Standard Errors for Fitted Coefficients

What a Coefficient Standard Error Measures

A fitted GLM coefficient is an estimate based on a particular sample of insurance experience. If a comparable sample were collected under the same conditions and the model were refitted, the resulting estimate would generally differ. A standard error summarizes the estimated sampling variability of that coefficient estimate under a stated model and covariance-estimation method.

This uncertainty concerns the estimated model parameter, not the spread of outcomes among policies. For example, a claim-count model may have substantial policy-to-policy variation even when its fitted rating coefficient is estimated precisely from extensive experience. Conversely, a coefficient may have a large standard error when the available data provide little information about it, even if individual outcomes are not especially variable.

Definition: Let $\hat{\boldsymbol{\beta}}$ be the vector of fitted GLM coefficients and let $\widehat{\operatorname{Cov}}(\hat{\boldsymbol{\beta}})$ be an estimated covariance matrix for those coefficient estimates. The estimated standard error of coefficient $\hat{\beta}_j$ is the square root of its estimated variance: $$ \operatorname{SE}(\hat{\beta}_j) = \sqrt{\widehat{\operatorname{Var}}(\hat{\beta}_j)} = \sqrt{\left[\widehat{\operatorname{Cov}}(\hat{\boldsymbol{\beta}})\right]_{jj}}. $$ An off-diagonal entry $\widehat{\operatorname{Cov}}(\hat{\beta}_j,\hat{\beta}_k)$ estimates how the two coefficient estimates vary together under the specified covariance method.

A coefficient and its standard error use the same coefficient scale. For a log-link model, a coefficient is on the linear-predictor, or log, scale, so its standard error is also on that scale. It is not a standard error for a policy's claim count, loss, or premium. Translating uncertainty from a coefficient to a multiplicative relativity is a separate calculation.

Why the Covariance Matrix Uses Information

For a likelihood-fitted GLM, the usual model-based covariance estimate is obtained from the curvature of the log-likelihood near its maximum. Write the coefficient vector as $\boldsymbol{\beta}$, the log-likelihood as $\ell(\boldsymbol{\beta})$, and the score as $U(\boldsymbol{\beta})=\partial\ell/\partial\boldsymbol{\beta}$. The Fisher information is the expected negative derivative of the score; the observed information uses the negative derivative evaluated at the fitted coefficients. Under regularity conditions and a sufficiently informative sample, either provides the local approximation underlying the usual standard errors.

Approximation: For an identifiable GLM fitted by maximum likelihood, the model-based covariance is estimated by the inverse information matrix evaluated at the fitted model. In common GLM notation this is $$ \widehat{\operatorname{Cov}}(\hat{\boldsymbol{\beta}}) \approx \hat{\phi}\,(X^\mathsf{T}WX)^{-1}, $$ where $X$ is the model matrix, $W$ is the fitted GLM weight matrix, and $\hat{\phi}$ is the estimated dispersion when the model uses one. For a Poisson model with its usual fixed dispersion of 1, the factor $\hat{\phi}$ is 1. The exact weights and dispersion convention depend on the GLM and software parameterization.

The approximation follows from a local expansion of the score. At the maximum-likelihood estimate, the score is zero. Near the true coefficient vector, a first-order Taylor expansion gives the fitted coefficient error approximately as the inverse information multiplied by the score at the true value. Under the likelihood assumptions, the score has covariance approximately equal to the information. Multiplying these matrices gives the inverse information as the approximate covariance of the estimate:

\begin{aligned} 0=U(\hat{\boldsymbol{\beta}}) &\approx U(\boldsymbol{\beta})- I(\boldsymbol{\beta})(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta}),\\ \hat{\boldsymbol{\beta}}-\boldsymbol{\beta} &\approx I(\boldsymbol{\beta})^{-1}U(\boldsymbol{\beta}),\\ \operatorname{Cov}(\hat{\boldsymbol{\beta}}) &\approx I^{-1}\operatorname{Cov}(U)I^{-1} =I^{-1}I I^{-1} =I^{-1}. \end{aligned}

This is a local, large-sample approximation, not a guarantee that the estimate is close to the true parameter in every sample. It relies on an appropriately specified likelihood, identifiable coefficients, and information that is not singular. Sparse data, near-collinearity, or a coefficient close to a boundary can make the approximation unreliable or the estimated standard error very large.

Worked Example: Reading a Coefficient Covariance Matrix

A hypothetical insurer fits a log-link frequency model with an intercept $\hat{\beta}_0= -3.20$ and a territory coefficient $\hat{\beta}_1=0.12$. The reported estimated covariance matrix, ordered as intercept then territory, is $$ \begin{pmatrix} 0.040 & -0.006\\ -0.006 & 0.010 \end{pmatrix}. $$ Find both coefficient standard errors and the correlation between the two coefficient estimates. State what the results describe.

Method. Take the square root of each diagonal variance to obtain its standard error. Use covariance divided by the product of the standard errors to obtain the correlation.

For the intercept, $$ \operatorname{SE}(\hat{\beta}_0)=\sqrt{0.040}=0.200. $$ For the territory coefficient, $$ \operatorname{SE}(\hat{\beta}_1)=\sqrt{0.010}=0.100. $$ The correlation is $$ \frac{\widehat{\operatorname{Cov}}(\hat{\beta}_0,\hat{\beta}_1)} {\operatorname{SE}(\hat{\beta}_0)\operatorname{SE}(\hat{\beta}_1)} = \frac{-0.006}{(0.200)(0.100)} =-0.300. $$

Thus, under the covariance method used for this output, the estimated standard errors are 0.200 and 0.100, and the coefficient estimates have estimated correlation $-0.300$. The negative correlation describes how the two estimates co-vary across repeated samples under the model approximation; it does not mean that policies with higher territory values have lower claim counts.

Check. The diagonal entries are variances, since $0.200^2=0.040$ and $0.100^2=0.010$. The covariance matrix is symmetric and the computed correlation lies between $-1$ and 1. A common mistake is to report 0.040 and 0.010 as standard errors; those are variances, so each must be square-rooted.

A Poisson Example With an Exposure Offset

For a Poisson frequency model, suppose independent claim counts $Y_j$ have means $\mu_j$, with $Y_j\sim\operatorname{Poisson}(\mu_j)$. A log-link model with exposure $e_j$ can be written as $\log(\mu_j)=\log(e_j)+\mathbf{x}_j^\mathsf{T}\boldsymbol{\beta}$. The exposure is known and enters as an offset; the information for the coefficients depends on the fitted means and predictor rows. The following grouped example makes that calculation explicit.

Worked Example: Standard Errors for a Two-Group Frequency Model

A hypothetical portfolio has two rating groups. Group 0 has 2,000 policy-years and 40 claims; group 1 has 1,000 policy-years and 30 claims. Assume independent Poisson claim counts and fit $\log(\mu_j)=\log(e_j)+\alpha+\beta x_j$, where $x_j=0$ for group 0 and $x_j=1$ for group 1. Find the fitted coefficients and their model-based standard errors.

Method and assumptions. Use the Poisson maximum-likelihood estimates for the two group frequencies, express them as $\alpha$ and $\beta$, then invert the fitted Fisher information. The exposure is measured in policy-years and is known. The fitted Poisson model assumes conditional independence and variance equal to the mean.

The fitted frequency in group 0 is $40/2000=0.020$ claims per policy-year. Since its linear predictor is $\alpha$, $$ \hat{\alpha}=\log(0.020)=-3.912023. $$ The fitted frequency in group 1 is $30/1000=0.030$. Its linear predictor is $\alpha+\beta$, so $$ \hat{\beta}=\log(0.030)-\log(0.020) =\log(1.5)=0.405465. $$

For a Poisson log-link model, each observation contributes fitted mean $\hat{\mu}_j$ times the outer product of its predictor row to the information. Aggregating within the two groups, the predictor rows are $(1,0)$ and $(1,1)$, and the fitted total means are 40 and 30. Therefore, $$ \widehat{I} = 40\begin{pmatrix}1&0\\0&0\end{pmatrix} + 30\begin{pmatrix}1&1\\1&1\end{pmatrix} = \begin{pmatrix}70&30\\30&30\end{pmatrix}. $$ The determinant is $70(30)-30(30)=1200$, so $$ \widehat{I}^{-1} = \frac{1}{1200} \begin{pmatrix}30&-30\\-30&70\end{pmatrix} = \begin{pmatrix} 0.025000&-0.025000\\ -0.025000&0.058333 \end{pmatrix}. $$ The Poisson dispersion is fixed at 1 here. Consequently, $$ \operatorname{SE}(\hat{\alpha})=\sqrt{0.025000}=0.1581, \qquad \operatorname{SE}(\hat{\beta})=\sqrt{0.058333}=0.2415. $$

The fitted intercept and group coefficient are on the log-frequency scale, and their standard errors are on that same scale. The negative covariance, $-0.025000$, reflects the fact that changing the fitted baseline can partly offset a change in the group difference. It is not a measure of how claim counts vary between individual policyholders.

Check. The implied group linear predictors are $-3.912023$ and $-3.912023+0.405465=-3.506558$. Exponentiating gives 0.020 and 0.030, respectively, as required. A common mistake is to calculate information using exposure alone; for this model the Poisson information contributions use fitted claim means, with the exposure already included in those means through the offset.

Model-Based and Robust Standard Errors

The inverse-information calculation is model-based: its validity depends on the assumed variance structure and other likelihood assumptions being suitable. If the mean model is retained but the variance or dependence assumptions are not trusted, an analyst may report a robust covariance estimate instead. A commonly used sandwich form is $A^{-1}BA^{-1}$, where $A$ measures the curvature of the estimating equations and $B$ estimates the variability of their contributions. For clustered data, $B$ must account for the specified clustering; treating dependent observations as independent does not make the result robust to that dependence.

A robust standard error is still about coefficient estimation uncertainty, but it answers that question using a different covariance estimate. It does not change the fitted coefficient itself. Nor should a larger robust standard error be interpreted as an increase in the underlying risk for an individual policy: it indicates that the coefficient's estimated sampling uncertainty is larger under the alternative covariance method.

Worked Example: Comparing Two Reported Standard Errors

A hypothetical insurer reports model-based and cluster-robust standard errors for two coefficients. For a continuous vehicle-age predictor, the coefficient estimate is $-0.030$, with model-based standard error $0.008$ and cluster-robust standard error $0.011$. For a territory indicator, the estimate is $0.200$, with corresponding standard errors $0.120$ and $0.150$. Compare the standard errors and explain the interpretation.

Method. Compare each robust standard error with its model-based counterpart. Since both estimates for a given coefficient come from the same fitted mean model, differences in the reported uncertainty arise from the covariance method, not a change in the coefficient estimate.

For vehicle age, the increase is $0.011-0.008=0.003$. Relative to the model-based standard error, this is $0.003/0.008=0.375$, or 37.5%. For territory, the increase is $0.150-0.120=0.030$, and the relative increase is $0.030/0.120=0.25$, or 25%.

The robust calculation therefore reports greater coefficient estimation uncertainty for both effects. It does not change the fitted estimates $-0.030$ and $0.200$, and it does not show that individual risks vary by 37.5% or 25%. Whether the robust results are appropriate depends on whether their assumptions match the data structure and the chosen clustering.

Check. The robust standard errors are larger in both comparisons, and the percentage calculations use the model-based standard error as the denominator. A common mistake is to compare the numerical estimates, $-0.030$ and $0.200$, with the standard errors as though they were all measures of the same quantity; estimates describe fitted effects, while standard errors describe their estimated sampling uncertainty.

Reading Output Carefully

When reading a GLM output table, identify the coefficient scale, the reported standard-error method, and the model assumptions before drawing conclusions. Standard errors can help describe how precisely coefficients are estimated, but the number alone does not establish that the model is well specified or that a coefficient is actuarially appropriate.

Do not confuse standard error with response variation. A standard error describes uncertainty in an estimated coefficient. Claim counts, losses, or premiums have their own distributions and variability.
Do not confuse variance with standard error. The diagonal of a covariance matrix contains variances; the standard errors are their square roots.
Do not ignore off-diagonal entries. Coefficient estimates may be correlated. A covariance matrix contains information that is not present in a list of standard errors alone.
Check what covariance estimator was used. Model-based and robust standard errors can differ even when the fitted coefficients are identical.
Remember the approximation. Standard errors based on local information are most dependable when the model is identifiable and the data provide adequate information for the fitted parameters.

Key takeaway: A GLM coefficient standard error is the square root of an estimated coefficient variance. It quantifies sampling uncertainty in that fitted coefficient under a stated model and covariance method; it is not a measure of policy-to-policy outcome variation.

Check Your Understanding

For each problem, focus on what the covariance information says about coefficient estimation.

A fitted model reports a coefficient variance of 0.0225. What is its standard error, and on what scale is it reported if the coefficient is on the log-frequency scale? (Answer: Standard error $=0.150$, on the log-frequency scale.)
A two-coefficient covariance matrix has diagonal entries 0.16 and 0.09, and off-diagonal entry 0.06. Find both standard errors and the correlation between estimates. (Answer: Standard errors $0.400$ and $0.300$; correlation $0.500$.)
For an independent Poisson log-link model, a fitted information matrix is $\begin{pmatrix}50&20\\20&20\end{pmatrix}$, with dispersion 1. Find the estimated variance and standard error of the second coefficient. (Answer: The inverse matrix has second diagonal entry $50/600=0.08333$; standard error $=0.2887$.)
A model-based standard error is 0.040 and a robust standard error is 0.050 for the same fitted coefficient. By what percentage did the standard error increase, and did the coefficient estimate necessarily change? (Answer: It increased by 25%; no, the coefficient estimate need not change.)
A colleague says a coefficient standard error of 0.18 means individual policies' claim frequencies typically differ by 0.18. Is that interpretation correct? (Answer: No. It measures estimated sampling uncertainty in the coefficient, not policy-to-policy variation in claim frequency.)

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