Frequency, Severity, and Variability

Why Separate Frequency, Severity, and Variability?

A loss estimate is more informative when it distinguishes how often losses occur from how large they are when they occur. A hypothetical insurer might have many small claims, or a few large claims, even if both portfolios have the same expected annual loss. The expected value alone does not reveal that difference.

Frequency and severity also have their own variability. Claim counts can fluctuate from year to year, and individual claim amounts can vary. Both sources contribute to the variability of total loss. This tutorial develops the basic definitions and calculations for a single period, such as one year, without discounting future payments. The treatment is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: Frequency is the number of losses arising in a specified period, represented here by the nonnegative integer-valued random variable $N$. Severity is the amount of an individual loss, represented by $X$, conditional on a loss occurring. Aggregate loss is the total amount of all losses in the period, represented by $S_N=\sum_{j=1}^{N}X_j$, with $S_N=0$ when $N=0$. Variability describes how dispersed possible outcomes are around their expected value; variance and standard deviation are common numerical summaries.

The distinction depends on the question being asked. The count $N$ answers “how many losses?”; an individual severity $X_j$ answers “how much was the $j$th loss?”; and $S_N$ answers “how much was lost in total?” For a defined period and exposure base, frequency can be summarized by its distribution, mean, and variance. Severity can be summarized by its distribution, mean, and variance, conditional on a claim occurring.

For the basic aggregate formulas below, assume the severities $X_1,X_2,\ldots$ are independent and identically distributed, each with mean $\mu_X=E[X]$ and variance $\sigma_X^2=\operatorname{Var}(X)$. Assume also that the severity sequence is independent of the count $N$, and that the relevant means and variances are finite. These are model assumptions, not universal truths: for example, claim sizes might depend on the number of claims, or claim severities might be dependent. In such cases, the formulas may need adjustment.

Describe Each Component Before Combining Them

A frequency model specifies probabilities for possible values of $N$. For instance, a binomial count can represent a fixed number of independent exposure units, each with the same probability of a loss. A Poisson count is another standard model for events occurring over a specified period. Choosing either model requires checking whether its assumptions are suitable for the situation; the model name alone does not make it appropriate.

A severity model describes the distribution of a loss given that one occurs. Its mean measures the average individual loss, while its variance indicates how much individual losses vary. A constant severity has zero variance: all covered losses have the same assumed amount. That simplifying assumption can help isolate the effect of frequency.

The period and exposure base must be consistent. An expected count of three claims per year should not be combined with a severity mean measured per month unless the time basis is reconciled. In the examples, every model describes one hypothetical year, and all amounts are in dollars.

Worked Example: Fixed Loss Amount With a Random Count

A hypothetical insurer covers five independent pieces of equipment for one year. Each piece has a probability of $0.20$ of producing one covered loss during the year, and there is at most one loss per piece. If a loss occurs, its amount is exactly \$3,000. Find the expected aggregate loss, its variance and standard deviation, and the probability of at least one loss. This is a one-year undiscounted calculation; no interest-rate or mortality assumption is used.

Method and frequency model. The count has a binomial distribution, $N\sim\operatorname{Binomial}(5,0.20)$, because there are five independent opportunities, each with the same loss probability. Therefore,

E[N]=5(0.20)=1,\qquad \operatorname{Var}(N)=5(0.20)(0.80)=0.80.

Severity model. Conditional on a loss, $X=3{,}000$ with probability 1. Thus $E[X]=3{,}000$ and $\operatorname{Var}(X)=0$. Aggregate loss is $S_N=3{,}000N$, so

E[S_N]=3{,}000E[N]=3{,}000(1)=\$3{,}000,

and

\operatorname{Var}(S_N)=3{,}000^2\operatorname{Var}(N) =9{,}000{,}000(0.80)=7{,}200{,}000\text{ dollars}^2.

The standard deviation is $\sqrt{7{,}200{,}000}=\$2{,}683.28$, rounded to the nearest cent. The probability of no losses is $0.80^5=0.32768$, so the probability of at least one loss is $1-0.32768=0.67232$, or $0.6723$ to four decimals.

Check and common mistake avoided. The expected aggregate loss equals one expected claim times \$3,000 per claim. Its standard deviation is not \$3,000: the claim amount is fixed, but the count varies. The variance calculation correctly scales the count variance by the square of the claim amount.

Expected Aggregate Loss and Its Variance

Let $\mu_X=E[X]$ and $\sigma_X^2=\operatorname{Var}(X)$. Under the independent, identically distributed severity assumptions stated above, the expected aggregate loss is

E[S_N]=E[N]\mu_X.

This says that the expected total is the expected number of losses multiplied by the expected amount per loss. For variability, the frequency and severity components both matter:

\operatorname{Var}(S_N) =E[N]\sigma_X^2+\operatorname{Var}(N)\mu_X^2.

The first term reflects variation in individual loss amounts across the expected number of claims. The second reflects variation in the count, with the average severity squared scaling that count variation. This expression is valid under the stated independence assumptions; it is not a general formula for every dependent count-and-severity model.

Derivation: Conditional on $N$, the sum $S_N$ contains $N$ independent severities. Therefore $E[S_N\mid N]=N\mu_X$ and $\operatorname{Var}(S_N\mid N)=N\sigma_X^2$. The law of total expectation gives $E[S_N]=E[N\mu_X]=E[N]\mu_X$. The law of total variance gives $$ \operatorname{Var}(S_N) =E[\operatorname{Var}(S_N\mid N)] +\operatorname{Var}(E[S_N\mid N]) =E[N]\sigma_X^2+\operatorname{Var}(N)\mu_X^2. $$

The formula can also be expressed using the severity second moment $E[X^2]=\sigma_X^2+\mu_X^2$:

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

This follows by substituting $\sigma_X^2=E[X^2]-\mu_X^2$ into the variance formula and collecting terms. In the special case $N\sim\operatorname{Poisson}(\lambda)$, $E[N]=\operatorname{Var}(N)=\lambda$, so the expression simplifies to $\operatorname{Var}(S_N)=\lambda E[X^2]$. The Poisson simplification relies on the Poisson count model; it is not the general aggregate variance formula.

Worked Example: Random Frequency and Random Severity

A hypothetical service-contract portfolio has a one-year claim count $N\sim\operatorname{Poisson}(2)$. Each claim is independently \$500 or \$1,500, each with probability $0.50$, and claim severities are independent of $N$. Find the expected aggregate loss, its variance and standard deviation, and the probability of no claims. This is an undiscounted one-year model; no interest-rate or mortality assumption is used.

Method and severity calculations. First calculate the severity mean and second moment:

E[X]=0.50(500)+0.50(1{,}500)=250+750=\$1{,}000,

$$ E[X^2]=0.50(500^2)+0.50(1{,}500^2) =0.50(250{,}000)+0.50(2{,}250{,}000) =1{,}250{,}000. $$

Thus $\operatorname{Var}(X)=E[X^2]-(E[X])^2=1{,}250{,}000-1{,}000{,}000=250{,}000$ dollars squared. For this Poisson model, $E[N]=\operatorname{Var}(N)=2$, so

E[S_N]=2(1{,}000)=\$2{,}000,

and

\operatorname{Var}(S_N) =2(250{,}000)+2(1{,}000^2) =500{,}000+2{,}000{,}000 =2{,}500{,}000\text{ dollars}^2.

The standard deviation is $\sqrt{2{,}500{,}000}=\$1{,}581.14$, rounded to the nearest cent. The Poisson probability of zero claims is $P(N=0)=e^{-2}=0.1353$ to four decimals.

Check and common mistake avoided. The Poisson shortcut gives $2E[X^2]=2(1{,}250{,}000)=2{,}500{,}000$, agreeing with the general formula. A tempting but incorrect expression is $E[N]E[X^2]+\operatorname{Var}(N)\mu_X^2$, which here gives $2(1{,}250{,}000)+2(1{,}000{,}000)=4{,}500{,}000$. It counts an extra $E[N]\mu_X^2$. The correct second-moment form subtracts that term through $(\operatorname{Var}(N)-E[N])\mu_X^2$.

How Frequency Variability Changes the Total

Two portfolios can have the same expected claim count and the same severity distribution but different aggregate variability if their count distributions differ. A fixed count removes count variation, but it does not remove severity variation. Conversely, if severity is fixed, all aggregate variability comes from frequency.

Worked Example: Same Expected Count, Different Count Variability

Compare two hypothetical one-year portfolios with the same independent severity model: each claim is \$500 or \$1,500 with equal probability. In Portfolio A, exactly two claims occur. In Portfolio B, $N\sim\operatorname{Poisson}(2)$. Find the expected aggregate loss and variance for each, then explain the difference. No discounting, interest-rate assumption, or mortality assumption is involved.

Method. From the severity calculation, $E[X]=\$1{,}000$ and $\operatorname{Var}(X)=250{,}000$ dollars squared. Portfolio A has $E[N]=2$ and $\operatorname{Var}(N)=0$, because its count is fixed. Therefore,

E[S_N]=2(1{,}000)=\$2{,}000,\qquad \operatorname{Var}(S_N)=2(250{,}000)+0(1{,}000^2)=500{,}000.

Its standard deviation is $\sqrt{500{,}000}=\$707.11$, rounded to the nearest cent. For Portfolio B, the Poisson count has $E[N]=\operatorname{Var}(N)=2$. Therefore,

E[S_N]=2(1{,}000)=\$2{,}000,\qquad \operatorname{Var}(S_N)=2(250{,}000)+2(1{,}000^2)=2{,}500{,}000.

Portfolio B's standard deviation is $\sqrt{2{,}500{,}000}=\$1{,}581.14$, rounded to the nearest cent. Both expected losses are \$2,000, but Portfolio B has greater variance because its count fluctuates as well as its severity. Portfolio A still has nonzero variance because the two claim amounts can differ.

Check and common mistake avoided. Equal expected aggregate losses do not imply equal variability. Treating both counts as fixed at their common mean would incorrectly give Portfolio B the variance of Portfolio A and omit the count-variation term.

Common Interpretation Traps

Keep the scale of each quantity clear. A count is measured in claims per period; severity and aggregate loss are measured in dollars; variance is measured in dollars squared. Standard deviation returns to dollars, making it easier to compare with a dollar-valued expected loss, though it describes dispersion rather than a guaranteed range.

Mixing frequency with severity. An expected count is not an expected loss amount. Multiply by expected severity to obtain expected aggregate loss under the stated model.
Ignoring conditional severity. Severity is the amount given that a loss occurs. A portfolio with no claims in a period has aggregate loss zero, not a severity observation of zero.
Using a variance formula without its assumptions. The compound formula assumes independent severities with a common distribution, independent of the count, and finite second moments.
Confusing second moment and variance. $E[X^2]$ is not $\operatorname{Var}(X)$; rather, $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. Substitution must preserve the subtraction in the correct equivalent aggregate formula.
Comparing models with different periods or exposure bases. A monthly count and an annual severity cannot be combined as though they described the same period without aligning the assumptions and units.

Key takeaway: Describe a loss model in three parts: the distribution of the count $N$, the distribution of an individual severity $X$, and the resulting aggregate loss $S_N$. Under the independent compound model, expected aggregate loss is $E[N]E[X]$, while aggregate variance includes both severity variation and frequency variation.

Check Your Understanding

For each hypothetical one-year model, identify the frequency and severity inputs and calculate the requested aggregate quantities. Round dollar standard deviations to the nearest cent and probabilities to four decimals where applicable.

A workshop insures four independent machines. Each has probability $0.10$ of one loss during the year, with at most one loss per machine; each loss is exactly \$10,000. Find expected aggregate loss and aggregate variance. (Answer: $E[S_N]=\$4{,}000$; $\operatorname{Var}(S_N)=36{,}000{,}000$ dollars squared.)
A hypothetical roadside-assistance program has $N\sim\operatorname{Poisson}(3)$ claims in one year, and every claim costs exactly \$400. Find expected aggregate loss, variance, standard deviation, and probability of no claims. (Answer: $E[S_N]=\$1{,}200$; variance $=480{,}000$ dollars squared; standard deviation $=\$692.82$; $P(N=0)=0.0498$.)
A one-year portfolio has exactly four claims. Each independent claim is \$200 with probability $0.75$, or \$800 with probability $0.25$. Find the severity mean and variance, then the aggregate mean and variance. (Answer: $E[X]=\$350$; $\operatorname{Var}(X)=67{,}500$ dollars squared; $E[S_N]=\$1{,}400$; $\operatorname{Var}(S_N)=270{,}000$ dollars squared.)
A hypothetical product has $N\sim\operatorname{Poisson}(1)$ claims per year. Each independent claim is \$1,000 or \$3,000 with equal probability. Find the expected aggregate loss, severity second moment, aggregate variance, and aggregate standard deviation. (Answer: $E[S_N]=\$2{,}000$; $E[X^2]=5{,}000{,}000$; $\operatorname{Var}(S_N)=5{,}000{,}000$ dollars squared; standard deviation $=\$2{,}236.07$.)
Compare two hypothetical one-year count models with fixed severity \$2,000 per claim: Model A has $N\sim\operatorname{Binomial}(10,0.10)$; Model B has $N\sim\operatorname{Poisson}(1)$. Calculate expected aggregate loss and variance under both, then state which has greater variance and by how much. (Answer: both expected losses are $\$2{,}000$; Model A variance $=3{,}600{,}000$ dollars squared; Model B variance $=4{,}000{,}000$ dollars squared; Model B exceeds Model A by $400{,}000$ dollars squared.)

What You'll Learn