Parametric Bootstrap From A Fitted Loss Model

Why Generate Bootstrap Samples From a Fitted Loss Model?

A severity sample contains only one realized set of losses. A bootstrap asks how a chosen statistic might vary across repeated samples. In a parametric bootstrap, the repeated samples are generated from a specified probability model fitted to the observed data, rather than by resampling the observed losses themselves. This is useful when a fitted severity distribution provides a reasonable description of the loss process, including loss values not present in the original sample.

The resulting distribution is conditional on the fitted model and its parameter values. It describes sampling variation under that plug-in model; it does not, by itself, account for uncertainty about which model is appropriate or for uncertainty in the fitted parameters. Those distinctions matter when interpreting a bootstrap standard error or interval.

Definition: Let $X_1,\ldots,X_n$ be observed independent severity losses, and let $F_{\hat{\theta}}$ be a fitted severity distribution with parameter estimate $\hat{\theta}$. A parametric bootstrap replicate is a new sample $X_1^*,\ldots,X_n^*$ generated independently from $F_{\hat{\theta}}$. For a statistic $T$, calculate $T^*=T(X_1^*,\ldots,X_n^*)$. Repeating this process $M$ times gives a simulated approximation to the sampling distribution of $T$ under the fitted model.

The sample size in each replicate is ordinarily the observed severity sample size $n$. For example, if 40 individual claim severities were used to fit a model, each replicate contains 40 simulated severities—not the portfolio’s number of policies. The statistic must also match the question: it could be a sample mean, a sample quantile, or a proportion of losses above a specified threshold.

The Plug-In Procedure

The fitted model is held fixed during this procedure. Generate losses from that model, apply the statistic to each replicate, and examine the resulting values. If the statistic is the sample mean, for instance, each replicate produces a different mean even though the generating distribution is unchanged.

Specify the observed statistic.
Identify the severity data, the sample size $n$, and the precise statistic $T$ whose sampling variability is of interest.

Fit the severity model.
Estimate its parameters from the observed data and write down the fitted distribution $F_{\hat{\theta}}$. The parameter values remain fixed for this plug-in procedure.

Generate and evaluate replicates.
For each $b=1,\ldots,M$, generate $n$ independent losses from $F_{\hat{\theta}}$ and calculate $T_b^*$.

Summarize the replicate distribution.
Use its center, spread, or quantiles to describe the statistic’s variation under the fitted model. More replicates generally reduce simulation noise in these summaries.

For replicate statistics $T_1^*,\ldots,T_M^*$, their average and sample standard deviation are

\bar{T}^*=\frac{1}{M}\sum_{b=1}^{M}T_b^*, \qquad \widehat{\operatorname{SE}}_{\mathrm{boot}}(T) = \sqrt{\frac{1}{M-1}\sum_{b=1}^{M}(T_b^*-\bar{T}^*)^2}.

The standard deviation summarizes the spread of the replicate statistic distribution. The average can be compared with the observed statistic as a diagnostic for the plug-in distribution’s center. With a small number of replicates, both summaries can be noisy; a few illustrative replicates are not a substitute for a sufficiently large simulation when precise results are needed.

Why the Replicate Distribution Represents Sampling Variation

The justification is the sampling model. Once the fitted parameters are held fixed, each simulated replicate is an independent sample of size $n$ from $F_{\hat{\theta}}$. Thus the distribution of $T^*$ is the sampling distribution of the statistic under that fitted model. The bootstrap estimates this distribution by Monte Carlo simulation.

For example, suppose the fitted severity model is exponential with mean $\hat{\theta}$. Its density is $f(x)=\hat{\theta}^{-1}e^{-x/\hat{\theta}}$ for $x\geq 0$. Each bootstrap loss has mean $\hat{\theta}$ and variance $\hat{\theta}^2$. Independence then gives the following result for the replicate sample mean:

E[\bar{X}^*]=\frac{1}{n}\sum_{i=1}^n E[X_i^*]=\hat{\theta}, \qquad \operatorname{Var}(\bar{X}^*)= \frac{1}{n^2}\sum_{i=1}^n\operatorname{Var}(X_i^*) =\frac{\hat{\theta}^2}{n}.

The simulated replicate means fluctuate around $\hat{\theta}$; their empirical standard deviation approximates $\hat{\theta}/\sqrt{n}$. This derivation explains the procedure for a sample mean under this model. For a statistic whose distribution is not available in a convenient form, simulation provides an empirical approximation instead.

Worked Example: Bootstrap Standard Error of an Exponential Sample Mean

A hypothetical severity sample, in dollars, is $500, 1{,}000, 1{,}500, 1{,}000$. Fit an exponential model by maximum likelihood, then use three specified bootstrap replicates to estimate the variability of the sample mean. Use the inverse transform $X^*=-\hat{\theta}\ln(1-U)$, with the uniforms shown below.

Fit the model. The observed sample size is $n=4$, and the total loss is $500+1{,}000+1{,}500+1{,}000=4{,}000$. For an exponential model with rate $\lambda$, the maximum-likelihood estimate is $n/\sum x_i=4/4{,}000=0.001$ per dollar. Equivalently, the fitted mean is $\hat{\theta}=1/\hat{\lambda}=\$1{,}000$.

Generate the replicates. Each row uses four independent uniforms and produces four losses. The resulting bootstrap means are:

Replicate	Uniform values	Generated losses (dollars, to cents)	Sample mean
1	0.2, 0.4, 0.6, 0.8	223.14, 510.83, 916.29, 1,609.44	814.93
2	0.1, 0.3, 0.5, 0.7	105.36, 356.67, 693.15, 1,203.97	589.79
3	0.25, 0.45, 0.65, 0.85	287.68, 597.84, 1,049.82, 1,897.12	958.12

For example, the first loss in replicate 1 is $-1{,}000\ln(0.8)=\$223.14$. Its replicate mean is $(223.14+510.83+916.29+1{,}609.44)/4=\$814.925$, or $\$814.93$ to the nearest cent. The other row means are calculated in the same way.

Summarize the replicate means. Their average is $(814.93+589.79+958.12)/3=\$787.61$. Using the unrounded replicate means, their sample standard deviation is approximately $\$185.68$. Thus the estimated bootstrap standard error from these three deliberately limited replicates is $\$185.68$. The theoretical standard deviation under the fitted model is $1{,}000/\sqrt{4}=\$500$; the difference illustrates how unstable a standard deviation based on only three replicates can be.

Check and common mistake. Each replicate has four losses, matching the observed severity sample size. Using $1/\hat{\theta}=0.001$ as the scale in the inverse transform would confuse the exponential rate with its mean and produce losses on the wrong scale.

Other Statistics and the Meaning of the Model

A bootstrap replicate statistic need not be a mean. For a sample quantile, first generate the losses, sort them, and apply the stated quantile convention. For an exceedance proportion, count the replicate losses meeting the specified threshold and divide by the replicate sample size. The statistic and its conventions must be fixed before comparing replicate values; different quantile conventions, for example, can produce different results in a small sample.

Worked Example: Replicate Distribution of a Sample Quantile

Suppose a hypothetical fitted exponential severity model has mean $\hat{\theta}=\$1{,}000$. For this example, define the empirical 80th percentile of a sample of size five by the nearest-rank rule: sort the losses and take the observation in position $\lceil 0.8n\rceil$. Four bootstrap replicates use the following uniforms:

Replicate 1: $U=(0.1,0.3,0.5,0.7,0.9)$ generates losses $\$105.36,\ \$356.67,\ \$693.15,\ \$1{,}203.97,\ \$2{,}302.59$. Since $\lceil0.8(5)\rceil=4$, the replicate 80th percentile is $\$1{,}203.97$.

Replicate 2: $U=(0.05,0.25,0.45,0.65,0.85)$ generates $\$51.29,\ \$287.68,\ \$597.84,\ \$1{,}049.82,\ \$1{,}897.12$. The fourth ordered loss is $\$1{,}049.82$.

Replicate 3: $U=(0.15,0.35,0.55,0.75,0.95)$ generates $\$162.52,\ \$430.78,\ \$798.51,\ \$1{,}386.29,\ \$2{,}995.73$. The fourth ordered loss is $\$1{,}386.29$.

Replicate 4: $U=(0.02,0.2,0.4,0.6,0.8)$ generates $\$20.20,\ \$223.14,\ \$510.83,\ \$916.29,\ \$1{,}609.44$. The fourth ordered loss is $\$916.29$.

The four replicate quantiles are therefore $\$1{,}203.97,\ \$1{,}049.82,\ \$1{,}386.29,\ \$916.29$. Their range is $\$1{,}386.29-\$916.29=\$470.00$, and their average is $\$1{,}139.09$. These summaries describe the spread and center of this small simulated replicate distribution; they are not precise estimates from only four replicates.

Check and common mistake. The nearest-rank position is the fourth observation, not $0.8$ times the largest loss or the average of the fourth and fifth losses. Changing the quantile definition mid-calculation would change the statistic being bootstrapped.

Worked Example: Exceedance Proportions Under a Pareto Severity Model

A hypothetical fitted Pareto severity model has minimum loss $x_m=\$1{,}000$ and shape $\alpha=2$, with $F_X(x)=1-(x_m/x)^\alpha$ for $x\geq x_m$. Use four bootstrap samples of size four to study the proportion of losses strictly greater than $\$2{,}000$. The inverse transform is $X^*=x_m(1-U)^{-1/\alpha}$.

For a threshold of $\$2{,}000$, the model’s exceedance probability is $S_X(2{,}000)=(1{,}000/2{,}000)^2=0.25$. The specified uniforms produce these replicate results:

Replicate 1, $U=(0.36,0.75,0.9375,0.84)$, gives losses $\$1{,}250,\ \$2{,}000,\ \$4{,}000,\ \$2{,}500$. Two losses are strictly greater than $\$2{,}000$, so the proportion is $2/4=0.50$.
Replicate 2, $U=(0.19,0.36,0.75,0.9375)$, gives losses $\$1{,}111.11,\ \$1{,}250,\ \$2{,}000,\ \$4{,}000$. One loss exceeds the threshold, so the proportion is $1/4=0.25$.
Replicate 3, $U=(0.51,0.75,0.84,0.96)$, gives losses $\$1{,}428.57,\ \$2{,}000,\ \$2{,}500,\ \$5{,}000$. Two losses exceed the threshold, so the proportion is $2/4=0.50$.
Replicate 4, $U=(0.36,0.51,0.64,0.75)$, gives losses $\$1{,}250,\ \$1{,}428.57,\ \$1{,}666.67,\ \$2{,}000$. No loss is strictly greater than the threshold, so the proportion is $0/4=0$.

The replicate proportions are $0.50, 0.25, 0.50, 0$. Their mean is $1.25/4=0.3125$, and their sample standard deviation is $\sqrt{0.171875/3}=0.2394$, to four decimal places. The replicate mean differs from the model probability $0.25$ because only four simulated samples were used. A loss equal to $\$2{,}000$ is not counted: the statistic was defined using a strict exceedance.

Check and common mistake. Apply the strict threshold to each generated loss before calculating the proportion. Counting the loss equal to $\$2{,}000$ as an exceedance would change the specified statistic.

What This Bootstrap Does—and Does Not—Measure

A parametric bootstrap is only as credible as the assumptions used to generate its samples. Its interpretation requires care:

Model adequacy: the fitted severity family should be a reasonable representation of the losses relevant to the statistic. A poor tail fit can make simulated tail summaries misleading.
Sampling assumptions: the procedure above assumes independent, identically distributed severities and uses the observed sample size. Dependence or systematically different risk groups require a model and simulation design that represent those features.
Observation mechanism: the description assumes fully observed severities. If the data are truncated or censored, simply generating complete losses does not reproduce the way the observed sample was collected.
Fixed fitted parameters: this plug-in procedure does not refit the model in each replicate. It therefore describes variation of the chosen statistic under the fixed fitted distribution, not the full variation of a statistic that depends on estimating model parameters.
Monte Carlo error: the replicate distribution is simulated, so its summaries vary from run to run. Use enough replicates for the desired precision and check that reported summaries are stable.

The distinction about refitting is especially important. If the target is an estimate obtained by fitting a parameter to each sample, holding the original fitted parameter fixed does not reproduce that estimator’s full repeated-sample process. The procedure must simulate data and re-estimate the parameter in every replicate to assess that additional estimation uncertainty.

Key takeaway: A parametric bootstrap generates new severity samples from a fitted model, applies the same stated statistic to each sample, and uses the replicate distribution to approximate sampling variation conditional on the model. State the sample size, fitted distribution, statistic, and whether parameters are held fixed; these choices determine what the results mean.

Check Your Understanding

For each question, use the stated model and statistic. Round probabilities to four decimal places and dollar amounts to the nearest cent where needed.

A fitted exponential severity model has mean $\$800$. One bootstrap replicate contains losses $\$100,\ \$300,\ \$700,\ \$1{,}100$. Calculate its sample mean. (Answer: $\$550.00$)
A fitted exponential model has mean $\$500$. For uniforms $0.2,\ 0.6,\ 0.8$, generate the losses using $X^*=-500\ln(1-U)$, then identify the largest. (Answer: $\$804.72$)
A fitted Pareto model has minimum loss $\$1{,}000$ and shape $\alpha=2.5$. Calculate the model probability that a loss strictly exceeds $\$2{,}000$. (Answer: $0.1768$)
Four bootstrap replicate statistics are $900,\ 1{,}100,\ 1{,}000,\ 1{,}400$ dollars. Calculate their average and their sample standard deviation. (Answer: average $\$1{,}100.00$; sample standard deviation $\$216.02$)
A severity model was fitted using six observed claims from a portfolio with 20 policies. Under the stated claim-level bootstrap procedure, how many severities should each replicate contain, and are the fitted parameters refitted? (Answer: six severities; no, the fitted parameters are held fixed.)

What You'll Learn