Bootstrap Intervals for Severity Summaries

Why Bootstrap Severity Summaries?

A severity estimate such as the average claim amount or a high claim quantile is calculated from a finite sample. A different sample of claims would generally produce a different estimate. The nonparametric bootstrap uses the observed complete claim amounts as a stand-in for the unknown severity distribution and repeatedly samples from them to approximate how much a summary can vary.

The resampling unit matters. When the target is a summary of individual claim severities, each bootstrap sample consists of the same number of complete claim amounts as the original sample, drawn with replacement. This is not the same procedure as resampling policy periods: periods can contain different numbers of claims, and resampling them answers a different uncertainty question.

Definition: Let $X_1,\ldots,X_n$ be an observed sample of complete claim amounts. A nonparametric bootstrap replicate is a sample of size $n$ drawn with replacement from the empirical distribution that assigns probability $1/n$ to each observed claim amount. Calculate the severity statistic of interest, such as the sample mean or sample quantile, for each replicate. Repeating this process $B$ times gives $B$ bootstrap estimates.

A claim amount can be selected more than once in a replicate, while another observed amount can be omitted. The original sample statistic is denoted by $\hat{\theta}$; the statistic from bootstrap replicate $b$ is denoted by $\hat{\theta}^{*(b)}$, for $b=1,\ldots,B$. In practice, the replicate samples are generated randomly. Small, explicitly listed sets of replicates below are for calculation practice; they are not a recommendation to use a small $B$ for an actual analysis.

Bootstrap Standard Errors and Intervals

The spread of the bootstrap estimates measures sampling variability as represented by the empirical distribution. The bootstrap standard error is the sample standard deviation of the $B$ replicate estimates:

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

Two commonly used intervals are the percentile interval and the basic bootstrap interval. Write $q^*_{\alpha}$ for the empirical $\alpha$-quantile of the bootstrap estimates, using the stated quantile convention. The percentile interval takes its endpoints directly from the bootstrap distribution. The basic interval reflects those endpoints around the original estimate.

\begin{aligned} \text{Percentile interval:}\quad &[q^*_{\alpha/2},q^*_{1-\alpha/2}],\\ \text{Basic interval:}\quad &[2\hat{\theta}-q^*_{1-\alpha/2},\;2\hat{\theta}-q^*_{\alpha/2}]. \end{aligned}

For example, a 95% interval uses $\alpha=0.05$. With many replicates, the empirical quantiles approximate the corresponding quantiles of the bootstrap distribution. A different software quantile convention or a different random set of replicates can give slightly different endpoints, especially when $B$ is small or the statistic has many ties. State the convention when a calculation depends on it.

Assumptions and scope: Claim-level resampling is appropriate when the observed complete claim amounts can reasonably be treated as independent, identically distributed observations from the severity population of interest. It estimates uncertainty in a statistic of that claim-severity distribution, conditional on the observed data and sample size. It does not automatically account for claim dependence, changes in the severity distribution over time, uncertainty in claim frequency, or missing, truncated, or censored claim amounts.

Why Resampling With Replacement Works

The bootstrap substitutes the empirical distribution for the unknown population distribution. Each draw from that empirical distribution has probability $1/n$ of selecting each observed claim amount. Thus, a bootstrap replicate imitates taking another sample of $n$ claims from a population represented by the observed data. Recalculating the statistic in each replicate approximates the variation that would result from repeating the sampling process.

For the sample mean, this idea also gives a short variance justification. Conditional on the observed data, one bootstrap draw has mean $\bar{X}$ and variance equal to the empirical variance with divisor $n$. The $n$ draws in a replicate are independent conditional on the data, so the variance of their average is the variance of one draw divided by $n$:

\operatorname{Var}^*(\bar{X}^*) = \frac{1}{n}\left(\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2\right).

This explains why bootstrap means vary around the original sample mean and why their spread can estimate the mean's sampling error. The formula is conditional on the empirical distribution; it is not a guarantee that the bootstrap captures every source of real-world variation.

Worked Example: Bootstrap Standard Error for a Mean

A hypothetical portfolio has three observed, complete claim amounts: $100, $200, and $300. Estimate the bootstrap standard error of the sample mean using the exact variance of one draw from the empirical distribution. Then form a normal-approximation interval using 1.96 bootstrap standard errors.

Method and assumptions. Treat the three claim amounts as independent observations from a common severity distribution. A bootstrap draw selects each of $100, $200, and $300 with probability $1/3$. The observed mean is

\bar{X}=\frac{100+200+300}{3}=\frac{600}{3}=\$200.

The variance of one empirical draw, using divisor $n$, is

\frac{(100-200)^2+(200-200)^2+(300-200)^2}{3} = \frac{10{,}000+0+10{,}000}{3} = \frac{20{,}000}{3} = \$6{,}666.67.

Each replicate mean averages three independent draws from this empirical distribution. Therefore,

\operatorname{Var}^*(\bar{X}^*) = \frac{6{,}666.67}{3} = \$2{,}222.22, \qquad \widehat{\operatorname{SE}}_{\mathrm{boot}} = \sqrt{2{,}222.22} = \$47.14.

The requested normal-approximation interval is

200\mathbin{\pm}1.96(47.14) = 200\mathbin{\pm}92.3944 = [\$107.61,\ \$292.39].

Check. The interval is centered at the observed mean of $200, and its half-width is $92.39 after rounding to the nearest cent. This calculation uses the normal approximation; it is not the same as taking bootstrap percentile endpoints. With only three observed claims, neither the approximation nor the empirical resampling distribution should be mistaken for precise information about the tails of the underlying severity distribution.

Common mistake avoided: Dividing the empirical variance by $n-1$ inside the conditional bootstrap-draw calculation would use a different variance convention. A bootstrap draw selects from the empirical distribution with probabilities $1/n$, whose variance uses divisor $n$. The replicate-mean variance then divides that quantity by $n$.

Percentile and Basic Intervals in Practice

A percentile interval is simple to interpret: its lower and upper endpoints are the corresponding lower and upper quantiles of the bootstrap estimates. A basic interval instead uses the bootstrap distribution to approximate how far the estimator may fall from the observed estimate, then reflects those deviations around $\hat{\theta}$. Neither construction removes bias or repairs a mismatch between the data and the assumed sampling unit.

Worked Example: Percentile and Basic Intervals for a Mean

Use the same three complete claim amounts, $100, $200, and $300. Suppose a small illustrative run produces the following eight bootstrap samples, each of size three. Use the nearest-rank empirical quantile convention: the $p$-quantile is the observation in position $\lceil 8p\rceil$ after sorting, with positions starting at 1. Calculate the 95% percentile and basic intervals.

Method. Calculate each replicate mean, sort the eight means, and take positions $\lceil8(0.025)\rceil=1$ and $\lceil8(0.975)\rceil=8$. The replicate calculations are:

\begin{array}{c|c} \text{Bootstrap sample}&\text{Replicate mean}\\ (100,100,200)&400/3=\$133.33\\ (100,200,300)&600/3=\$200.00\\ (300,200,200)&700/3=\$233.33\\ (100,300,300)&700/3=\$233.33\\ (200,200,200)&600/3=\$200.00\\ (100,100,100)&300/3=\$100.00\\ (300,300,300)&900/3=\$300.00\\ (200,300,300)&800/3=\$266.67 \end{array}

In increasing order, the replicate means are $100.00, $133.33, $200.00, $200.00, $233.33, $233.33, $266.67, and $300.00. The first and eighth order statistics are therefore $q^*_{0.025}=\$100$ and $q^*_{0.975}=\$300$. The 95% percentile interval is $[\$100,\$300]$.

The original estimate is $\hat{\theta}=\$200$. Reflecting the endpoints around it gives the basic interval:

[2(200)-300,\;2(200)-100]=[\$100,\$300].

Check. Here both intervals have the same endpoints because the two bootstrap quantiles are equally spaced around the observed mean. This coincidence is not a general property. Eight replicates are deliberately too few for a useful operational 95% interval: the nearest-rank rule forces the endpoints to the minimum and maximum. A real analysis should use many more replicates, check Monte Carlo stability, and report an interval suited to its data and purpose.

Common mistake avoided: The percentile interval uses the bootstrap quantiles as written. For the basic interval, the upper bootstrap quantile becomes the lower endpoint after reflection, and vice versa. Using the same endpoint order for both formulas can produce reversed or incorrect limits.

Bootstrapping a Severity Quantile

Bootstrap resampling applies to quantiles as well as means. First define the empirical quantile convention, then apply that same definition to the original data and to every bootstrap replicate. For a sample of size $n$, one common convention is the order statistic $X_{(\lceil np\rceil)}$ for quantile level $p$, where $X_{(j)}$ is the $j$th smallest observation. Other conventions interpolate between observations, so the convention can affect results in small samples.

Worked Example: A Bootstrap Interval for an Upper Quantile

A hypothetical set of four complete claims, in dollars, is $20,40,60,80$. Estimate its empirical 75th percentile using the order-statistic convention $X_{(\lceil np\rceil)}$. Then use the eight listed bootstrap replicates below to form a 95% percentile interval for that quantile, using the nearest-rank convention for the eight replicate statistics.

Original estimate. Here $n=4$ and $\lceil4(0.75)\rceil=3$. The ordered observations are $20,40,60,80$, so the empirical 75th percentile is $X_{(3)}=\$60$.

For each bootstrap sample, sort its four values and select the third value. The resulting replicate quantiles are:

\begin{array}{c|c} \text{Bootstrap sample}&\text{75th-percentile estimate}\\ (20,20,40,80)&\$40\\ (20,40,60,80)&\$60\\ (40,60,60,80)&\$60\\ (20,60,80,80)&\$80\\ (20,20,20,60)&\$20\\ (40,40,60,60)&\$60\\ (40,60,80,80)&\$80\\ (20,40,40,60)&\$40 \end{array}

Sorted, the eight replicate quantiles are $20,40,40,60,60,60,80,80$. The 2.5% and 97.5% nearest-rank positions are 1 and 8, so the 95% percentile interval is $[\$20,\$80]$.

Check. The original empirical estimate, $60, lies within the interval, and the two endpoints are observed claim amounts. With this tiny resampling run, the interval is coarse and sensitive to individual claims and ties. The calculation demonstrates the method, not a reliable tail-quantile assessment.

Common mistake avoided: Do not calculate the third order statistic of the original sample and reuse it as every replicate's estimate. Each resampled sample must be sorted independently, and its own third order statistic must be selected.

Claims Are Not Policy Periods

Resampling complete claim amounts treats claims as the observations. Resampling policy periods treats periods as observations, with each selected period carrying its own set of claims. These procedures generally target different summaries and represent different sampling assumptions. If the question concerns a claim-weighted severity mean, each claim contributes one amount to that mean. If the question concerns variation among annual or policy-period experience, the period may be the relevant unit; merely resampling individual claims does not reproduce variation between periods.

Before resampling, also check that the data contain complete amounts appropriate to the severity definition. A recorded payment limited by a policy maximum is not necessarily the same as a complete underlying loss. Similarly, a sample observed only above a deductible does not automatically represent the full severity distribution. Earlier material on censored and truncated claims explains why those observation mechanisms need separate treatment; resampling the recorded values alone does not undo them.

Worked Example: Claim Resampling Versus Period Resampling

A hypothetical insurer observes two policy periods. Period A has two complete claims of $100 and $300. Period B has four complete claims, each of $500. Calculate the claim-weighted mean and the unweighted mean of the two period means. Identify what each mean weights equally.

Claim-weighted mean. There are six claims, with total amount

$$ 100+300+500+500+500+500=2{,}400. $$

Thus the mean across claims is $2{,}400/6=\$400$. It gives each claim equal weight. Period A's mean is $(100+300)/2=\$200$, while Period B's mean is $(500+500+500+500)/4=\$500$.

Period-weighted mean. Giving each of the two periods equal weight gives

\frac{200+500}{2}=\$350.

Check. The two answers differ: the claim-weighted mean is $400, while the period-weighted mean is $350. Period B has more claims, so it contributes more to the claim-weighted mean, but each period contributes equally to the period-weighted mean. Resampling claims is aligned with uncertainty in a claim-level severity summary under an independent-claim assumption; resampling periods is not an interchangeable shortcut for that calculation.

Common mistake avoided: Treating all six claims as if they represented six equally sized policy periods would change the weighting. Always identify whether the target statistic weights claims equally or periods equally before choosing what to resample.

Common Traps and Practical Checks

Resampling without replacement. That selects a subset rather than generating bootstrap samples from the empirical distribution. A bootstrap sample has size $n$ and is drawn with replacement.
Changing the sample size. For the claim-level procedure described here, every replicate contains exactly $n$ claims. A variable number of claims belongs to a different model or resampling design.
Confusing the original estimate with the bootstrap average. The bootstrap estimates describe resampling variability; their average $\bar{\theta}^*$ need not equal the original estimate exactly in a finite simulation.
Assuming a bootstrap repairs problematic data. Censoring, truncation, inflation adjustments, and dependence require appropriate handling before or within an analysis. Resampling cannot recover information that was not observed.
Reporting unstable endpoints as precise. A small $B$, especially for a high severity quantile, can make interval endpoints depend heavily on a few replicates. Increase $B$ and assess whether the endpoints have stabilized.

Key takeaway: To bootstrap a severity summary, resample the complete claim amounts with replacement, keeping the replicate sample size equal to the original claim sample size. Recalculate the chosen statistic in every replicate, then use the replicate distribution for a standard error or interval. The resampling unit must match the uncertainty question.

Check Your Understanding

For each problem, use the stated quantile convention and show the resampling unit that matches the target.

A complete claim sample is $2,4,6,8$. Using the empirical distribution with divisor $n$, calculate the exact bootstrap standard error of the sample mean. (Answer: $\$1.1180$)
For a sample $50,100,150$, eight bootstrap mean estimates are $66.67,100,133.33,116.67,100,50,150,133.33$. Using nearest-rank quantiles, with position $\lceil Bp\rceil$, find the 90% percentile interval. (Answer: $[\$50,\$150]$)
The ordered claim amounts are $10,20,30,40,50$. Using $X_{(\lceil np\rceil)}$, find the empirical 80th percentile. (Answer: $\$40$)
Two periods contain claims $(100,300)$ and $(500,500,500,500)$. Find the claim-weighted mean and the equally weighted mean of period means. (Answer: claim-weighted mean $\$400$; period-weighted mean $\$350$)
A 95% interval is based on $B=6$ bootstrap quantile estimates. Under the nearest-rank convention, which order statistics supply the endpoints? (Answer: the 1st and 6th order statistics)

What You'll Learn