Delta-Method Uncertainty for Relativities

Why a Relativity Has Its Own Uncertainty

In a GLM with a log link, a fitted coefficient is on the log scale, while a rating relativity is usually reported on the multiplicative scale. If $\hat{\beta}$ estimates a coefficient $\beta$, the corresponding fitted relativity is $R=\exp(\hat{\beta})$. The uncertainty reported for $\hat{\beta}$ cannot simply be copied as the uncertainty for $R$: exponentiation changes the scale and the size of the estimation error.

The delta method gives a first-order approximation to the variance of a smooth transformation of an estimator. Here, the transformation is the exponential function. This tutorial derives and applies the resulting standard error for a single exponentiated coefficient. The method is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Suppose a log-link GLM has coefficient $\beta$, with estimate $\hat{\beta}$ and estimated standard error $s_{\hat{\beta}}$. The fitted multiplicative relativity is $\hat{R}=\exp(\hat{\beta})$. The first-order delta-method standard error for this relativity is $$ s_{\hat{R}} \approx \left|\frac{d}{d\beta}\exp(\beta)\bigg|_{\beta=\hat{\beta}}\right|s_{\hat{\beta}} =\exp(\hat{\beta})s_{\hat{\beta}}. $$ This is an approximation to the standard deviation of the fitted relativity, not an exact standard error in every finite sample.

Deriving the Delta-Method Formula

Assume that $\hat{\beta}$ is approximately unbiased for $\beta$ and has an approximately normal sampling distribution with variance $V_{\hat{\beta}}$. More generally, the usual large-sample GLM result is that the estimator is approximately normal around the true coefficient, with a variance estimated from the fitted model. Let $g(\beta)=\exp(\beta)$. A first-order Taylor expansion around $\beta$ gives

g(\hat{\beta}) \approx g(\beta)+g'(\beta)(\hat{\beta}-\beta).

The first term on the right is constant, so it contributes no variance. The random term is the coefficient estimation error multiplied by $g'(\beta)$. Therefore,

\operatorname{Var}\bigl(g(\hat{\beta})\bigr) \approx \bigl(g'(\beta)\bigr)^2\operatorname{Var}(\hat{\beta}).

For $g(\beta)=\exp(\beta)$, the derivative is $g'(\beta)=\exp(\beta)$. Replacing the unknown $\beta$ and variance with their fitted estimates gives the practical formula:

\widehat{\operatorname{Var}}(\hat{R}) \approx \exp(2\hat{\beta})\,s_{\hat{\beta}}^2, \qquad s_{\hat{R}} \approx \exp(\hat{\beta})s_{\hat{\beta}}.

The derivative is positive, so the absolute-value bars do not change the result. The variance formula squares both the relativity and the coefficient standard error; the standard-error formula multiplies them without squaring. Keeping those two formulas distinct is essential.

Interpretation: To first order, the relativity’s standard error as a proportion of the relativity equals the coefficient’s standard error: $$ \frac{s_{\hat{R}}}{\hat{R}}\approx s_{\hat{\beta}}. $$ Thus, if $s_{\hat{\beta}}=0.08$, the relativity’s approximate standard error is about 8% of its fitted value. This is a local, first-order statement, not a guarantee that the full sampling distribution is symmetric or that the approximation is accurate in a small sample.

Applying the Formula

The input $s_{\hat{\beta}}$ should be the standard error for the specific coefficient being exponentiated. If model output provides a coefficient covariance matrix, the coefficient’s variance is the corresponding diagonal entry, and its standard error is the square root of that entry. For one coefficient, an off-diagonal covariance with another coefficient is not part of this calculation.

A log-link coefficient is interpreted relative to the model’s coding and predictor scale. The transformation here gives the multiplicative factor for a one-unit increase in the coded predictor, or for the specified category relative to its reference when the coefficient is a categorical contrast. The delta-method calculation does not determine what comparison the coefficient represents; that interpretation comes from the model specification.

Worked Example: Frequency Relativity Above One

A hypothetical insurer fits a Poisson frequency GLM with a log link and a known exposure offset. For one binary territory indicator, the fitted coefficient is $\hat{\beta}=0.20$, with estimated standard error $s_{\hat{\beta}}=0.08$. The response is a claim count, modeled as Poisson conditional on the predictors and exposure; no mortality or interest assumption is involved. Calculate the fitted territory relativity and its delta-method standard error.

Step 1: Identify the transformation. Under the log link, the coefficient corresponds to a multiplicative frequency relativity, so $\hat{R}=\exp(\hat{\beta})$.

Step 2: Calculate the fitted relativity. $$ \hat{R}=\exp(0.20)=1.2214028. $$ The territory’s fitted frequency is therefore about 1.2214 times the reference territory’s fitted frequency, holding the other model terms and exposure treatment fixed.

Step 3: Transform the uncertainty. $$ s_{\hat{R}}\approx \exp(0.20)(0.08) =1.2214028(0.08) =0.0977122. $$ The approximate standard error of the relativity is 0.0977.

Step 4: Check the scale. $$ \frac{s_{\hat{R}}}{\hat{R}} =\frac{0.0977122}{1.2214028} =0.0800. $$ This matches the coefficient standard error of 0.08, as the relative-standard-error relationship predicts. Rounded at the final step to four decimal places, the result is $\hat{R}=1.2214$ and $s_{\hat{R}}\approx0.0977$.

Method and trap avoided. The delta method multiplies the coefficient standard error by the fitted relativity. A common error is to report 0.08 itself as the relativity’s standard error, even though 0.08 is measured on the coefficient scale.

Worked Example: Relativity Below One

A hypothetical insurer uses a Poisson frequency GLM with a log link and a known exposure offset to estimate the effect of an optional safety feature. Claim counts are modeled as Poisson conditional on the predictors and exposure. The fitted coefficient for having the feature is $\hat{\beta}=-0.25$, with standard error $0.10$. Find the fitted relativity and its approximate standard error. No mortality or interest assumption is involved.

Step 1: Exponentiate the coefficient. $$ \hat{R}=\exp(-0.25)=0.7788008. $$ The fitted frequency for policies with the feature is about 0.7788 times the fitted frequency for the reference group, conditional on the other predictors.

Step 2: Apply the derivative at the fitted coefficient. Since the derivative of $\exp(\beta)$ is $\exp(\beta)$, $$ s_{\hat{R}}\approx \exp(-0.25)(0.10) =0.7788008(0.10) =0.0778801. $$

Step 3: Check the proportional scale. $$ \frac{0.0778801}{0.7788008}=0.1000. $$ The approximate standard error is 10% of the fitted relativity, matching the coefficient standard error. Rounded only at the final step to four decimal places, $\hat{R}=0.7788$ and $s_{\hat{R}}\approx0.0779$.

Method and trap avoided. A negative coefficient produces a relativity below one, but its delta-method standard error remains positive. Do not carry the negative sign of $\hat{\beta}$ into the standard error; the derivative $\exp(\hat{\beta})$ is positive.

Using a Covariance Matrix

In fitted GLM output, uncertainty is often summarized by an estimated covariance matrix for the coefficient vector. For a single exponentiated coefficient, use the square root of that coefficient’s diagonal variance, then apply the scalar formula. Do not confuse a variance entry with a standard error, and do not use a covariance between two different coefficients as though it were the variance of the coefficient of interest.

Worked Example: Extracting the Coefficient Standard Error

A hypothetical insurer fits a Poisson frequency GLM with a log link and known exposure offsets. Claim counts are conditionally Poisson. The fitted coefficient for a vehicle-use indicator is $\hat{\beta}=0.40$. The relevant portion of the fitted coefficient covariance matrix, in the order intercept then indicator, is

\widehat{\operatorname{Cov}}(\hat{\boldsymbol{\beta}}) = \begin{pmatrix} 0.0100 & -0.0020\\ -0.0020 & 0.0144 \end{pmatrix}.

Calculate the indicator’s fitted relativity and delta-method standard error. No mortality or interest assumption is involved.

Step 1: Select the correct variance. The indicator coefficient is the second coefficient, so its estimated variance is the second diagonal entry, $0.0144$. The covariance $-0.0020$ with the intercept is not the indicator coefficient’s variance.

Step 2: Convert the variance to a coefficient standard error. $$ s_{\hat{\beta}}=\sqrt{0.0144}=0.12. $$

Step 3: Calculate the relativity. $$ \hat{R}=\exp(0.40)=1.4918247. $$

Step 4: Calculate its standard error and verify with the variance formula. $$ s_{\hat{R}}\approx 1.4918247(0.12)=0.1790190, $$ and $$ \widehat{\operatorname{Var}}(\hat{R}) \approx (1.4918247)^2(0.0144) =0.0320478, \qquad \sqrt{0.0320478}=0.1790190. $$ The two calculations agree. Rounded at the final step to four decimal places, $\hat{R}=1.4918$ and $s_{\hat{R}}\approx0.1790$.

Method and trap avoided. First take the square root of the variance-matrix diagonal entry to obtain the coefficient standard error. Multiplying the relativity by $0.0144$ directly would mix a variance with a standard error and give the wrong result.

Approximation Limits and Exam Traps

The delta method is a large-sample, first-order approximation. It is most useful when the coefficient estimator is reasonably well behaved and its standard error is not so large that the exponential transformation makes the sampling distribution strongly asymmetric. Because exponentiation is nonlinear, an approximately normal coefficient estimator does not imply an exactly normal relativity estimator. In particular, the fitted relativity is positive, while a normal approximation on the relativity scale can assign probability to negative values. Treat the delta-method standard error as a local measure of variability, not as a complete description of the relativity’s distribution.

The method also depends on using the correct coefficient. A categorical coefficient may represent a level relative to a reference under a particular contrast coding; a continuous coefficient may represent a one-unit predictor change. Exponentiating the wrong coefficient, or misreading the predictor unit, gives the wrong relativity before uncertainty is even calculated. The standard error transformation cannot correct an interpretation error.

Confirm the coefficient’s meaning.
Check the log link, predictor coding, and unit or reference comparison represented by the coefficient.

Obtain its standard error.
Use the reported coefficient standard error or take the square root of the appropriate diagonal variance entry.

Exponentiate the estimate.
Compute $\hat{R}=\exp(\hat{\beta})$.

Transform the standard error.
Compute $s_{\hat{R}}\approx\hat{R}s_{\hat{\beta}}$, then check that $s_{\hat{R}}/\hat{R}\approx s_{\hat{\beta}}$.

A frequent exam error is to apply the variance formula but label the result as a standard error. For example, if $\hat{R}=1.20$ and $s_{\hat{\beta}}=0.10$, then the approximate relativity variance is $1.20^2(0.10^2)=0.0144$, while its standard error is $\sqrt{0.0144}=0.12$. Another error is to use $\exp(s_{\hat{\beta}})$, which is not the delta-method standard error. The derivative is evaluated at the coefficient estimate; the rule is multiplication by $\exp(\hat{\beta})$, not exponentiation of the standard error.

Key takeaway: For a single log-link coefficient, exponentiate the estimate to obtain the relativity and multiply that relativity by the coefficient standard error to obtain its first-order delta-method standard error. Keep the variance and standard error formulas separate, and interpret the coefficient before transforming it.

Check Your Understanding

For each problem, calculate the fitted relativity and its delta-method standard error. Assume the stated coefficient standard errors are already obtained from the fitted GLM. No mortality or interest assumptions apply.

A hypothetical Poisson frequency model with a log link and known exposure offset gives $\hat{\beta}=0.10$ and $s_{\hat{\beta}}=0.05$ for a district indicator. Find $\hat{R}$ and $s_{\hat{R}}$. (Answer: $\hat{R}=1.1052$, $s_{\hat{R}}\approx0.0553$.)
A hypothetical Poisson frequency model with a log link gives $\hat{\beta}=-0.40$ and $s_{\hat{\beta}}=0.15$ for a policy feature. Find the relativity and standard error. (Answer: $\hat{R}=0.6703$, $s_{\hat{R}}\approx0.1005$.)
A fitted log-link GLM gives $\hat{\beta}=0.25$, and the coefficient’s diagonal variance entry in the covariance matrix is $0.0100$. Find the coefficient standard error first, then the relativity standard error. (Answer: $s_{\hat{\beta}}=0.1000$, $\hat{R}=1.2840$, $s_{\hat{R}}\approx0.1284$.)
A hypothetical Poisson frequency model gives $\hat{\beta}=0.06$ and $s_{\hat{\beta}}=0.20$ for a territory contrast. Calculate the two fitted-scale quantities and state the relative-standard-error check. (Answer: $\hat{R}=1.0618$, $s_{\hat{R}}\approx0.2124$, and $s_{\hat{R}}/\hat{R}\approx0.2000$.)
A hypothetical Poisson frequency model with a log link gives $\hat{\beta}=0.30$, with estimated coefficient variance $0.0025$. Calculate the fitted relativity and its approximate standard error. (Answer: $s_{\hat{\beta}}=0.0500$, $\hat{R}=1.3499$, $s_{\hat{R}}\approx0.0675$.)

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