Compound Binomial And Negative Binomial Loss Simulation

Introduction

A compound loss simulation has two distinct random ingredients: how many claims occur and how large each claim is. For a binomial frequency model, the claim count is limited by a fixed number of trials. For a negative binomial model, the count is tied to the number of trials needed to achieve a specified number of successes. The simulation must preserve that distinction: frequency parameters generate the count, while severity parameters generate claim amounts.

This tutorial develops both algorithms and shows how to combine them. The settings and values are hypothetical. The methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $N$ be a claim count and let $X_1,X_2,\ldots$ be independent, identically distributed claim severities, independent of $N$. The aggregate loss is $$ S_N=\sum_{j=1}^{N}X_j, $$ with $S_0=0$. The distribution used to generate $N$ is the frequency model; the distribution used to generate each $X_j$ is the severity model.

Simulating a Binomial Claim Count

Suppose a hypothetical portfolio has $m$ independent policies, each of which can produce at most one claim during the period. If each policy has claim probability $p$, then the number of claims has distribution $N\sim\operatorname{Binomial}(m,p)$. The assumptions are a fixed number $m$ of trials, the same claim probability $p$ on every trial, and independence among trials.

The most direct simulation is to generate one independent uniform random number for each policy. A value at or below $p$ represents a claim; a value above $p$ represents no claim. Count the claims to obtain $N$. Equivalently, one may generate $N$ directly from the binomial inverse CDF, but the trial method makes the fixed-trial interpretation explicit.

N=\sum_{k=1}^{m} I_k,\qquad I_k= \begin{cases} 1,&U_k\le p,\\ 0,&U_k>p, \end{cases} \qquad U_k\sim\operatorname{Uniform}(0,1).

This model is not appropriate when a policy can generate multiple claims in the period, or when the trial probabilities differ without an appropriate modification. Once $N$ has been generated, generate exactly $N$ severities from the severity model. Do not generate severities for policies that did not produce a claim.

Worked Example: A Binomial Portfolio With Discrete Severities

A hypothetical portfolio contains $m=5$ policies. Each has claim probability $p=0.20$, independently. Conditional on a claim, its severity is $X=\$100$ with probability $0.75$ and $X=\$400$ with probability $0.25$. Use the frequency uniforms $0.08,0.44,0.19,0.91,0.03$, followed by severity uniforms $0.62,0.91,0.18$, to simulate aggregate loss.

Method: Bernoulli trials for frequency, followed by inverse-CDF severity generation. Compare each frequency uniform with $p=0.20$:

$0.08\le0.20$: claim.
$0.44>0.20$: no claim.
$0.19\le0.20$: claim.
$0.91>0.20$: no claim.
$0.03\le0.20$: claim.

Thus $N=3$. For severity, a uniform value at or below $0.75$ selects $\$100$, and a value above $0.75$ selects $\$400$. The three severity uniforms give $0.62\mapsto\$100$, $0.91\mapsto\$400$, and $0.18\mapsto\$100$. Therefore

S_3=\$100+\$400+\$100=\$600.

The simulated count is three and the simulated aggregate is $\$600$. The common mistake is to compare severity uniforms with the frequency probability $0.20$. That probability is used only to decide whether each policy generates a claim; the severity cutoff is $0.75$. As a separate check on the distinction, the expected severity is $0.75(\$100)+0.25(\$400)=\$175$, so the model’s expected aggregate is $5(0.20)(\$175)=\$175$. The realized $\$600$ is one simulated outcome, not that expectation.

Simulating a Negative Binomial Claim Count

“Negative binomial” has more than one convention in common use. Here $N$ is specifically the number of failures before the $r$th success in independent trials, each with success probability $p$. Thus $r$ is a positive integer and $0<p<1$. Under this convention, a success is not itself a claim: the claim count is the number of failures encountered before the stopping success.

Definition: With the failures-before-$r$-successes convention, $$ P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n,\qquad n=0,1,2,\ldots. $$ Here $p$ is the trial success probability and $r$ is the required number of successes. Neither parameter describes claim severity.

There are two useful simulation methods. The sequential method generates one uniform per trial and continues until the $r$th success. The inverse-CDF method first generates one uniform and selects the smallest $n$ whose cumulative probability is at least that uniform. The second method can be convenient when a single uniform is supplied or when a distribution table is available.

For inverse-CDF simulation, the first probability and a recurrence are

P(N=0)=p^r,\qquad P(N=n+1)=P(N=n)\frac{n+r}{n+1}(1-p).

The recurrence follows by dividing the probability at $n+1$ by the probability at $n$: the ratio of the binomial coefficients is $(n+r)/(n+1)$, and the remaining factor is $1-p$. Accumulate the probabilities until the cumulative total reaches the uniform draw. This recurrence computes probabilities for the chosen convention; it does not remove the need to specify that convention.

Worked Example: Sequential Negative Binomial Simulation

A hypothetical portfolio uses a negative binomial frequency model with $r=3$ and $p=0.60$, where $N$ counts failures before the third success. Use trial uniforms $0.72,0.21,0.48,0.83,0.59$. Each claim severity is exponential with mean $\$500$. Use severity uniforms $0.20,0.60$ for the claims that occur.

Method: sequential Bernoulli trials, then inverse-transform severities. A uniform at or below $0.60$ is a success; a value above $0.60$ is a failure. The sequence is $0.72$ failure, $0.21$ success, $0.48$ success, $0.83$ failure, and $0.59$ success. The third success occurs on the fifth trial, after two failures. Therefore $N=2$.

For an exponential severity with mean $500$, the CDF is $F_X(x)=1-e^{-x/500}$ for $x\ge0$. Setting $u=F_X(x)$ and solving for $x$ gives $x=-500\ln(1-u)$. The two severities are

X_1=-500\ln(0.80)=\$111.5718\ldots,\qquad X_2=-500\ln(0.40)=\$458.1454\ldots.

Adding the unrounded values gives $\$569.7172\ldots$, so the simulated aggregate, rounded to the nearest cent, is $\$569.72$. The common mistake is to stop after three trials because $r=3$. The process stops at the third success, not the third trial; here the third success occurs on trial five.

Inverse-CDF Simulation and the Compound-Loss Procedure

For a discrete random variable, inverse-CDF simulation uses the rule $N=\min\{n:F_N(n)\ge u\}$, where $u$ is a uniform draw. This convention assigns a draw exactly on a cumulative-probability boundary to the value at that boundary. Since exact boundary hits have probability zero for an ideal continuous uniform generator, other consistent endpoint conventions do not change the theoretical distribution.

The same logic generates discrete severities: select the smallest severity value whose cumulative probability is at least the severity uniform. For a continuous severity with an invertible CDF, solve $F_X(x)=u$. In every case, use fresh independent uniforms for the frequency and severity stages.

Specify the frequency model.
Record its parameters and convention: for example, $m,p$ for a binomial count, or $r,p$ for failures before $r$ successes.

Generate the count.
Use $m$ Bernoulli trials for a binomial count, or continue negative binomial trials until $r$ successes. In an inverse-CDF method, use the frequency CDF.

Generate the severities.
For each $j=1,\ldots,N$, use an independent uniform and the severity distribution’s inverse CDF or discrete cumulative probabilities.

Sum the claim amounts.
Set $S_N=0$ when $N=0$; otherwise add the $N$ generated severities. Keep the frequency and severity inputs identified separately.

Worked Example: Negative Binomial Inverse CDF

A hypothetical aggregate-loss model has $r=2$, $p=0.50$, and defines $N$ as failures before the second success. Use frequency uniform $u=0.70$. Conditional severities are $\$250$ with probability $0.60$ and $\$1{,}000$ with probability $0.40$. Use severity uniforms $0.11,0.77,0.52$.

Method: negative binomial inverse CDF followed by discrete severity inverse CDF. First calculate the count probabilities using the stated convention:

P(N=0)=0.50^2=0.25,\quad P(N=1)=2(0.50)^2(0.50)=0.25,\quad P(N=2)=3(0.50)^2(0.50)^2=0.1875,\quad P(N=3)=4(0.50)^2(0.50)^3=0.125.

The cumulative probabilities are $F_N(0)=0.25$, $F_N(1)=0.50$, $F_N(2)=0.6875$, and $F_N(3)=0.8125$. Since $0.6875<0.70\le0.8125$, the inverse-CDF count is $N=3$.

The severity cutoff is $0.60$: uniforms $0.11$ and $0.52$ select $\$250$, while $0.77$ selects $\$1{,}000$. Therefore

S_3=\$250+\$1{,}000+\$250=\$1{,}500.

The result is a count of three and aggregate loss of $\$1{,}500$. A common error is to use $0.70$, the frequency uniform, as a severity cutoff. It determines the count only; the severity uniforms must be compared with the severity model’s cumulative probability $0.60$.

Why Frequency and Severity Must Stay Separate

The separation is not merely a bookkeeping preference. Under the independence and identical-distribution assumptions above, let a claim severity have mean $\mu_X$ and variance $\sigma_X^2$, both finite. Conditional on $N=n$, the aggregate is a sum of $n$ independent severities, so its conditional mean is $n\mu_X$ and its conditional variance is $n\sigma_X^2$. Taking expectations and applying the law of total variance gives

E[S_N]=E[N]\mu_X,\qquad \operatorname{Var}(S_N)=E[N]\sigma_X^2+\operatorname{Var}(N)\mu_X^2.

For the binomial model, $E[N]=mp$ and $\operatorname{Var}(N)=mp(1-p)$. For the failures-before-$r$-successes negative binomial convention, $E[N]=r(1-p)/p$ and $\operatorname{Var}(N)=r(1-p)/p^2$. These expressions make the parameter roles explicit: $m,p$ or $r,p$ determine frequency, whereas $\mu_X,\sigma_X^2$ come from severity. The formulas require the stated independence and finite severity moments; dependence between count and severities would require a different calculation.

The negative binomial convention deserves particular care in implementation. Some software defines its variable as the number of trials required to reach $r$ successes, which equals the failures count plus $r$. If a model instead counts total trials, its values and formulas differ. Always inspect what the simulated count represents before treating it as the number of claims.

Key takeaway: Generate the count with frequency parameters, generate exactly that many independent severities with severity parameters, and sum them. For negative binomial simulation, state whether the count is failures or total trials. A correct aggregate simulation depends on all three choices being explicit.

Check Your Understanding

For each problem, apply the stated negative binomial convention where relevant and keep frequency uniforms separate from severity uniforms.

A binomial model has $m=4$ and $p=0.25$. The frequency uniforms are $0.10,0.31,0.24,0.90$. Severities are $\$80$ or $\$200$, each with probability $0.50$, and the two severity uniforms are $0.20,0.80$. Find the simulated count and aggregate. (Answer: $N=2,\ S_N=\$280$)
A negative binomial model counts failures before $r=2$ successes, with $p=0.40$. The trial uniforms are $0.65,0.20,0.91,0.39$, and each claim is a fixed $\$300$. Find the count and aggregate. (Answer: $N=2,\ S_N=\$600$)
For $r=1,p=0.50$, simulate failures before the first success by inverse CDF using $u=0.60$. Each severity is exponential with mean $\$100$; the severity uniform for the resulting claim is $0.50$. Give the count and aggregate to the nearest cent. (Answer: $N=1,\ S_N=\$69.31$)
A binomial model has $m=6,p=0.10$, with frequency uniforms $0.03,0.11,0.07,0.90,0.09,0.40$. Severity is $\$50$ with probability $0.70$ and $\$150$ with probability $0.30$; the severity uniforms are $0.20,0.80,0.60$. Find the count and aggregate. (Answer: $N=3,\ S_N=\$250$)
A binomial model has $m=3,p=0.40$, with frequency uniforms $0.20,0.70,0.40$. Severity is $\$100$ with probability $0.25$ and $\$300$ with probability $0.75$; severity uniforms are $0.10,0.90$. Find the realized aggregate and the model’s expected aggregate. (Answer: realized $S_N=\$400$; $E[S_N]=\$300$)

What You'll Learn