Why Overdispersion Changes Inference
A generalized linear model (GLM) specifies how the response mean depends on predictors and, through a variance function, how variability relates to that mean. If the observed responses vary more than the model's nominal variance allows, the coefficient estimates may still describe the fitted mean relationship, but their usual standard errors can be too small. That can make confidence intervals too narrow and tests too optimistic.
The central task is to separate two questions: is the mean structure appropriate, and is the assumed variance scale appropriate? A scale adjustment addresses the second question. It does not repair a wrong link, an omitted predictor, or a pattern of dependence between observations.
For the scale adjustment described here, assume observations are independent conditional on their predictors, the mean model and variance function are suitable, and the extra variability can reasonably be represented by one common multiplier \(\phi\). These assumptions matter: a single multiplier cannot describe every form of dependence or every change in variability.
Estimating a Common Dispersion Scale
A common estimate uses Pearson's chi-square statistic, the sum of squared Pearson residuals. If the model has \(n\) observations and \(p\) estimated regression coefficients, its residual degrees of freedom are \(n-p\). Under the stated common-scale assumptions, the statistic is approximately \(\phi(n-p)\), which motivates dividing it by the residual degrees of freedom.
This is an estimate, not a proof that overdispersion exists. Sampling variation can put the ratio above one, and residual patterns can signal that a single scale factor is not the right explanation. A deviance divided by residual degrees of freedom is also sometimes used as a rough scale estimate, but it is not identical to the Pearson estimate and its suitability depends on the model and data.
Worked Example: Adjusting a Coefficient Test
A hypothetical Poisson count model for monthly service requests has \(n=83\) independent observations and \(p=3\) fitted coefficients, including the intercept. Its Pearson statistic is \(Q_P=120\). A predictor's estimated coefficient is \(0.24\), with conventional model-based standard error \(0.08\). Estimate the scale, adjust the standard error, and compare the two-sided 5% Wald test with the unadjusted test.
Method. Estimate the common dispersion from Pearson's statistic, then multiply the conventional standard error by \(\sqrt{\hat{\phi}}\). This method addresses excess variance; it does not change the estimated coefficient.
Step 1: Residual degrees of freedom and scale. The residual degrees of freedom are \(83-3=80\). Thus \(\hat{\phi}=120/80=1.5\).
Step 2: Adjust the standard error. The adjusted standard error is \(0.08\sqrt{1.5}\). Since \(\sqrt{1.5}\approx1.2247\), this is \(0.09798\), or \(0.098\) to three decimals.
Step 3: Calculate the test statistics. Without adjustment, the Wald statistic is \(0.24/0.08=3.00\). With adjustment, it is \(0.24/0.09798\approx2.45\). Using a large-sample standard normal reference, the two-sided p-values are approximately \(0.0027\) and \(0.0143\), respectively.
Step 4: Interpret and check. Both tests reject at the 5% level, but the adjusted evidence is weaker. The adjustment has left the estimate \(0.24\) unchanged and increased its standard error. The common mistake is to multiply the coefficient by \(\hat{\phi}\); it is the variance that is multiplied by \(\hat{\phi}\), so the standard error is multiplied by its square root.
Why the Standard Error Is Multiplied by the Square Root
The reason follows from how information changes when the variance is inflated. For a GLM with the same mean structure and variance function, multiplying every response variance by \(\phi\) divides the corresponding information weights by \(\phi\). In the usual local large-sample approximation, the coefficient covariance matrix is the inverse of the weighted information matrix.
Let \(W\) denote the information-weight matrix when the scale is one. The usual approximate coefficient covariance matrix is \((X^{mathsf T}WX)^{-1}\), where \(X\) is the design matrix.
With variance \(\phi V(\mu_i)\), the information weights are divided by \(\phi\). The information matrix becomes \(X^{mathsf T}WX/\phi\).
\((X^{mathsf T}WX/\phi)^{-1}=\phi(X^{mathsf T}WX)^{-1}\). Thus the coefficient covariance matrix is multiplied by \(\phi\).
Taking the square root of a coefficient variance multiplies its standard error by \(\sqrt{\phi}\). Replacing \(\phi\) with \(\hat{\phi}\) gives the estimated scale adjustment.
This argument is an approximate large-sample justification; it is not a guarantee that adjusted tests have an exact small-sample reference distribution. When dispersion is estimated, a normal or chi-square reference may be replaced in practice by a \(t\) or \(F\) approximation using residual degrees of freedom. The exact choice depends on the model and inferential procedure.
Comparing Nested Models with Adjusted Dispersion
For nested mean models, a deviance difference measures the improvement in fit from adding terms. With a fully specified likelihood and its assumptions satisfied, a likelihood-ratio statistic can have an approximate chi-square reference distribution. Under a quasi-likelihood model, however, there is no fully specified likelihood from which to obtain that likelihood-ratio test. A common alternative is an approximate dispersion-adjusted \(F\) comparison.
Here \(D_{\text{reduced}}\) and \(D_{\text{full}}\) are the residual deviances for nested reduced and full models, \(q\) is the number of added coefficients, and \(p_{\text{full}}\) is the number of coefficients in the full model. This is an approximate procedure: its appropriateness depends on the model, the dispersion estimate, and the data. It should not be described as an exact likelihood-ratio test.
Worked Example: Testing Added Predictors with Excess Dispersion
A hypothetical quasi-Poisson model predicts the number of equipment alerts per facility. The reduced model has residual deviance \(146\). Adding two predictors gives a full-model residual deviance of \(128\), with 48 residual degrees of freedom. The full model's Pearson statistic is \(72\). Use the approximate dispersion-adjusted comparison and state its limitation.
Method. Estimate dispersion from the full model and divide the average deviance reduction per added coefficient by that estimate.
Step 1: Estimate dispersion. The full-model residual degrees of freedom are 48, so \(\hat{\phi}=72/48=1.5\).
Step 2: Find the deviance difference. The reduction is \(146-128=18\). There are \(q=2\) added coefficients, so the reduction per coefficient is \(18/2=9\).
Step 3: Calculate the adjusted statistic. \(F\approx9/1.5=6\). The approximate reference distribution is \(F_{2,48}\). Its upper-tail probability is approximately \(0.0047\).
Step 4: Interpret and check. Under this approximate quasi-likelihood comparison, the added predictors improve the mean fit at the 5% level. The common mistake is to compare 18 directly with a chi-square distribution with 2 degrees of freedom, ignoring the estimated overdispersion. This calculation is not an exact likelihood-ratio test: the quasi-likelihood specification does not provide a full likelihood for that test.
Quasi-Likelihood and What It Does Not Change
Quasi-likelihood specifies a relationship between the mean and variance without requiring a complete probability distribution for the response. For a quasi-Poisson model, for example, the mean follows the chosen link and predictor structure, while the variance is \(\phi\mu_i\). The Poisson model is the special case with \(\phi=1\).
When the mean structure and variance function are retained and \(\phi\) is a common multiplier, the quasi-Poisson coefficient estimates are the same as the Poisson estimates. The multiplier changes the estimated uncertainty, not the fitted mean. A quasi-binomial model makes the analogous variance adjustment for binomial responses. It does not turn an inadequate mean model into an adequate one.
Worked Example: Same Fitted Mean, Wider Uncertainty
A hypothetical quasi-Poisson model for claims counts per policy group has a log link and fitted linear predictor \(\hat{\eta}=1.1+0.2x\). At \(x=2\), the estimated standard error of the linear predictor under scale one is \(0.12\). Suppose a Pearson estimate gives \(\hat{\phi}=2\). Find the fitted mean, the adjusted standard error, and a 95% Wald interval for the mean using a normal approximation on the link scale.
Method. Evaluate the mean from the unchanged predictor, scale the linear-predictor standard error by \(\sqrt{\hat{\phi}}\), and transform the interval endpoints back using the exponential link inverse.
Step 1: Calculate the linear predictor and mean. At \(x=2\), \(\hat{\eta}=1.1+0.2(2)=1.5\). The fitted mean is \(\hat{\mu}=\exp(1.5)\approx4.4817\) claims per policy group.
Step 2: Adjust the standard error. The adjusted standard error is \(0.12\sqrt{2}\approx0.1697\).
Step 3: Form the interval on the link scale. Using \(1.96\) for a 95% normal interval, the margin is \(1.96(0.1697)\approx0.3326\). The endpoints are \(1.5-0.3326=1.1674\) and \(1.5+0.3326=1.8326\).
Step 4: Transform and check. Exponentiating gives an approximate interval \((\exp(1.1674),\exp(1.8326))=(3.213,6.250)\) claims per policy group, to three decimals. The fitted mean remains about \(4.4817\); the wider interval reflects the larger estimated variance. The common mistake is to multiply the fitted mean by \(\hat{\phi}\). Dispersion changes the variance and uncertainty, not the mean calculation.
Choosing a Remedy and Avoiding the Main Trap
A scale adjustment is appropriate only when a common variance multiplier is a reasonable description of the excess variation. It can make standard errors and tests more cautious while preserving a plausible mean model. Quasi-likelihood formalizes this mean-and-variance approach when a full response distribution is not specified.
Overdispersion can also arise because the model's assumptions do not capture important features of the data. Examples include omitted structure in the mean, dependence among observations, or variability that changes in a way a single \(\phi\) cannot represent. In those cases, merely inflating standard errors can leave the model's predictions or interpretation inadequate. Investigate the scientific structure and residual evidence before treating one scale estimate as a complete remedy.
- Scale adjustment: retains the fitted mean and inflates coefficient variances by an estimated common factor.
- Quasi-likelihood: specifies mean and variance relationships without asserting a full probability distribution; likelihood-ratio tests are not automatically available.
- Mean-model revision: is needed when the predictor or link structure does not capture the mean pattern. Changing only \(\phi\) cannot correct that pattern.
- Dependence or changing variance: may require a model that represents those features rather than one common scale multiplier.
Check Your Understanding
For each problem, state the method and keep the mean and variance roles separate.
- A Poisson model has 62 observations, 2 fitted coefficients, and Pearson statistic \(Q_P=90\). What is the Pearson estimate of dispersion? (Answer: \(\hat{\phi}=90/(62-2)=1.5\).)
- A coefficient estimate is \(0.30\), its conventional standard error is \(0.10\), and \(\hat{\phi}=2.25\). Find the adjusted standard error and adjusted Wald statistic. (Answer: adjusted standard error \(0.15\); adjusted Wald statistic \(0.30/0.15=2.00\).)
- Nested models have residual deviances 154 and 136. The full model adds 3 coefficients, has 40 residual degrees of freedom, and has Pearson statistic 60. Calculate the approximate dispersion-adjusted \(F\) statistic and its degrees of freedom. (Answer: \(\hat{\phi}=60/40=1.5\), \(F\approx[(154-136)/3]/1.5=4\), with degrees of freedom \((3,40)\).)
- A quasi-Poisson log-link model has predictor \(\hat{\eta}=0.4+0.15x\), with \(\hat{\phi}=1.8\). At \(x=4\), find the fitted mean and variance. (Answer: \(\hat{\eta}=1\), fitted mean \(e^1\approx2.7183\), and variance \(1.8(2.7183)\approx4.8930\).)
- A quasi-Poisson fit has an estimated dispersion above one, but its residuals also show a systematic pattern across a predictor. Is multiplying standard errors by \(\sqrt{\hat{\phi}}\) enough to address both findings? (Answer: No. It adjusts uncertainty for a common variance inflation; it does not correct a systematic mean-pattern problem.)