Why Measure Influence?
A policy can have an unusual outcome without substantially changing a fitted rate model. Conversely, a policy with an outcome that is not especially extreme may affect the fitted coefficients because its rating variables place it in a part of the predictor space with little data. Influence describes how much a fitted model changes when an observation is removed. Cook’s distance combines the observation’s effect on the fitted coefficients into one measure.
In ratemaking, this is useful as a screening diagnostic. A large value may identify a policy worth checking for data errors, unusual but valid experience, or a model structure that does not adequately represent the risk. It does not by itself show that the policy is erroneous or that it should be excluded. The measure is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
A Practical GLM Approximation
Refitting the model once for every observation can be computationally demanding, so GLM diagnostics often use a one-step approximation. To state it precisely, suppose the observations are independent conditional on the model, the design matrix is fixed, the GLM mean and variance functions are correctly specified for the calculation, and the fitted model has a finite, nonsingular coefficient estimate. Write the response mean as \(\mu_i\), the variance as \(\operatorname{Var}(Y_i)=\phi V(\mu_i)/a_i\), and the linear predictor as \(\eta_i=\mathbf{x}_i^{\mathsf T}\boldsymbol\beta\). Here \(\phi\) is the dispersion parameter, \(a_i\) is a prior weight, and \(V(\mu_i)\) is the GLM variance function. For a Poisson model with its usual dispersion convention, \(\phi=1\); for other models, it may be estimated.
Let \(h_{ii}\) be the \(i\)th diagonal element of the GLM weighted hat matrix, measuring the observation’s leverage. Define the unscaled Pearson residual as \(r_{P,i}=\sqrt{a_i}(y_i-\hat\mu_i)/\sqrt{V(\hat\mu_i)}\); the term “unscaled” means that \(\phi\) is not included in its denominator. With this convention, the common approximate Cook’s distance is
The formula is an approximation to the coefficient-change measure, not a replacement for deletion and refitting when an exact assessment is needed. Its usual local derivation relies on the weighted least-squares representation used in fitting a GLM. It can be less reliable when deleting an observation would lead to a large nonlinear change in the fitted model, when a coefficient estimate becomes unstable, or when the model assumptions are poor.
Why the Approximation Has This Form
A short derivation explains the roles of the residual and leverage. Let \(A\) denote the full-data information matrix without the dispersion factor, and let \(\mathbf{x}_i\) be the design row for observation \(i\). In the local weighted least-squares representation, write \(w_i=a_i(d\mu_i/d\eta_i)^2/V(\mu_i)\). Then \(h_{ii}=w_i\mathbf{x}_i^{\mathsf T}A^{-1}\mathbf{x}_i\). The observation’s contribution to the score is proportional to \(\mathbf{x}_i a_i(d\mu_i/d\eta_i)(y_i-\mu_i)/V(\mu_i)\).
At the full-data estimate the total score is zero. Removing observation \(i\) therefore leaves a score equal to the negative of its individual contribution. A first-order Taylor expansion of the deleted score around the full-data estimate gives the coefficient change. Accounting for the fact that the deleted information matrix omits observation \(i\)’s contribution introduces the leverage adjustment \(1-h_{ii}\). Thus, up to sign, the one-step coefficient change is approximately
For a GLM with dispersion \(\phi\), the coefficient covariance is approximately \(\phi A^{-1}\). Substituting the one-step change into the covariance-scaled squared distance gives a factor \(\mathbf{x}_i^{\mathsf T}A^{-1}\mathbf{x}_i\). Using \(h_{ii}=w_i\mathbf{x}_i^{\mathsf T}A^{-1}\mathbf{x}_i\), the remaining response factor becomes \(a_i(y_i-\hat\mu_i)^2/[\phi V(\hat\mu_i)]=r_{P,i}^2/\phi\). Dividing by \(p\) yields the approximation above. This derivation is local: it explains the diagnostic formula but does not make it an exact substitute for refitting.
Calculating and Interpreting Cook’s Distance
Worked Example: Screening Three Policies
A hypothetical claim-frequency GLM has \(p=4\) fitted coefficients and dispersion estimate \(\phi=1.1\). Three policies have the following unscaled Pearson residuals and leverages: Policy A has \(r_{P,A}=-1.5\) and \(h_{AA}=0.08\); Policy B has \(r_{P,B}=3.0\) and \(h_{BB}=0.12\); Policy C has \(r_{P,C}=2.2\) and \(h_{CC}=0.20\). Calculate the approximate Cook’s distances, rank the policies, and identify the common error avoided.
Method. Use \(D_i\approx r_{P,i}^2h_{ii}/[p\phi(1-h_{ii})^2]\). The sign of a residual does not affect this measure because it is squared.
Policy A. The numerator is \((-1.5)^2(0.08)=2.25(0.08)=0.18\). The denominator is \(4(1.1)(1-0.08)^2=4.4(0.92)^2=4.4(0.8464)=3.72416\). Therefore \(D_A\approx0.18/3.72416=0.0483\).
Policy B. The numerator is \(3.0^2(0.12)=9.00(0.12)=1.08\). The denominator is \(4(1.1)(1-0.12)^2=4.4(0.88)^2=4.4(0.7744)=3.40736\). Therefore \(D_B\approx1.08/3.40736=0.3170\).
Policy C. The numerator is \(2.2^2(0.20)=4.84(0.20)=0.968\). The denominator is \(4(1.1)(1-0.20)^2=4.4(0.80)^2=4.4(0.64)=2.816\). Therefore \(D_C\approx0.968/2.816=0.3438\).
The ranking is Policy C, Policy B, then Policy A. Policy C has the largest approximate influence despite having a smaller residual than Policy B, because its leverage is higher. The common mistake is to use \(1-h_{ii}\) rather than \((1-h_{ii})^2\) in the denominator, or to round an intermediate denominator too early. Keeping the stated denominator gives Policy B’s four-decimal result as \(0.3170\). These values screen policies; they do not establish that any policy is invalid.
Worked Example: Computing a Deletion-Based Distance
A hypothetical two-coefficient GLM has a full-data estimated coefficient covariance matrix \(\widehat{\operatorname{Cov}}(\hat{\boldsymbol\beta})=\begin{pmatrix}0.25&0\\0&0.04\end{pmatrix}\). After deleting one policy and refitting, the full-data estimate minus the deleted-data estimate is \(\boldsymbol\Delta=(0.5,-0.2)^{\mathsf T}\). Calculate the deletion-based Cook’s distance.
Method. Here \(p=2\). The inverse covariance matrix is \(\begin{pmatrix}4&0\\0&25\end{pmatrix}\), since \(1/0.25=4\) and \(1/0.04=25\). The quadratic form is \(\boldsymbol\Delta^{\mathsf T}\widehat{\operatorname{Cov}}(\hat{\boldsymbol\beta})^{-1}\boldsymbol\Delta =(0.5)^2(4)+(-0.2)^2(25)=1+1=2\).
Dividing by \(p=2\) gives \(D_i=2/2=1.0000\). The squared coefficient changes are weighted differently because the two coefficients have different uncertainty scales. This is why simply adding the raw squared coefficient differences would not calculate Cook’s distance. The result is exact for the stated deletion refit and covariance definition; it need not equal the one-step approximation exactly.
Screening Is Not a Decision Rule
A screening threshold is sometimes set for a particular analysis to make review manageable. For example, \(4/n\), where \(n\) is the number of observations, is used as a rough flag in some settings. It is a heuristic, not a universal GLM cutoff or a formal test. Interpretation depends on the model, sample size, leverage distribution, dispersion treatment, and the purpose of the analysis. A value above a chosen screening level means “inspect this observation,” not “remove this observation.”
A useful review asks whether the record is correct, whether its rating variables were coded as intended, and whether the fitted model represents the policy’s risk adequately. If the observation is valid, it may be important to retain it: its influence can reveal that a particular rating class is poorly supported or that the model is sensitive to a genuine part of the portfolio. Any refit or model change should be assessed on its merits, rather than made solely to reduce Cook’s distance.
Record the coefficient count, dispersion convention, residuals, and leverage values used in the diagnostic.
Use the deletion-based definition for a refit assessment or the one-step formula as a screening approximation. Confirm whether residuals are dispersion-scaled.
Check the data and model context. Do not delete a valid policy just because its diagnostic value is large.
Common Calculation Traps
- Forgetting the square on the leverage adjustment. The approximation has \((1-h_{ii})^2\), not \(1-h_{ii}\).
- Mixing residual conventions. The formula shown uses an unscaled Pearson residual and divides by \(\phi\). If a residual is already divided by \(\sqrt{\phi}\), dividing by \(\phi\) again would count dispersion twice.
- Ignoring the coefficient count. The quadratic distance is divided by \(p\), the number of fitted coefficients, including the intercept when one is fitted.
- Treating a large value as proof of bad data. Influence is a reason to investigate, not evidence that an observation should automatically be excluded.
Check Your Understanding
For the first three questions, use \(D_i\approx r_{P,i}^2h_{ii}/[p\phi(1-h_{ii})^2]\) with unscaled Pearson residuals. Round Cook’s distances to four decimal places.
- A hypothetical severity GLM has \(p=3\), \(\phi=1.2\), \(r_{P,i}=2.0\), and \(h_{ii}=0.15\). Calculate the approximate Cook’s distance. (Answer: \(0.2307\))
- A fitted frequency model has \(p=5\), \(\phi=1.0\), \(r_{P,i}=-1.5\), and \(h_{ii}=0.20\). Calculate the approximate Cook’s distance and explain why the residual’s negative sign does not reduce it. (Answer: \(0.1406\); the residual is squared.)
- In a hypothetical model with \(n=50\), an analyst uses \(4/n\) only as a screening heuristic. The model has \(p=4\), \(\phi=1.1\). Policy A has \(r_P=1.0,h=0.10\); Policy B has \(r_P=2.0,h=0.15\). Calculate both approximate distances and identify which exceeds the heuristic level. (Answer: \(4/n=0.0800\); \(D_A=0.0281\), \(D_B=0.1887\); only Policy B exceeds it.)
- A two-coefficient refit gives \(\boldsymbol\Delta=(0.5,-0.2)^{\mathsf T}\) and full-fit coefficient covariance \(\begin{pmatrix}0.25&0\\0&0.04\end{pmatrix}\). Calculate the deletion-based Cook’s distance. (Answer: \(1.0000\))
- For an approximate GLM Cook’s distance, \(p=3\), \(\phi=1\), \(h_{ii}=0.25\), and the desired distance is \(0.20\). What is the magnitude of the unscaled Pearson residual that gives this distance? (Answer: \(|r_{P,i}|=\sqrt{0.20(3)(1-0.25)^2/0.25}=\sqrt{1.35}\approx1.1619\))