Worked Chain: Negative Binomial Frequency With Coinsurance

Building the Frequency-Severity Chain

For a compound loss model, the aggregate payment depends on two separate sources of uncertainty: how many claims occur and how much is paid on each claim. Coinsurance changes the payment amount for each claim, but it does not change the claim count. Keeping those parts distinct is essential when calculating the aggregate variance.

This tutorial uses a negative binomial model for claim frequency, a gamma model for gross claim severity, and a fixed coinsurance percentage applied to every claim. All scenarios and values below are hypothetical. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Model and notation: Let $N$ be the number of claims, and let $X_1,X_2,\ldots$ be independent, identically distributed gross claim amounts, independent of $N$. Under coinsurance rate $c$, where $0<c\leq 1$, the payment on a claim is $Y=cX$. The aggregate payment is $S=\sum_{j=1}^{N}Y_j$, with $S=0$ when $N=0$.

Use the following negative binomial convention throughout: $N$ is the number of failures before the $r$th success in independent trials, each with success probability $p$. Thus $0<p<1$, $r>0$ (an integer in this failures-before-$r$-successes construction), and

P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n,\qquad n=0,1,2,\ldots,

with

E[N]=\frac{r(1-p)}{p}, \qquad \operatorname{Var}(N)=\frac{r(1-p)}{p^2}.

Some sources use a different negative binomial convention or parameter meaning. Therefore, the probability model and the meaning of $p$ must be checked before applying these formulas.

Gamma severity convention: Write $X\sim\operatorname{Gamma}(\alpha,\theta)$, where $\alpha>0$ is the shape and $\theta>0$ is the scale. Then $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$. Under fixed coinsurance $c$, $Y=cX$, so $E[Y]=c\alpha\theta$ and $\operatorname{Var}(Y)=c^2\alpha\theta^2$.

Payment Moments Under Coinsurance

Coinsurance is a multiplicative coverage modification: the insurer pays the fraction $c$ of each gross loss. For a random variable $X$ with finite second moment, multiplying by a constant gives $E[cX]=cE[X]$ and $\operatorname{Var}(cX)=c^2\operatorname{Var}(X)$. The square in the variance formula is important. The second raw moment of the payment is

E[Y^2]=\operatorname{Var}(Y)+(E[Y])^2 =c^2\left(\operatorname{Var}(X)+(E[X])^2\right) =c^2E[X^2].

For gamma severity, $E[X^2]=\alpha\theta^2+(\alpha\theta)^2=\alpha(\alpha+1)\theta^2$. Therefore $E[Y^2]=c^2\alpha(\alpha+1)\theta^2$. These moments describe the payment on one claim; they are not yet the aggregate moments.

Worked Example: Coinsurance and Negative Binomial Aggregate Moments

A hypothetical insurer has claim count $N$ equal to the number of failures before the eighth success, with $p=0.5$. Gross claim amounts are independent with $X\sim\operatorname{Gamma}(2,250)$, using the shape-scale convention, and the insurer pays 80% of each claim. Find the payment mean and variance, the aggregate mean and variance, and a normal approximation to $P(S>5000)$. Assume claim amounts are independent of the claim count.

Method. First calculate the one-claim payment moments. Then combine them with the frequency moments. Finally, standardize the threshold using a normal distribution with the calculated aggregate mean and variance.

1. Frequency moments. Here $r=8$ and $p=0.5$, so

E[N]=\frac{8(1-0.5)}{0.5}=8, \qquad \operatorname{Var}(N)=\frac{8(1-0.5)}{(0.5)^2}=16.

2. Payment moments. The gross gamma mean is $E[X]=2(250)=500$, and its variance is $\operatorname{Var}(X)=2(250)^2=125{,}000$. With $c=0.8$,

E[Y]=0.8(500)=400, \qquad \operatorname{Var}(Y)=(0.8)^2(125{,}000)=80{,}000.

The payment second moment, needed as a check, is $E[Y^2]=80{,}000+400^2=240{,}000$.

3. Aggregate moments. The aggregate mean is $E[S]=E[N]E[Y]=8(400)=3{,}200$. The aggregate variance is

\operatorname{Var}(S) =8(80{,}000)+16(400)^2 =640{,}000+2{,}560{,}000 =3{,}200{,}000.

4. Approximate probability. Use $S$ approximately normal with mean $3{,}200$ and variance $3{,}200{,}000$. Its standard deviation is $\sqrt{3{,}200{,}000}\approx1{,}788.854$. Hence

z=\frac{5{,}000-3{,}200}{1{,}788.854}\approx1.0062, \qquad P(S>5{,}000)\approx1-\Phi(1.0062)=0.1571.

The result is an approximation, not an exact compound-distribution probability. The check $E[Y^2]=240{,}000$ also gives $\operatorname{Var}(S)=8(240{,}000)+(16-8)(400)^2=3{,}200{,}000$, agreeing with the calculation above.

Why the Aggregate Variance Has Two Terms

Conditional on $N=n$, the aggregate is a sum of $n$ independent payment amounts. Therefore $E[S\mid N]=NE[Y]$ and $\operatorname{Var}(S\mid N)=N\operatorname{Var}(Y)$. Applying the law of total variance separates variation within a fixed claim count from variation in the claim count itself.

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N])\\ &=E[N]\operatorname{Var}(Y) +\operatorname{Var}(N)(E[Y])^2. \end{aligned}

The first term reflects random payment sizes given the number of claims. The second reflects the variability in how many payments occur. In particular, it is not enough to multiply the payment variance by the expected claim count; that would omit the second source of uncertainty.

Aggregate-moment result: For independent claim count $N$ and independent, identically distributed payments $Y_j$, with finite second moments, $$ E[S]=E[N]E[Y], \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2. $$ For the negative binomial convention used here, substitute $E[N]=r(1-p)/p$ and $\operatorname{Var}(N)=r(1-p)/p^2$.

An equivalent form is $\operatorname{Var}(S)=E[N]E[Y^2]+(\operatorname{Var}(N)-E[N])(E[Y])^2$. This follows by replacing $\operatorname{Var}(Y)$ with $E[Y^2]-(E[Y])^2$. Either form is valid; the first often makes the two sources of variation easier to interpret.

Further Worked Chains

Worked Example: A Different Negative Binomial Parameter

A hypothetical benefit program has claim count equal to the number of failures before the fifth success, with success probability $p=0.4$. Gross claims follow $X\sim\operatorname{Gamma}(3,100)$, and the insurer pays 60% of each claim. Find the aggregate mean and variance, then approximate $P(S>2500)$. Assume independence between claim count and severities.

Frequency. With $r=5$ and $p=0.4$,

E[N]=\frac{5(0.6)}{0.4}=7.5, \qquad \operatorname{Var}(N)=\frac{5(0.6)}{(0.4)^2}=18.75.

Payment. The gross mean is $3(100)=300$ and the gross variance is $3(100)^2=30{,}000$. Thus $E[Y]=0.6(300)=180$, $\operatorname{Var}(Y)=(0.6)^2(30{,}000)=10{,}800$, and $(E[Y])^2=32{,}400$.

Aggregate moments.

$$ E[S]=7.5(180)=1{,}350, $$

and

\operatorname{Var}(S) =7.5(10{,}800)+18.75(32{,}400) =81{,}000+607{,}500 =688{,}500.

The standard deviation is $\sqrt{688{,}500}\approx829.759$. For the requested upper tail,

z=\frac{2{,}500-1{,}350}{829.759}\approx1.3859, \qquad P(S>2{,}500)\approx1-\Phi(1.3859)=0.0829.

The parameter check is important: $p=0.4$ is the success probability in the stated convention, so the expected number of failures is $7.5$, not $rp=2$. The aggregate variance includes both $81{,}000$ from payment-size variation and $607{,}500$ from frequency variation.

Worked Example: Checking a Lower-Tail Approximation

A hypothetical insurer counts claims as failures before the fourth success, with $p=0.25$. Claim amounts are independent with $X\sim\operatorname{Gamma}(2,150)$, and the insurer pays half of each claim. Find the aggregate moments and approximate $P(S\leq 3000)$.

Frequency. Here $r=4$, so $E[N]=4(0.75)/0.25=12$ and $\operatorname{Var}(N)=4(0.75)/(0.25)^2=48$.

Payment. The gross mean is $2(150)=300$, and the gross variance is $2(150)^2=45{,}000$. Therefore $E[Y]=0.5(300)=150$ and $\operatorname{Var}(Y)=(0.5)^2(45{,}000)=11{,}250$. The squared payment mean is $150^2=22{,}500$.

Aggregate moments.

$$ E[S]=12(150)=1{,}800, $$

\operatorname{Var}(S) =12(11{,}250)+48(22{,}500) =135{,}000+1{,}080{,}000 =1{,}215{,}000.

Thus the standard deviation is $\sqrt{1{,}215{,}000}\approx1{,}102.270$. Standardizing the lower-tail threshold gives

z=\frac{3{,}000-1{,}800}{1{,}102.270}\approx1.0887, \qquad P(S\leq3{,}000)\approx\Phi(1.0887)=0.8618.

As a check, the mean and variance are positive, and the threshold is above the mean, so a probability above one-half is consistent with the fitted normal approximation.

Normal Approximation and Its Limits

Once the aggregate mean $\mu_S$ and standard deviation $\sigma_S$ are calculated, the normal approximation replaces $S$ by a normal random variable with the same mean and variance. For a threshold $x$, the calculation is

P(S>x)\approx1-\Phi\left(\frac{x-\mu_S}{\sigma_S}\right), \qquad P(S\leq x)\approx\Phi\left(\frac{x-\mu_S}{\sigma_S}\right).

Here $\Phi$ is the standard normal cumulative distribution function. This approximation can be useful when direct compound-distribution calculations are impractical, but matching the first two moments does not make the aggregate distribution exactly normal. A negative binomial count can produce substantial frequency variation, and gamma payments are right-skewed. In particular, the true aggregate has a probability mass at zero equal to $P(N=0)=p^r$, whereas a normal distribution has no point mass. Consider whether that feature or strong skewness could materially affect the requested probability.

For continuous claim amounts and a positive threshold, there is generally no need for a discrete continuity correction: the positive aggregate has a continuous distribution, although it has an atom at zero. A normal approximation is not a substitute for an exact or recursive method when high precision is needed or the threshold lies in a region where the approximation is poor.

Common Exam Trap: Scaling Variance Incorrectly

A frequent error is to multiply the gross variance by the coinsurance percentage instead of by its square. In the first worked example, this incorrect calculation would give

\operatorname{Var}(Y)\ \text{(incorrect)} =0.8(125{,}000)=100{,}000.

It would then produce the incorrect aggregate variance $8(100{,}000)+16(400)^2=3{,}360{,}000$. Since $Y=0.8X$, variance scales by the square of the constant:

\operatorname{Var}(Y)=0.8^2(125{,}000)=80{,}000, \qquad \operatorname{Var}(S)=8(80{,}000)+16(400)^2=3{,}200{,}000.

A separate trap is to use a negative binomial formula without checking its convention. For failures before the $r$th success, the mean is $r(1-p)/p$, not $rp$. Another is to omit the frequency-variance contribution $\operatorname{Var}(N)(E[Y])^2$. Both errors can substantially distort the aggregate result.

Key takeaway: Calculate frequency moments from the explicitly stated negative binomial convention, scale severity variance by $c^2$, and combine the two sources of aggregate variation. Only then standardize a threshold for a normal approximation.

Check Your Understanding

For each question, assume claim count is independent of the claim severities and use the failures-before-$r$-successes convention for the negative binomial count.

A hypothetical insurer has $r=3$, $p=0.5$, gross severity $X\sim\operatorname{Gamma}(2,50)$, and coinsurance $c=0.8$. Find the mean and variance of one payment. (Answer: $E[Y]=80$, $\operatorname{Var}(Y)=3{,}200$.)
Let $r=6$, $p=0.4$, $X\sim\operatorname{Gamma}(2,100)$, and $c=0.5$. Find the aggregate mean and variance. (Answer: $E[S]=900$, $\operatorname{Var}(S)=270{,}000$.)
Let $r=4$, $p=0.5$, $X\sim\operatorname{Gamma}(1,200)$, and $c=0.75$. Find the aggregate mean and variance. (Answer: $E[S]=600$, $\operatorname{Var}(S)=270{,}000$.)
Let $r=5$, $p=0.5$, $X\sim\operatorname{Gamma}(2,100)$, and $c=0.5$. Use a normal approximation to find $P(S>1000)$. (Answer: $0.0787$.)
Let $r=3$, $p=0.25$, $X\sim\operatorname{Gamma}(2,50)$, and $c=0.8$. Find the aggregate moments and use a normal approximation for $P(S\leq1000)$. (Answer: $E[S]=720$, $\operatorname{Var}(S)=259{,}200$, and $P(S\leq1000)\approx0.7088$.)

What You'll Learn