Repeated-Sample Simulation For Actuarial Estimators

Why Simulate Repeated Datasets?

A loss model describes the random observations that could arise in a portfolio. An estimator is a rule applied to those observations to estimate a quantity of interest. Even when the data-generating model is fixed, a different sample would generally produce a different estimate. Repeated-sample simulation makes that variation visible: generate many independent datasets of the same design, calculate the estimator in each one, and examine the resulting estimates.

The object of interest is the estimator’s sampling distribution, not the distribution of individual claim amounts. For example, a severity model may produce individual claims with a highly skewed distribution, while the sample mean of a fixed number of claims has its own distribution. The latter is what repeated-sample simulation approximates when the estimator is the sample mean.

Definition: Let $D$ denote a dataset generated under a specified model and sampling design, and let $\hat{\theta}=T(D)$ be an estimator computed from that dataset. The sampling distribution of $\hat{\theta}$ is the probability distribution of $T(D)$ over all datasets that could be generated under the same model and design. Repeated-sample simulation approximates this distribution by generating $R$ independent datasets $D_1,\ldots,D_R$ and calculating $\hat{\theta}_r=T(D_r)$ for each replication $r$.

A simulation study must specify both the data-generating model and the estimator. The model includes its parameter values, independence assumptions, and sample size or exposure. The estimator must then be applied in the same way to every generated dataset. If the aim is to study finite-sample behavior under a known model, the model parameters are held fixed across replications; it is the generated observations, not the model, that change.

What the Repeated Estimates Tell Us

The $R$ simulated estimates can be summarized to approximate features of the estimator’s sampling distribution. If the target is $\theta$, the average simulated estimate is an approximation to the estimator’s expected value. Its difference from $\theta$ indicates simulated bias. The spread of the estimates describes sampling variability.

\bar{\hat{\theta}}_{\mathrm{sim}}=\frac{1}{R}\sum_{r=1}^{R}\hat{\theta}_r,\qquad \widehat{\operatorname{Bias}}_{\mathrm{sim}}=\bar{\hat{\theta}}_{\mathrm{sim}}-\theta

The variance of the finite set of simulated estimates, using divisor $R$, is a direct Monte Carlo approximation to the sampling variance. A variance estimate with divisor $R-1$ is also often reported as an unbiased sample variance for the unknown variance of the simulated-estimate distribution. These are distinct summaries, so state which divisor is being used.

\widehat{\operatorname{Var}}_{\mathrm{sim}}(\hat{\theta}) =\frac{1}{R}\sum_{r=1}^{R}(\hat{\theta}_r-\bar{\hat{\theta}}_{\mathrm{sim}})^2

The square root of a variance estimate is a simulated standard deviation of the estimator. It is not the standard deviation of individual claims. Also, the simulated standard deviation of the estimates is not the same as the Monte Carlo standard error of their simulated average. Under independent replications, that latter standard error is approximately the simulated standard deviation divided by $\sqrt{R}$.

A Short Justification: The Sample Mean

Suppose each dataset consists of $n$ independent observations $X_1,\ldots,X_n$, each with mean $\mu$ and finite variance $\sigma^2$. The estimator is $\bar{X}=(X_1+\cdots+X_n)/n$. Linearity of expectation gives $E[\bar{X}]=\mu$. Independence makes the covariance terms zero when expanding the variance of a sum, so

\operatorname{Var}(\bar{X}) =\frac{1}{n^2}\operatorname{Var}\left(\sum_{j=1}^{n}X_j\right) =\frac{1}{n^2}\sum_{j=1}^{n}\operatorname{Var}(X_j) =\frac{n\sigma^2}{n^2} =\frac{\sigma^2}{n}.

This result provides a useful check on a repeated-sample simulation of a sample mean: across many replications, the average should be near $\mu$, and the variance should be near $\sigma^2/n$. Neither comparison is guaranteed to be close in a small simulation. The simulated summaries themselves vary because only $R$ datasets were generated.

Worked Repeated-Sample Simulations

Worked Example: Estimating a Claim Probability

A hypothetical warranty model has independent claim indicators $I_j\sim\operatorname{Bernoulli}(p)$, with $p=0.40$. Each simulated dataset has $n=4$ warranties, and the estimator is $\hat{p}=\sum_{j=1}^{4}I_j/4$. Use the following five rows of simulated uniform variates, with a claim recorded when $U<0.40$, to approximate the sampling mean and variance of $\hat p$.

Method. Apply the same threshold rule to every uniform variate, count the claims in each dataset, and divide by four. Independent uniform draws generate independent Bernoulli outcomes with probability $0.40$.

Replication	Simulated uniform variates	Claim count	$\hat p$
1	0.12, 0.55, 0.71, 0.93	1	0.25
2	0.08, 0.24, 0.61, 0.82	2	0.50
3	0.11, 0.33, 0.47, 0.75	2	0.50
4	0.03, 0.18, 0.27, 0.62	3	0.75
5	0.43, 0.51, 0.66, 0.97	0	0.00

The five estimates sum to $0.25+0.50+0.50+0.75+0.00=2.00$, so their simulated mean is $2.00/5=0.40$. The deviations from $0.40$ are $-0.15,0.10,0.10,0.35,-0.40$. Their squared deviations sum to $0.0225+0.0100+0.0100+0.1225+0.1600=0.3250$. Thus, using divisor $R=5$, the simulated variance is $0.3250/5=0.0650$.

As a check, the exact estimator mean is $p=0.40$, and its exact variance is $p(1-p)/n=(0.40)(0.60)/4=0.0600$. The simulated mean matches exactly in this small run, while the simulated variance is close but not exact. Common mistake avoided: do not treat the five estimates as five individual warranty outcomes. Each estimate is based on four warranties, and the sampling variation being summarized is variation across the five dataset-level estimates.

Worked Example: Sampling Distribution of an Exponential Severity Mean

Suppose individual hypothetical claim severities are independent exponential random variables with mean $\theta=100$ dollars. Each dataset contains $n=3$ claims, and the estimator is the sample mean. Four independently simulated datasets produce severities $(20,40,60)$, $(40,80,120)$, $(60,120,180)$, and $(80,160,240)$, all in dollars. Calculate the simulated mean and variance of the estimator using divisor $R$, then compare them with the model results.

Method. Recompute the sample mean separately for each three-claim dataset. For an exponential distribution with mean $100$, the variance is $100^2=10{,}000$ dollars squared. By the sample-mean result, the estimator’s theoretical mean is $100$ dollars and its theoretical variance is $10{,}000/3$ dollars squared.

The four estimates are $120/3=40$, $240/3=80$, $360/3=120$, and $480/3=160$ dollars. Their sum is $400$, so their simulated mean is $400/4=\$100$. The deviations from this mean are $-60,-20,20,60$ dollars. The squared deviations sum to $3{,}600+400+400+3{,}600=8{,}000$ dollars squared, giving a simulated variance of $8{,}000/4=2{,}000$ dollars squared.

The exact variance is $10{,}000/3=\$3{,}333.33$ squared, whereas the four-replication approximation is $2{,}000$ dollars squared. This difference is plausible with only four datasets; it does not change the theoretical result. Common mistake avoided: the exponential severity variance is $10{,}000$, but the variance of the mean of three independent claims is one-third of that amount.

Worked Example: Simulating a Compound Poisson Pure-Premium Estimator

A hypothetical portfolio model has independent policy claim counts $N\sim\operatorname{Poisson}(\lambda)$, with $\lambda=1$ claim per policy. Claim severities are independent exponential amounts with mean $\theta=\$100$, independent of counts. For a dataset of $m=2$ policies, estimate expected aggregate loss per policy by $\hat{\mu}=(S_1+S_2)/2$, where $S_j$ is the total loss for policy $j$. Four simulated datasets have the following policy-level counts and severities:

Replication 1: counts $(0,1)$; the one claim is $\$50$.
Replication 2: counts $(2,1)$; severities are $\$40,\$60$ for the first policy and $\$80$ for the second.
Replication 3: counts $(1,1)$; severities are $\$100$ and $\$200$.
Replication 4: counts $(0,0)$; there are no severities.

Method. Sum all claim payments within each policy, then average the two policy totals. The four aggregate totals are $\$50$, $\$180$, $\$300$, and $\$0$. The corresponding estimator values are therefore $\$25,\$90,\$150,\$0$ per policy.

The simulated mean is $(25+90+150+0)/4=\$66.25$. The deviations from $\$66.25$ are $-\$41.25,\$23.75,\$83.75,-\$66.25$. Their squared values are $1{,}701.5625$, $564.0625$, $7{,}014.0625$, and $4{,}389.0625$ dollars squared. The sum is $13{,}668.75$, so the simulated variance with divisor four is $13{,}668.75/4=\$3{,}417.1875$ squared.

The model’s expected aggregate loss per policy is $\lambda E[X]=1(100)=\$100$. For a compound Poisson total, $\operatorname{Var}(S)=\lambda E[X^2]$. Here $E[X^2]=\operatorname{Var}(X)+(E[X])^2=10{,}000+10{,}000=20{,}000$, so the variance of the two-policy average is $20{,}000/2=\$10{,}000$ squared. The four simulated datasets give a much smaller variance estimate, illustrating the instability of very few replications. Common mistake avoided: the target is the average aggregate loss per policy, not the average severity among observed claims; zero-claim policies remain in the denominator.

Practical Design and Interpretation

A reliable repeated-sample study follows a consistent sequence. First, define the estimand, the data-generating model, and the sampling design. Next, generate one complete dataset under that model and compute the estimator exactly as specified. Repeat this process independently $R$ times. Finally, summarize the estimates and compare the results with theoretical values when available.

Specify the target and model.
State the quantity being estimated, the model distributions and parameters, independence assumptions, and the dataset size or exposure.

Generate a full dataset.
Simulate all required observations for one replication. Preserve dependencies or shared model assumptions specified by the problem.

Recompute the estimator.
Apply the same estimator to that dataset, including its treatment of zero claims or other special outcomes.

Repeat and summarize.
Generate independent datasets, retain one estimate per replication, and report the number of replications and the summary convention used.

Increasing $R$ generally reduces Monte Carlo noise in summaries of the sampling distribution, but does not repair a misspecified data-generating model, an incorrect estimator, or a coding error. Likewise, a single simulated run is not evidence that an estimator is biased or unbiased. Compare simulated patterns with known theoretical properties and inspect the implementation carefully.

The most important distinction is between two levels of variation. Within one dataset, losses or claim counts vary according to the model. Across repeated datasets, the estimator varies because it is recalculated from different observations. The sampling distribution concerns the second level. This simulation approach is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, state the requested result and keep the simulated-estimate variation distinct from individual-observation variation.

A Bernoulli model has $p=0.30$, and each dataset has $n=5$. Four simulated datasets have claim counts $1,2,0,3$. Find the average of the four estimates and their variance using divisor $R=4$. (Answer: mean $0.3000$; variance $0.0500$.)
Individual severities are exponential with mean $\$200$. A dataset contains four independent claims, and the estimator is the sample mean. Find the estimator’s exact expected value, variance, and standard deviation. (Answer: mean $\$200$; variance $\$10{,}000$ dollars squared; standard deviation $\$100$.)
A Poisson claim-count model has rate $\lambda=0.5$ per policy. Three simulated datasets each cover ten policies and have claim counts $3,4,6$. For the rate estimator, calculate the average of the three estimates and the variance across them using divisor $R=3$. (Answer: mean $0.4333$ claims per policy; variance $0.0156$ squared claims per policy.)
For one policy, aggregate loss is compound Poisson with $\lambda=2$ and exponential severities with mean $\$50$. A portfolio estimator is the average aggregate loss over $m=4$ independent policies. Find its expected value and variance. (Answer: mean $\$100$; variance $\$2{,}500$ dollars squared.)
Four simulated estimates of a target $\theta=100$ are $80,100,120,140$. Find their simulated bias, their variance using divisor $R=4$, and the Monte Carlo standard error of their average using that variance. (Answer: bias $10$; variance $500$; Monte Carlo standard error $\sqrt{125}\approx11.18$.)

What You'll Learn