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Quasi-Binomial and Quasi-Poisson · Tutorial 714 of 981

What Quasi-GLM Inference Does Not Provide

Understand what quasi-GLMs can support, why they do not supply ordinary likelihood-based comparisons, and how to distinguish quasi-inference from likelihood inference.

Intermediate 11 min read

What You'll Learn

  • Distinguish a quasi-GLM’s mean–variance specification from a full probability model
  • See how different response distributions can share the same quasi-Poisson moments
  • Explain why ordinary AIC and likelihood-ratio tests are not supplied by quasi-GLMs
  • Recognize quasi-likelihood and quasi-deviance as criteria that are not automatically log-likelihoods
  • Interpret quasi-Wald and approximate F inference without calling it likelihood inference
  • Identify why predictive probabilities require additional distributional assumptions

What a Quasi-GLM Specifies

A quasi-binomial or quasi-Poisson model can estimate regression effects and quantify their uncertainty without specifying a complete probability distribution for each response. That distinction matters when choosing how to compare models or describe uncertainty: some familiar GLM outputs require a likelihood, while quasi-inference relies on the specified mean and variance relationships.

For independent responses \(Y_i\), a quasi-GLM specifies a relationship between the mean \(\mu_i=E[Y_i]\) and predictors, usually through a link function, and a variance function. In the quasi-Poisson case, the variance is \(\phi\mu_i\). In the quasi-binomial case, for a count \(Y_i\) out of \(m_i\) trials with success probability \(p_i\), the mean is \(m_i p_i\) and the variance is \(\phi m_i p_i(1-p_i)\). These statements constrain moments; they do not, by themselves, say what probability \(Y_i\) assigns to each possible value.

Definition: A full probability model specifies a probability mass function or density for the response, including how probability is assigned across all possible outcomes. A quasi-GLM specification supplies a mean model and a variance relationship, but does not generally determine that mass function or density. A quasi-likelihood criterion may be constructed from the mean and variance; it is not automatically the log-likelihood of a probability model.

This is not a claim that quasi-GLMs provide no inference. Their estimating equations can yield regression estimates and, with appropriate variance estimation and assumptions, Wald tests, quasi-score tests, or approximate F tests. The limitation is narrower: without a specified probability model, they do not supply the ordinary likelihood and its likelihood-based comparisons.

Why Moments Do Not Determine a Likelihood

The essential issue is that a mean and variance need not determine a unique distribution. The following example verifies this directly for a quasi-Poisson mean–variance relationship.

Worked Example: Two Count Distributions With the Same Quasi-Poisson Moments

Suppose the specified mean is \(\mu=2\), and the quasi-Poisson dispersion is \(\phi=2\). The variance specification is therefore \(\operatorname{Var}(Y)=\phi\mu=4\). Consider two possible distributions for \(Y\).

Distribution A: \(P(Y=0)=1/2\) and \(P(Y=4)=1/2\). Its mean is \(E[Y]=0(1/2)+4(1/2)=2\). Its second moment is \(E[Y^2]=0^2(1/2)+4^2(1/2)=8\), so its variance is \(8-2^2=4\).

Distribution B: \(P(Y=0)=2/5\), \(P(Y=2)=1/3\), and \(P(Y=5)=4/15\). These probabilities sum to \(2/5+1/3+4/15=6/15+5/15+4/15=1\). Its mean is \(0(2/5)+2(1/3)+5(4/15)=2/3+4/3=2\). Its second moment is \(0^2(2/5)+2^2(1/3)+5^2(4/15)=4/3+20/3=8\), so its variance is again \(8-2^2=4\).

The distributions are different: for example, Distribution A assigns positive probability to 4, while Distribution B assigns none. Yet both satisfy the same quasi-Poisson mean and variance. Thus those moment specifications alone cannot identify a unique probability mass function, and cannot identify a unique likelihood for the response.

Result: A mean–variance specification does not, in general, determine a likelihood. In the example, two distinct probability mass functions satisfy the same specified moments. Because a likelihood is built from the probability of the observed outcomes under a specified distribution, those moments alone cannot select which likelihood to use. \(\square\)

The same distinction applies to quasi-binomial models. The binomial distribution is one possible full distribution, with variance \(m_i p_i(1-p_i)\); allowing an extra factor \(\phi\) changes the variance specification, but does not itself define a new probability mass function. A quasi-binomial model should therefore not be read as a complete distributional description merely because its mean resembles the binomial mean.

Quasi-Likelihood Is Not Necessarily Log-Likelihood

A quasi-likelihood criterion can be constructed from the specified variance function. For a response with mean \(\mu\), variance \(\phi V(\mu)\), fixed \(\phi>0\), and observed value \(y\), one common definition is a function \(Q(y;\mu)\) whose derivative with respect to \(\mu\) is \(\partial Q(y;\mu)/\partial\mu=(y-\mu)/\{\phi V(\mu)\}\). On an interval where \(V(t)>0\), this can be obtained by integration from any reference point \(\mu_0\) in the interval:

$$ Q(y;\mu)=\int_{\mu_0}^{\mu}\frac{y-t}{\phi V(t)}\,dt, \qquad \frac{\partial Q(y;\mu)}{\partial\mu} =\frac{y-\mu}{\phi V(\mu)}. $$

The derivative has the form used to construct quasi-score estimating equations. But the integration defines a criterion as a function of the mean; it does not establish that \(\exp\{Q(y;\mu)\}\) is a probability mass or density as a function of \(y\). In particular, the mean–variance specification does not provide the response probabilities needed to check that a proposed mass function sums to one or a density integrates to one. Some particular quasi-likelihood criteria can coincide with a log-likelihood up to terms that do not depend on the model parameters under additional distributional assumptions. That coincidence must be justified by the specified probability model; it is not guaranteed by the quasi-GLM specification alone.

Result: Integrating the quasi-score derivative gives a criterion in \(\mu\), but a likelihood requires a normalized probability model for the observed response. The derivative condition alone does not supply that model. The two distinct distributions in the preceding example satisfy the same moment specification, confirming that the missing information cannot be recovered from the quasi-likelihood criterion alone. \(\square\)

This distinction also explains why a quasi-deviance is not automatically a likelihood-ratio statistic. Deviance-like quantities can be formed from a variance function and used to summarize lack of fit or to construct approximate tests. But the fact that a quantity is called a deviance does not make it a difference between maximized log-likelihoods. That interpretation requires an actual likelihood.

Which Likelihood-Based Quantities Are Unavailable?

For a specified likelihood \(L(\theta)\), the log-likelihood is \(\ell(\theta)=\log L(\theta)\). AIC is conventionally defined as \(-2\ell(\hat{\theta})+2k\), where \(\hat{\theta}\) maximizes the likelihood and \(k\) is the number of estimated parameters. A quasi-GLM’s mean and variance specification does not supply this maximized log-likelihood. Therefore, an ordinary AIC for that quasi-model is not defined by the quasi specification.

There are criteria sometimes called QAIC or quasi-AIC. Their definitions require additional conventions and assumptions, and they are not the ordinary AIC obtained from a specified quasi-GLM likelihood. If such a criterion is used, its definition and justification need to be stated; it should not be presented simply as the quasi-model’s AIC.

A likelihood-ratio test also requires likelihoods. For nested models fitted to the same data, its statistic is \(2\{\ell(\hat{\theta}_{\mathrm{full}})-\ell(\hat{\theta}_{\mathrm{reduced}})\}\). Under regularity conditions and the null hypothesis, this statistic has an asymptotic chi-square reference distribution with degrees of freedom given by the difference in the numbers of free parameters. Without specified likelihoods, the quasi-GLM does not provide this statistic or its likelihood-ratio reference distribution. A difference in quasi-deviances is not, solely by being a difference, a likelihood-ratio statistic.

Worked Example: A Deviance Difference Is Not Automatically an LR Test

Suppose two observed counts are \(y_1=0\) and \(y_2=4\). Under an intercept-only Poisson mean model, the fitted mean for each count is \(\hat{\mu}=2\). The Poisson-form deviance contribution for a count \(y\) is \(2\{y\log(y/\hat{\mu})-(y-\hat{\mu})\}\), with the \(y\log(y/\hat{\mu})\) term taken as zero when \(y=0\).

For \(y_1=0\), the contribution is \(2\{0-(0-2)\}=4\). For \(y_2=4\), it is \(2\{4\log(4/2)-(4-2)\}=8\log 2-4\). The total is therefore \(4+(8\log 2-4)=8\log 2\), approximately \(5.545\).

This calculation gives a Poisson-form deviance for the fitted means. If the model is being used only as a quasi-Poisson specification, that number alone does not establish a log-likelihood difference: the quasi specification has not asserted the Poisson mass function. A likelihood-ratio test would require a specified likelihood for both models, fitted under the appropriate conditions. The deviance calculation is not a substitute for that requirement.

What Quasi-Inference Does Provide

The absence of a full likelihood does not erase the inferences available from the mean model. The earlier material on quasi-score equations and covariance scaling develops how coefficient estimates and their estimated uncertainty are obtained. In suitable large samples, those estimates can support quasi-Wald tests and confidence intervals. With estimated dispersion, approximate F procedures may be used where their conditions are appropriate. These are quasi-inference procedures, not likelihood-ratio tests under another name.

Worked Example: A Quasi-Wald Test Is Not a Likelihood-Ratio Test

Suppose a fitted quasi-Poisson regression estimates a coefficient as \(\hat{\beta}=0.40\). Its unscaled standard error is \(0.10\), and the dispersion value used to scale its covariance is \(\phi=2\). The scaled standard error is \(\sqrt{2}(0.10)\), which is approximately \(0.1414\).

For the null hypothesis \(H_0:\beta=0\), the quasi-Wald statistic is \(z=0.40/0.1414\approx2.83\). Using a standard normal reference distribution gives an approximate two-sided \(p\)-value of \(0.0047\). This calculation uses the coefficient estimate and its scaled uncertainty; it does not require a likelihood-ratio statistic.

The result is conditional on the stated large-sample normal approximation and the specified dispersion treatment. It does not provide an AIC for this model, nor does it become a likelihood-ratio test because its \(p\)-value is small. A predictive probability for a future count also cannot be obtained from this mean, variance, and standard error alone: that probability requires a full predictive distribution or further distributional assumptions.

The practical boundary is therefore clear. Quasi-GLMs can support inference about regression means and effects using estimating-equation methods and an appropriate covariance estimate. They do not, without extra assumptions, determine outcome probabilities, tail probabilities, or the likelihood-based quantities that depend on those probabilities. If the scientific question requires such quantities, a full response distribution must be specified and assessed.

Common Mistake

Wrong reasoning: “The software reports a deviance for my quasi-Poisson fit, so I can subtract deviances for nested models, call the result a likelihood-ratio statistic, and use the usual chi-square reference distribution.”

Correction: A deviance-like quantity can be calculated from a quasi model, but a likelihood-ratio statistic is defined using maximized log-likelihoods. A quasi mean–variance specification does not supply those likelihoods. Use a justified quasi-inference procedure, such as an appropriate Wald or approximate F procedure, or specify a full probability model if likelihood-based comparison is required. Do not label a quasi-deviance comparison an LR test unless the likelihood interpretation and its assumptions have actually been established.

Key takeaway: Quasi-binomial and quasi-Poisson specifications support mean-based estimating equations and associated quasi-inference. They do not, by themselves, define a full likelihood. Ordinary AIC and likelihood-ratio tests therefore do not follow from the quasi specification; using them requires additional distributional assumptions.

Check Your Understanding

Check that you can distinguish moment-based inference from likelihood-based inference.

  1. Why can two different response distributions satisfy the same quasi-Poisson mean and variance?
  2. What additional object, beyond a mean and variance relationship, is needed to define a likelihood?
  3. Why does a quasi-deviance difference not automatically have the interpretation of a likelihood-ratio statistic?
  4. What information is used in a quasi-Wald test that is not the maximized log-likelihood?
  5. Why does a quasi-GLM alone not determine a predictive probability for a future count?
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