Claim Counts and Severity via Conditioning

Why a Random Claim Count Changes Aggregate Variability

An insurer’s total claim amount over a period depends on two sources of uncertainty: how many claims occur and how large those claims are. Even if the average claim size is known, the total can vary because the count varies. Conversely, even with a fixed number of claims, the total can vary because individual claim amounts differ.

A useful model separates these two sources. Let $N$ be the number of claims in the period, let $X_j$ be the amount of the $j$th claim, and let $S_N$ be the aggregate claim amount. The claim amounts might represent hypothetical payments under a specified policy; the calculations below do not assume data from any actual insurer.

Definition: The aggregate claim amount is the random sum $S_N=\sum_{j=1}^{N}X_j$. When $N=0$, the sum is empty and $S_N=0$. In the standard iid severity model, the severities $X_1,X_2,\ldots$ have a common distribution, are independent of one another, and are independent of $N$. Write $\mu=E[X_1]$ and $\sigma^2=\operatorname{Var}(X_1)$.

Assume that $N$ is a nonnegative integer-valued random variable, and that $E[N]$ and $\operatorname{Var}(N)$ are finite. Also assume $E[X_1]$ and $\operatorname{Var}(X_1)$ are finite. Under these conditions, conditioning on the claim count gives the aggregate moments.

Deriving the Mean and Variance

If the claim count is known to be $n$, then the aggregate amount is the sum of exactly $n$ severities. Independence and identical distribution imply that its conditional mean is $n\mu$, while its conditional variance is $n\sigma^2$. For $n=0$, both quantities are zero, consistent with the empty-sum definition.

E[S_N\mid N]=N\mu, \qquad \operatorname{Var}(S_N\mid N)=N\sigma^2.

The law of total expectation gives the mean. For the variance, the law of total variance says that the unconditional variance is the expected conditional variance plus the variance of the conditional mean. The first term captures variation in severities when the count is fixed; the second captures variation in the count itself.

Condition on the count.
For a fixed count $N=n$, add the means of $n$ severities to obtain $n\mu$. Independence makes the variance of their sum $n\sigma^2$.

Average the conditional mean.
Apply total expectation: $E[S_N]=E[E[S_N\mid N]]=E[N\mu]=E[N]\mu$.

Combine the two variance components.
Apply total variance: $\operatorname{Var}(S_N)=E[N\sigma^2]+\operatorname{Var}(N\mu)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2$.

Theorem: For a random claim count $N$ independent of iid severities with mean $\mu$ and variance $\sigma^2$, $$ E[S_N]=E[N]\mu, \qquad \operatorname{Var}(S_N)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2. $$ Both terms in the variance are needed in general. The result requires the stated independence and finite-moment assumptions.

The variance formula can also be written using the severity second moment. Since $E[X_1^2]=\sigma^2+\mu^2$, algebra gives

\operatorname{Var}(S_N) =E[N]E[X_1^2]+\bigl(\operatorname{Var}(N)-E[N]\bigr)\mu^2.

This form is sometimes convenient when a problem supplies $E[X_1^2]$ rather than the severity variance. For a Poisson count with mean $\lambda$, $\operatorname{Var}(N)=E[N]=\lambda$, so it simplifies to $\operatorname{Var}(S_N)=\lambda E[X_1^2]$. This simplification depends on the Poisson count model; it is not the general formula.

Worked Applications

Worked Example: A Binomial Claim Count and Two-Point Severity

For a hypothetical portfolio, suppose there are four independent exposure units, each of which produces one claim during the year with probability $0.20$. Thus $N\sim\operatorname{Binomial}(4,0.20)$. Assume the claim payments are independent of the count and of one another. Each payment is $1,000 or $2,000 with probability $0.50$ each. Find the mean and variance of the annual aggregate payment.

Method. Calculate the count and severity moments separately, then substitute them into the random-sum formulas.

For the binomial count,

E[N]=4(0.20)=0.80, \qquad \operatorname{Var}(N)=4(0.20)(0.80)=0.64.

For one claim payment,

\mu=0.50(1000)+0.50(2000)=1500,

and

E[X_1^2]=0.50(1000^2)+0.50(2000^2) =500{,}000+2{,}000{,}000=2{,}500{,}000.

Therefore $\sigma^2=E[X_1^2]-\mu^2=2{,}500{,}000-2{,}250{,}000=250{,}000$. The aggregate mean is

$$ E[S_N]=0.80(1500)=1200. $$

The aggregate variance is

\operatorname{Var}(S_N) =0.80(250{,}000)+0.64(1500^2) =200{,}000+1{,}440{,}000 =1{,}640{,}000.

Check. The mean is below one average claim payment because the expected number of claims is only $0.80$. The variance includes both the conditional severity component, $200{,}000$, and the claim-count component, $1{,}440{,}000$. Omitting the latter would substantially understate the aggregate variance.

Worked Example: Poisson Frequency and Exponential Severity

Suppose the number of claims in a hypothetical six-month period has distribution $N\sim\operatorname{Poisson}(3)$. Conditional on a claim occurring, its payment is exponentially distributed with mean $400, independently across claims and independently of $N$. Find the expected aggregate payment and its variance.

Method. Use the Poisson mean and variance, together with the mean and variance of the exponential severity. The exponential parameterization here is by mean: the rate is $1/400$ per dollar.

For $N\sim\operatorname{Poisson}(3)$, $E[N]=3$ and $\operatorname{Var}(N)=3$. For an exponential random variable with mean $400, $\mu=400$ and $\sigma^2=400^2=160{,}000$. Hence

$$ E[S_N]=3(400)=1200, $$

and

\operatorname{Var}(S_N) =3(160{,}000)+3(400^2) =480{,}000+480{,}000 =960{,}000.

Check. The exponential second moment is $E[X_1^2]=\sigma^2+\mu^2=160{,}000+160{,}000=320{,}000$. The Poisson special case gives $\operatorname{Var}(S_N)=3(320{,}000)=960{,}000$, agreeing with the calculation. This check relies on the assumed Poisson count and the independence of count and severity.

Worked Example: A Discrete Claim Count With Unequal Severity Probabilities

In a hypothetical coverage pool, the claim count is $0$, $1$, or $3$, with respective probabilities $0.20$, $0.50$, and $0.30$. Claim payments are iid, independent of the count, and equal to $500, $1,000, or $2,500 with probabilities $0.50$, $0.30$, and $0.20$, respectively. Find the aggregate mean and variance.

Method. Compute the first two moments of the count directly from its probability distribution, and compute the severity variance from its second moment.

The count mean is

$$ E[N]=0(0.20)+1(0.50)+3(0.30)=0.50+0.90=1.40. $$

Its second moment is

$$ E[N^2]=0^2(0.20)+1^2(0.50)+3^2(0.30) =0.50+2.70=3.20, $$

so $\operatorname{Var}(N)=3.20-(1.40)^2=3.20-1.96=1.24$.

For a single payment,

\mu=500(0.50)+1000(0.30)+2500(0.20) =250+300+500=1050,

and

E[X_1^2]=500^2(0.50)+1000^2(0.30)+2500^2(0.20) =125{,}000+300{,}000+1{,}250{,}000=1{,}675{,}000.

Thus $\sigma^2=1{,}675{,}000-1050^2=1{,}675{,}000-1{,}102{,}500=572{,}500$. The aggregate mean is

$$ E[S_N]=1.40(1050)=1470. $$

The aggregate variance is

\begin{aligned} \operatorname{Var}(S_N) &=1.40(572{,}500)+1.24(1{,}102{,}500)\\ &=801{,}500+1{,}367{,}100 =2{,}168{,}600. \end{aligned}

Check. The second-moment form gives $E[S_N^2]=1.40(1{,}675{,}000)+(3.20-1.40)(1{,}102{,}500)=2{,}345{,}000+1{,}984{,}500=4{,}329{,}500$. Subtracting $E[S_N]^2=1470^2=2{,}160{,}900$ gives $4{,}329{,}500-2{,}160{,}900=2{,}168{,}600$, as required.

Common Mistake: Treating the Count as Fixed

A frequent error is to use only $E[N]\sigma^2$ for the aggregate variance. That is the expected variance of the aggregate conditional on the count, not its full unconditional variance. The missing term is $\operatorname{Var}(N)\mu^2$, which measures the additional fluctuation caused by having more or fewer claims than expected.

Exam trap: In the first worked example, the incomplete calculation $\operatorname{Var}(S_N)=0.80(250{,}000)=200{,}000$ leaves out claim-count variation. The correction is $200{,}000+0.64(1500^2)=1{,}640{,}000$. Do not confuse conditional variance given $N$ with unconditional variance.

Another assumption to check is independence. If claim severities depend on the count, or if severities are dependent across claims, the conditional mean or variance need not be $N\mu$ or $N\sigma^2$. In such a model, conditioning is still a useful method, but the conditional moments must be derived from the specified dependence structure rather than copied from the iid formula.

For a standard iid random-sum problem, a reliable sequence is: identify the count distribution; find $E[N]$ and $\operatorname{Var}(N)$; find the severity mean and variance; verify independence; then calculate the aggregate moments. This method is aligned with the SOA Exam P (Probability) syllabus.

Check Your Understanding

For each problem, assume severities are independent of the count and iid unless stated otherwise.

A count has distribution $N\sim\operatorname{Binomial}(5,0.10)$. Each claim payment is a fixed $2,000. Find the aggregate mean and variance. (Answer: $E[S_N]=\$1{,}000$, $\operatorname{Var}(S_N)=\$1{,}800{,}000$)
A claim count is Poisson with mean $4$. Each severity is $100 or $300 with equal probability. Find the aggregate mean and variance. (Answer: $E[S_N]=\$800$, $\operatorname{Var}(S_N)=\$200{,}000$)
The count is $0$, $2$, or $4$ with probabilities $0.25$, $0.50$, or $0.25$. Each severity is $1$, $3$, or $5$ units with equal probability. Find the aggregate mean and variance in squared units. (Answer: $E[S_N]=6$ units, $\operatorname{Var}(S_N)=70/3$ squared units)
A count is $1$ with probability $0.40$ and $3$ with probability $0.60$. Each severity is $500 with probability $0.75$ and $2,000 with probability $0.25$. Find the aggregate mean and variance. (Answer: $E[S_N]=\$1{,}925$, $\operatorname{Var}(S_N)=\$1{,}663{,}125$)

What You'll Learn