Separating Frequency And Severity Contributions

Why Separate Aggregate Variance?

Aggregate loss varies for two distinct reasons. The number of claims can change from one period to another, and, even with a fixed number of claims, the amounts paid can differ. Separating these sources of variation helps explain what drives an aggregate portfolio’s volatility and how a change in a frequency or severity assumption affects its variance.

Let $N$ be the number of claims during a specified period, and let $X_1,X_2,\ldots$ be the claim severities. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. The standard decomposition applies when $N$ is independent of the severities, the severities are independent and identically distributed, and the relevant second moments are finite.

Definition: Write $\mu=E[X]$ and $\sigma^2=\operatorname{Var}(X)$ for the mean and variance of one claim severity. Under the assumptions above, the severity contribution to aggregate variance is $E[N]\sigma^2$, and the frequency contribution is $\operatorname{Var}(N)\mu^2$. Their sum is the aggregate variance.

Deriving the Two Contributions

Condition on the observed claim count. If $N=n$, aggregate loss is a sum of $n$ independent severities. Its conditional mean is $n\mu$, and its conditional variance is $n\sigma^2$. The law of total variance separates variability within each fixed-count scenario from variability between scenarios with different counts.

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

The first term averages the conditional severity variance across the possible claim counts. With $n$ claims, that variance is $n\sigma^2$; averaging over $N$ gives $E[N]\sigma^2$. The second term measures how much the conditional mean aggregate loss $N\mu$ changes as the claim count changes. Since multiplying a random variable by $\mu$ multiplies its variance by $\mu^2$, this term is $\operatorname{Var}(N)\mu^2$.

The decomposition also gives $E[S]=E[N]\mu$. The mean uses the expected claim count and mean severity; the variance adds two separate sources of fluctuation. Both contributions have squared currency units: $\sigma^2$ is in squared currency, while $\mu^2$ is in squared currency and claim-count variance is dimensionless.

Find the severity moments.
Calculate $\mu=E[X]$ and $\sigma^2=\operatorname{Var}(X)$ from the claim-severity model.

Find the frequency moments.
Calculate $E[N]$ and $\operatorname{Var}(N)$ from the claim-count model.

Calculate and interpret the terms.
Use $E[N]\sigma^2$ for severity variation and $\operatorname{Var}(N)\mu^2$ for claim-count variation; check that they sum to $\operatorname{Var}(S)$.

Worked Examples

Worked Example: A Poisson Count and Two Severity Amounts

A hypothetical portfolio has a Poisson claim count with parameter $\lambda=4$ during the period. Claim severities are independent of the count and of one another. Each claim is \$1,000 with probability $0.75$ and \$5,000 with probability $0.25$. Find the expected aggregate loss, both variance contributions, and the share of aggregate variance due to each source. Work in thousands of dollars.

Method and severity moments. Measure a severity in thousands of dollars, so $X=1$ with probability $0.75$ and $X=5$ with probability $0.25$. Then

\begin{aligned} \mu=E[X]&=0.75(1)+0.25(5)=2,\\ E[X^2]&=0.75(1^2)+0.25(5^2)=0.75+6.25=7,\\ \sigma^2&=E[X^2]-\mu^2=7-2^2=3. \end{aligned}

For a Poisson random variable with parameter $\lambda$, $E[N]=\operatorname{Var}(N)=\lambda$. Here both moments equal $4$. Thus the expected aggregate loss is $E[S]=4(2)=8$, or \$8,000.

Severity contribution: $E[N]\sigma^2=4(3)=12$ thousand-dollar-squared units. Frequency contribution: $\operatorname{Var}(N)\mu^2=4(2^2)=16$ thousand-dollar-squared units. Therefore $\operatorname{Var}(S)=12+16=28$ thousand-dollar-squared units.

The severity share is $12/28=0.428571\ldots$, or $42.8571\%$; the frequency share is $16/28=0.571428\ldots$, or $57.1429\%$. The shares sum to $100\%$, and the two contributions sum to the calculated total variance. The method avoids treating the Poisson count as fixed: doing so would omit the frequency contribution.

Worked Example: Binomial Frequency With Variable Severities

A hypothetical program can generate at most 10 claims in a period. Each opportunity independently produces a claim with probability $0.2$, so $N\sim\operatorname{Binomial}(10,0.2)$. Independently, each claim is \$2,000 or \$6,000 with equal probability. Find the two contributions and the aggregate variance, working in thousands of dollars.

Method and severity moments. The mean severity is $\mu=0.5(2)+0.5(6)=4$. Its second moment is $E[X^2]=0.5(2^2)+0.5(6^2)=2+18=20$, so $\sigma^2=20-4^2=4$.

For a binomial count with parameters $n$ and $p$, $E[N]=np$ and $\operatorname{Var}(N)=np(1-p)$. Consequently, $E[N]=10(0.2)=2$, and $\operatorname{Var}(N)=10(0.2)(0.8)=1.6$. The expected aggregate loss is $2(4)=8$, or \$8,000.

The severity contribution is $2(4)=8$ thousand-dollar-squared units. The frequency contribution is $1.6(4^2)=1.6(16)=25.6$ thousand-dollar-squared units. Hence $\operatorname{Var}(S)=8+25.6=33.6$ thousand-dollar-squared units.

The severity share is $8/33.6=0.238095\ldots$, or $23.8095\%$, and the frequency share is $25.6/33.6=0.761904\ldots$, or $76.1905\%$. The contributions check because $8+25.6=33.6$. A common error is to use $np$ as the binomial variance; the correct variance is $np(1-p)$, which is $1.6$, not $2$.

Worked Example: Comparing Frequency Models With the Same Mean

Consider the same hypothetical severity model for each of three portfolios: an independent claim severity of \$1,000 or \$3,000, each with probability $0.5$. Compare aggregate variance for (a) Poisson frequency with $\lambda=2$, (b) binomial frequency with $n=8$ and $p=0.25$, and (c) a negative binomial count $N$ equal to the number of failures before $r=3$ successes, with success probability $p=0.6$. In every case, assume the count and severities are independent. Work in thousands of dollars.

Method and common severity moments. The severity mean is $\mu=0.5(1)+0.5(3)=2$. The second moment is $E[X^2]=0.5(1^2)+0.5(3^2)=5$, so $\sigma^2=5-2^2=1$. All three frequency models have mean $2$, so each has expected aggregate loss $2(2)=4$, or \$4,000. In every portfolio, the severity contribution is $E[N]\sigma^2=2(1)=2$ thousand-dollar-squared units.

(a) Poisson frequency. A Poisson count with $\lambda=2$ has variance $2$. The frequency contribution is $2(2^2)=8$, so the total aggregate variance is $2+8=10$ thousand-dollar-squared units.

(b) Binomial frequency. The mean is $8(0.25)=2$, and the variance is $8(0.25)(0.75)=1.5$. The frequency contribution is $1.5(2^2)=6$, so the total is $2+6=8$ thousand-dollar-squared units.

(c) Negative binomial frequency. Under the stated failures-before-$r$-successes parameterization, $E[N]=r(1-p)/p$ and $\operatorname{Var}(N)=r(1-p)/p^2$. Thus $E[N]=3(0.4)/0.6=2$, while $\operatorname{Var}(N)=3(0.4)/(0.6^2)=1.2/0.36=10/3$. The frequency contribution is $(10/3)(2^2)=40/3$. The total is $2+40/3=46/3=15.3333\ldots$ thousand-dollar-squared units.

The means and severity contributions match, but the frequency contributions differ because the count variances are $1.5$, $2$, and $10/3$. The aggregate variances are therefore $8$, $10$, and $46/3$ thousand-dollar-squared units, respectively. This comparison avoids assuming that equal expected claim counts imply equal aggregate variances; the dispersion of the count also matters.

Interpreting the Contributions Carefully

A contribution is an additive part of the aggregate variance, not a separate variance for a different aggregate portfolio. Its proportion of total variance can be useful for comparison: divide each contribution by their sum. A larger frequency share indicates that variability in claim count accounts for a larger portion of total variance under the specified model. It does not mean that individual claim amounts are irrelevant; the frequency term itself depends on the mean severity $\mu$.

Some limiting cases provide useful checks. If the claim count is fixed at $n$, then $\operatorname{Var}(N)=0$, so only the severity contribution remains: $\operatorname{Var}(S)=n\sigma^2$. If severity is deterministic, then $\sigma^2=0$, so all aggregate variance comes from the claim count: $\operatorname{Var}(S)=\operatorname{Var}(N)\mu^2$. If both are deterministic, aggregate variance is zero.

The formula relies on the stated independence and identical-severity assumptions. If severity depends on the count, or claim severities are dependent on one another, the conditional mean or variance used in the derivation may change, and this two-term formula need not apply. Always check the model assumptions before using it. A second frequent trap is confusing $E[X^2]$ with $\operatorname{Var}(X)$: the severity contribution uses $\sigma^2=E[X^2]-\mu^2$, not the second moment alone.

Key takeaway: For an independent, identically distributed severity model with finite second moments, aggregate variance is $E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2$. The first term reflects claim-amount variation at a given count; the second reflects variation in the count itself.

Check Your Understanding

For each problem, identify the frequency and severity contributions before giving the requested aggregate result. Monetary calculations are in thousands of dollars unless stated otherwise.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(3)$. Each independent claim is \$2,000 with probability $0.75$ and \$6,000 with probability $0.25$. Find the expected aggregate loss and both variance contributions. (Answer: Expected loss $=\$9,000$; severity contribution $=9$, frequency contribution $=27$, and aggregate variance $=36$ thousand-dollar-squared units.)
A program has $N\sim\operatorname{Binomial}(8,0.25)$, and every claim is exactly \$5,000. Find the aggregate variance and identify its source. (Answer: Severity contribution $=0$; frequency contribution and aggregate variance $=37.5$ thousand-dollar-squared units.)
A count follows the negative binomial model in which $N$ is the number of failures before $r=3$ successes, with success probability $p=0.6$. Independent severities are \$1,000 or \$5,000 with equal probability. Find both contributions and the total. (Answer: Severity contribution $=8$, frequency contribution $=30$, and aggregate variance $=38$ thousand-dollar-squared units.)
Compare two hypothetical portfolios with the same severity mean of \$2,000 and severity variance of $5$ thousand-dollar-squared units. Portfolio A has $N\sim\operatorname{Poisson}(3)$; Portfolio B has $N\sim\operatorname{Binomial}(12,0.25)$. Find each aggregate variance. (Answer: A has severity contribution $15$, frequency contribution $12$, and total $27$; B has severity contribution $15$, frequency contribution $9$, and total $24$, all in thousand-dollar-squared units.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(5)$, and the independent severity has mean \$4,000 and variance $9$ thousand-dollar-squared units. Find the aggregate variance and the percentage of it due to each contribution. (Answer: Severity contribution $=45$, frequency contribution $=80$, total $=125$ thousand-dollar-squared units; severity share $=36.0000\%$, frequency share $=64.0000\%$.)

What You'll Learn