Plug-In Aggregate Loss Moments

Why Aggregate-Loss Moments Need Both Models

An aggregate loss over a fixed period depends on two sources of variation: how many payments occur and how large those payments are. A frequency model describes the claim count; a severity model describes the payment amount for a claim. Combining fitted moments from both models gives a quick summary of aggregate risk without requiring the full aggregate-loss distribution.

The word plug-in means that estimated model parameters, and the moments calculated from them, are substituted into the aggregate formulas as if they were fixed. The resulting mean and variance describe process variation under the fitted models. They do not, by themselves, add uncertainty about the estimated parameters.

Definition: Let $N$ be the number of covered payments during the period, and let $Y_i^P$ be the amount of the $i$th payment. The aggregate payment is $$ S=\sum_{i=1}^{N}Y_i^P, $$ where the sum is defined to be zero when $N=0$. In the standard compound model, the payment amounts are mutually independent and identically distributed, and the entire sequence of payment amounts is independent of $N$. Write $m=E[Y^P]$ and $\sigma^2=\operatorname{Var}(Y^P)$, where $Y^P$ has the common payment distribution.

The mutual-independence condition matters: identical marginal distributions alone do not imply that the variance of a sum is the number of terms times the individual variance. For example, if every payment in a period were the same random amount, payments would be perfectly dependent. The standard compound-model variance formula would then not apply.

Use payment severity after any applicable coverage terms have been applied. A deductible, limit, or coinsurance arrangement can change the distribution—and therefore the moments—of $Y^P$. The frequency moments must refer to the count of payments represented by that payment model.

The Plug-In Moment Formulas

Suppose the fitted frequency model gives $E[N]$ and $\operatorname{Var}(N)$, while the fitted payment model gives $m$ and $\sigma^2$. Under the standard compound-model assumptions, the aggregate moments are:

Theorem: For mutually independent, identically distributed payment amounts that are independent of $N$, $$ E[S]=E[N]m \qquad\text{and}\qquad \operatorname{Var}(S)=E[N]\sigma^2+\operatorname{Var}(N)m^2. $$

The first term in the variance, $E[N]\sigma^2$, is the contribution from variation in payment sizes when the expected number of payments is taken into account. The second term, $\operatorname{Var}(N)m^2$, is the contribution from variation in the count. In particular, frequency variability remains important even if the payment amounts themselves are constant.

Why the Variance Formula Holds

Condition on the claim count. If $N=n$, the aggregate is a sum of $n$ mutually independent payments. Its conditional mean is $nm$, and its conditional variance is $n\sigma^2$. The variance decomposition then combines the average conditional variance with the variance of the conditional mean.

E[S\mid N=n]=nm,\qquad \operatorname{Var}(S\mid N=n)=n\sigma^2

\begin{aligned} E[S] &=E\!\left[E[S\mid N]\right] =E[Nm] =E[N]m,\\ \operatorname{Var}(S) &=E\!\left[\operatorname{Var}(S\mid N)\right] +\operatorname{Var}\!\left(E[S\mid N]\right)\\ &=E[N\sigma^2]+\operatorname{Var}(Nm)\\ &=E[N]\sigma^2+\operatorname{Var}(N)m^2. \end{aligned}

This derivation uses both parts of the payment assumption: given the count, the payments are mutually independent with the same variance, and their distribution does not change with the count. If payment amounts are dependent, the conditional variance generally includes covariance terms and need not equal $n\sigma^2$.

For plug-in work, replace the moments in the formula with fitted moments. For example, if the estimated frequency mean is $\widehat{E[N]}$, use that estimate in both the aggregate mean and the severity-variance contribution. The symbol $\widehat{\ }$ can help distinguish estimated inputs from population quantities; it does not change the algebra.

Worked Example: Poisson Frequency and Fitted Payment Moments

A hypothetical portfolio has a fitted Poisson payment count with parameter $\widehat{\lambda}=4$. Its fitted payment distribution has mean $\hat m=\$2{,}500$ and standard deviation $\hat\sigma=\$1{,}500$. Calculate the plug-in aggregate mean, variance, and standard deviation. State which source of variability contributes more to the aggregate variance.

Method and assumptions. For a Poisson count, $E[N]=\operatorname{Var}(N)=\lambda$, so both fitted frequency moments equal 4. Assume the payments are mutually independent and identically distributed, and independent of the count.

First square the fitted payment standard deviation to obtain the payment variance:

\hat\sigma^2=(\$1{,}500)^2=\$2{,}250{,}000, \qquad \hat m^2=(\$2{,}500)^2=\$6{,}250{,}000.

The aggregate mean is:

\widehat{E[S]}=4(\$2{,}500)=\$10{,}000.

The severity-variability contribution to the variance is $4(\$2{,}250{,}000)=\$9{,}000{,}000$. The count-variability contribution is $4(\$6{,}250{,}000)=\$25{,}000{,}000$. Therefore:

\widehat{\operatorname{Var}}(S) =\$9{,}000{,}000+\$25{,}000{,}000 =\$34{,}000{,}000.

The aggregate standard deviation is $\sqrt{\$34{,}000{,}000}=\$5{,}830.951\ldots$, or \$5,830.95 to the nearest cent. The count-variability contribution is larger: \$25 million versus \$9 million. As a check, the aggregate mean is four times the mean payment, and both variance contributions have units of dollars squared.

Common mistake avoided. Using only $E[N]\hat\sigma^2=\$9{,}000{,}000$ omits the count-variability term. That is not the compound Poisson aggregate variance; the correction is to add $\operatorname{Var}(N)\hat m^2=\$25{,}000{,}000$.

How Frequency Overdispersion Changes Aggregate Risk

A frequency distribution is overdispersed relative to a Poisson distribution with the same mean when its variance exceeds its mean. In the aggregate variance formula, holding the frequency mean and payment distribution fixed, increasing $\operatorname{Var}(N)$ increases aggregate variance by the increase in $\operatorname{Var}(N)$ multiplied by $m^2$. The aggregate mean does not change under that comparison.

For common frequency models, the relevant moments are:

Frequency model	Mean	Variance
Poisson with parameter $\lambda$	$\lambda$	$\lambda$
Binomial with parameters $n,p$	$np$	$np(1-p)$
Negative binomial, parameterized by its fitted mean and variance	$\mu_N$	$\sigma_N^2$, as specified by the fitted model

For a negative binomial model, parameter conventions vary, so use the mean and variance implied by the stated parameterization rather than assuming a particular convention. The plug-in aggregate formula needs those two moments; it does not require the frequency probability mass function.

Worked Example: Binomial Frequency With Limited Count Variability

A hypothetical group of eight units each has an estimated probability $p=0.25$ of producing one covered payment during the period, independently of the other units. Each payment has fitted mean $\$1{,}200$ and variance $\$160{,}000$. Calculate the aggregate mean and variance.

Method and assumptions. The count is binomial with $n=8$ and $p=0.25$. Assume all payment amounts are mutually independent, identically distributed, and independent of the count. The fitted frequency moments are:

E[N]=np=8(0.25)=2, \qquad \operatorname{Var}(N)=np(1-p)=8(0.25)(0.75)=1.5.

The aggregate mean is:

\widehat{E[S]}=2(\$1{,}200)=\$2{,}400.

For the variance, the severity contribution is $2(\$160{,}000)=\$320{,}000$. The frequency contribution is $1.5(\$1{,}200)^2=1.5(\$1{,}440{,}000)=\$2{,}160{,}000$. Thus:

\widehat{\operatorname{Var}}(S) =\$320{,}000+\$2{,}160{,}000 =\$2{,}480{,}000.

The aggregate standard deviation is $\sqrt{\$2{,}480{,}000}=\$1{,}574.801\ldots$, or \$1,574.80 to the nearest cent. The variance is positive and larger than the severity-only contribution, as expected because count variability adds a second nonnegative term.

Common mistake avoided. Substituting $np=2$ for both frequency moments would incorrectly treat the binomial count as Poisson. Here the binomial variance is $1.5$, not 2, because each of the eight units can generate at most one payment.

Worked Example: Overdispersed Frequency and Severity Variability

A hypothetical fitted negative binomial count model has mean 6 and variance 15. The fitted payment distribution has mean $\$800$ and standard deviation $\$600$. Calculate the aggregate mean and variance, then compare the variance with that under Poisson frequency having the same mean and the same payment model.

Method and assumptions. Use the supplied negative binomial moments directly; no further parameterization is needed. Assume payments are mutually independent and identically distributed, and independent of the count. The payment variance is $(\$600)^2=\$360{,}000$, and the squared payment mean is $(\$800)^2=\$640{,}000$.

For the negative binomial model, the aggregate mean is $6(\$800)=\$4{,}800$. Its severity contribution is $6(\$360{,}000)=\$2{,}160{,}000$, while its frequency contribution is $15(\$640{,}000)=\$9{,}600{,}000$. Therefore:

\widehat{\operatorname{Var}}(S) =\$2{,}160{,}000+\$9{,}600{,}000 =\$11{,}760{,}000.

With Poisson frequency of mean 6, the frequency variance is also 6. The mean remains $\$4{,}800$, and the aggregate variance becomes:

\operatorname{Var}(S_{\text{Poisson}}) =6(\$360{,}000)+6(\$640{,}000) =\$2{,}160{,}000+\$3{,}840{,}000 =\$6{,}000{,}000.

The negative binomial fit has the same aggregate mean but a larger variance because its count variance, 15, exceeds the Poisson variance of 6. The aggregate standard deviation under the negative binomial fit is $\sqrt{\$11{,}760{,}000}=\$3{,}429.285\ldots$, or \$3,429.29 to the nearest cent. The comparison isolates the effect of frequency overdispersion because the count means and payment moments are held fixed.

Common mistake avoided. Replacing the supplied negative binomial variance of 15 with its mean of 6 understates the frequency contribution. The variance term must use $\operatorname{Var}(N)$, not $E[N]$, unless the fitted frequency model is Poisson.

Interpreting Results and Avoiding Scope Errors

The decomposition is useful because it shows whether aggregate variability is driven more by payment sizes or by the number of payments. It also helps check a model comparison: if two frequency fits have equal means and use the same payment model, they give the same aggregate mean, while the fit with the larger count variance gives the larger aggregate variance.

These are moment calculations, not a complete description of the aggregate distribution. Two aggregate models can share a mean and variance but have different probabilities of extreme outcomes. Aggregate-tail measures and distribution calculations require additional methods. Also, plug-in moments treat fitted inputs as fixed. If the purpose is to include uncertainty in estimated parameters, that additional source of uncertainty must be handled separately; it is not captured merely by substituting estimates in the formulas.

Key takeaway: Under a compound model with mutually independent, identically distributed payments independent of the count, aggregate mean equals frequency mean times payment mean. Aggregate variance is the sum of a severity term, $E[N]\operatorname{Var}(Y^P)$, and a frequency term, $\operatorname{Var}(N)(E[Y^P])^2$. Use the fitted variance of the count, especially when frequency is overdispersed.

Check Your Understanding

For each problem, assume payment amounts are mutually independent and identically distributed, and independent of the claim count. Use the stated fitted moments as fixed plug-in inputs.

A fitted Poisson count has parameter 2. Each payment has mean $\$300$ and variance $\$40{,}000$. Find the aggregate mean and variance. (Answer: Mean $\$600$; variance $\$260{,}000$.)
A fitted binomial count has parameters $n=8$ and $p=0.25$. Payments have mean $\$1{,}200$ and variance $\$160{,}000$. Find the aggregate mean and variance. (Answer: Mean $\$2{,}400$; variance $\$2{,}480{,}000$.)
A fitted frequency model has mean 5 and variance 11. Payments have mean $\$500$ and variance $\$90{,}000$. Find the aggregate mean and variance. (Answer: Mean $\$2{,}500$; variance $\$3{,}200{,}000$.)
A fitted Poisson count has parameter 3.5. A payment has mean $\$400$ and variance $\$90{,}000$. Calculate the aggregate mean and variance, showing the separate variance contributions. (Answer: Mean $\$1{,}400$; severity contribution $\$315{,}000$; frequency contribution $\$560{,}000$; total variance $\$875{,}000$.)
Two fitted frequency models both have mean 4. Model A has variance 4, while Model B has variance 10. Both use payment moments $E[Y^P]=\$600$ and $\operatorname{Var}(Y^P)=\$160{,}000$. Compare their aggregate means and variances. (Answer: Both means are $\$2{,}400$; Model A variance is $\$2{,}080{,}000$, and Model B variance is $\$4{,}240{,}000$.)

Frequency model	Mean	Variance
Poisson with parameter \(\lambda\)	\(\lambda\)	\(\lambda\)
Binomial with parameters \(n,p\)	\(np\)	\(np(1-p)\)
Negative binomial, parameterized by its fitted mean and variance	\(\mu_N\)	\(\sigma_N^2\), as specified by the fitted model

What You'll Learn