Frequency-Severity Variance Decomposition

Why Separate Frequency and Severity?

An aggregate claim amount varies for two distinct reasons: the number of claims can change, and the amounts paid on those claims can change. Separating these sources helps identify what drives conditional aggregate-loss variance and, in a credibility model, which part contributes to expected process variance. The distinction depends on the frequency distribution: a Poisson count has a special relationship between its mean and variance, but the general decomposition does not require Poisson frequency.

Let \(\Theta\) denote a risk's latent risk parameter, \(N\) its claim count in a period, and \(X_1,X_2,\ldots\) the severities of its claims. Define the aggregate loss as \(S=\sum_{j=1}^{N}X_j\), with \(S=0\) when \(N=0\). The calculations below concern losses before any deductible, limit, or other payment adjustment.

Definition: Conditional on \(\Theta=\theta\), write \(m(\theta)=E[N\mid\Theta=\theta]\), \(v_N(\theta)=\operatorname{Var}(N\mid\Theta=\theta)\), \(\mu(\theta)=E[X_j\mid\Theta=\theta]\), and \(\sigma^2(\theta)=\operatorname{Var}(X_j\mid\Theta=\theta)\). The frequency-severity decomposition assumes that, given \(\theta\), the severities are identically distributed and independent of one another, and the count is independent of the severity sequence. Under these assumptions, the conditional aggregate variance is \(m(\theta)\sigma^2(\theta)+v_N(\theta)\mu^2(\theta)\).

The first term, \(m(\theta)\sigma^2(\theta)\), is the contribution from random claim sizes at the expected number of claims. The second term, \(v_N(\theta)\mu^2(\theta)\), is the contribution from a random claim count, scaled by the squared conditional mean severity. The squared mean matters: a fluctuation of one claim in the count changes expected aggregate loss by approximately one mean claim amount.

Deriving the Conditional Formula

Condition on the claim count as well as on the risk parameter. Given \(N=n\) and \(\Theta=\theta\), the aggregate is a sum of \(n\) independent severities. Its conditional mean is \(n\mu(\theta)\), and its conditional variance is \(n\sigma^2(\theta)\). Apply the law of total variance, first conditioning on \(N\) while holding \(\theta\) fixed.

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

This derivation shows why the two terms add: the law of total variance separates variation remaining after the count is known from variation in the conditional mean caused by the count. Conditional independence is essential to the first calculation, since it makes the variance of the sum of \(n\) severities equal to \(n\sigma^2(\theta)\). If severities are dependent, or their distribution changes with the claim's position or count, the formula may need additional covariance or conditional-distribution terms.

Worked Example: A Poisson Count With Two Possible Claim Sizes

For one hypothetical risk, conditional on its risk type, the annual claim count is Poisson with mean \(3\). Each claim is \(1{,}000\) dollars with probability \(0.75\) and \(3{,}000\) dollars with probability \(0.25\), independently of the count and other claim sizes. Find the expected aggregate and its variance, and separate the two variance contributions.

Method. Calculate the conditional severity mean and variance, then use the general random-sum decomposition. Because the count is Poisson, its variance equals its mean, but the two components should still be calculated separately.

The conditional mean severity is \( \mu=0.75(1{,}000)+0.25(3{,}000)=750+750=1{,}500 \) dollars. The second moment is \( E[X^2]=0.75(1{,}000^2)+0.25(3{,}000^2)=750{,}000+2{,}250{,}000=3{,}000{,}000 \) dollars squared.

Therefore, the severity variance is \( \sigma^2=E[X^2]-\mu^2=3{,}000{,}000-1{,}500^2=3{,}000{,}000-2{,}250{,}000=750{,}000 \) dollars squared. For a Poisson count, \(m=v_N=3\). The expected aggregate is \(E[S]=E[N]\mu=3(1{,}500)=4{,}500\) dollars.

The claim-size contribution to variance is \(m\sigma^2=3(750{,}000)=2{,}250{,}000\) dollars squared. The claim-count contribution is \(v_N\mu^2=3(2{,}250{,}000)=6{,}750{,}000\) dollars squared. Thus \(\operatorname{Var}(S)=2{,}250{,}000+6{,}750{,}000=9{,}000{,}000\) dollars squared.

Check. For Poisson frequency, the compound-Poisson variance is \(E[N]E[X^2]=3(3{,}000{,}000)=9{,}000{,}000\), matching the decomposition. The common mistake avoided here is to treat all of \(E[N]E[X^2]\) as the claim-size contribution; the count contribution is substantial too.

General Frequency Versus Poisson Frequency

The general formula uses both the conditional mean and conditional variance of the count. It does not permit replacing the count variance with the count mean unless the conditional count distribution supports that equality. In particular, fixed counts have zero count variance, while overdispersed counts can have variance greater than their mean.

Worked Example: A Non-Poisson Count

For a hypothetical coverage, conditional on risk type, \(N\) is equally likely to be \(1\) or \(3\). Claim severities are conditionally independent of \(N\), with mean \(400\) dollars and variance \(40{,}000\) dollars squared. Find the expected aggregate and aggregate variance.

Method. Find the count mean and variance directly from its two-point distribution, then apply the general formula. Do not impose the Poisson equality between count mean and variance.

The count mean is \(E[N]=(1+3)/2=2\). Its second moment is \(E[N^2]=(1^2+3^2)/2=(1+9)/2=5\), so \(\operatorname{Var}(N)=5-2^2=1\).

The expected aggregate is \(E[S]=E[N]E[X]=2(400)=800\) dollars. The claim-size contribution is \(2(40{,}000)=80{,}000\) dollars squared. The claim-count contribution is \(1(400^2)=160{,}000\) dollars squared. Therefore, \(\operatorname{Var}(S)=80{,}000+160{,}000=240{,}000\) dollars squared.

Check. The formula specialized to Poisson frequency would incorrectly use count variance \(2\), producing \(80{,}000+320{,}000=400{,}000\). The actual count variance is \(1\), so the correct answer is \(240{,}000\) dollars squared.

Expected Process Variance Across Risks

In the Bühlmann setting, a risk parameter can vary across risks. The expected process variance is the population expectation of the conditional variance of an observation, here the aggregate loss. Thus it averages the within-risk frequency and severity contributions over the distribution of \(\Theta\):

\begin{aligned} \text{Expected process variance} &=E[\operatorname{Var}(S\mid\Theta)]\\ &=E[m(\Theta)\sigma^2(\Theta)] +E[v_N(\Theta)\mu^2(\Theta)]. \end{aligned}

Both terms are inside the expectation over risks. The first averages conditional claim-size variation, weighted by conditional expected count. The second averages conditional count variation, weighted by squared conditional mean severity. This quantity is not, by itself, the full unconditional variance of aggregate losses across risks. The total variance also includes variation in conditional aggregate means, \(E[S\mid\Theta]=m(\Theta)\mu(\Theta)\):

\operatorname{Var}(S) =E[\operatorname{Var}(S\mid\Theta)] +\operatorname{Var}(m(\Theta)\mu(\Theta)).

Consequently, heterogeneity in expected frequency or expected severity can increase the total variance without being part of expected process variance. This distinction is important when interpreting the within-risk and between-risk components in credibility theory.

Worked Example: Averaging Process Variance Over Two Risk Types

A hypothetical portfolio has two risk types, each with probability \(0.5\). Conditional on type A, claim count is Poisson with mean \(2\), and severity has mean \(100\) dollars and variance \(900\) dollars squared. Conditional on type B, count is Poisson with mean \(5\), and severity has mean \(200\) dollars and variance \(3{,}600\) dollars squared. Find expected process variance, separating frequency and severity contributions. Also find the between-risk variance of conditional aggregate means and the total aggregate variance.

Method. For each type, use \(v_N=m=\lambda\). Then average the two conditional variance components using the type probabilities. Separately calculate each type's conditional aggregate mean for the between-risk component.

For type A, the severity contribution is \(2(900)=1{,}800\), and the frequency contribution is \(2(100^2)=20{,}000\), both in dollars squared. Its conditional aggregate variance is \(1{,}800+20{,}000=21{,}800\). Its conditional aggregate mean is \(2(100)=200\) dollars.

For type B, the severity contribution is \(5(3{,}600)=18{,}000\), and the frequency contribution is \(5(200^2)=200{,}000\), both in dollars squared. Its conditional aggregate variance is \(18{,}000+200{,}000=218{,}000\). Its conditional aggregate mean is \(5(200)=1{,}000\) dollars.

The expected severity contribution is \(0.5(1{,}800)+0.5(18{,}000)=900+9{,}000=9{,}900\) dollars squared. The expected frequency contribution is \(0.5(20{,}000)+0.5(200{,}000)=10{,}000+100{,}000=110{,}000\) dollars squared. Therefore, expected process variance is \(9{,}900+110{,}000=119{,}900\) dollars squared.

The conditional aggregate mean is \(200\) or \(1{,}000\), each with probability \(0.5\). Its population mean is \(0.5(200)+0.5(1{,}000)=600\) dollars, and its variance is \(0.5(200-600)^2+0.5(1{,}000-600)^2=0.5(160{,}000)+0.5(160{,}000)=160{,}000\) dollars squared. Thus total aggregate variance is \(119{,}900+160{,}000=279{,}900\) dollars squared.

Check. Expected process variance is the average of the two conditional variances: \(0.5(21{,}800)+0.5(218{,}000)=119{,}900\). The additional \(160{,}000\) is between-risk variation in conditional means, not another process-variance term.

Using the Decomposition Correctly

When the conditional count is Poisson with mean \(\lambda(\theta)\), its variance is also \(\lambda(\theta)\). The two-term expression then simplifies:

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

The simplification follows from \(E[X^2\mid\theta]=\sigma^2(\theta)+\mu^2(\theta)\). It is a useful check, but it should not obscure the separate sources of variance. For non-Poisson frequency, the general form remains \(m\sigma^2+v_N\mu^2\). For expected process variance, average these conditional contributions over risk types; do not add between-risk variance unless the question asks for total unconditional variance.

A frequent exam trap is to confuse the count mean and variance, or to include only the severity term when the count is random. Another is to use \(E[N]\operatorname{Var}(X)\) as the whole aggregate variance. That omits \( \operatorname{Var}(N)\mu^2 \), which is zero only when frequency is fixed. These results are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Condition on the risk type.
Identify the conditional count mean and variance and the conditional severity mean and variance.

Check the assumptions.
Confirm that severities are conditionally iid and independent of the count. If not, the simple random-sum formula may not apply.

Compute both contributions.
Use \(m\sigma^2\) for claim-size variation and \(v_N\mu^2\) for claim-count variation.

Average only when requested.
Expected process variance averages the conditional variance. Total unconditional variance also includes variance in the conditional aggregate means.

Key takeaway: Given a risk type, aggregate-loss variance is the sum of a severity term, \(E[N\mid\theta]\operatorname{Var}(X\mid\theta)\), and a frequency term, \(\operatorname{Var}(N\mid\theta)E[X\mid\theta]^2\). Expected process variance is the population average of these two terms.

Check Your Understanding

For each problem, use conditional frequency and severity moments and keep process variance distinct from between-risk variation.

A hypothetical policy has Poisson claim count with mean \(4\). Each claim is \(50\) dollars with probability \(0.6\) and \(150\) dollars with probability \(0.4\). Find the severity and frequency contributions to aggregate variance and their sum. (Answer: severity contribution \(9{,}600\), frequency contribution \(32{,}400\), aggregate variance \(42{,}000\) dollars squared.)
A hypothetical contract always has exactly \(3\) claims in the period. Conditional severities are independent with mean \(250\) dollars and variance \(10{,}000\) dollars squared. Find the aggregate variance and explain which contribution is zero. (Answer: \(30{,}000\) dollars squared; frequency contribution is zero.)
For a hypothetical risk type, claim count has probabilities \(0.25, 0.50, 0.25\) at counts \(0,2,4\), respectively. Severity has mean \(80\) dollars and variance \(400\) dollars squared. Find the count mean and variance, then aggregate variance. (Answer: count mean \(2\), count variance \(2\), aggregate variance \(13{,}600\) dollars squared.)
A portfolio has two equally likely risk types. Type A has Poisson count mean \(1\), severity mean \(100\) dollars, and severity variance \(900\) dollars squared. Type B has Poisson count mean \(3\), severity mean \(200\) dollars, and severity variance \(1{,}600\) dollars squared. Find expected process variance, with its two contributions separated. (Answer: expected severity contribution \(2{,}850\), expected frequency contribution \(65{,}000\), expected process variance \(67{,}850\) dollars squared.)
For a hypothetical risk, the conditional count mean is \(5\), its conditional variance is \(8\), and severity has conditional mean \(120\) dollars and variance \(2{,}500\) dollars squared. Find the conditional aggregate variance, and state why using a Poisson formula would be incorrect. (Answer: \(127{,}700\) dollars squared; the count variance is \(8\), not the mean \(5\).)

What You'll Learn