phi=D_F/10=2\), and the full model adds \(q=2\) coefficients. The scaled deviance comparison is

F=\frac{(D_R-D_F)/q}{\hat\phi} =\frac{(30-20)/2}{2}=2.5.

Under the usual approximate nested quasi-GLM procedure, this is compared with an \(F\) reference distribution with 2 and 10 degrees of freedom. The calculation is not a likelihood-ratio test: labeling the deviance difference 10 as a chi-square statistic with 2 degrees of freedom would require likelihood justification that the quasi specification alone does not supply.

Common Mistakes and a Practical Check

Multiplying coefficients by dispersion. A common scalar in the quasi-score cancels from its zero. Dispersion scales the usual model-based covariance, not the fitted coefficient vector.
Dropping trial totals. A proportion must retain its binomial denominator, either explicitly or through the appropriate weight. Equal proportions from unequal numbers of trials do not generally contribute equally.
Calling a deviance comparison a likelihood-ratio test. Quasi-likelihood does not automatically provide a full likelihood. Use an appropriate quasi-GLM inference procedure and state its approximate assumptions.
Confusing a mean estimate with its uncertainty. A larger standard error means less precision under the specified calculation; it does not mean the fitted mean or effect estimate has been multiplied.
Changing weights or offsets while attributing the result to dispersion. A changed data representation, trial total, exposure offset, or observation weight can change the estimating equation and hence the fitted means. A scalar dispersion adjustment alone cannot explain that change.

A useful diagnostic is to ask four separate questions: Did the mean model or data representation change? Were trial totals, offsets, and weights carried through correctly? Is the reported quantity a coefficient, a fitted mean, or an uncertainty measure? Does the proposed test require a likelihood that the model does not specify? Keeping these questions separate prevents most interpretive errors.

Check Your Understanding

Use the distinctions in this tutorial to assess each statement.

A quasi-Poisson fit has the same mean model and data as a Poisson fit, but its estimated dispersion is larger. Which quantities can change, and which remain unchanged under the common-scalar quasi-score setup?
Two groups have the same observed success proportion, but one has ten times as many trials. Why should their binomial contributions not be treated as equally informative?
A researcher reports a chi-square likelihood-ratio test after fitting two quasi-Poisson models. What assumption or justification is missing?
If the dispersion estimate is multiplied by 9 and the fitted means and weight matrix are unchanged, by what factor does the model-based standard error change?
Can a fitted mean change after altering trial weights even though the dispersion parameter is unchanged? Explain why.