How Frequency And Severity Variability Shape Stop-Loss Risk

Why Equal Means Do Not Mean Equal Stop-Loss Risk

Two aggregate loss models can have the same expected total loss yet behave quite differently above a deductible. One model may produce a comparatively stable amount, while another produces many small totals and occasional large ones. A stop-loss expectation measures the average amount by which aggregate loss exceeds a chosen deductible, so it responds to the shape of the distribution, not just its mean.

Here, $S$ denotes aggregate loss and $d\geq 0$ is a stop-loss deductible. There is no discounting in the examples. The comparison is about the distribution of aggregate loss, not the pricing of a hypothetical reinsurance contract.

Definition: The stop-loss expectation of an aggregate loss $S$ at deductible $d$ is $\Pi_S(d)=E[(S-d)_+]$, where $(z)_+=\max(z,0)$. Equivalently, $\Pi_S(d)=E[S-d\mid S>d]P(S>d)$ when $P(S>d)>0$. It is the expected amount above the deductible, counting zero when aggregate loss does not exceed $d$.

The deductible is applied to the aggregate $S$. Thus, this quantity is generally not the same as adding up the stop-loss amounts for individual claims. A comparison must use the same deductible and the same definition of aggregate loss in each model.

Frequency and Severity Both Contribute to Aggregate Variability

Consider a compound aggregate model $S=\sum_{j=1}^{N}X_j$, where $N$ is a nonnegative integer-valued claim count and the claim severities $X_j$ are independent and identically distributed nonnegative random variables, independent of $N$. Assume the relevant moments are finite. Write $m=E[X]$, $s^2=\operatorname{Var}(X)$, and suppose $E[N]$ and $\operatorname{Var}(N)$ are finite. Conditioning on $N$ gives the standard moment formulas:

E[S]=E[N]m,\qquad \operatorname{Var}(S)=E[N]s^2+\operatorname{Var}(N)m^2.

The first variance term reflects variation in individual claim sizes across the expected number of claims. The second reflects variation in how many claims occur. In particular, holding $E[N]$ and $m$ fixed holds the aggregate mean fixed, but a change in either severity variance or frequency variance can change aggregate variance.

Important limitation: The compound variance formula describes the first two moments; it does not by itself determine a stop-loss expectation. Stop-loss values depend on the distribution above the deductible. A larger variance does not guarantee a larger stop-loss value at every deductible.

Why Convex Spreading Raises Stop-Loss Values

A precise way to express “more variable” for this comparison is convex order. For random variables $S_A$ and $S_B$ with finite means, $S_A$ is smaller than $S_B$ in convex order if $E[\phi(S_A)]\leq E[\phi(S_B)]$ for every convex function $\phi$ for which the expectations exist. If their means are equal and this ordering holds, then the stop-loss expectation for $S_B$ is at least as large at every deductible.

E[S_A]=E[S_B]\ \text{and}\ S_A\leq_{\mathrm{cx}}S_B \quad\Longrightarrow\quad E[(S_A-d)_+]\leq E[(S_B-d)_+]\quad\text{for every }d\geq0.

The justification is direct: for a fixed $d$, the function $\phi_d(s)=(s-d)_+$ is convex in $s$, since it is the maximum of the two affine functions $0$ and $s-d$. Applying the definition of convex order to this particular convex function gives the stated inequality. Equal means alone do not establish convex order, and neither does a comparison of variances alone.

Fix the deductible and aggregate definition.
Confirm that the two models measure the same kind of total loss and use the same $d$.

Compare means and moments.
Use the frequency-severity formulas to identify which source of variability changes.

Evaluate the stop-loss expectation.
Use the aggregate distribution or an appropriate transform; do not infer its value from variance alone.

State the strength of the conclusion.
A result at one deductible is a point comparison. A claim for all deductibles requires a distributional ordering such as convex order.

Worked Comparisons

Worked Example: Changing Frequency Variability With a Fixed Claim Size

A hypothetical portfolio has deterministic claim severity $X=\$10$. Compare two annual claim-count models: Model A has exactly one claim; Model B has zero claims or two claims, each with probability $0.5$. Find each aggregate mean, variance, and stop-loss expectation at $d=\$12$.

Method and assumptions. Severity is exactly $\$10$ per claim, the count and severity are independent, and the aggregate is $S=\sum_{j=1}^N X_j$. No interest or mortality assumptions are needed. In both models, $E[N]=1$, so $E[S]=1(\$10)=\$10$.

Model A. Since $N=1$ with certainty, $S_A=\$10$ with certainty. Therefore, $\operatorname{Var}(S_A)=0$, and $E[(S_A-\$12)_+]=(\$10-\$12)_+=\$0.00$.

Model B. The aggregate is $\$0$ with probability $0.5$ and $\$20$ with probability $0.5$. Its mean is $0.5(\$0)+0.5(\$20)=\$10$. Its variance is $0.5(\$0-\$10)^2+0.5(\$20-\$10)^2=0.5(100)+0.5(100)=100$ dollars squared.

At $d=\$12$, the excess is $\$0$ when $S_B=\$0$ and $\$8$ when $S_B=\$20$. Thus, $E[(S_B-\$12)_+]=0.5(\$0)+0.5(\$8)=\$4.00$.

Check and exam trap. Both means are $\$10$, but the stop-loss expectations are not equal. The common mistake is to conclude that equal expected claim counts and equal severities imply equal aggregate risk; Model B’s more variable count changes the aggregate distribution.

Worked Example: Changing Severity Variability at a Fixed Claim Count

A hypothetical contract has exactly one claim, so aggregate loss equals severity. Compare a constant severity of $\$100$ with an exponential severity having mean $\$100$, at deductible $d=\$150$. Find the stop-loss expectation for each model.

Method and assumptions. The constant model has $X=\$100$. In the second model, $X$ is exponentially distributed with rate $1/100$ per dollar, so $E[X]=\$100$ and $\operatorname{Var}(X)=(\$100)^2$. There is no discounting. Since $N=1$, $S=X$.

For the constant model, $(\$100-\$150)_+=\$0$, so the stop-loss expectation is $\$0.00$.

For the exponential model, the density is $f_X(x)=(1/100)e^{-x/100}$ for $x\geq0$. The stop-loss expectation is

\begin{aligned} E[(X-150)_+] &=\int_{150}^{\infty}(x-150)\frac{1}{100}e^{-x/100}\,dx\\ &=100e^{-150/100}\\ &=100e^{-1.5}\\ &\approx \$22.31. \end{aligned}

The integration result follows by substituting $y=x-150$: the integral becomes $e^{-1.5}\int_0^\infty y(1/100)e^{-y/100}\,dy$, and that latter integral is the mean $\$100$ of an exponential variable with rate $1/100$. The final value is rounded to the nearest cent.

Check and exam trap. Both severities have mean $\$100$, but the variable severity has a positive chance of exceeding $\$150$. Do not substitute the mean into the stop-loss function: in general $E[(X-d)_+]\neq(E[X]-d)_+$.

Worked Example: Equal Means and Variances, Different Stop-Loss Values

Two hypothetical aggregate models have the following distributions. Model A is $\$0$ or $\$20$, each with probability $0.5$. Model B is $\$0$ with probability $0.2$, $\$5$ with probability $0.5$, and $\$25$ with probability $0.3$. Compare their means, variances, and stop-loss expectations at $d=\$15$.

Method. For each discrete distribution, calculate the mean, then the variance as $E[S^2]-(E[S])^2$, and calculate the excess separately at each possible aggregate amount.

For Model A, $E[S_A]=0.5(0)+0.5(20)=\$10$. Its second moment is $E[S_A^2]=0.5(0^2)+0.5(20^2)=200$ dollars squared, so $\operatorname{Var}(S_A)=200-10^2=100$ dollars squared. At $d=\$15$, only the $\$20$ outcome exceeds the deductible, giving $E[(S_A-15)_+]=0.5(5)=\$2.50$.

For Model B, $E[S_B]=0.2(0)+0.5(5)+0.3(25)=0+2.5+7.5=\$10$. Its second moment is $0.2(0^2)+0.5(5^2)+0.3(25^2)=0+12.5+187.5=200$ dollars squared. Hence $\operatorname{Var}(S_B)=200-10^2=100$ dollars squared. Only the $\$25$ outcome exceeds $\$15$, so $E[(S_B-15)_+]=0.3(10)=\$3.00$.

Check and exam trap. The means and variances agree, but the stop-loss values differ. A variance calculation cannot replace a tail calculation: the two distributions place their probability differently relative to the deductible. Both stop-loss values use the same deductible and are reported to the nearest cent.

Interpreting the Comparison

A stop-loss expectation combines the probability of exceeding the deductible with the size of the excess when it is exceeded. A model can have a larger probability of crossing $d$ but smaller average excess conditional on crossing it; the product of those effects determines the expectation. This is why a comparison at one deductible need not hold at another.

The compound variance formula is still useful. It shows whether variability is coming from frequency, severity, or both, and it can flag a calculation that is inconsistent with the model. But it is a diagnostic moment calculation, not a substitute for evaluating the aggregate distribution or its stop-loss transform. When convex order is established, it provides the stronger all-deductible comparison; otherwise, compute the stop-loss values at the deductibles of interest.

Key takeaway: Equal aggregate means do not imply equal stop-loss expectations. Frequency and severity variability affect the aggregate distribution, while the stop-loss expectation depends on how that distribution lies above the deductible. Use convex order for a valid all-deductible conclusion; do not use variance alone to claim one.

Check Your Understanding

For each question, use the aggregate distribution or the stated compound model. Report monetary answers to the nearest cent unless the answer is an exact moment.

Model A has aggregate loss $\$12$ with certainty. Model B has aggregate loss $\$0$ or $\$24$, each with probability $0.5$. Find both means and stop-loss expectations at $d=\$15$. (Answer: Both means are $\$12$; the stop-loss expectations are $\$0.00$ and $\$4.50$, respectively.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ and deterministic severity $X=\$5$ per claim, independently of $N$. Find $E[S]$, $\operatorname{Var}(S)$, and $E[(S-\$6)_+]$. (Answer: $E[S]=\$10$, $\operatorname{Var}(S)=50$ dollars squared, and the stop-loss expectation is $4+8e^{-2}\approx\$5.08$.)
A single claim is either a constant $\$10$ or is $\$0$ and $\$20$ with equal probabilities. Compare the stop-loss expectations at $d=\$12$. (Answer: The constant model gives $\$0.00$; the two-point model gives $\$4.00$.)
Severity is deterministically $\$3$. In Model A, $N=2$ with certainty. In Model B, $N=0$ or $N=4$, each with probability $0.5$. Find each aggregate mean, variance, and stop-loss expectation at $d=\$7$. (Answer: Both means are $\$6$; the variances are $0$ and $36$ dollars squared; the stop-loss expectations are $\$0.00$ and $\$2.50$, respectively.)
Model A has aggregate loss $\$0$ or $\$20$, each with probability $0.5$. Model B has losses $\$0,\$5,\$25$ with probabilities $0.2,0.5,0.3$. At $d=\$15$, compare the means, variances, and stop-loss expectations. (Answer: Both means are $\$10$, both variances are $100$ dollars squared, and the stop-loss expectations are $\$2.50$ and $\$3.00$, respectively.)

What You'll Learn