Exam Workshop: Compare Pricing Across Severity Models

Why Compare Prices Across Severity Models?

A reinsurance layer can have very different prices under two severity models even when both models have the same expected claim size. The reason is that a layer does not respond equally to every part of the loss distribution: an attachment point filters out smaller losses, and a limit caps the payment on larger ones. Pricing therefore depends on the probability mass in the part of the severity distribution that the layer actually covers.

This comparison is useful when alternative fitted models provide plausible descriptions of the same claim data. The objective is not to declare one model universally correct. It is to calculate each indicated price on the same claim-count, coverage, and loading bases, then identify which assumptions explain the difference. The analysis below treats fitted parameters as fixed; it does not include uncertainty in those parameter estimates.

Definition: For loss $X$, attachment $d$, and layer width $L$, the payment per loss to a layer attaching at $d$ and covering the next $L$ dollars is $$ Y=\min\bigl((X-d)_+,L\bigr), \qquad (z)_+=\max(z,0). $$ Thus $Y=0$ when $X\le d$, $Y=X-d$ when $d<X<d+L$, and $Y=L$ when $X\ge d+L$. The layer payment is not the full loss above the attachment: it is capped at the layer width.

Expected Payment From a Layer

For a nonnegative continuous loss with survival function $S_X(x)=P(X>x)$, a capped layer payment can be represented as accumulated indicators: $Y$ is the length of the portion of the interval from 0 to $L$ for which $X>d+t$. Taking expectations gives the standard tail-integral formula.

E[Y]=\int_0^L S_X(d+t)\,dt =\int_d^{d+L}S_X(x)\,dx.

The integral uses the survival function over the layer itself. It is not enough to compare $E[X]$, or even the probability of exceeding the attachment: the expected payment also depends on how far losses continue into the layer. For a finite layer, another useful identity is the difference of two unlimited-layer expectations:

$$ E[Y]=E[(X-d)_+]-E[(X-d-L)_+]. $$

This follows because the first term includes all loss above $d$, while the second removes the part above $d+L$. The difference leaves precisely the payment capped at $L$. For a capped payment, its second moment is

E[Y^2]=2\int_0^L tS_X(d+t)\,dt.

To see why, use the tail-integral identity $E[Y^2]=\int_0^\infty 2tP(Y>t)\,dt$. Since $0\le Y\le L$, the integrand is zero for $t\ge L$, and for $0\le t<L$, $P(Y>t)=S_X(d+t)$. This gives the displayed formula.

Two Fitted Models for a Consistent Comparison

The examples use the following hypothetical fitted severity models. Both have mean loss of $1,000, so differences in layer prices are not caused by differences in the average severity.

Model	Survival function	Parameters and mean
Exponential	$S_X(x)=e^{-x/\mu}$, $x\ge0$	$\mu=1{,}000$; $E[X]=\mu=1{,}000$
Lomax (Pareto II)	$S_X(x)=(1+x/\theta)^{-\alpha}$, $x\ge0$	$\theta=2{,}000,\ \alpha=3$; $E[X]=\theta/(\alpha-1)=1{,}000$

The exponential model has a tail that declines exponentially. The Lomax model has a heavier, polynomially declining tail. With these parameters the Lomax survival probability is below the exponential survival probability at some moderate loss amounts and above it farther into the tail. Therefore, a price comparison can depend on the layer’s attachment and width, not just on the two models’ common mean.

Useful formulas: For an exponential model with mean $\mu$, $$ E[Y]=\mu\left(e^{-d/\mu}-e^{-(d+L)/\mu}\right). $$ For a Lomax model with $\alpha>1$, $$ E[Y]=\frac{\theta}{\alpha-1} \left[ \left(1+\frac{d}{\theta}\right)^{1-\alpha} - \left(1+\frac{d+L}{\theta}\right)^{1-\alpha} \right]. $$ These are layer-payment formulas, not formulas for an uncapped excess above $d$.

Worked Example: Price a High Layer Under Both Models

A hypothetical reinsurer covers a $1,000 layer attaching at $3,000. The fitted severity models are the exponential and Lomax models specified above. There are 100 expected claims, and the indicated gross premium is defined to be 115% of expected ceded claims. Calculate the premium under each model.

Method. Calculate the expected payment per claim for the capped layer, multiply by the expected claim count, and then apply the stated 15% loading. The loading is applied to expected aggregate claim cost, not to the severity mean.

Exponential model. Here $\mu=1{,}000$, $d/\mu=3$, and $(d+L)/\mu=4$:

E[Y] =1{,}000(e^{-3}-e^{-4}) =1{,}000(0.0497871-0.0183156) =\$31.47.

The expected ceded claims are $100(31.4714)=\$3{,}147.14$. The indicated gross premium is $1.15(3{,}147.14)=\$3{,}619.21$.

Lomax model. Here $\theta=2{,}000$, $\alpha=3$, $d/\theta=1.5$, and $(d+L)/\theta=2$:

\begin{aligned} E[Y] &=\frac{2{,}000}{2}\left[(2.5)^{-2}-(3)^{-2}\right]\\ &=1{,}000(0.160000-0.111111)\\ &=\$48.89. \end{aligned}

The expected ceded claims are $100(48.8889)=\$4{,}888.89$. The indicated gross premium is $1.15(4{,}888.89)=\$5{,}622.22$.

Check and trap avoided. The Lomax indication exceeds the exponential indication by $\$5{,}622.22-\$3{,}619.21=\$2{,}003.01$. The calculation applies the same layer and claim count to both models. A common mistake is to use the uncapped expected excess above $3,000. For example, the Lomax expectation above the attachment is $\$160$ per claim, not the correct capped layer payment of $\$48.89$. The finite layer requires subtracting the expected excess above $4,000 as well.

From Per-Claim Prices to Aggregate Cost

Let $N$ be the number of claims, and let the layer payments $Y_1,Y_2,\ldots$ be independent and identically distributed, independent of $N$. The annual aggregate ceded cost is $Z=\sum_{j=1}^{N}Y_j$. If $N$ is Poisson with mean $\lambda$, then

E[Z]=\lambda E[Y], \qquad \operatorname{Var}(Z)=\lambda E[Y^2].

The variance expression uses the second moment of the payment, not its variance. Conditional on $N$, the aggregate has mean $NE[Y]$ and variance $N\operatorname{Var}(Y)$. Applying the law of total variance gives

\begin{aligned} \operatorname{Var}(Z) &=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2\\ &=\lambda\operatorname{Var}(Y)+\lambda(E[Y])^2\\ &=\lambda E[Y^2], \end{aligned}

because a Poisson variable has both mean and variance $\lambda$. If the claim count is fixed at $n$ instead, the variance is $n\operatorname{Var}(Y)$. The Poisson formula should not be used automatically for a fixed-count portfolio or for a frequency model with a different variance.

Worked Example: Compare Annual Aggregate Moments

For the same hypothetical $1,000 layer attaching at $3,000, suppose annual claim count is Poisson with mean $\lambda=100$, independent of severities. Calculate the aggregate mean and variance under both models.

Method. Use $E[Z]=\lambda E[Y]$ and $\operatorname{Var}(Z)=\lambda E[Y^2]$. First calculate the per-claim second moment from the capped-layer integral.

Exponential model. With $d=3{,}000$, $L=1{,}000$, and $\mu=1{,}000$:

\begin{aligned} E[Y^2] &=2\int_0^{1{,}000}t e^{-(3{,}000+t)/1{,}000}\,dt\\ &=2(1{,}000)^2e^{-3}\left[1-2e^{-1}\right]\\ &=26{,}311.58. \end{aligned}

The aggregate mean is $100(31.4714)=\$3{,}147.14$. Its variance is $100(26{,}311.58)=2{,}631{,}158$ dollars squared. As a check, the per-claim variance is $26{,}311.58-(31.4714)^2=25{,}321.13$, which is positive.

Lomax model. Set $a=1+d/\theta=2.5$ and $b=1+(d+L)/\theta=3$. Substituting $z=1+(d+t)/\theta$ in the second-moment integral gives

\begin{aligned} E[Y^2] &=2\theta^2\int_a^b(z-a)z^{-3}\,dz\\ &=\theta^2\left(\frac{1}{a}-\frac{2}{b}+\frac{a}{b^2}\right)\\ &=4{,}000{,}000(0.4-0.666667+0.277778)\\ &=44{,}444.44. \end{aligned}

The aggregate mean is $100(48.8889)=\$4{,}888.89$, and the aggregate variance is $100(44{,}444.44)=4{,}444{,}444$ dollars squared. The per-claim variance check is $44{,}444.44-(48.8889)^2=42{,}054.32>0$.

Check and trap avoided. The Lomax model gives both the higher aggregate mean and the higher aggregate variance for this layer. Using $100\operatorname{Var}(Y)$ for the aggregate variance would omit the variation in the Poisson claim count; the correct Poisson calculation is $100E[Y^2]$.

Attachment Choice Reveals Tail Sensitivity

Comparing models at more than one attachment helps explain a price difference. For a fixed layer width, increasing the attachment moves the covered interval farther into the severity tail. If one model assigns relatively more probability to large losses, its layer price can become increasingly high relative to the other model at higher attachments. This is a model-specific comparison, not a rule that every heavy-tailed model must price every layer higher.

Worked Example: Test How the Difference Changes by Layer

Using the same two fitted models, compare the expected payment per claim for a $1,000 layer attaching at $2,000 with the payment for the $1,000 layer attaching at $3,000. State how the model difference changes across the two layers.

Method. Apply each model’s capped-layer formula separately at each attachment; do not infer the new price by multiplying the old price by a common factor.

Layer attaching at $2,000. Under the exponential model:

E[Y]=1{,}000(e^{-2}-e^{-3}) =1{,}000(0.1353353-0.0497871) =\$85.55.

Under the Lomax model:

E[Y]=1{,}000\left[(2)^{-2}-(2.5)^{-2}\right] =1{,}000(0.25-0.16) =\$90.00.

The Lomax amount is $\$90.00-\$85.55=\$4.45$ higher, or about $4.45/85.55=5.20\%$ above the exponential amount.

Layer attaching at $3,000. From the earlier layer calculations, the exponential payment is $\$31.47$ and the Lomax payment is $\$48.89$. The difference is $\$17.42$, about $17.42/31.47=55.35\%$ of the exponential amount.

Check and trap avoided. The higher layer shows a much larger relative price difference for these particular models and parameters. The absolute expected payment falls for each model as the layer moves up, but the model indications do not fall at the same rate. A common mistake is to conclude from equal means that the layer prices should be similar; the calculations show that equal overall means do not imply equal tail probabilities or layer costs.

Exam Checks for a Model Comparison

Before interpreting a comparison, verify that both indications use the same loss basis, attachment, layer width, claim-count basis, and pricing adjustment. The fitted severity model changes; the contract definition should not. Also distinguish expected cost from a tail measure: aggregate mean and variance are not themselves a high quantile or a probability of exceeding a threshold.

Fix the contract transformation.
Write the payment per loss, including both attachment and cap, before substituting a severity model.

Calculate each model’s layer expectation.
Integrate its survival function over the covered severity interval.

Apply frequency and pricing bases consistently.
Use the stated claim-count model for aggregate costs, then apply only the stated loading or other adjustment.

Explain the difference in context.
Identify which part of the severity tail drives the comparison and avoid attributing the result to the mean alone.

Key takeaway: Reinsurance prices across severity models are comparable only when the coverage transformation and claim-count basis match. Layer survival integrals show what severity region drives expected cost; compound-model moments then translate that per-claim cost into aggregate results. This treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated continuous severity model and layer definition. Round monetary answers to the nearest cent.

A fitted exponential severity model has mean $\mu=\$500$. A hypothetical layer attaches at $\$1{,}000$ and has width $\$500$. Find its expected payment per claim. (Answer: $\$42.77$)
A fitted Lomax model has $\theta=\$1{,}000$ and $\alpha=3$. A layer attaches at $\$1{,}000$ and has width $\$1{,}000$. Find its expected payment per claim. (Answer: $\$69.44$)
Annual claim count is Poisson with mean 40. The expected payment per claim to a specified layer is $\$25$. Find expected annual aggregate layer cost. (Answer: $\$1{,}000.00$)
Annual claim count is Poisson with mean 20, and a specified layer payment $Y$ has $E[Y^2]=900$ dollars squared. Find the annual aggregate cost variance. (Answer: $18{,}000$ dollars squared)
A Lomax severity model has $\theta=\$1{,}000$ and $\alpha=2$. A layer attaches at $\$2{,}000$ and has width $\$1{,}000$. Find its expected payment per claim. (Answer: $\$83.33$)

Model	Survival function	Parameters and mean
Exponential	\(S_X(x)=e^{-x/\mu}\), \(x\ge0\)	\(\mu=1{,}000\); \(E[X]=\mu=1{,}000\)
Lomax (Pareto II)	\(S_X(x)=(1+x/\theta)^{-\alpha}\), \(x\ge0\)	\(\theta=2{,}000,\ \alpha=3\); \(E[X]=\theta/(\alpha-1)=1{,}000\)

What You'll Learn