Worked Layer-Loss Calculations For Common Severity Distributions

Layer Payments and Severity Models

For an excess layer, pricing begins with the payment generated by one ground-up loss. The layer’s expected payment depends on the severity distribution, the attachment point, and the limit. A loss just above the attachment may produce only a small payment; a sufficiently large loss exhausts the limit. Calculating the expected payment therefore requires more than multiplying the limit by the probability that a loss reaches the layer.

This tutorial calculates per-occurrence expected payments for exponential, Pareto, and lognormal severities. These calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. They provide a severity-level building block; they do not by themselves include a claim frequency assumption or calculate an annual aggregate loss.

Definition: Let $X\ge 0$ be a ground-up loss, let $a\ge 0$ be a layer’s attachment point, and let $h>0$ be its limit. The payment from the layer is $Y=\min((X-a)_+,h)$, where $(z)_+=\max(z,0)$. Equivalently, $Y=(X-a)_+-(X-a-h)_+$. The layer covers losses above $a$, up to the endpoint $a+h$.

Write $C(d)=E[(X-d)_+]$ for the expected excess loss above threshold $d$. This is an expected amount per ground-up loss, not a probability. The expected payment for the layer is $E[Y]=C(a)-C(a+h)$. Calculating the two excess expectations and subtracting them is often easier than integrating the layer payment directly.

Why the Excess-Loss Difference Works

The identity follows by checking what happens at each possible loss amount. If $X\le a$, both excess terms are zero. If $a<X\le a+h$, the first term is $X-a$ and the second is zero. If $X>a+h$, their difference is $(X-a)-(X-a-h)=h$. These are exactly the three amounts paid by the layer.

\min((X-a)_+,h)=(X-a)_+-(X-a-h)_+

Taking expectations on both sides gives $E[Y]=C(a)-C(a+h)$. For a nonnegative loss, the tail-integral identity also gives $C(d)=\int_d^\infty S_X(x)\,dx$. To see why, for each fixed loss value $X$, the amount $(X-d)_+$ is the length of the interval from $d$ to $X$, when $X>d$. Integrating the indicator that $X$ exceeds each point in that interval and then taking expectations yields the survival-function integral. This identity assumes a nonnegative severity; it applies to the continuous distributions used here.

Exponential Severity

Suppose $X$ has an exponential distribution with mean $\theta>0$. Its survival function is $S_X(x)=e^{-x/\theta}$ for $x\ge 0$. Integrating the tail above $d$ gives $C(d)=\theta e^{-d/\theta}$. Thus, for an attachment $a$ and limit $h$, the expected layer payment is the difference between the excess expectations at $a$ and $a+h$.

C(d)=\int_d^\infty e^{-x/\theta}\,dx =\theta e^{-d/\theta}, \qquad E[Y]=\theta\left(e^{-a/\theta}-e^{-(a+h)/\theta}\right).

The exponential model is memoryless: conditional on exceeding a threshold, the remaining expected amount above that threshold is $\theta$. This property makes the excess formula especially direct. It does not mean that every loss reaching a finite layer pays the full limit.

Worked Example: Exponential Severity in a Finite Layer

A hypothetical property severity model is exponential with mean $\theta=\$60{,}000$. Calculate the expected payment per occurrence for a layer attaching at $a=\$30{,}000$ with limit $h=\$30{,}000$. Use dollars throughout and round the final payment to the nearest cent.

Method. The exponential excess formula gives $E[Y]=\theta(e^{-a/\theta}-e^{-(a+h)/\theta})$. The layer endpoint is $a+h=\$60{,}000$.

Substitute the values: $a/\theta=30{,}000/60{,}000=0.5$, and $(a+h)/\theta=60{,}000/60{,}000=1$. Therefore, $e^{-0.5}=0.6065306597$ and $e^{-1}=0.3678794412$.

The difference is $0.6065306597-0.3678794412=0.2386512185$. Multiplying by the mean gives $\$60{,}000(0.2386512185)=\$14{,}319.0731\ldots$. Rounded only at the final step, the expected payment is \$14,319.07 per occurrence.

Check. The expected payment is positive and below the \$30,000 layer limit. A common mistake is to calculate $hP(X>a)=\$30{,}000e^{-0.5}=\$18{,}195.92$, as if every loss entering the layer exhausted it. That overstates the payment: losses between \$30,000 and \$60,000 pay only the amount above \$30,000. Subtracting the excess above the endpoint correctly removes the part beyond the layer.

Pareto Severity

Consider a Pareto Type I severity with scale $b>0$ and shape $\alpha>1$. Its survival function is $S_X(x)=(b/x)^\alpha$ for $x\ge b$, and $X\ge b$. The condition $\alpha>1$ ensures a finite mean. For thresholds $d\ge b$, integrating the survival function gives a convenient closed form for the expected excess. If $d<b$, this form does not apply directly because the survival function is 1 between 0 and $b$; the integral must include that interval as well.

C(d)=\int_d^\infty \left(\frac{b}{x}\right)^\alpha dx =\frac{b^\alpha d^{1-\alpha}}{\alpha-1} =\frac{b}{\alpha-1}\left(\frac{b}{d}\right)^{\alpha-1}, \qquad d\ge b.

For a layer with $a\ge b$, both $a$ and $a+h$ satisfy the formula’s condition, so its expected payment is $C(a)-C(a+h)$. A smaller shape parameter means a heavier Pareto tail, provided the parameters are compared on a meaningful common basis. The expected cost of a particular finite layer still depends on where that layer sits relative to the severity scale.

Worked Example: Pareto Severity Above Its Scale

For a hypothetical liability portfolio, suppose ground-up severity is Pareto Type I with scale $b=\$10{,}000$ and shape $\alpha=2.5$. Find the expected payment per occurrence for a layer attaching at \$20,000 with a \$20,000 limit. Both thresholds are at or above $b$. Round the final payment to the nearest cent.

Method. Apply the Pareto excess formula at the attachment $d=\$20{,}000$ and endpoint $d=\$40{,}000$. Since $\alpha-1=1.5$, the coefficient $b/(\alpha-1)$ is $\$10{,}000/1.5=\$6{,}666.6667$.

At the attachment, $b/d=10{,}000/20{,}000=0.5$, so $C(20{,}000)=\$6{,}666.6667(0.5)^{1.5}$. Since $(0.5)^{1.5}=0.35355339$, this is approximately $\$2{,}357.02$.

At the endpoint, $b/d=10{,}000/40{,}000=0.25$, and $(0.25)^{1.5}=0.125$. Thus $C(40{,}000)=\$6{,}666.6667(0.125)=\$833.3333$. Subtracting gives $\$2{,}357.0226-\$833.3333=\$1{,}523.6893\ldots$, or \$1,523.69 per occurrence to the nearest cent.

Check. The result is positive and less than the \$20,000 limit. The formula is valid at both thresholds because each is at least the \$10,000 scale. Using the formula at an attachment below $b$ without adding the survival integral from that attachment to $b$ would be a condition error.

Lognormal Severity

Suppose $X$ is lognormal, meaning $\ln X\sim N(\mu,\sigma^2)$, with $\sigma>0$. In this notation the second parameter of the normal distribution is the variance; here $\sigma^2$ is the variance of $\ln X$. Let $\Phi$ denote the standard normal cumulative distribution function. For $d>0$, the expected excess has a closed form involving two normal probabilities. The first term is a truncated first moment of $X$; the second accounts for the threshold amount on the event that the loss exceeds $d$.

C(d)=E[X\,1_{\{X>d\}}]-dP(X>d) =e^{\mu+\sigma^2/2} \Phi\left(\frac{\mu+\sigma^2-\ln d}{\sigma}\right) -d\Phi\left(\frac{\mu-\ln d}{\sigma}\right).

To justify the truncated-moment term, write $Z=(\ln X-\mu)/\sigma$, so $Z\sim N(0,1)$ and $X=e^{\mu+\sigma Z}$. The event $X>d$ is $Z>c$, where $c=(\ln d-\mu)/\sigma$. Completing the square in the standard normal density gives $e^{\sigma z}\phi(z)=e^{\sigma^2/2}\phi(z-\sigma)$, where $\phi$ is the standard normal density. Therefore, $E[X1_{\{X>d\}}]=e^{\mu+\sigma^2/2}\Phi(\sigma-c)$, which is the first term in the formula. Also, $P(X>d)=\Phi(-c)=\Phi((\mu-\ln d)/\sigma)$. Subtracting $dP(X>d)$ yields $C(d)$. For a layer, subtract $C(a+h)$ from $C(a)$; use the formula only at positive thresholds.

Worked Example: Lognormal Severity in a Finite Layer

A hypothetical commercial property severity model has $\ln X\sim N(\ln 50,0.25)$, where $X$ is measured in thousands of dollars. Thus the lognormal median is \$50,000 and $\sigma=0.5$. Calculate the expected payment per occurrence for a layer attaching at \$25,000 with a \$25,000 limit. Work in thousands of dollars, use unrounded normal probabilities in the calculations, and round the final answer to the nearest \$10.

Method. The layer endpoint is \$50,000. Evaluate the lognormal excess formula at $d=25$ and $d=50$, in thousands, then take their difference. The mean severity is $e^{\mu+\sigma^2/2}=e^{\ln 50+0.125}=56.6574$ thousand dollars (shown to four decimal places).

At $d=25$, the two standard normal arguments are $(\ln 50+0.25-\ln 25)/0.5=1.8863$ and $(\ln 50-\ln 25)/0.5=1.3863$. Using the unrounded normal probabilities (approximately $\Phi(1.8863)=0.9704$ and $\Phi(1.3863)=0.9172$), the excess expectation is approximately $C(25)=32.0495$ thousand dollars.

At $d=50$, the arguments are $0.5$ and $0$. Using $\Phi(0.5)=0.6914625$ and $\Phi(0)=0.5$, $C(50)=56.6574(0.6914625)-50(0.5)$, with the mean and probabilities evaluated at full precision, is approximately $14.1765$ thousand dollars.

The layer expectation is $C(25)-C(50)=32.0495-14.1765=17.8730$ thousand dollars. Rounded to the nearest \$10, the expected payment is approximately \$17,870 per occurrence.

Check. The payment is positive and below the \$25,000 limit. The excess at the endpoint is subtracted because losses above \$50,000 exceed this layer; they do not increase its payment beyond the limit. Rounding the normal probabilities too early can shift the final result, so retain calculator precision until the final rounding.

Comparing Tail Behavior Across Models

Distribution labels alone do not determine the cost of every layer. The attachment and endpoint matter, and distributions with the same mean can place different amounts of expected loss in a particular band. The following comparison holds the mean at \$20,000 and considers a high layer from \$100,000 to \$200,000. All amounts in the calculation are in thousands of dollars.

Worked Example: Equal Means, Different High-Layer Costs

Compare the expected payment per occurrence for a hypothetical layer attaching at \$100,000 with a \$100,000 limit under two severities: (1) exponential with mean \$20,000; and (2) Pareto Type I with scale $b=\$10{,}000$ and shape $\alpha=2$. The Pareto mean is $\alpha b/(\alpha-1)=2(10{,}000)=\$20{,}000$, so both models have the same mean. Report final amounts to the nearest cent.

Exponential calculation. In thousands, $\theta=20$, $a=100$, and $a+h=200$. Therefore, $E[Y]=20(e^{-5}-e^{-10})$. Using $e^{-5}=0.0067379470$ and $e^{-10}=0.0000453999$, the difference is $0.0066925471$. The expected payment is $20(0.0066925471)=0.1338509$ thousand dollars, or \$133.85.

Pareto calculation. For $\alpha=2$, the excess formula is $C(d)=b^2/d$ when $d\ge b$. In thousands, $b=10$, so $C(100)=10^2/100=1$ and $C(200)=10^2/200=0.5$. The layer expectation is $1-0.5=0.5$ thousand dollars, or \$500.00.

Check and interpretation. Both calculations use the same attachment, limit, and mean, yet the Pareto model produces the larger expected payment for this high layer. Its heavier tail places relatively more expected loss far above the mean. This comparison is specific to the parameters and layer shown; it is not a rule that one distribution must cost more at every attachment.

Practical Checks and Common Traps

These formulas are most useful when paired with a few basic checks. The excess expectation $C(d)$ should not increase as $d$ increases, because raising the threshold cannot increase the amount above it. Consequently $C(a)-C(a+h)$ must be nonnegative. The layer payment also cannot exceed $h$ for any individual loss, so its expected payment cannot exceed the limit.

A particularly common error is to calculate $hP(X>a)$, treating every loss that reaches the layer as a full-limit claim. That quantity counts the limit for losses that may only penetrate the layer slightly. Another error is to use the Pareto formula below its scale without accounting for the interval on which its survival function is 1. For the lognormal formula, the arguments of the two normal probabilities differ: the truncated first moment uses $\mu+\sigma^2$, while the exceedance probability uses $\mu$. Mixing these arguments changes the result.

Key takeaway: For a layer attaching at $a$ with limit $h$, calculate the expected excess above $a$, subtract the expected excess above $a+h$, and check that the result is between zero and $h$. Confirm the severity model’s assumptions and the formula’s threshold conditions before substituting values.

Check Your Understanding

For each problem, calculate a per-occurrence expected layer payment. Round dollar amounts to the nearest cent unless the answer is given as an exact expression.

A hypothetical exponential severity has mean \$40,000. A layer attaches at \$20,000 and has a \$20,000 limit. Find its expected payment. (Answer: \$9,546.05)
A hypothetical Pareto Type I severity has scale $b=\$8,000$ and shape $\alpha=3$. Find the expected payment for a layer attaching at \$16,000 with a \$16,000 limit. (Answer: \$750.00)
A hypothetical exponential severity has mean \$25,000. Calculate the expected payment for a layer attaching at zero with a \$50,000 limit. (Answer: \$21,616.62)
A hypothetical Pareto Type I severity has scale $b=\$5,000$ and shape $\alpha=2$. Calculate the expected payment for a layer attaching at \$10,000 with a \$10,000 limit. (Answer: \$1,250.00)
For a hypothetical lognormal severity with $\ln X\sim N(\ln 12,1)$, where $X$ is in thousands of dollars, calculate the expected payment for a layer attaching at 12 with a 12-thousand-dollar limit. Express the answer using $\Phi$, without numerical rounding. (Answer: $12e^{1/2}\{\Phi(1)-\Phi(1-\ln 2)\}-6+24\Phi(-\ln 2)$ thousand dollars, approximately \$4,230)

What You'll Learn