Nonparametric Bootstrap Resampling Of Claim-Level Data

Why Resample Claims?

A claim sample is only one realization of the underlying severity process. Its sample mean, upper quantile, or fitted severity parameter would generally change if a different set of claims had been observed. The nonparametric bootstrap uses the observed claim amounts to approximate that repeated-sample variation, without requiring a named parametric severity distribution for the resampling step.

The key operation is to draw claim amounts with replacement from the observed sample. Each bootstrap replicate contains the same number of claims as the original sample. A claim amount can appear more than once in a replicate, and another observed amount can be absent. Those repetitions and omissions are essential: sampling without replacement would simply rearrange the original data and would not create a useful distribution of statistics.

Definition: Given observed claim amounts $x_1,\ldots,x_n$, a nonparametric bootstrap replicate is a sample of size $n$ drawn independently with replacement from those $n$ observed values. Recalculate the statistic of interest from each replicate. The resulting collection of statistics is the bootstrap distribution of that statistic, conditional on the observed data.

This procedure treats the observed claims as the support of an estimated severity distribution. It is a claim-level procedure: it resamples claim amounts, not policy counts or whole policy records. Its basic interpretation assumes the observed amounts are a representative sample of independent, identically distributed claims from the population of interest. If claims are dependent, systematically sampled, truncated, or censored, resampling the recorded amounts as if they were independent complete claims may not represent the actual sampling process.

The Resampling Procedure

Let $F$ be the unknown distribution of claim severity. The empirical distribution $F_n$ assigns probability $1/n$ to each observed claim amount, counting repeated observed values with their multiplicities. In each bootstrap replicate, draw $n$ independent values from $F_n$, calculate the chosen statistic, and repeat the process $B$ times. The $B$ resulting statistics approximate the sampling distribution of the statistic.

Fix the observed sample.
Record the $n$ claim amounts and identify the statistic to investigate, such as a mean, quantile, or fitted parameter.

Draw a replicate of size $n$.
Select $n$ observed claim amounts independently and with replacement. Each draw has probability $1/n$ of selecting each sample position.

Recalculate the statistic.
Use the replicate, rather than the original data, to compute the mean, quantile, or model estimate.

Repeat and inspect the results.
Repeat the draw and calculation $B$ times. The empirical distribution of the $B$ statistics describes the bootstrap approximation to sampling variation.

Why the Empirical Distribution Is Resampled

The bootstrap substitutes $F_n$ for the unknown population distribution $F$. This makes it possible to imitate repeated sampling using only the observed data. Conditional on the data, every bootstrap draw has mean equal to the observed sample mean. Consequently, the bootstrap mean statistic is centered at the original sample mean in expectation.

E^*[X_j^*]=\frac{1}{n}\sum_{i=1}^{n}x_i=\bar{x} \qquad\Longrightarrow\qquad E^*[\bar{X}^*]=\bar{x}.

Here $X_j^*$ denotes a bootstrap draw, $\bar{X}^*$ the mean in one bootstrap replicate, and $E^*$ expectation over the resampling while holding the observed data fixed. The equality follows because each of the $n$ observed amounts is equally likely to be selected on any draw, and the mean of $n$ draws is the average of their means. This does not say that $\bar{x}$ equals the unknown population mean; it explains why the bootstrap distribution is centered around the estimate from the observed sample.

For a statistic that is not a mean, there is no general rule that the bootstrap distribution has a particular center or shape. The practical method is to recalculate the actual statistic in every replicate. That matters especially for quantiles and fitted parameters, whose behavior can be sensitive to a small sample or to a few large claims.

Worked Example: Bootstrap Distribution of a Claim Mean

A hypothetical insurer observes two complete claim amounts, $120$ and $300$ dollars. Enumerate every ordered bootstrap replicate of size two and find the bootstrap distribution of the sample mean. Use exact enumeration rather than random simulation.

Method. Each draw selects either $120$ or $300$, each with probability $1/2$. There are $2^2=4$ equally likely ordered replicates:

(120,120),\quad(120,300),\quad(300,120),\quad(300,300).

Recalculate the mean in each replicate. The four means are, respectively,

\frac{120+120}{2}=120,\quad \frac{120+300}{2}=210,\quad \frac{300+120}{2}=210,\quad \frac{300+300}{2}=300.

Thus the bootstrap distribution assigns probability $1/4$ to $120$, probability $2/4=1/2$ to $210$, and probability $1/4$ to $300$. The original sample mean is $(120+300)/2=210$, which is also the mean of these four bootstrap means: $(120+210+210+300)/4=210$.

Check. The probabilities sum to $1/4+1/2+1/4=1$, and the bootstrap distribution is centered at the observed mean as expected. A common mistake is to draw two claims without replacement: that produces only $(120,300)$ and $(300,120)$, both with mean $210$, and incorrectly removes all resampling variation.

Using Bootstrap Replicates for Quantiles

A sample quantile is computed from the ordered values in a sample. To keep the convention unambiguous, this tutorial uses the empirical inverse-CDF quantile: for probability $p$, the quantile is the observation in position $\lceil np\rceil$ after sorting the $n$ values, where $\lceil\cdot\rceil$ is the ceiling function. In every replicate, sort that replicate and apply the same convention. Small samples can produce a discrete bootstrap distribution with substantial probability at only a few values.

Worked Example: Resampling an Upper Claim Quantile

A hypothetical sample of three claim amounts is $40$, $70$, and $130$ dollars. Estimate the $75$th percentile using the stated empirical inverse-CDF convention, and find the exact bootstrap distribution of that quantile.

Original estimate. The position is $\lceil 3(0.75)\rceil=\lceil2.25\rceil=3$. The third ordered observation is $130$, so the sample $75$th percentile is $130$ dollars.

Bootstrap calculation. Each replicate has three draws with replacement, so there are $3^3=27$ equally likely ordered replicates. For this sample, the $75$th percentile is the third ordered value: the maximum of the three draws. The maximum is $40$ only when all three draws are $40$, which occurs once. It is $70$ when all three draws come from $\{40,70\}$, but not all are $40$; there are $2^3-1^3=7$ such replicates. The remaining $27-1-7=19$ replicates have maximum $130$.

P^*(Q_{0.75}^*=40)=\frac{1}{27},\qquad P^*(Q_{0.75}^*=70)=\frac{7}{27},\qquad P^*(Q_{0.75}^*=130)=\frac{19}{27}.

Check. The counts $1+7+19=27$, so the probabilities sum to one. The bootstrap distribution is coarse because there are only three distinct observed values and three claims per replicate. A common mistake is to calculate the quantile position as $np=2.25$ and then use a non-integer position without stating an interpolation rule. Under the convention specified here, the position is rounded up to $3$.

Refitting a Severity Parameter

The same resampling logic applies when the statistic is an estimated model parameter. The model is fitted anew in each replicate; the original fitted parameter is not held fixed. This is distinct from drawing new claims from a fitted parametric severity model: nonparametric resampling draws only from the observed claim amounts.

Worked Example: Bootstrap Replicates of an Exponential Mean

Suppose three hypothetical, complete claim amounts are $100$, $200$, and $300$ dollars. Assume for fitting purposes that claims are independent and identically distributed exponential amounts with density $f_X(x)=\theta^{-1}e^{-x/\theta}$ for $x>0$, where $\theta$ is the mean severity. For four specified bootstrap replicates, calculate the maximum likelihood estimate of $\theta$, and also the corresponding rate estimate $\lambda=1/\theta$.

Method and original fit. For complete exponential observations $x_1,\ldots,x_n$, the log likelihood is $\ell(\theta)=-n\log\theta-\sum_i x_i/\theta$. Setting its derivative equal to zero gives $-n/\theta+\sum_i x_i/\theta^2=0$, hence $\hat{\theta}=\sum_i x_i/n=\bar{x}$. The original fit is $(100+200+300)/3=200$ dollars, so $\hat{\lambda}=1/200=0.005$ per dollar.

Refit each replicate. The specified resamples, shown by their claim amounts, are $(100,100,300)$, $(200,300,300)$, $(100,200,200)$, and $(100,200,300)$. For each one, the exponential mean estimate is its sample average, and the rate estimate is its reciprocal.

Replicate	Sum of amounts	$\hat{\theta}^*$	$\hat{\lambda}^*$
$(100,100,300)$	$500$	$500/3$ dollars	$3/500=0.006$ per dollar
$(200,300,300)$	$800$	$800/3$ dollars	$3/800=0.00375$ per dollar
$(100,200,200)$	$500$	$500/3$ dollars	$3/500=0.006$ per dollar
$(100,200,300)$	$600$	$200$ dollars	$1/200=0.005$ per dollar

Check. Each replicate contains exactly three claims, and each fitted mean equals its replicate total divided by three. Each rate is the reciprocal of that mean. These four specified replicates illustrate refitting; four replicates are far too few for a stable assessment of uncertainty. A common mistake is to resample the data but report the original estimate $200$ for every replicate, which discards the parameter variation the procedure is intended to reveal.

Interpreting the Results and Avoiding Misuse

With a sufficiently large $B$, the bootstrap statistics form an empirical approximation to the statistic’s sampling distribution under the empirical model. Their spread and shape can be inspected to see how sensitive the estimate is to resampling the observed claims. Increasing $B$ makes the Monte Carlo representation of this bootstrap distribution less noisy, but does not add information to the original claim sample or correct a mismatch between the sample and the target population.

Keep the replicate size fixed. For claim-level resampling of $n$ observed amounts, every replicate has $n$ draws. Changing the count changes the sampling procedure.
Resample with replacement. Without replacement, the sample values are merely reordered, so symmetric statistics such as the mean do not vary.
Recompute the target statistic. Apply the same definition and estimation method to every replicate, including the same quantile convention or model-fitting assumptions.
Respect the data structure. The basic procedure treats claim amounts as independent observations from one severity population. A claim-level resample is not automatically appropriate when observations are dependent or represent a different sampling design.
Do not confuse bootstrap and parametric simulation. Nonparametric resampling selects observed values; a parametric simulation generates new values from a specified fitted distribution.

Key takeaway: A nonparametric claim-level bootstrap repeatedly samples $n$ observed claim amounts with replacement and recalculates the statistic of interest. It approximates repeated-sample variation by using the empirical distribution in place of the unknown severity distribution, subject to the assumptions represented by the observed sample.

Check Your Understanding

For each problem, use resampling with replacement and preserve the original sample size. The stated quantile convention is the empirical inverse-CDF convention.

A claim sample is $10$ and $50$ dollars. What is the probability that a bootstrap replicate of size two contains two $10$-dollar claims, and what is its mean? (Answer: Probability $1/4$; mean $\$10$.)
A sample contains $20$, $40$, and $80$ dollars. A specified bootstrap replicate is $(80,20,80)$. Find its mean and its empirical $75$th percentile. (Answer: Mean $\$60$; $75$th percentile $\$80$.)
For the sample $30$, $60$, and $90$ dollars, what is the probability that a bootstrap replicate of size three has maximum $60$? (Answer: $7/27$.)
For complete exponential claim data $50$, $100$, and $150$ dollars, a bootstrap replicate is $(50,50,150)$. Find the exponential maximum likelihood estimate of the mean parameter and the corresponding rate. (Answer: $\hat{\theta}=250/3$ dollars; $\hat{\lambda}=3/250=0.012$ per dollar.)
An analyst has $n=40$ independent, complete claim amounts and wants a nonparametric claim-level bootstrap distribution of the sample mean. State the replicate size and how each replicate is generated. (Answer: Each replicate has 40 claim amounts, drawn independently with replacement from the 40 observed amounts; calculate the mean of each replicate.)

Replicate	Sum of amounts	\(\hat{\theta}^*\)	\(\hat{\lambda}^*\)
\((100,100,300)\)	\(500\)	\(500/3\) dollars	\(3/500=0.006\) per dollar
\((200,300,300)\)	\(800\)	\(800/3\) dollars	\(3/800=0.00375\) per dollar
\((100,200,200)\)	\(500\)	\(500/3\) dollars	\(3/500=0.006\) per dollar
\((100,200,300)\)	\(600\)	\(200\) dollars	\(1/200=0.005\) per dollar

What You'll Learn