Deriving Aggregate Means and Variances

Why Aggregate Moments Need Both Frequency and Severity

An aggregate loss is the total paid across a random number of claims. Its expected value depends on both how many claims occur, on average, and how much is paid per claim, on average. Its variance also reflects two sources of uncertainty: variation in individual payments and variation in the number of claims. Coverage terms such as deductibles, limits, coinsurance, and reinsurance change the payment distribution, so the relevant severity moments are those of the payment after the contract terms have been applied.

The calculations below use a random-sum model. The claim count is independent of the claim payments, and the payments are independent and identically distributed. A payment may be zero; this is often the simplest way to represent a claim that produces no payment under the contract. If claim count and severity are dependent, or if payment amounts are not identically distributed, the formulas in this tutorial do not automatically apply.

Definition: Let $N$ be the number of claims and let $Y_1,Y_2,\ldots$ be independent, identically distributed payments per claim, independent of $N$. Define the aggregate payment by $S=\sum_{j=1}^{N}Y_j$, with $S=0$ when $N=0$. Write $m_1=E[Y]$ and $m_2=E[Y^2]$, where $Y$ has the common distribution of the $Y_j$. Assume the moments used below are finite.

The Mean and Variance Formulas

For any frequency distribution with finite mean and variance, the aggregate moments are

$$ E[S]=E[N]m_1 $$

and

\operatorname{Var}(S)=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)m_1^2 =E[N]m_2+\bigl(\operatorname{Var}(N)-E[N]\bigr)m_1^2.

The first variance form separates the two sources of variability. The term $E[N]\operatorname{Var}(Y)$ is the contribution from variation in payment size; $\operatorname{Var}(N)m_1^2$ is the contribution from variation in claim count. The second form is convenient when the severity calculation gives $m_1$ and $m_2$ directly. The equality follows from $\operatorname{Var}(Y)=m_2-m_1^2$.

Condition on the count.
For a fixed count $N=n$, the aggregate is the sum of $n$ independent payments. Its mean is $nm_1$, and its variance is $n\operatorname{Var}(Y)$.

Average the conditional mean.
The law of total expectation gives $E[S]=E[E[S\mid N]]=E[Nm_1]=E[N]m_1$.

Combine the two variance sources.
The law of total variance gives $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. Substituting the fixed-count variance and mean yields $E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)m_1^2$.

This derivation shows why the frequency variance cannot generally be ignored. Two portfolios can have the same expected claim count and the same payment distribution, yet have different aggregate variances if their claim-count variances differ.

Useful special case: If $N\sim\operatorname{Poisson}(\lambda)$, then $E[N]=\operatorname{Var}(N)=\lambda$, so $\operatorname{Var}(S)=\lambda m_2$. This simplification depends on Poisson frequency. It is not a general compound-model variance formula.

Find Payment Moments After Applying the Contract

Let $X$ denote the loss on a claim and let $Y$ be the insurer’s payment under the specified contract. First translate the contract into a payment map $Y=g(X)$. Then calculate $m_1=E[g(X)]$ and $m_2=E[g(X)^2]$ using the distribution of $X$. In particular, a deductible can create a probability mass at zero, and that mass contributes zero to both payment moments.

For example, with an ordinary deductible $d$ and no upper limit, the payment per loss is $Y=(X-d)_+=\max(X-d,0)$. If a limit $u$ caps payment above the deductible, the payment map is $Y=\min((X-d)_+,u)$. If a coinsurance percentage $c$ applies to that covered amount, the payment becomes $Y=c\min((X-d)_+,u)$. The order and meaning of contract terms must be read from the stated payment map; do not assume that a limit applies to the same quantity as the deductible.

Worked Example: Poisson Frequency With a Deductible

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$ claims in a year. Claim losses are independent of $N$, and each loss is $100 with probability 0.75 or $300 with probability 0.25. The policy has a $100 ordinary deductible and no limit. Find the mean and variance of annual insurer payments.

Method. Transform each loss to the insurer’s payment, find its first two moments, and use the Poisson aggregate formulas. This avoids treating the original loss as though it were the payment.

For loss $X=$100$, the payment is $(100-100)_+=$0$. For loss $X=$300$, it is $(300-100)_+=$200$. Thus $Y$ is $0 with probability 0.75 and $200 with probability 0.25.

The payment moments are

m_1=E[Y]=0(0.75)+200(0.25)=$50, \qquad m_2=E[Y^2]=0^2(0.75)+200^2(0.25)=$10{,}000.

The annual mean is $E[S]=4(50)=$200$. Since the frequency is Poisson with parameter 4, its variance is also 4, and the aggregate variance is

\operatorname{Var}(S)=4(10{,}000)=$40{,}000.

As a check, $\operatorname{Var}(Y)=10{,}000-50^2=7{,}500$, so the general formula gives $4(7{,}500)+4(50^2)=30{,}000+10{,}000=$40{,}000$, the same result. The variance is in dollars squared; the standard deviation, if needed, is $\sqrt{40{,}000}=$200$.

Frequency Variability Changes the Aggregate Variance

The mean aggregate payment is the product of the mean count and mean payment under the stated independence assumptions. The variance is more sensitive to the frequency model. For a fixed claim count, $\operatorname{Var}(N)=0$, so the count contributes no additional variability. For a frequency distribution with variance greater than its mean, the count-variation term can make the aggregate variance substantially larger than the Poisson case with the same mean.

Worked Example: A Limit, Coinsurance, and Binomial Frequency

A hypothetical insurer has $N\sim\operatorname{Binomial}(5,0.4)$ claims. Each loss, independently of the count and other losses, is $100 with probability 0.5 or $500 with probability 0.5. The policy applies a $100 deductible, caps the amount above the deductible at $200, and pays 60% of that capped amount. Find the annual mean and variance of insurer payments.

Method. Apply the contract terms in their stated order to obtain the payment per loss. Then use the binomial mean and variance, not the Poisson special case.

For a $100 loss, the amount above the deductible is $0, so the insurer pays $0. For a $500 loss, the amount above the deductible is $400, the cap reduces it to $200, and 60% coinsurance gives a payment of $0.60(200)=$120$. Thus $Y$ equals $0 or $120, each with probability 0.5.

Therefore,

m_1=0(0.5)+120(0.5)=$60, \qquad m_2=0^2(0.5)+120^2(0.5)=$7{,}200.

For $N\sim\operatorname{Binomial}(5,0.4)$, $E[N]=5(0.4)=2$ and $\operatorname{Var}(N)=5(0.4)(0.6)=1.2$. The aggregate mean is $2(60)=$120$. Also, $\operatorname{Var}(Y)=7{,}200-60^2=3{,}600$, so

\operatorname{Var}(S)=2(3{,}600)+1.2(60^2) =7{,}200+4{,}320 =$11{,}520.

As a check using the second-moment form, $2(7{,}200)+(1.2-2)(60^2)=14{,}400-2{,}880=$11{,}520$. The common error is to use $\operatorname{Var}(S)=E[N]m_2$, which would give $14,400. That shortcut is valid for Poisson frequency, not for this binomial model.

Worked Example: Negative Binomial Frequency With a Per-Claim Limit

A hypothetical portfolio has a negative binomial claim count $N$, defined as the number of failures before $r=2$ successes in independent Bernoulli trials with success probability $p=0.5$. Thus $E[N]=r(1-p)/p=2$ and $\operatorname{Var}(N)=r(1-p)/p^2=4$. Claim losses are independent of $N$; each is $200 or $400 with probability 0.5 each. The insurer pays the loss above a $100 deductible, subject to a maximum payment of $150 per claim. Find the aggregate mean and variance.

Method. Calculate the payment outcomes and moments, then use the general random-sum formula with the negative binomial frequency moments. The cap is on the insurer’s payment, not on the original loss.

For $X=$200$, the amount above the deductible is $100, below the $150 maximum, so $Y=$100$. For $X=$400$, the amount above the deductible is $300, capped at $150, so $Y=$150$. Each payment occurs with probability 0.5. Hence

m_1=100(0.5)+150(0.5)=$125, \qquad m_2=100^2(0.5)+150^2(0.5)=$16{,}250.

The aggregate mean is $E[S]=2(125)=$250$. The payment variance is $16{,}250-125^2=16{,}250-15{,}625=625$. Therefore,

\operatorname{Var}(S)=2(625)+4(125^2) =1{,}250+62{,}500 =$63{,}750.

Using the second-moment form gives $2(16{,}250)+(4-2)(15{,}625)=32{,}500+31{,}250=$63{,}750$, confirming the result. Substituting the Poisson shortcut would instead produce $2(16{,}250)=$32{,}500$, which understates the variance because this frequency has variance 4, greater than its mean of 2.

A Reliable Calculation Sequence

A consistent sequence helps prevent errors in multi-part problems. The frequency model determines $E[N]$ and $\operatorname{Var}(N)$; the contract and loss distribution determine $m_1$ and $m_2$. Keep those calculations separate until the final substitution.

Write the aggregate as $S=\sum_{j=1}^{N}Y_j$ and state that frequency and payments are independent, with payments identically distributed.
Translate all coverage terms into one payment per loss $Y=g(X)$, preserving the stated order of deductible, limit, coinsurance, and reinsurance operations.
List the possible payment values and probabilities, including any zero-payment outcome. Calculate $m_1=E[Y]$ and $m_2=E[Y^2]$.
Calculate $E[N]$ and $\operatorname{Var}(N)$ from the specified frequency model.
Substitute into $E[S]=E[N]m_1$ and $\operatorname{Var}(S)=E[N]m_2+(\operatorname{Var}(N)-E[N])m_1^2$. Check the variance using the equivalent form when practical.

When a contract includes reinsurance, define whose payment is being measured before calculating moments. A ceded payment and a retained payment generally have different payment distributions and therefore different $m_1$ and $m_2$. Apply the same aggregate formulas to either quantity only after identifying its per-claim payment map.

Key takeaway: Aggregate mean uses frequency mean times payment mean. Aggregate variance requires both payment moments and the frequency variance. Apply coverage modifications to each loss first, and use the Poisson shortcut only when the claim count is Poisson. These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the payment distribution after the stated contract terms and the given frequency moments.

A hypothetical insurer has Poisson claim frequency with parameter 3. Each claim produces a payment of $100 with probability 0.4 and $0 otherwise, independently of frequency. Find the aggregate mean and variance. (Answer: mean $120; variance $12,000.)
A portfolio has $N\sim\operatorname{Binomial}(8,0.25)$. Each claim payment is $0 or $200, each with probability 0.5, independently. Find the aggregate mean and variance. (Answer: mean $200; variance $35,000.)
A frequency model has $E[N]=5$ and $\operatorname{Var}(N)=8$. Payments are $0, $50, or $150 with probabilities 0.2, 0.5, or 0.3. Find $m_1$, $m_2$, and the aggregate mean and variance. (Answer: $m_1=$70$, $m_2=$8{,}000$; mean $350; variance $54,700.)
A hypothetical portfolio has Poisson frequency with parameter 4. Losses are $100, $250, or $500 with probabilities 0.2, 0.5, or 0.3. The insurer pays the amount above a $100 deductible, subject to a $250 per-claim maximum. Find the aggregate mean and variance. (Answer: mean $600; variance $120,000.)

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