Why Assess Dispersion?
A GLM specifies both a mean structure and a relationship between the mean and the response variance. For example, a Poisson frequency model assumes that the conditional variance equals the conditional mean. Even if fitted means appear reasonable, observed variation may be larger or smaller than that assumption predicts. The Pearson chi-square statistic provides an aggregate check of this variation relative to the model’s residual degrees of freedom.
This assessment is useful because uncertainty estimates rely on a variance assumption. If the response varies more than the assumed model allows, standard errors calculated under that assumption can be too small. If it varies less, they can be too large. The dispersion calculation summarizes this issue; it does not, by itself, explain its cause or establish that the model’s mean structure is adequate.
The variance model determines the denominator in each Pearson residual. For a Poisson response, \(V(\mu)=\mu\), so \(r_{Pi}=(y_i-\hat{\mu}_i)/\sqrt{\hat{\mu}_i}\). For a binomial response recorded as a number of successes \(y_i\) out of \(m_i\) trials, the variance of the count is \(m_i\hat{p}_i(1-\hat{p}_i)\), where \(\hat{p}_i\) is the fitted success probability. Use the variance for the response as recorded, not a variance for a different scale.
The residual degrees of freedom account approximately for the parameters used to fit the model. In an ordinary unweighted GLM with \(n\) independent response observations and design-matrix rank \(p\), \(\nu=n-p\). If observations are aggregated, weighted, or otherwise dependent, the relevant units and degrees of freedom depend on the model and data construction; simply counting rows may not be justified. In particular, do not substitute the number of predictor columns if some are linearly dependent: use the fitted design-matrix rank.
Interpreting the Dispersion Estimate
| Result | Interpretation | Typical implication |
|---|---|---|
| \(\hat{\phi}\) near 1 | Aggregate residual variation is broadly consistent with the assumed variance scale. | No strong aggregate evidence of a different scale, subject to the model and approximation. |
| \(\hat{\phi}\) greater than 1 | Overdispersion: residual variation is greater than the unit-dispersion model predicts. | Under a scalar dispersion adjustment, standard errors are multiplied by \(\sqrt{\hat{\phi}}\). |
| \(\hat{\phi}\) less than 1 | Underdispersion: residual variation is less than the unit-dispersion model predicts. | A scalar adjustment would reduce standard errors by \(\sqrt{\hat{\phi}}\), if that model is appropriate. |
“Near 1” is not a universal numerical cutoff. The size of the data set, the response distribution, leverage, and the way the model was fitted all affect how much the ratio may vary. A chi-square reference can help calibrate the comparison, but the ratio is an approximate diagnostic, not a verdict on every aspect of fit.
Why Compare the Statistic With Residual Degrees of Freedom?
Under a correctly specified mean and variance model with unit dispersion, the residuals, after fitting parameters, have less variation available than they would if the means were known in advance. The residual degrees of freedom provide the corresponding approximate scale for their squared standardized values. This explains why comparing \(X_P^2\) with \(n\), rather than with \(n-p\), generally fails to account for estimation.
Each Pearson residual divides its observed-minus-fitted difference by the model-based standard deviation for that observation.
The Pearson chi-square is the total squared standardized discrepancy: \(X_P^2=\sum_i r_{Pi}^2\).
Estimating \(p\) independent parameters uses degrees of freedom. In a large-sample, correctly specified GLM, the residual sum of squares is often approximately chi-square with \(\nu=n-p\) degrees of freedom.
A chi-square variable with \(\nu\) degrees of freedom has expectation \(\nu\). Thus \(X_P^2/\nu\) estimates the residual variance scale relative to the assumed scale.
The last step gives a short justification for the dispersion estimate. If the residual statistic is approximately distributed as \(\phi\chi^2_\nu\), its expected value is approximately \(\phi\nu\). Dividing the observed statistic by \(\nu\) therefore estimates \(\phi\). When \(\phi=1\), the assumed variance scale is supported at this aggregate level; values above or below 1 indicate greater or lesser variation, respectively.
The chi-square approximation is most useful as a large-sample guide under a suitable mean and variance specification and independent observational units. Discrete responses, small expected counts, influential observations, dependence, or misspecification can weaken it. Also, a large ratio does not identify a single cause: the variance function may be inadequate, the mean structure may be wrong, or the data may have features not represented by the model. Treat the ratio as a signal to investigate, not as proof of a particular explanation.
Worked Examples
Worked Example: Assessing a Poisson Frequency Model
A hypothetical insurer fits a Poisson claim-frequency GLM to \(n=120\) independent policy observations. The design matrix has rank \(p=5\). The sum of the squared unscaled Pearson residuals is \(X_P^2=161\). Assess dispersion and compare the result with the unit-dispersion assumption.
Method. For a Poisson model, each contribution is \((y_i-\hat{\mu}_i)^2/\hat{\mu}_i\), and the given sum is the Pearson chi-square. First calculate the residual degrees of freedom:
Then calculate the Pearson dispersion estimate:
Since \(1.400>1\), the residual variation is larger than the Poisson unit-dispersion assumption predicts. A scalar dispersion adjustment, if appropriate, would multiply standard errors by \(\sqrt{1.400}=1.1832\), approximately. As a rough reference, a chi-square variable with 115 degrees of freedom has standard deviation \(\sqrt{2(115)}=\sqrt{230}\approx15.17\). The observed statistic exceeds its expected value of 115 by \(161-115=46\), or about \(46/15.17=3.03\) such standard deviations. This is a diagnostic comparison, not an exact test calculation.
Check. The ratio is \(161/115=1.400\), not \(161/120\): the denominator accounts for the five fitted dimensions. The calculation indicates overdispersion relative to the fitted Poisson variance scale, but does not establish its cause. A common mistake is to call the model wrong solely from this ratio; the ratio alone cannot distinguish a variance problem from other sources of residual variation.
Worked Example: Recognizing Underdispersion in Grouped Binomial Data
A hypothetical equipment insurer models the number of units with a specified event in each of eight independent groups. Each group contains 20 units, and the fitted event probability is \(0.5\) for every group. The observed counts are \(11,9,11,9,11,9,11,9\). The fitted model has design-matrix rank 3. Calculate the dispersion estimate and interpret it.
Method. For a binomial count with \(m_i=20\) and \(\hat{p}_i=0.5\), the fitted mean is \(20(0.5)=10\), and the fitted variance is \(20(0.5)(1-0.5)=5\). Each observed count differs from its fitted mean by either \(1\) or \(-1\). Thus, for every group, the Pearson contribution is
There are eight groups, so \(X_P^2=8(0.2)=1.6\). The residual degrees of freedom are \(\nu=8-3=5\), giving
This is underdispersion relative to the fitted binomial variance scale. A scalar adjustment would multiply standard errors by \(\sqrt{0.32}\approx0.5657\), if the assumptions for that adjustment were justified. It would not change the observed counts, fitted probabilities, or the interpretation of the data.
Check. Each group contributes \(0.2\), so the total is \(1.6\); dividing by five residual degrees of freedom gives \(0.32\). A common mistake is to use the number of trials, 20, as the number of observations for residual degrees of freedom. Here the responses are eight grouped counts, and the stated model rank is three, so the stated residual degrees of freedom are \(8-3=5\).
Worked Example: Adjusting a Coefficient Standard Error
A hypothetical claim-severity model uses \(n=50\) independent observations and has design-matrix rank \(p=4\). Its Pearson chi-square is 69. A log-link coefficient for a rating factor is estimated as \(0.30\), with conventional unit-dispersion standard error \(0.08\). Under a scalar quasi-likelihood dispersion adjustment, calculate the adjusted standard error and an approximate 95% interval for the coefficient and its relativity.
Method. First find the residual degrees of freedom and dispersion estimate:
In a scalar dispersion model, the variance of an estimator is multiplied by \(\hat{\phi}\); its standard error is therefore multiplied by the square root, not by \(\hat{\phi}\) itself:
Using the normal critical value \(1.96\), the approximate coefficient interval is
Exponentiating the endpoints gives the interval on the multiplicative relativity scale: \((e^{0.1079},e^{0.4921})\approx(1.114,1.636)\). The point estimate remains \(e^{0.30}\approx1.350\); the scalar adjustment changes the estimated uncertainty, not the fitted coefficient or relativity.
Check. The dispersion estimate is 1.50 and the standard-error multiplier is \(\sqrt{1.50}\approx1.2247\). A common mistake is multiplying the standard error by 1.50, which would give 0.1200. That scales the variance factor as though it were a standard-error factor and overstates the adjustment.
What a Dispersion Adjustment Does—and Does Not Do
If a GLM is extended to use variance \(\operatorname{Var}(Y_i\mid x_i)=\phi V(\mu_i)\) with a single common dispersion parameter, an estimated \(\phi\) can be used to scale estimated variances and standard errors. In common quasi-likelihood applications, multiplying all variances by the same constant leaves the fitted mean estimates unchanged, while their estimated uncertainty changes. The exact estimation and inferential procedure still depends on the model; do not assume every GLM software fit uses the same adjustment automatically.
For overdispersion, multiplying standard errors by \(\sqrt{\hat{\phi}}\) generally widens confidence intervals and reduces the magnitude of Wald statistics for fixed coefficient estimates. For underdispersion, the same formula gives a multiplier below one, but a smaller standard error should not be accepted mechanically. First consider whether the variance model, sampling design, independence assumption, and definition of the observational units support that interpretation.
Most importantly, scalar rescaling cannot repair an incorrect mean structure. It does not correct biased fitted means, omitted effects, an unsuitable link, or a response distribution that fails in ways a single scale factor cannot represent. Nor does one aggregate number show where residual discrepancies occur. Use it to assess overall dispersion, and keep the conclusion appropriately narrow: it describes total standardized residual variation relative to the available residual degrees of freedom.
Check Your Understanding
Work through each calculation, then compare your result with the answer at the end of the question.
- A hypothetical Poisson frequency GLM has 90 independent observations, design-matrix rank 6, and Pearson chi-square 100.8. Find the residual degrees of freedom, dispersion estimate, and standard-error multiplier under a scalar adjustment. (Answer: \(\nu=84\), \(\hat{\phi}=1.200\), multiplier \(\sqrt{1.2}=1.0954\); overdispersion.)
- A binomial model has 12 independent grouped-count observations and design-matrix rank 4. Its Pearson chi-square is 6.4. Calculate the dispersion estimate and state whether it indicates overdispersion or underdispersion. (Answer: \(\nu=8\), \(\hat{\phi}=0.800\); underdispersion.)
- A hypothetical GLM has 75 independent observations, rank 5, Pearson chi-square 112, and a coefficient’s unit-dispersion standard error is 0.06. Find the adjusted standard error under a scalar dispersion model. (Answer: \(\nu=70\), \(\hat{\phi}=1.600\), adjusted standard error \(0.06\sqrt{1.6}=0.0759\), approximately.)
- A model has \(X_P^2=58\) and residual degrees of freedom 50. Interpret the ratio and state one conclusion that cannot be drawn from it alone. (Answer: \(\hat{\phi}=58/50=1.16\), indicating modest aggregate overdispersion relative to unit dispersion; the ratio alone cannot identify its cause or prove which part of the model is misspecified.)