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Detecting Overdispersion · Tutorial 670 of 981

Influential Cases and False Dispersion Signals

Learn to distinguish a broad dispersion pattern from a signal driven by one unusual response or an influential design point.

Intermediate 10 min read

What You'll Learn

  • Separate a large residual contribution from genuine evidence of broad extra variation
  • Recalculate dispersion diagnostics after deleting and refitting a case
  • Use leverage and standardized Pearson residuals to identify influential design points
  • Interpret case-deletion coefficient changes alongside dispersion ratios
  • Avoid treating a single influential observation as proof of overdispersion

Why Inspect Influence Before Interpreting Dispersion

A Pearson dispersion ratio can look large because many observations vary more than the fitted GLM predicts. It can also look large because one observation contributes an unusually large residual. Those situations call for different interpretations: broad extra variation may affect uncertainty throughout the data, while a single unusual case may instead point to a data issue, a local model failure, or an influential observation.

The total statistic alone does not say how its evidence is distributed. A useful diagnostic therefore combines residual inspection with case deletion: remove a case, refit the model, and examine how the fitted coefficients and dispersion diagnostic change. A case can matter because its response is unusual, because its design point has high leverage, or because both features occur together.

Definition: A case is influential for a fitted GLM when deleting it and refitting produces a meaningful change in fitted coefficients, fitted means, or a diagnostic of interest. A case-deletion comparison compares the original fit with the fit obtained after removing one observation. It is a diagnostic comparison, not by itself a formal test of overdispersion.

Residual Contributions and Case Deletion

For independent observations with fitted means \(\hat{\mu}_i\) and model variances \(V(\hat{\mu}_i)\), the Pearson residual and Pearson statistic are

$$ r_{Pi}=\frac{y_i-\hat{\mu}_i}{\sqrt{V(\hat{\mu}_i)}}, \qquad X_P^2=\sum_{i=1}^{n} r_{Pi}^2. $$

For grouped binomial data with \(y_i\) successes out of \(m_i\) trials, \(V(\hat{\mu}_i)=m_i\hat{p}_i(1-\hat{p}_i)\). For Poisson counts, \(V(\hat{\mu}_i)=\hat{\mu}_i\). Each squared Pearson residual is a nonnegative contribution to the total. A large contribution identifies a case worth investigating, but it does not establish why that case is unusual.

There are two distinct deletion calculations. One can omit a contribution while keeping the original fitted values fixed, or remove the case and refit the model. The first calculation is a simple accounting check. The second is the relevant assessment of influence, since deleting a case may change every fitted mean and therefore every remaining residual.

Result: If fitted values are held fixed, the Pearson statistic after omitting case \(i\) is exactly \(X_P^2-r_{Pi}^2\). After refitting without that case, however, the new statistic is generally not \(X_P^2-r_{Pi}^2\).

To verify the fixed-fit identity, write the original statistic as \(X_P^2=r_{Pi}^2+\sum_{j\ne i}r_{Pj}^2\). Omitting only the \(i\)th term leaves \(\sum_{j\ne i}r_{Pj}^2=X_P^2-r_{Pi}^2\). Under refitting, the remaining residuals become \(r_{Pj}^{(-i)}\), calculated using new fitted means \(\hat{\mu}_j^{(-i)}\). Their sum of squares need not equal the old remaining sum. This distinction is essential when a case has substantial influence.

Worked Example: One Group Drives a Binomial Dispersion Signal

Suppose four independent groups each contain 10 trials, with observed successes \(5,5,5,10\). Fit an intercept-only binomial model, so the common success probability is estimated by the pooled proportion.

The full-data estimate is \(\hat p=(5+5+5+10)/40=25/40=0.625\). Each group has fitted count \(10(0.625)=6.25\) and variance \(10(0.625)(0.375)=2.34375=75/32\).

For each of the first three groups, the squared Pearson residual is \((5-6.25)^2/2.34375=(25/16)/(75/32)=2/3\), which is approximately \(0.6667\). For the fourth group, it is \((10-6.25)^2/2.34375=(225/16)/(75/32)=6\). Thus the exact statistic is

$$ X_P^2=3\left(\frac{2}{3}\right)+6=8. $$

The fitted model has one parameter, so the residual degrees of freedom are \(4-1=3\), and the dispersion ratio is \(8/3\approx2.67\). The fourth group contributes 6 of the total 8 units. That concentration suggests checking whether the apparent signal depends on this group.

Now delete the fourth group and refit. The estimate becomes \(\hat p^{(-4)}=15/30=0.5\). Each remaining group has five fitted successes, so all three Pearson residuals are zero. The refitted statistic is \(0\), with \(3-1=2\) residual degrees of freedom. This stark change shows that the original diagnostic was dominated by one group, not that the other groups independently displayed extra variation. The unusual group still needs contextual investigation; deletion does not establish that it is erroneous or justify ignoring it.

Leverage and the Residual That Looks Too Small

A case can influence a fit even when its raw residual is not large. Leverage describes how unusual a design point is relative to the other design points, in the geometry of the fitted model. In a GLM, the working-weight hat matrix is

$$ H=W^{1/2}X(X^\mathsf{T}WX)^{-1}X^\mathsf{T}W^{1/2}, $$

where \(X\) is the model matrix and \(W\) is the diagonal matrix of fitted working weights. This expression assumes \(X^\mathsf{T}WX\) is invertible. The diagonal entry \(h_{ii}\) is the leverage for case \(i\); large values indicate that the case has substantial potential to affect the fitted values.

In a linearized GLM fit, the weighted working residuals are approximately projected by \(I-H\). If the standardized working errors have covariance matrix \(I\), the covariance of these residuals is \(I-H\), since \(H\) is symmetric and idempotent: \((I-H)(I-H)^\mathsf{T}=I-H\). In particular, the approximate residual variance for case \(i\) is \(1-h_{ii}\). This motivates examining Pearson residuals adjusted by \(\sqrt{1-h_{ii}}\), especially when leverage differs substantially across cases. The approximation is a diagnostic guide, not an exact finite-sample distributional result for every GLM.

A high-leverage observation can pull the fitted mean toward itself, reducing its own raw residual while materially changing the fitted coefficients. Conversely, a large residual at an ordinary design point may have limited effect on the coefficient estimates. Residual size and leverage therefore answer different questions and should be considered together.

Worked Example: A High-Exposure Poisson Case

Consider three independent Poisson counts with exposures \(e=(1,1,10)\). Fit a common rate \(\lambda\), using \(\log(e_i)\) as an offset, to observed counts \(y=(1,1,20)\). The fitted rate is total count divided by total exposure: \(\hat\lambda=22/12=11/6\).

The fitted means are \(11/6,11/6,55/3\). The Pearson contributions for the first two cases are each \((1-11/6)^2/(11/6)=25/66\). The third contribution is \((20-55/3)^2/(55/3)=5/33\). Hence \(X_P^2=25/66+25/66+5/33=10/11\), and the dispersion ratio is \((10/11)/2=5/11\).

For this intercept-only rate model, the working weights are the fitted Poisson means, proportional to exposure. Thus the leverage values are \(1/12,1/12,10/12=5/6\). The high-exposure observation has high leverage even though its Pearson residual is only \((5/3)/\sqrt{55/3}\), approximately \(0.389\). Its leverage-adjusted value is approximately \(0.389/\sqrt{1/6}=0.953\).

Delete the high-exposure case and refit. The remaining estimated rate is \(2/2=1\), rather than \(11/6\); the log-rate coefficient changes from \(\log(11/6)\) to \(\log(1)=0\). The two remaining counts exactly equal their refitted means, so their refitted Pearson statistic is zero. This example is not evidence of overdispersion: the full-data dispersion ratio is below one. It illustrates a separate point: a case can substantially affect a fitted coefficient because of its design, even when its own residual does not look extreme.

Approximating the Effect of Deleting a Case

A refit is the clearest way to assess influence, but the score equation explains why unusual cases can matter. Let \(U(\beta)=\sum_i u_i(\beta)\) be the full-model score, with a finite maximum-likelihood estimate \(\hat\beta\) satisfying \(U(\hat\beta)=0\). Let \(U_{-i}(\beta)=U(\beta)-u_i(\beta)\) be the score after case \(i\) is removed, and suppose the deleted fit \(\hat\beta_{-i}\) exists.

Result: If the score is differentiable, the deleted-fit information matrix \(J_{-i}(\hat\beta)=-\partial U_{-i}(\beta)/\partial\beta^\mathsf{T}\vert_{\hat\beta}\) is nonsingular, and the deletion change is small enough for a first-order Taylor approximation to be informative, then
$$ \hat\beta_{-i}-\hat\beta \;\approx\; -J_{-i}(\hat\beta)^{-1}u_i(\hat\beta). $$

To derive this approximation, set \(\delta=\hat\beta_{-i}-\hat\beta\). Taylor expansion of the deleted score around \(\hat\beta\) gives \(0=U_{-i}(\hat\beta_{-i})=U_{-i}(\hat\beta)-J_{-i}(\hat\beta)\delta+R(\delta)\), where differentiability implies \(R(\delta)=o(\|\delta\|)\) as \(\delta\) tends to zero. Since \(U(\hat\beta)=0\), the deleted score at the original estimate is \(U_{-i}(\hat\beta)=U(\hat\beta)-u_i(\hat\beta)=-u_i(\hat\beta)\). Substitution yields \(0=-u_i(\hat\beta)-J_{-i}(\hat\beta)\delta+o(\|\delta\|)\). Ignoring only this higher-order remainder and multiplying by the inverse information matrix gives the stated first-order approximation. It is not an exact deletion formula; a full refit is preferable when the change is large, the information matrix is poorly conditioned, or the fit is near a boundary. \(\square\)

The expression combines the case's score contribution with the information available to estimate the coefficients. A large case score can be influential, but so can a design that makes the information matrix sensitive in a particular direction. In practice, compare the approximate signal with an actual refit rather than treating the approximation as a replacement.

Worked Example: Distinguishing a Large Residual From a Broad Pattern

Suppose an intercept-only Poisson model is fitted to counts \(2,2,2,10\). The fitted mean is \(\hat\mu=(2+2+2+10)/4=4\). The first three Pearson contributions are each \((2-4)^2/4=1\), while the last is \((10-4)^2/4=9\). Thus \(X_P^2=3(1)+9=12\), with \(4-1=3\) residual degrees of freedom and dispersion ratio \(12/3=4\).

The final observation accounts for \(9/12=75\%\) of the statistic. Removing it and refitting gives \(\hat\mu^{(-4)}=6/3=2\). The remaining observations equal their refitted means, so \(X_{P,-4}^2=0\), with two residual degrees of freedom. The original ratio is therefore highly sensitive to one response. This does not prove the count of 10 is a recording error: it could reflect an omitted predictor, dependence, latent heterogeneity, or genuine unusual variation. The diagnostic identifies concentration of the signal and motivates investigation; it does not identify the cause by itself.

Interpreting the Diagnostic Pattern

A few practical checks help keep the interpretation disciplined:

  • Inspect the contribution distribution. List or plot squared Pearson residuals to see whether the total is broadly shared or dominated by a small number of cases.
  • Check both response and design. Review unusual outcomes, data collection, exposures, and predictor combinations. A high residual and a high-leverage design point are distinct features.
  • Refit after deletion. Recalculate coefficients, fitted means, residuals, and the dispersion ratio using the reduced data. Do not subtract one old residual contribution and call that a refitted analysis.
  • Track degrees of freedom. The residual degrees of freedom generally change when a case is removed. Compare the recomputed ratios with that change in view.
  • Retain the scientific context. Do not discard a case simply because its removal improves a diagnostic. Establish whether it is erroneous, outside the target population, or a valid observation that the model fails to describe.

A common mistake is to reason, “The dispersion ratio is high, so the model has broad extra variation,” without checking which observations generated the statistic. The correction is to examine case-level residual contributions and conduct refitted deletion comparisons. A second mistake is to conclude that a high-leverage case is harmless because its raw residual is small. Leverage can pull the fit toward the case; check the leverage-adjusted residual and the change in coefficients after deletion. Neither check alone proves a cause, and deleting a case is a sensitivity analysis rather than a routine data-cleaning rule.

Key takeaway: An apparent dispersion signal is more persuasive as evidence of widespread extra variation when it is not dominated by a few cases and remains similar under reasonable refitted case-deletion checks. Large changes instead flag sensitivity that needs explanation before drawing a dispersion conclusion.

Check Your Understanding

Use these questions to test whether you can separate residual size, leverage, and influence.

  1. Why is \(X_P^2-r_{Pi}^2\) not generally the Pearson statistic from a model refitted without case \(i\)?
  2. In the four-group binomial example, what fraction of the original Pearson statistic came from the group with 10 successes?
  3. How can an observation have a modest raw Pearson residual and still noticeably affect fitted coefficients?
  4. What information does a case-deletion comparison provide that inspection of the full-data dispersion ratio does not?
  5. Why is deleting a case because it raises the dispersion ratio not, by itself, a defensible analysis decision?
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