Bootstrap Uncertainty For Nonlinear Actuarial Functionals

Why Transform the Parameters in Every Replicate?

A bootstrap analysis often begins with a fitted severity model, but the actuarial quantity of interest may not be one of the model parameters. A premium calculation, for example, may depend on the expected payment after a deductible or limit. Such a quantity is a functional of the fitted model: it is calculated by applying a specified rule to the model parameters. If that rule is nonlinear, uncertainty in the parameters does not pass through it as though it were a linear transformation.

The key procedure is simple: in each bootstrap replicate, refit the model and calculate the target functional using that replicate's fitted parameters. The resulting replicate functional values approximate the sampling distribution of the target. This tutorial focuses on that transformation and its uncertainty, not on how to generate or refit the bootstrap samples.

Definition: Let $\hat{\theta}$ be a fitted parameter vector and let $g(\theta)$ be the actuarial functional of interest. The original estimate is $g(\hat{\theta})$. In bootstrap replicate $b$, refit the model to obtain $\hat{\theta}^{(b)}$, then calculate $T^{(b)}=g(\hat{\theta}^{(b)})$. For $B$ replicates, the bootstrap mean is $\bar{T}_{\ast}=B^{-1}\sum_{b=1}^{B}T^{(b)}$, and the bootstrap standard error is the sample standard deviation of the $T^{(b)}$ values.

The superscript $(b)$ identifies a bootstrap replicate; it is not an exponent. Each replicate value must be calculated using the parameters refitted in that replicate. Reusing the original fitted parameters would produce the same functional value every time and would omit the parameter uncertainty the bootstrap is intended to represent.

Limited Expected Value Under an Exponential Model

Suppose a hypothetical ground-up loss $X$ has an exponential distribution with mean parameter $\theta>0$, so $S_X(x)=e^{-x/\theta}$ for $x\geq 0$. For a payment capped at $u>0$, the limited expected value is $E[\min(X,u)]$. Under these assumptions,

g(\theta)=E[\min(X,u)] =\int_0^u S_X(x)\,dx =\theta\left(1-e^{-u/\theta}\right).

The survival integral follows because, for any nonnegative random variable $Z$, $E[Z]=\int_0^\infty P(Z>x)\,dx$. For $Z=\min(X,u)$, the probability $P(Z>x)$ equals $S_X(x)$ for $0\leq x<u$, and is zero for $x\geq u$. Integrating the exponential survival function from zero to $u$ gives the displayed formula.

Worked Example: Averaging Limited Expected Values

A hypothetical severity model is exponential with mean parameter $\theta$, and the target is the expected payment per loss capped at $u=\$50$. Two bootstrap replicates produce refitted estimates $\hat{\theta}^{(1)}=\$50$ and $\hat{\theta}^{(2)}=\$100$. Calculate the mean replicate functional and compare it with the functional evaluated at the mean fitted parameter.

Method. Evaluate $g(\theta)=\theta(1-e^{-50/\theta})$ separately at each replicate estimate.

For replicate 1, \[ g(50)=50(1-e^{-1})=50(1-0.3678794)=\$31.6060. \] For replicate 2, \[ g(100)=100(1-e^{-0.5})=100(1-0.6065307)=\$39.3469. \] Therefore the mean of the two replicate functional values is \[ \bar{T}_{\ast}=\frac{31.6060+39.3469}{2}=\$35.47645\approx \$35.48. \]

The mean fitted parameter is $(50+100)/2=\$75$. Evaluating the functional at this mean gives \[ g(75)=75(1-e^{-50/75}) =75(1-e^{-2/3}) =75(1-0.5134171) =\$36.4937\approx \$36.49. \] Thus the mean replicate functional is approximately $\$35.48$, whereas the functional at the mean fitted parameter is approximately $\$36.49$. The two operations do not give the same answer. Common mistake avoided: averaging the fitted parameters first and treating that as the bootstrap average of the target. The target must be recalculated in every replicate.

Bootstrap Mean, Standard Error, and Bias

Given the replicate functional values $T^{(1)},\ldots,T^{(B)}$, their average describes the center of the bootstrap distribution. Their sample standard deviation estimates the standard error of the original functional estimate. The denominator in the sample variance is $B-1$, not $B$.

\bar{T}_{\ast}=\frac{1}{B}\sum_{b=1}^{B}T^{(b)}, \qquad \widehat{\operatorname{SE}}_{\ast}\{g(\hat{\theta})\} = \sqrt{\frac{1}{B-1}\sum_{b=1}^{B} \left(T^{(b)}-\bar{T}_{\ast}\right)^2}.

A common bootstrap estimate of bias is the bootstrap mean minus the original plug-in estimate: \[ \widehat{\operatorname{Bias}}_{\ast} =\bar{T}_{\ast}-g(\hat{\theta}). \] These quantities answer different questions. The standard error describes the spread of replicate functional values. The estimated bias describes the difference between their average and the original estimate. A small bootstrap standard error does not imply that this estimated bias is zero.

Worked Example: Deductible-Adjusted Mean and Bootstrap Standard Error

Assume hypothetical claim severity $X$ is exponential with mean $\theta$. A policy pays per loss $Y^L=(X-d)_+$, where $d=\$50$ and $(z)_+=\max(z,0)$. Three bootstrap replicates give refitted estimates $\hat{\theta}^{(1)}=\$40$, $\hat{\theta}^{(2)}=\$60$, and $\hat{\theta}^{(3)}=\$100$. Calculate the replicate expected payments, their bootstrap mean, and their sample standard error.

Method. For an exponential severity, $E[(X-d)_+]=\int_d^\infty S_X(x)\,dx=\theta e^{-d/\theta}$. Apply this formula to each refitted mean parameter.

The three replicate functional values are \[ T^{(1)}=40e^{-50/40}=40e^{-1.25}=\$11.4602, \] \[ T^{(2)}=60e^{-50/60}=60e^{-5/6}=\$26.0759, \] \[ T^{(3)}=100e^{-50/100}=100e^{-0.5}=\$60.6531. \] Their mean is \[ \bar{T}_{\ast} =\frac{11.4602+26.0759+60.6531}{3} =\$32.7297 \approx \$32.73. \]

The deviations from the mean, using the displayed values, are approximately $-21.2695$, $-6.6538$, and $27.9234$. Their squares are approximately $452.3927$, $44.2734$, and $779.7134$. Thus \[ \widehat{\operatorname{SE}}_{\ast} =\sqrt{\frac{452.3927+44.2734+779.7134}{3-1}} =\sqrt{638.1898} \approx \$25.26. \] The squared deviations sum to $1{,}276.3795$, so the calculation uses the sample-variance denominator $2$. Common mistake avoided: dividing by $B$ instead of $B-1$, or calculating the standard error from the parameter estimates rather than from the deductible-adjusted functional values.

Why Nonlinear Transformations Change the Bootstrap Distribution

In general, a nonlinear function does not commute with averaging: the average of $g(\hat{\theta}^{(b)})$ need not equal $g$ of the average $\hat{\theta}^{(b)}$. The limited expected value above provides a precise illustration. For fixed $u>0$, its second derivative with respect to $\theta$ is \[ g''(\theta)=-\frac{u^2}{\theta^3}e^{-u/\theta}<0. \] Therefore $g$ is concave for positive $\theta$. Jensen's inequality implies that, for positive replicate estimates, \[ \frac{1}{B}\sum_{b=1}^{B}g(\hat{\theta}^{(b)}) \leq g\left(\frac{1}{B}\sum_{b=1}^{B}\hat{\theta}^{(b)}\right). \] The inequality can be strict when the replicate parameters vary. This explains the direction of the difference in the first worked example; it is not a rounding discrepancy.

For other targets the direction of the difference may be different, or the function may change curvature over the relevant parameter range. The general procedure does not depend on knowing the direction: calculate the target separately for every replicate, then summarize the resulting values.

Deductibles and Payment Limits

Coverage modifications must be incorporated into the functional in the same order they apply to the loss. Suppose, hypothetically, the payment per loss is $Y^L=\min((X-d)_+,u)$, where $d$ is a deductible and $u$ is the maximum payment after the deductible. Under the exponential model with mean $\theta$, the expected payment is \[ E[Y^L]=\int_d^{d+u}S_X(x)\,dx =\theta\left(e^{-d/\theta}-e^{-(d+u)/\theta}\right). \] The interval of integration begins at the deductible and ends at the ground-up loss amount $d+u$ that exhausts the payment limit.

Worked Example: A Deductible Followed by a Payment Limit

A hypothetical policy has payment per loss $Y^L=\min((X-\$20)_+,\$30)$. Assume ground-up severity $X$ is exponential with mean parameter $\theta$. Three bootstrap replicates produce refitted estimates $\$40$, $\$60$, and $\$100$. Calculate the expected payment in each replicate, the bootstrap mean, and the bootstrap standard error.

Method. Use $E[Y^L]=\theta(e^{-20/\theta}-e^{-50/\theta})$ in each replicate. The upper integration endpoint is $d+u=\$50$, not the payment limit $\$30$.

For $\theta=\$40$, \[ T^{(1)}=40(e^{-20/40}-e^{-50/40}) =40(e^{-0.5}-e^{-1.25}) =40(0.6065307-0.2865048) =\$12.8010. \] For $\theta=\$60$, \[ T^{(2)}=60(e^{-20/60}-e^{-50/60}) =60(e^{-1/3}-e^{-5/6}) =60(0.7165313-0.4345982) =\$16.9160. \] For $\theta=\$100$, \[ T^{(3)}=100(e^{-0.2}-e^{-0.5}) =100(0.8187308-0.6065307) =\$21.2200. \]

The bootstrap mean is \[ \bar{T}_{\ast} =\frac{12.8010+16.9160+21.2200}{3} =\$16.9790 \approx \$16.98. \] The deviations from the mean are approximately $-4.1780$, $-0.0630$, and $4.2410$. Their squared values are approximately $17.4555$, $0.0040$, and $17.9861$. Therefore \[ \widehat{\operatorname{SE}}_{\ast} =\sqrt{\frac{17.4555+0.0040+17.9861}{2}} =\sqrt{17.7228} \approx \$4.21. \] The mean lies between the smallest and largest replicate payments, as it should. Common mistake avoided: treating the payment limit as the upper ground-up loss amount. Here the policy pays at most $\$30$, after the first $\$20$ of loss, so the ground-up endpoint is $\$50$.

Practical Checks and Exam Traps

The functional must match the quantity being estimated. A limited expected value $E[\min(X,u)]$ is not the same as an expected payment above a deductible, $E[(X-d)_+]$, or a deductible-and-limit payment, $E[\min((X-d)_+,u)]$. Write the payment definition first, then derive or select its expected value. This guards against applying a limit to the wrong amount or subtracting a deductible after the expected value has already been calculated.

Keep the model assumptions and units consistent across replicates. In the exponential examples, $\theta$, $d$, and $u$ are all measured in dollars, so ratios such as $d/\theta$ are dimensionless. If losses are in thousands of dollars, all monetary parameters and reported functional values must use that same unit. Round the final reported summaries, not each intermediate functional value; early rounding can slightly alter the bootstrap mean and standard error.

A finite number of replicates also introduces simulation variability in the summaries. Increasing $B$ generally stabilizes the estimated bootstrap mean and standard error, but does not change the rule for calculating $T^{(b)}$. The bootstrap standard error describes estimated sampling uncertainty in the target; it is not the standard error of the Monte Carlo average $\bar{T}_{\ast}$. These answer different questions.

Key takeaway: Refit the model in every bootstrap replicate, calculate the full actuarial functional from that replicate's fitted parameters, and summarize those functional values. Do not substitute the functional evaluated at an average parameter for the average of the replicate functionals.

Check Your Understanding

Assume exponential severity wherever a model is specified. Calculate the requested quantities from the replicate functional values, retaining unrounded values until the final answer.

For limited expected value $E[\min(X,30)]$, two bootstrap mean estimates are $\theta=\$30$ and $\theta=\$60$. Find the mean of the two replicate functional values. (Answer: approximately $\$21.29$)
Three bootstrap replicate values of a target functional are $\$8$, $\$10$, and $\$18$. Calculate their mean and sample standard error. (Answer: mean $\$12.00$; standard error approximately $\$5.29$)
A policy pays per loss $Y^L=(X-\$40)_+$. A bootstrap replicate has fitted exponential mean $\theta=\$80$. Calculate the replicate expected payment. (Answer: approximately $\$48.52$)
A policy pays $Y^L=\min((X-\$10)_+,\$20)$. Two bootstrap replicates have exponential mean estimates $\theta=\$50$ and $\theta=\$100$. Calculate the mean replicate expected payment. (Answer: approximately $\$14.95$)
For $E[\min(X,\$40)]$, the original fitted exponential mean is $\hat{\theta}=\$60$, and two bootstrap replicates have fitted means $\$40$ and $\$80$. Calculate the mean replicate functional and the functional at the mean replicate parameter. (Answer: approximately $\$28.38$ and $\$29.19$, respectively)

What You'll Learn