Likelihood Criteria Under Overdispersion

Why Dispersion Changes Model Comparisons

A candidate rate model can have an appropriate mean structure and still understate the variability of observed claims. For example, a Poisson GLM assumes that, conditional on the predictors, the count variance equals the count mean. If unmodeled heterogeneity or dependence makes observed counts more variable, that assumption may fail. In that situation, a small reduction in deviance can look like stronger evidence for a more complex model than the data justify.

The remedy is not to multiply a Poisson likelihood by a dispersion factor and call the result a new Poisson likelihood. A Poisson distribution has variance equal to its mean. Instead, distinguish two approaches: use a quasi-likelihood method with an explicitly stated dispersion adjustment, or specify a full probability model whose variance structure includes dispersion and compare its actual likelihood.

Definition: A model is overdispersed relative to its stated variance function when the observed variation is systematically greater than that function predicts. In a Poisson mean model, the conditional variance assumption is \(\operatorname{Var}(Y_i\mid x_i)=\mu_i\). A quasi-Poisson specification instead states \(\operatorname{Var}(Y_i\mid x_i)=\phi\mu_i\), where \(\phi\) is a dispersion parameter. For \(\phi>1\), it allows extra variance, but it does not define a Poisson probability distribution when \(\phi\ne1\).

A full likelihood specifies a probability distribution for every observation, including its variance and any dispersion parameters. AIC is calculated from that likelihood, and its parameter count includes every parameter estimated by maximum likelihood. Quasi-likelihood specifies a mean-variance relationship but need not specify a full probability distribution. Its criteria are useful when defined consistently, but they are not ordinary likelihood or AIC comparisons.

Estimating Dispersion for a Quasi-Poisson Mean Model

For independent observations with fitted means \(\hat\mu_i\), a common diagnostic estimate is the Pearson dispersion estimate. Let \(n\) be the number of observations and \(p\) the number of fitted mean-model coefficients. The residual degrees of freedom are \(n-p\). For the quasi-Poisson variance function \(\phi\mu_i\), the Pearson statistic is divided by those degrees of freedom to estimate \(\phi\).

X^2_P=\sum_{i=1}^{n}\frac{(y_i-\hat\mu_i)^2}{\hat\mu_i}, \qquad \hat\phi_P=\frac{X^2_P}{n-p}.

This estimate is a diagnostic and a working scale estimate, not proof that the quasi-Poisson model is a full probability distribution. In particular, the Poisson deviance remains the deviance from a Poisson likelihood; dividing it by \(\hat\phi_P\) produces a dispersion-adjusted comparison, not a new Poisson log-likelihood.

Worked Example: Diagnosing Extra-Poisson Variation

A hypothetical insurer fits a Poisson mean model to four independent policy-group claim counts. The observed counts are \(0,8,10,2\), and the fitted Poisson means are \(2,4,6,4\). The mean model has two estimated coefficients. Estimate the Pearson dispersion and interpret it.

Method. Calculate each squared Pearson residual, sum them, and divide by the residual degrees of freedom. The Poisson variance used for this diagnostic is the fitted mean.

The four contributions are:

\frac{(0-2)^2}{2}=2,\qquad \frac{(8-4)^2}{4}=4,\qquad \frac{(10-6)^2}{6}=\frac{8}{3},\qquad \frac{(2-4)^2}{4}=1.

Thus \(X^2_P=2+4+8/3+1=29/3=9.6667\). The residual degrees of freedom are \(n-p=4-2=2\), so:

\hat\phi_P=\frac{9.6667}{2}=4.8333.

The diagnostic indicates substantial extra-Poisson variation in this small hypothetical data set. It does not establish that the true dispersion is exactly 4.8333; with only two residual degrees of freedom, the estimate is particularly uncertain. Common mistake avoided: interpreting \(\hat\phi_P\) as a multiplier that turns the fitted Poisson distribution into a valid Poisson distribution with variance \(4.8333\mu_i\). It does not.

Quasi-Likelihood Comparisons: Use the Scale Consistently

When candidate mean models use the same response, the same quasi-Poisson variance function, and the same observations, a dispersion-adjusted deviance criterion can be useful. One explicit convention is to rank candidates by \(D/\hat\phi+2K\), where \(D\) is the Poisson deviance, \(K\) is the number of estimated mean-model coefficients, and \(\hat\phi\) is one common dispersion estimate obtained from a prespecified reference or global model. Lower scores are preferred. This is a defined quasi-likelihood criterion, not ordinary AIC.

Using a common scale matters. If each candidate has its own dispersion estimate, simply dividing each deviance by its own estimate can produce a comparison that depends on arbitrary additive constants in the quasi-likelihood. If dispersion is to vary across candidate models, use a suitable full likelihood or a carefully specified quasi-likelihood criterion; do not treat candidate-specific rescaled Poisson deviances as ordinary likelihoods.

Derivation: For a fixed Poisson response and fixed data, the Poisson deviance for a fitted mean model is \(D=2(\ell_{\mathrm{sat}}-\ell_{\mathrm{fit}})\), where \(\ell_{\mathrm{sat}}\) is the log-likelihood of the saturated model. In a deviance-based quasi-likelihood comparison with one common \(\hat\phi\), the fit term is scaled by \(1/\hat\phi\). Since \(\ell_{\mathrm{sat}}\) is the same for all candidates on the same observations, differences in scaled deviance are proportional to differences in their scaled fit terms. Adding \(2K\) penalizes additional fitted mean coefficients. This justifies the stated ranking convention, but does not turn the scaled score into a likelihood or make it an ordinary AIC.

Worked Example: Ranking Mean Models With a Common Dispersion Estimate

A hypothetical insurer compares two quasi-Poisson claim-frequency mean models on identical observations. A prespecified reference model gives a common dispersion estimate \(\hat\phi=2.0\). Candidate A has Poisson deviance \(D_A=24\) and \(K_A=3\) mean coefficients. Candidate B has \(D_B=18\) and \(K_B=5\). Use the deviance-based quasi-likelihood criterion \(D/\hat\phi+2K\).

Method. Apply the same reference dispersion to both models, then add twice the number of mean coefficients.

Q_A=\frac{24}{2.0}+2(3)=12+6=18, \qquad Q_B=\frac{18}{2.0}+2(5)=9+10=19.

Candidate A has the lower score, 18 versus 19, so it is preferred under this stated criterion. The result reflects the trade-off between Candidate B’s deviance reduction of \(24-18=6\) and its two additional coefficients, which add \(2(5-3)=4\) penalty units after the common dispersion adjustment.

If the analyst incorrectly ignores overdispersion and compares \(D+2K\), the scores are \(24+6=30\) for A and \(18+10=28\) for B, so the ranking reverses. That unscaled calculation treats Poisson-level variation as appropriate. Common mistake avoided: reporting the reversal as an ordinary AIC result. These are quasi-Poisson mean models; the scaled score is meaningful only under the stated convention and common scale.

Full Likelihoods Can Model Dispersion Directly

A negative binomial model is one full-likelihood alternative for overdispersed counts. Under the NB2 parameterization, its conditional mean is \(\mu_i\) and its variance is \(\mu_i+\alpha\mu_i^2\), with \(\alpha>0\). Equivalently, the size parameter is \(r=1/\alpha\). The probability of count \(y_i\) is:

P(Y_i=y_i)= \frac{\Gamma(y_i+r)}{\Gamma(r)y_i!} \left(\frac{r}{r+\mu_i}\right)^r \left(\frac{\mu_i}{r+\mu_i}\right)^{y_i}.

This specifies a probability distribution, unlike a quasi-Poisson mean-variance relationship. A Poisson model and a negative binomial model can be compared by AIC when they are fitted to the same response observations using proper likelihoods. The Poisson model has no freely estimated dispersion parameter; the negative binomial model estimates its extra parameter \(\alpha\), which must be counted in AIC. The AIC is \(-2\ell+2K\), where \(\ell\) is the maximized log-likelihood and \(K\) counts all estimated parameters in that likelihood.

AIC comparisons require comparable likelihoods for the same observed data and response scale. They do not require the candidate distributions to have identical variance assumptions; the purpose may be to compare those assumptions. They do require actual likelihoods. A quasi-Poisson fit cannot be assigned an ordinary AIC just by inserting its coefficient estimates into a Poisson likelihood and scaling the result.

Worked Example: AIC for Two Full-Likelihood Models

For the same hypothetical claim-count observations, an insurer fits a Poisson mean model with two estimated coefficients and a negative binomial model with the same two mean coefficients plus an estimated \(\alpha\). The maximized log-likelihoods are \(-114.2\) for Poisson and \(-101.6\) for negative binomial. Calculate AIC and select the lower-AIC model.

Method. Both models are stipulated to be full likelihoods on the identical counts. Count the mean coefficients and any estimated dispersion parameter before applying AIC.

For Poisson, \(K_P=2\), so \(AIC_P=-2(-114.2)+2(2)=228.4+4=232.4\). For negative binomial, \(K_{NB}=2+1=3\), so \(AIC_{NB}=-2(-101.6)+2(3)=203.2+6=209.2\).

The negative binomial AIC is lower by \(232.4-209.2=23.2\), so it is preferred among these two fitted full-likelihood candidates by AIC. Its better likelihood more than offsets the penalty for estimating \(\alpha\). Common mistake avoided: omitting the dispersion parameter from \(K_{NB}\), which would under-penalize the negative binomial model.

A Second Full-Likelihood Illustration: An Estimated Normal Scale

The same distinction appears in a normal model with an unknown variance. Suppose independent responses satisfy \(Y_i\sim N(\mu_i,\sigma^2)\), with a common variance \(\sigma^2\) estimated from the data. This is a full likelihood: \(\sigma^2\) is part of the model and is estimated by maximum likelihood. For a fitted mean structure with residual sum of squares \(SSE\), the maximum-likelihood estimate is \(SSE/n\), not the unbiased estimate \(SSE/(n-p)\).

Worked Example: AIC With an Estimated Variance Parameter

A hypothetical insurer models a continuous claim-cost measure using independent normal observations. There are \(n=8\) observations. Model A has two mean coefficients and \(SSE_A=12\); Model B has three mean coefficients and \(SSE_B=6\). Each model estimates one common normal variance. Calculate AIC using maximum likelihood, counting the variance parameter.

Method. For each model, estimate \(\sigma^2\) as \(SSE/n\), substitute into the maximized normal log-likelihood, and count the mean coefficients plus one variance parameter in AIC.

For Model A, \(\hat\sigma_A^2=12/8=1.5\), and \(K_A=2+1=3\). Its maximized log-likelihood is:

\ell_A=-\frac{8}{2}\left[\log(2\pi)+1+\log(1.5)\right] =-4(1.8379+1+0.4055)=-12.9736.

Therefore \(AIC_A=-2(-12.9736)+2(3)=25.9472+6=31.9472\).

For Model B, \(\hat\sigma_B^2=6/8=0.75\), and \(K_B=3+1=4\). Its maximized log-likelihood is:

\ell_B=-4\left[\log(2\pi)+1+\log(0.75)\right] =-4(1.8379+1-0.2877)=-10.2008.

Thus \(AIC_B=-2(-10.2008)+2(4)=20.4016+8=28.4016\). Model B has the lower AIC. The calculation uses the maximum-likelihood variance estimate in the likelihood and counts it as an estimated parameter. Common mistake avoided: using the residual-degrees-of-freedom variance estimate inside the maximized likelihood or omitting the variance parameter from the AIC penalty.

Practical Checks Before Ranking Candidates

Before reporting a likelihood-based ranking, check what the criterion actually measures. Overdispersion is a warning that a Poisson variance assumption may understate uncertainty, not automatic evidence that one particular alternative model is correct. A negative binomial model, a quasi-Poisson model, or a revised mean structure answers a different modeling question; each requires assumptions that should be stated.

For ordinary AIC: confirm that each candidate has a specified, normalized likelihood for the same response observations. Include every estimated likelihood parameter, including dispersion parameters.
For quasi-likelihood: name the quasi criterion, state how its dispersion is estimated, and use a consistent scale across candidates when the comparison requires one. Do not label the result ordinary AIC.
For Poisson deviance: do not interpret a deviance difference as having its usual Poisson likelihood meaning when the variance assumption is materially wrong. A dispersion-adjusted quasi comparison is not a replacement Poisson likelihood-ratio test.
For interpretation: a lower score ranks candidates only under the selected criterion and its assumptions. It does not by itself establish that the fitted rates are stable, adequate for implementation, or causally interpretable.

Key takeaway: Overdispersion changes the uncertainty attached to model fit. Use a clearly defined quasi-likelihood comparison with an appropriate, consistently applied scale, or use a full likelihood whose distribution explicitly models dispersion. Never present a scaled Poisson deviance as an ordinary Poisson likelihood or AIC.

Check Your Understanding

For each item, identify the appropriate calculation or distinction before checking the final answer.

A quasi-Poisson reference model has Pearson statistic \(18\) and residual degrees of freedom \(6\). What is its Pearson dispersion estimate? (Answer: \(\hat\phi_P=3.000\).)
Two quasi-Poisson mean models use the same observations and a common reference dispersion \(\hat\phi=1.5\). Model A has deviance \(21\) and three mean coefficients; Model B has deviance \(16\) and five. Using \(D/\hat\phi+2K\), which ranks lower? (Answer: A scores \(21/1.5+6=20\); B scores \(16/1.5+10=20.6667\); Model A ranks lower.)
For \(n=10\) independent normal observations, Model A has \(SSE=20\) and two mean coefficients; Model B has \(SSE=12\) and three. Both estimate one common variance. Using the maximum-likelihood variance estimate and counting it in AIC, which model has lower AIC? (Answer: Model A has AIC approximately \(41.31\); Model B has AIC approximately \(38.20\); Model B ranks lower.)
Two full-likelihood count models use identical observations. A Poisson model has \(-2\ell=88\) and two estimated parameters. A negative binomial model has \(-2\ell=80\), two mean coefficients, and one estimated dispersion parameter. Which has lower AIC? (Answer: Poisson AIC \(=88+2(2)=92\); negative binomial AIC \(=80+2(3)=86\), so negative binomial ranks lower.)
An analyst multiplies a fitted Poisson likelihood’s deviance by \(1/2\) because the estimated quasi-Poisson dispersion is \(2\), then calls the result an ordinary Poisson AIC. Is that interpretation valid? (Answer: No. A dispersion-adjusted quasi criterion may be defined, but scaling the Poisson deviance does not create a Poisson likelihood or ordinary AIC.)

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