Why Simulate Pricing and Capital Indications?
A pricing model can specify claim frequency, claim severity, exposure, and coverage terms without yielding a convenient closed-form distribution for annual covered losses. Simulation combines those ingredients directly: generate a claim count, generate claim amounts, apply the contract, and add the resulting payments. Repeating that process produces a sample of annual aggregate losses that can support an expected-cost estimate and tail-risk measures such as VaR and TVaR.
Simulation is a numerical method, not a substitute for a correctly specified model. A sound indication therefore has two parts: a clear description of the model and calculation steps, and checks that expose implementation errors. In particular, exposure must scale the frequency model, coverage must be applied claim by claim before aggregation, and estimates from a finite set of trials must be reported as estimates rather than exact model values.
Set Up the Simulation Before Drawing Outcomes
For a basic compound Poisson model, suppose the number of claims in one policy-year follows a Poisson distribution with mean \(\lambda\), and claim severities are independent and identically distributed as \(X\), independently of \(N\). For exposure \(e\) policy-years, use \(N\sim\operatorname{Poisson}(e\lambda)\), provided the exposure is homogeneous and frequency is assumed proportional to exposure. If exposures differ in risk characteristics, use appropriate group-specific frequency parameters rather than scaling every policy by one common rate.
For a deductible \(d\) and a maximum payment per claim \(u\), where \(u\) is a cap on the insurer's payment, the payment per loss is \(Y^L=\min((X-d)_+,u)\), where \((z)_+=\max(z,0)\). Apply this transformation to each simulated gross claim, then sum the claim payments to obtain the annual aggregate. Do not apply the deductible or per-claim limit to the annual aggregate unless the contract specifically defines an aggregate feature.
A practical simulation records the model inputs and the random-number mapping. For each trial, generate the claim count from the exposure-adjusted frequency distribution; generate that many severities from the fitted severity distribution; apply the coverage transformation separately to each severity; and sum the payments. Keep one aggregate result per trial. The sample mean estimates expected aggregate covered cost, while the ordered aggregate results support empirical risk measures.
Record the pricing period, exposure, fitted frequency and severity distributions, their parameters, and any independence assumptions.
Draw a count using the frequency model adjusted for exposure. For each claim, draw a gross severity using the fitted severity model.
Transform each gross severity into its covered payment, preserving the distinction between per-claim terms and aggregate terms.
Calculate the sample mean and empirical tail measures, and compare the mean with an analytical expectation when that expectation is available.
Expected Cost as a Validation Check
Under the compound Poisson assumptions above, if the mean covered payment \(E[Y^L]\) is finite, then \(E[L]=e\lambda E[Y^L]\). This identity is valuable even when the aggregate distribution is difficult to calculate directly: the simulated mean should be reasonably consistent with this model expectation, allowing for Monte Carlo variation.
To justify the middle equality, condition on a fixed count \(N=n\). Independence and identical distribution of the covered payments give \(E[\sum_{j=1}^{n}Y_j^L\mid N=n]=nE[Y^L]\). Taking expectations over \(N\) gives \(E[N]E[Y^L]\), and a Poisson count with mean \(e\lambda\) has \(E[N]=e\lambda\). If frequency and severity are dependent, or if the portfolio is not represented by a homogeneous Poisson model, this calculation may not apply in this form.
Worked Example: Generating Covered Losses From Uniform Draws
Method and assumptions. Consider a hypothetical policy with exposure \(e=2\) policy-years. The fitted annual claim frequency is Poisson with \(\lambda=0.5\) claims per policy-year, so each trial has \(N\sim\operatorname{Poisson}(1)\). Gross severity is independent of frequency and has the discrete distribution \(P(X= \$500)=0.50\), \(P(X=\$1{,}500)=0.30\), and \(P(X=\$3{,}000)=0.20\). The policy has deductible \(d=\$250\) and per-claim payment limit \(u=\$1{,}000\). There is no discounting or mortality assumption: the model covers one annual short-term pricing period.
For the count simulation, the Poisson(1) cumulative probabilities are \(P(N=0)=0.3679\), \(P(N\leq1)=0.7358\), \(P(N\leq2)=0.9197\), and \(P(N\leq3)=0.9810\). Thus a uniform draw below 0.3679 gives zero claims; a draw from 0.3679 up to 0.7358 gives one; from 0.7358 up to 0.9197 gives two; and from 0.9197 up to 0.9810 gives three.
For severity, use a uniform draw below 0.50 for \(\$500\), from 0.50 up to 0.80 for \(\$1{,}500\), and from 0.80 to 1 for \(\$3{,}000\). The covered payments are, respectively, \(\min((500-250)_+,1000)=\$250\), \(\min((1500-250)_+,1000)=\$1{,}000\), and \(\min((3000-250)_+,1000)=\$1{,}000\).
Use the following five sets of draws. The severity draws are listed only for claims generated in that trial.
| Trial | Frequency uniform | Severity uniforms | Claim count | Aggregate covered loss |
|---|---|---|---|---|
| 1 | 0.20 | None | 0 | $0 |
| 2 | 0.50 | 0.60 | 1 | $1,000 |
| 3 | 0.80 | 0.20, 0.90 | 2 | $250 + $1,000 = $1,250 |
| 4 | 0.95 | 0.40, 0.70, 0.95 | 3 | $250 + $1,000 + $1,000 = $2,250 |
| 5 | 0.70 | 0.85 | 1 | $1,000 |
The five aggregate results in increasing order are \(\$0,\$1{,}000,\$1{,}000,\$1{,}250,\$2{,}250\). Their mean is \((0+1000+1000+1250+2250)/5=\$1{,}100\). Using the nearest-rank empirical quantile convention, the 80th percentile is the observation at rank \(\lceil5(0.80)\rceil=4\), so empirical VaR at 80% is \(\$1{,}250\). The upper 20% contains one of the five equally weighted trials, so empirical TVaR at 80% is \(\$2{,}250\). These are demonstrations of the calculation, not stable estimates: five trials are far too few for a dependable tail indication.
Common mistake avoided: The deductible is subtracted from each gross claim, and the payment cap is applied to each resulting claim payment. Applying a single deductible to the total of all claims would describe a different contract and would produce different simulated losses.
From Aggregate Outcomes to Pricing Measures
For \(M\) simulated aggregate losses \(L_1,\ldots,L_M\), the sample mean \(\bar L=M^{-1}\sum_{m=1}^{M}L_m\) estimates expected covered cost for the modeled portfolio and period. Divide by the corresponding number of policy-years to obtain a pure premium per policy-year, if that is the intended pricing basis. The denominator must match the exposure represented by the simulation.
For empirical VaR, sort the trial outcomes from smallest to largest and use the stated quantile convention. This tutorial uses the nearest-rank convention: for confidence level \(\alpha\), the empirical VaR is the observation at rank \(\lceil M\alpha\rceil\). Empirical TVaR is the average of the upper \(1-\alpha\) fraction of the empirical distribution. If that fraction cuts through an observation's probability mass, include the appropriate fraction of that observation rather than silently changing the tail probability.
The convention matters when outcomes are discrete or the simulation count is small. Other empirical quantile conventions can interpolate between order statistics and may yield different reported VaR values. Whichever convention is used, state it and apply it consistently. For capital or pricing decisions, also state whether the risk measure is calculated on an aggregate portfolio, per policy, or per exposure unit.
Worked Example: Combining Exposure Groups for Expected Cost
Method and assumptions. A hypothetical portfolio contains 10 policy-years in Group A and 5 policy-years in Group B. The fitted claim counts are Poisson with annual means \(\lambda_A=0.20\) and \(\lambda_B=0.40\) per policy-year, respectively. In both groups, gross severity is \(\$500\) or \(\$1{,}500\), each with probability 0.50, independently across claims and of frequency. Each claim has a deductible \(d=\$100\) and payment limit \(u=\$1{,}000\). The annual horizon is one year; no discounting or mortality assumption is used.
The covered payment for a \(\$500\) claim is \(\min((500-100)_+,1000)=\$400\). For a \(\$1{,}500\) claim it is \(\min((1500-100)_+,1000)=\$1{,}000\). Therefore, \(E[Y^L]=0.50(400)+0.50(1000)=\$700\).
For Group A, expected claim count is \(10(0.20)=2\), so expected covered cost is \(2(\$700)=\$1{,}400\). For Group B, expected claim count is \(5(0.40)=2\), so expected covered cost is \(2(\$700)=\$1{,}400\). Total expected portfolio cost is \(\$1{,}400+\$1{,}400=\$2{,}800\), over \(10+5=15\) policy-years. The indicated pure premium is \(\$2{,}800/15=\$186.666\ldots\), or \(\$186.67\) per policy-year to the nearest cent.
As a check, the group-specific formula gives \(10(0.20)(700)+5(0.40)(700)=1{,}400+1{,}400=\$2{,}800\), agreeing with the claim-count calculation. In a simulation, Group A and Group B must use their respective exposure-adjusted frequency distributions. Pooling the exposures and applying one unweighted average frequency rate would not preserve the stated group model.
Common mistake avoided: Exposure scales expected claim count, not claim severity. Multiplying severity by the number of policy-years and then also multiplying by frequency would count exposure twice.
Worked Example: Turning Simulated Tail Risk Into a Load
Method and assumptions. Suppose eight simulated annual aggregate losses for a hypothetical eight-policy-year portfolio, generated from a fitted frequency-severity model with claim-level coverage applied, are \(\$0,\$400,\$700,\$900,\$1{,}200,\$1{,}600,\$2{,}500,\$4{,}000\). The underlying model uses one-year aggregate losses and assumes independent frequency and severity as specified in its fit; the simulation output is provided here for the risk-measure calculation. No discounting or mortality assumption applies. Use the nearest-rank convention for VaR and define a capital measure as \(\operatorname{TVaR}_{0.75}-E[L]\). For illustration only, the pricing decision adds 5% of that capital measure as a portfolio risk load.
The sample mean is \((0+400+700+900+1200+1600+2500+4000)/8=11300/8=\$1{,}412.50\). At \(\alpha=0.75\), the VaR rank is \(\lceil8(0.75)\rceil=6\). The sixth ordered result is \(\$1{,}600\), so empirical \(\operatorname{VaR}_{0.75}=\$1{,}600\).
The upper 25% of eight equally weighted results contains the two largest losses, \(\$2{,}500\) and \(\$4{,}000\). Thus \(\operatorname{TVaR}_{0.75}=(2500+4000)/2=\$3{,}250\). The capital measure is \(\$3{,}250-\$1{,}412.50=\$1{,}837.50\). The illustrative 5% risk load is \(0.05(\$1{,}837.50)=\$91.875\), which rounds to \(\$91.88\) for reporting. Retaining the unrounded amount, the total indicated portfolio premium is \(\$1{,}412.50+\$91.875=\$1{,}504.375\), or \(\$1{,}504.38\) to the nearest cent. Per policy-year, this is \(\$1{,}504.375/8=\$188.046875\), or \(\$188.05\).
This load is a stated decision rule, not a universal formula for a risk load. A pricing indication should identify the chosen risk measure, confidence level, capital-to-premium factor, and allocation basis. This example also illustrates why rounding should be delayed: rounding the portfolio load before adding it to the expected cost can change the final cent.
Common mistake avoided: VaR is not the average of losses exceeding the quantile; that tail average is TVaR. Here VaR is \(\$1{,}600\), while the upper-quarter average is \(\$3{,}250\).
Implementation Checks and Exam Traps
A simulation can produce plausible-looking results while implementing the wrong model. Before interpreting an indication, check the following items:
- Exposure basis: Confirm that frequency is adjusted to the simulated policy-years and that results are divided by the matching exposure when reporting a per-unit premium.
- Coverage order: Apply each deductible, limit, coinsurance percentage, or other per-claim term as defined by the contract before summing claim payments.
- Frequency-severity structure: Check that the severity draws are generated once per simulated claim, not once per policy or once per simulation trial.
- Risk-measure convention: State the confidence level and empirical quantile convention. Handle fractional tail mass consistently for TVaR.
- Precision and stability: Keep unrounded trial totals through aggregation, and recognize that a small simulation can provide a noisy estimate, especially in the tail.
A further useful check is the expected-cost identity \(E[L]=e\lambda E[Y^L]\) under the homogeneous compound Poisson assumptions. A large discrepancy between the simulated mean and this benchmark can indicate an exposure, coverage, or random-number mapping error. Agreement is not proof that the entire model is appropriate: an incorrect severity tail can still have a reasonable mean but materially distort VaR and TVaR.
Check Your Understanding
For each problem, use the stated model and conventions. Amounts are hypothetical, and any indicated rounding is to the nearest cent.
- A block has 4 policy-years, with annual frequency Poisson at \(\lambda=0.30\) per policy-year. Severity is \(\$1{,}000\) with probability 0.40 and \(\$2{,}000\) with probability 0.60. Each claim has deductible \(d=\$200\) and per-claim payment limit \(u=\$1{,}200\). Find expected aggregate covered cost. (Answer: \(\$1{,}248\))
- For 3 policy-years, frequency is Poisson with \(\lambda=0.25\) per policy-year. Severity is equally likely to be \(\$400\) or \(\$900\). The deductible is \(d=\$100\), and the payment limit is \(u=\$700\). Find the expected aggregate covered cost. (Answer: \(\$375\))
- Six equally weighted simulated aggregate losses, already sorted, are \(\$0,\$300,\$500,\$800,\$1{,}200,\$2{,}000\). Using nearest-rank VaR and the integrated upper-tail mean definition of TVaR, find VaR and TVaR at \(\alpha=2/3\). (Answer: VaR \(=\$800\); TVaR \(=\$1{,}600\))
- Eight simulated aggregate losses for a four-policy-year portfolio are \(\$0,\$200,\$400,\$600,\$800,\$1{,}000,\$1{,}400,\$2{,}000\). Find the sample mean aggregate cost and the sample pure premium per policy-year. (Answer: mean aggregate cost \(=\$800\); pure premium \(=\$200\) per policy-year)
- A hypothetical gross claim is \(\$1{,}000\), subject to 10% severity trend before coverage is applied. The deductible is \(\$200\), and the per-claim payment limit is \(\$1{,}000\). Find the covered payment after trending the loss and applying the terms. (Answer: \(\$900\))