Standard Errors and Confidence Intervals for Fitted Parameters

Why fitted parameters need uncertainty measures

A fitted parameter is an estimate, not a known constant. A frequency estimate such as an annual claim rate, or a severity estimate such as a mean claim size, will vary if the data collection is repeated. A standard error describes the scale of this sampling variation; an approximate confidence interval uses it to express uncertainty about the underlying parameter. These tools are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

For likelihood-based fitting, the key idea is that a sharply peaked log-likelihood indicates a precisely estimated parameter, whereas a flatter peak indicates greater uncertainty. The curvature at the maximum gives a practical estimate of that precision. The resulting standard errors and confidence intervals are generally large-sample approximations, so their assumptions and limitations matter.

Definition: For a parameter vector $\boldsymbol{\theta}$ and log-likelihood $\ell(\boldsymbol{\theta})$, the observed information matrix at the maximum likelihood estimate $\widehat{\boldsymbol{\theta}}$ is $$ \mathbf{J}(\widehat{\boldsymbol{\theta}})=-\left.\frac{\partial^2\ell(\boldsymbol{\theta})}{\partial\boldsymbol{\theta}\,\partial\boldsymbol{\theta}^{\mathsf T}}\right|_{\boldsymbol{\theta}=\widehat{\boldsymbol{\theta}}}. $$ If the model is regular, the estimate is an interior maximum, and the sample is sufficiently large, its estimated covariance matrix is approximately $\mathbf{J}(\widehat{\boldsymbol{\theta}})^{-1}$. For one parameter $\theta$, this reduces to $\widehat{\operatorname{Var}}(\widehat{\theta})\approx 1/J(\widehat{\theta})$, and the standard error is the square root of this estimated variance.

From curvature to a confidence interval

Suppose the true parameter is near the fitted value. A second-order Taylor approximation to the log-likelihood near its maximum is

\ell(\theta)\approx \ell(\widehat{\theta})-\frac{1}{2}J(\widehat{\theta})(\theta-\widehat{\theta})^2.

The first derivative is zero at an interior maximum. The negative second derivative determines how rapidly the log-likelihood falls on either side of that maximum. Under standard large-sample likelihood theory, this local quadratic shape corresponds to an approximately normal sampling distribution for the estimate, with variance $1/J(\widehat{\theta})$. This explains why greater curvature means a smaller standard error.

For a scalar parameter, an approximate two-sided $100(1-\alpha)\%$ Wald confidence interval is the estimate plus or minus a standard-normal critical value times its standard error. For a 95% interval, the critical value is approximately 1.96. The interval is approximate, not a statement that the fixed parameter itself is random: under repeated sampling, intervals constructed by this procedure cover the true parameter at about the stated rate when the approximation is adequate.

\widehat{\theta}\ \pm\ z_{1-\alpha/2}\operatorname{SE}(\widehat{\theta}), \qquad \operatorname{SE}(\widehat{\theta})\approx\sqrt{\frac{1}{J(\widehat{\theta})}}.

For several parameters, use the inverse information matrix for the estimated covariance matrix. The standard error of a particular fitted component is the square root of the corresponding diagonal entry. Off-diagonal entries represent estimated covariances between fitted parameters; they cannot generally be discarded when uncertainty in a function of several parameters is being calculated.

Worked Example: A Bernoulli Claim-Frequency Parameter

A hypothetical portfolio records whether each of 100 independent, equally exposed policy-years has at least one claim. There are 20 policy-years with a claim. Assume each indicator is Bernoulli with claim-occurrence probability $p$. Find an approximate 95% confidence interval for $p$ using likelihood curvature. Give probabilities to four decimal places.

Method and fitted value. The likelihood, apart from a factor that does not involve $p$, is $L(p)=p^{20}(1-p)^{80}$. Its log-likelihood is $\ell(p)=20\ln p+80\ln(1-p)$. Setting the score to zero gives $20/p-80/(1-p)=0$, so $\widehat{p}=20/100=0.20$.

Curvature and standard error. The second derivative is $\ell''(p)=-20/p^2-80/(1-p)^2$. At $p=0.20$, the observed information is

J(0.20)=\frac{20}{0.20^2}+\frac{80}{0.80^2}=500+125=625.

Thus the estimated variance is $1/625=0.0016$, and the standard error is $\sqrt{0.0016}=0.04$. The 95% margin is $1.96(0.04)=0.0784$. The interval is $0.20\pm0.0784$, or $(0.1216,0.2784)$.

Check and trap avoided. The interval is centered on the fitted probability, and both endpoints lie between zero and one. A common error is to use the curvature itself as the variance, giving 625 rather than its reciprocal. Information measures precision; it is the inverse information that estimates variance.

Transformed parameters: the delta method

Actuarial work often needs a function of fitted parameters rather than the parameters separately. Examples include a mean expressed as the reciprocal of a rate, a transformed scale parameter, or expected aggregate loss formed from frequency and severity. The delta method approximates the uncertainty of such a function by linearizing it near the fitted values.

Theorem: Let $\widehat{\boldsymbol{\theta}}$ have estimated covariance matrix $\widehat{\boldsymbol{\Sigma}}$, and let $g(\boldsymbol{\theta})$ be differentiable near the fitted value. Then $$ \widehat{\operatorname{Var}}\{g(\widehat{\boldsymbol{\theta}})\}\approx \nabla g(\widehat{\boldsymbol{\theta}})^{\mathsf T} \widehat{\boldsymbol{\Sigma}}\, \nabla g(\widehat{\boldsymbol{\theta}}). $$ For one parameter, this is approximately $[g'(\widehat{\theta})]^2\widehat{\operatorname{Var}}(\widehat{\theta})$.

To see why, write the first-order Taylor approximation $g(\widehat{\boldsymbol{\theta}})\approx g(\boldsymbol{\theta})+ \nabla g(\boldsymbol{\theta})^{\mathsf T}(\widehat{\boldsymbol{\theta}}-\boldsymbol{\theta})$. Taking the variance of the right-hand side gives the gradient–covariance–gradient expression. This is a local approximation: it can be poor when the function is strongly curved across the likely range of estimates.

Worked Example: A Severity Rate and Its Approximate Interval

A hypothetical sample contains 25 complete claim severities, measured in thousands of dollars, with total severity 100. Assume independent observations from an exponential distribution with rate $\lambda$ per thousand dollars. The fitted rate is $\widehat{\lambda}=25/100=0.25$. Find its likelihood-based standard error and an approximate 95% confidence interval. Round the final values to three decimals.

Method. For complete exponential observations with sample size $n$ and total $T$, the log-likelihood is $\ell(\lambda)=n\ln\lambda-\lambda T$. Here $n=25$ and $T=100$, so $\ell(\lambda)=25\ln\lambda-100\lambda$. The second derivative is $\ell''(\lambda)=-25/\lambda^2$, giving observed information at the fitted rate of

J(0.25)=\frac{25}{0.25^2}=\frac{25}{0.0625}=400.

The estimated variance is $1/400=0.0025$, and the standard error is $\sqrt{0.0025}=0.05$ per thousand dollars. The 95% margin is $1.96(0.05)=0.098$. Therefore the interval is $0.25\pm0.098=(0.152,0.348)$ per thousand dollars, to three decimals.

Check and trap avoided. The fitted rate is the reciprocal of the observed sample mean of 4 thousand dollars, as expected for this likelihood. The interval remains positive. Do not calculate the standard error as $J$, or as $1/J$ without taking its square root: $1/J$ is the estimated variance, while the standard error is its square root.

Combining frequency and severity uncertainty

A common target is the expected aggregate loss per exposure period. Under a model in which the claim count $N$ has mean $\lambda$, individual claim severities have mean $\mu$, and count and severities are independent, the expected aggregate loss is $g(\lambda,\mu)=\lambda\mu$. If the two parameters are estimated from independent data, their estimated covariance is zero. If they are not independent, the covariance term must be retained.

Worked Example: Uncertainty in Expected Annual Loss

A hypothetical portfolio has 80 policy-years of exposure and 16 claims. Assume the annual claim count follows a Poisson model with rate $\lambda$ per policy-year. Separately, 625 complete claim severities are modeled as independent exponential amounts with mean $\mu$ dollars. The frequency and severity samples are independent, and their fitted values are $\widehat{\lambda}=0.20$ and $\widehat{\mu}=5{,}000$. Find a delta-method 95% confidence interval for expected annual aggregate loss per policy-year. Report dollars to the cent.

Method and component variances. The fitted expected loss is $\widehat{g}=\widehat{\lambda}\widehat{\mu}=0.20(5{,}000)=\$1{,}000$. For the Poisson rate based on 80 policy-years, the information is $80/\widehat{\lambda}=80/0.20=400$, so $\widehat{\operatorname{Var}}(\widehat{\lambda})=1/400=0.0025$. For an exponential mean, the information from 625 complete severities is $625/\widehat{\mu}^2$. Thus it is $625/5{,}000^2=0.000025$, giving $\widehat{\operatorname{Var}}(\widehat{\mu})=40{,}000$.

The gradient of $g(\lambda,\mu)=\lambda\mu$ is $(\mu,\lambda)$. Independence makes the covariance term zero, so

\widehat{\operatorname{Var}}(\widehat{g}) =5{,}000^2(0.0025)+0.20^2(40{,}000) =62{,}500+1{,}600 =64{,}100.

The standard error is $\sqrt{64{,}100}=\$253.18$ (rounded to cents). The margin is $1.96(\$253.18)=\$496.23$, to cents. The approximate 95% interval is therefore $\$1{,}000.00\pm\$496.23=(\$503.77,\$1{,}496.23)$.

Check and trap avoided. The fitted expected loss is the product of the two fitted means, and the interval is centered on that estimate. A frequent mistake is to vary only the frequency estimate or only the severity estimate. Both contribute: the delta variance includes the squared sensitivity to each parameter multiplied by that parameter’s variance.

When the approximation needs caution

Likelihood-based standard errors are most defensible when the model is identifiable, the true parameter is not on a boundary, the likelihood is sufficiently smooth, and the sample contains enough information for the local normal approximation to be reasonable. In small samples, with sparse counts, or with highly skewed estimates, a Wald interval may have poor coverage. It can also produce impossible values, such as a negative interval endpoint for a positive parameter.

Parameterization matters as well. Curvature is measured in the units of the parameter being fitted, so a standard error for a rate is not a standard error for its reciprocal mean. For a transformed quantity, apply the delta method to the correct function and use the full covariance matrix when parameters are correlated. Always check the estimate, variance, standard error, interval endpoints, and units separately.

Key takeaway: Invert the likelihood curvature to estimate variance, take its square root for the standard error, and multiply by the appropriate normal critical value for a Wald interval. For a derived quantity, propagate uncertainty using its gradient and the fitted-parameter covariance matrix.

Check Your Understanding

For each problem, use the stated large-sample normal approximation. Round final probabilities to four decimals and money amounts to cents where applicable.

A hypothetical group of 200 independent policy-years has 50 years with at least one claim. Treat each indicator as Bernoulli with probability $p$. Use likelihood curvature to find the standard error of $\widehat p$ and its approximate 95% interval. (Answer: standard error $0.0306$; interval $(0.1900, 0.3100)$.)
Independent observations follow $N(\mu,900)$, where the second argument is the variance. For $n=36$, the fitted mean is 120. Find its standard error and approximate 95% interval. (Answer: standard error $5$; interval $(110.20,129.80)$.)
A fitted exponential severity rate is $\widehat{\lambda}=0.5$ per thousand dollars, based on 100 complete claims. Using observed information $n/\widehat{\lambda}^{\,2}$, find its standard error and approximate 95% interval. (Answer: standard error $0.05$; interval $(0.402,0.598)$ per thousand dollars.)
Independent frequency and severity fits have $\widehat{\lambda}=0.10$, $\widehat{\operatorname{Var}}(\widehat{\lambda})=0.0004$, $\widehat{\mu}=\$8{,}000$, and $\widehat{\operatorname{Var}}(\widehat{\mu})=160{,}000$. Use the delta method for $g=\lambda\mu$ and give an approximate 95% interval. (Answer: estimate $\$800.00$; standard error $\$164.92$; interval $(\$476.75,\$1{,}123.25)$.)
A scalar fitted parameter is $\widehat{\beta}=\ln(1.2)$, with estimated variance $0.01$. For $g(\beta)=e^\beta$, use the delta method to find the approximate standard error and 95% interval for $g(\beta)$. (Answer: estimate $1.2$; standard error $0.12$; interval $(0.9648,1.4352)$.)

What You'll Learn