Testing Dispersion Parameters in GLMs

Why Test Dispersion?

In a generalized linear model, the dispersion parameter controls the scale of variation around the mean. Testing it can be meaningful when the assumed distribution includes a freely estimable dispersion parameter. It is not automatically meaningful just because a fitted model has a residual-based estimate of variability: some common GLMs fix dispersion as part of the distribution.

The central inferential question is whether the dispersion is a specified positive value, while allowing for the fact that regression coefficients may be known or may need to be estimated. That distinction changes the information available for a test. It also matters whether the null value is an interior point of a genuine likelihood model. If it is not, familiar chi-square reference distributions may fail.

Hypotheses and Regularity

Definition: Let $\phi$ denote an estimable GLM dispersion parameter, with parameter space contained in $(0,\infty)$. A two-sided test of a specified value $\phi_0$ is $$ H_0:\phi=\phi_0 \qquad\text{versus}\qquad H_1:\phi\ne\phi_0, $$ where $\phi_0>0$. Regression coefficients $\beta$ are held fixed if they are known; if they are unknown, they are nuisance parameters that must be estimated under both hypotheses.

For a standard exponential-dispersion model, a typical density or mass function has the form $$ f(y_i;\theta_i,\phi) =\exp\left\{\frac{y_i\theta_i-b(\theta_i)}{a(\phi)} +c(y_i,\phi)\right\}. $$ In the usual GLM formulation, $\theta_i$ is linked to the mean $\mu_i=E[Y_i]$, and the variance is $\operatorname{Var}(Y_i)=\phi V(\mu_i)$, possibly with a known observation weight included. The precise likelihood and the role of $\phi$ depend on the chosen distribution.

A standard chi-square test requires more than a parameter with the name “dispersion.” The null value should be an interior point of an identifiable, differentiable likelihood model; the true nuisance parameters should also be interior; and the relevant information matrix, including the information for $\phi$ after accounting for nuisance parameters, must be nonsingular. Independent observations and the usual large-sample regularity conditions for likelihood inference are assumed. In particular, the sample size must grow in a way that supplies information about $\phi$.

Known and Unknown Regression Coefficients

Write the parameter as $(\beta,\phi)$, and let $U_\beta$ and $U_\phi$ be the corresponding likelihood scores. When $\beta$ is known, the information for testing $\phi$ is simply $I_{\phi\phi}$. When $\beta$ is unknown, its estimation uses information that may overlap with information about $\phi$. The appropriate information for a dispersion score test is then the efficient information $$ I_{\phi\phi\cdot\beta} =I_{\phi\phi}-I_{\phi\beta}I_{\beta\beta}^{-1}I_{\beta\phi}, $$ where the information blocks are evaluated under the null and $I_{\beta\beta}$ is invertible. The corresponding efficient score is $$ U_{\phi\cdot\beta} =U_\phi-I_{\phi\beta}I_{\beta\beta}^{-1}U_\beta. $$

For the score test, these quantities are evaluated at $(\hat\beta_0,\phi_0)$, where $\hat\beta_0$ is the restricted maximum-likelihood estimate under $H_0$. The restricted fit satisfies $U_\beta(\hat\beta_0,\phi_0)=0$, so the efficient score there equals $U_\phi(\hat\beta_0,\phi_0)$. The information still needs the nuisance adjustment unless the parameters are information-orthogonal, meaning $I_{\phi\beta}=0$.

Worked Example: Known Means in a Normal Model

Suppose $Y_1,\ldots,Y_{10}$ are independent normal observations with known means $\mu_i$ and common variance $\phi$. The observed sum of squared deviations from those known means is $S=\sum_{i=1}^{10}(y_i-\mu_i)^2=20$. Test $H_0:\phi=1.5$ against $H_1:\phi\ne1.5$ by the likelihood-ratio statistic.

Apart from a constant that does not depend on $\phi$, the log-likelihood is $$ \ell(\phi)=-\frac{10}{2}\log(\phi)-\frac{20}{2\phi}. $$ Differentiating gives $\ell'(\phi)=-10/(2\phi)+20/(2\phi^2)$, so the unrestricted maximum occurs at $\hat\phi=S/10=2$. The likelihood-ratio statistic is $2\{\ell(\hat\phi)-\ell(\phi_0)\}$, which simplifies to $$ \Lambda =10\left\{\frac{\hat\phi}{\phi_0}-1-\log\left(\frac{\hat\phi}{\phi_0}\right)\right\} =10\left\{\frac{2}{1.5}-1-\log\left(\frac{2}{1.5}\right)\right\}. $$

Since $2/1.5=4/3$ and $\log(4/3)\approx0.287682$, the statistic is $10(1/3-0.287682)\approx0.457$. Under the regular null, its large-sample reference distribution is $\chi^2_1$. The calculation uses known means; it does not account for estimating regression coefficients.

Likelihood-Ratio Calibration

With unknown regression coefficients, the likelihood-ratio test compares the maximized likelihood under $H_0$ with the maximized likelihood under the unrestricted alternative. Both fits estimate $\beta$; only the null fit fixes $\phi$ at $\phi_0$. Under the regularity conditions above, this is a one-constraint test, so Wilks’ theorem gives a $\chi^2_1$ limiting reference distribution.

Theorem: Suppose the model is identifiable and regular in a neighborhood of the true parameter $(\beta_0,\phi_0)$, with $\phi_0$ interior, finite nonsingular information, and positive efficient information $I_{\phi\phi\cdot\beta}$. For independent observations satisfying the usual likelihood asymptotics, the likelihood-ratio, efficient score, and Wald statistics for $H_0:\phi=\phi_0$ each converge in distribution to $\chi^2_1$.

Derivation. Under these conditions, the local log-likelihood has a quadratic expansion. After maximizing over the nuisance coefficient displacement, its curvature in the $\phi$ direction is the Schur complement $I_{\phi\phi\cdot\beta}$. The remaining local score is $U_{\phi\cdot\beta}$. The efficient-score central limit theorem gives $$ \frac{U_{\phi\cdot\beta}}{\sqrt{I_{\phi\phi\cdot\beta}}} \xrightarrow{d}N(0,1) $$ under $H_0$, with information and score evaluated at the null fit (or their asymptotically equivalent values). The quadratic likelihood expansion therefore gives the likelihood-ratio statistic as $$ \Lambda =\frac{U_{\phi\cdot\beta}^{\,2}}{I_{\phi\phi\cdot\beta}}+o_p(1). $$ The leading term is the square of a quantity converging to $N(0,1)$, and hence converges to $\chi^2_1$. The efficient score statistic has that same leading term. The Wald statistic has the same limit because the unrestricted estimator is asymptotically normal with covariance given by the inverse information. Thus all three limits are $\chi^2_1$. $\square$

This result is asymptotic: it does not say the three tests give identical finite-sample answers. It also does not license using $\chi^2_1$ when the stated conditions fail.

Worked Example: Unknown Intercept in a Normal Model

Consider $Y_i=\beta+\varepsilon_i$, where the independent errors satisfy $\varepsilon_i\sim N(0,\phi)$. The five observations are $1,2,3,4,5$. Test $H_0:\phi=1.5$ against a two-sided alternative, treating the intercept $\beta$ as unknown.

The unrestricted estimate of the intercept is the sample mean, $\hat\beta=3$. The residual sum of squares is $$ \mathrm{RSS}=(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-3)^2 =4+1+0+1+4=10. $$ The unrestricted variance estimate is $\hat\phi=\mathrm{RSS}/5=2$. Profiling the normal log-likelihood over the intercept replaces the sum of squares about the known mean in the preceding calculation by RSS. Thus $$ \Lambda =5\left\{\frac{2}{1.5}-1-\log\left(\frac{2}{1.5}\right)\right\} =5(0.333333-0.287682) \approx0.228. $$

For comparison, the score at the restricted fit is $U_\phi=-5/(2\phi_0)+\mathrm{RSS}/(2\phi_0^2)$, which at $\phi_0=1.5$ equals $-5/3+10/4.5\approx0.556$. In the normal intercept model, the intercept and variance are information-orthogonal: the cross-information is zero because the fitted residuals sum to zero. Hence $I_{\phi\phi\cdot\beta}=5/(2(1.5)^2)\approx1.111$, and the score statistic is $0.556^2/1.111\approx0.278$. Both statistics have an asymptotic $\chi^2_1$ reference distribution under regularity, although their finite-sample values differ.

When Ordinary Chi-Square Calibration Is Inappropriate

The chi-square conclusion is conditional on having a regular likelihood parameter to test. Several important situations do not meet that condition.

Dispersion is fixed by the distribution. In the ordinary Poisson and binomial GLMs, $\phi=1$ is fixed. There is no freely estimated dispersion parameter in that likelihood, so a likelihood-ratio test of $\phi=1$ against a free $\phi$ is not defined within that model.
A quasi-likelihood supplies no full likelihood. A quasi-Poisson model may specify $\operatorname{Var}(Y_i)=\phi\mu_i$, but that variance specification alone does not define a likelihood for the observations. A likelihood-ratio statistic cannot be obtained from it without adding a distributional model.
The null is on a boundary. If the parameter space restricts $\phi$ to values at or above $\phi_0$, then testing $\phi=\phi_0$ places the null on its boundary. The regular interior-point argument for $\chi^2_1$ no longer applies; the limiting law depends on the boundary structure and other model conditions.
The parameter is not identified under the null. If the data distribution under $H_0$ does not depend on $\phi$, or the information for its estimable direction vanishes, the ordinary one-parameter Wilks result does not apply.

A residual-based dispersion estimate can still be useful descriptively or for a separately justified approximation. But an estimate or diagnostic is not, by itself, evidence that a regular likelihood-ratio test exists. First identify the model’s likelihood and parameter space; then determine whether the null is an interior, identifiable value.

Worked Example: A Poisson Dispersion Diagnostic Is Not a Likelihood-Ratio Test

Suppose four independent counts are $0,1,2,3$, and an intercept-only Poisson GLM is fitted. The maximum-likelihood estimate of the common mean is $\hat\mu=(0+1+2+3)/4=1.5$. The Pearson statistic is $$ X^2=\sum_{i=1}^4\frac{(y_i-\hat\mu)^2}{\hat\mu} =\frac{2.25+0.25+0.25+2.25}{1.5} =\frac{5}{1.5} \approx3.333. $$ There are $4-1=3$ residual degrees of freedom, so the Pearson dispersion ratio is $3.333/3\approx1.111$.

This ratio is a diagnostic summary. In the ordinary Poisson likelihood, the variance equals the mean and $\phi$ is fixed at $1$; it is not a fitted likelihood parameter. Therefore $1.111$ is not an estimate from an unrestricted Poisson likelihood, and comparing it to $1$ does not produce a Poisson likelihood-ratio test for dispersion. A model with an additional dispersion parameter would need to be specified before such a likelihood test could be considered.

Common Mistake

Common mistake: “The dispersion estimate is above its null value, so I can test it with a $\chi^2_1$ likelihood-ratio statistic.” The reasoning skips two checks: whether dispersion is a free parameter in the stated likelihood, and whether the null value is an interior identifiable point. If either check fails, the ordinary $\chi^2_1$ calibration is not justified. Establish the likelihood and its parameter space first; if the model is quasi-likelihood only, do not label a variance-based comparison a likelihood-ratio test.

When a regular dispersion test is available, the main practical distinction is whether regression coefficients are known. Known coefficients call for the ordinary information in the dispersion direction. Unknown coefficients call for nuisance-adjusted information and fits under the null and alternative. In either case, report the hypothesis, the fitted model, the test statistic, and why its reference distribution is appropriate.

Check Your Understanding

Use the model and regularity conditions, not just the name of a dispersion estimate, to answer these questions.

In a two-sided test of $\phi=\phi_0$, what must be true of $\phi_0$ for the ordinary interior-point chi-square argument to apply?
When $\beta$ is unknown, what information replaces $I_{\phi\phi}$ in the efficient score test, and what does the adjustment account for?
In the normal intercept example, why is the score statistic computed using residuals from the restricted fit rather than deviations from a known mean?
Why does a Pearson dispersion ratio from an ordinary Poisson GLM not, by itself, define a likelihood-ratio test for $\phi$?
Name one condition that can make the ordinary $\chi^2_1$ calibration inappropriate, even if a likelihood has been specified.