Worked Case: Pareto Fitting Through Per-Claim Reinsurance

Introduction

A reinsurance calculation can be numerically correct at each step and still produce the wrong answer if the data and payment amounts use different bases. A tail model fitted only to claims above a threshold describes the severity of those claims, not the probability that an arbitrary claim reaches the threshold. That distinction matters when moving from a fitted Pareto tail to an expected payment per claim or an expected layer cost.

This worked case fits a Pareto tail from threshold-exceedance counts, estimates payments above a deductible, and prices a per-claim excess-of-loss layer. The data and products below are hypothetical. Amounts are nominal dollars; there is no discounting or mortality assumption. The calculation is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Suppose claim severity $X$ has a Pareto tail above threshold $u$, with shape parameter $\alpha>0$. Its conditional survival function is $$ P(X>x\mid X>u)=\left(\frac{u}{x}\right)^\alpha,\qquad x\ge u. $$ If $q=P(X>u)$, the unconditional survival function in the modeled tail is $$ S_X(x)=P(X>x)=q\left(\frac{u}{x}\right)^\alpha,\qquad x\ge u. $$ The tail model does not specify the severity distribution below $u$. Calculations that depend only on payments above $u$ can nevertheless be made from this tail and $q$.

Fitting the Tail From Limited Data

When the available data contain counts at two thresholds, the ratio of those counts can identify the shape parameter without requiring every individual claim amount. For thresholds $u$ and $ku$, where $k>1$, the Pareto model gives

P(X>ku\mid X>u)=k^{-\alpha}.

If $n_u$ claims exceed $u$ and $n_{ku}$ of those exceed $ku$, the empirical conditional exceedance proportion is $n_{ku}/n_u$. Equating it to the model probability gives the grouped-data estimate

\widehat{\alpha}=-\frac{\ln(n_{ku}/n_u)}{\ln k}.

This is also the maximum-likelihood estimate when the grouped observation is treated as a binomial count: conditional on exceeding $u$, each claim either exceeds $ku$ or does not. The threshold probability $q$ is estimated separately from the proportion of all claims that exceed $u$. In practice, both estimates have sampling uncertainty; the worked calculations below use plug-in estimates and do not add a parameter-uncertainty adjustment.

Worked Example: Fitting a Pareto Tail From Two Threshold Counts

A hypothetical portfolio has 10,000 observed claims. Of these, 800 exceed $1,000, and 200 exceed $2,000. Fit the Pareto tail above $1,000 using these grouped counts. State both the tail shape estimate and the estimated proportion of all claims above $1,000.

Method. Use the conditional exceedance proportion among claims above $1,000 to estimate $\alpha$, and use the full claim count to estimate $q$. The threshold ratio is $k=2$.

First, estimate the probability of exceeding $2,000 conditional on exceeding $1,000:

\widehat{P}(X>2{,}000\mid X>1{,}000)=\frac{200}{800}=0.25.

Under the Pareto tail model, this probability is $2^{-\alpha}$. Therefore,

2^{-\widehat{\alpha}}=0.25=\frac{1}{4}=2^{-2}, \qquad \widehat{\alpha}=2.

Next, estimate the probability that an arbitrary claim exceeds the tail threshold:

\widehat{q}=\frac{800}{10{,}000}=0.08.

Thus, for $x\ge \$1{,}000$, the fitted unconditional survival function is

\widehat{S}_X(x)=0.08\left(\frac{1{,}000}{x}\right)^2.

Check. At $2,000, the fitted unconditional probability is $0.08(1/2)^2=0.02$, corresponding to $0.02(10{,}000)=200$ claims, as observed. The common mistake is to use $200/10{,}000$ as the conditional probability for fitting $\alpha$; that mixes the full claim population with the claims selected above $1,000.

Expected Payments Above a Deductible

For a per-claim deductible $d$, the insurer’s payment before any upper limit is $(X-d)_+=\max(X-d,0)$. If $d\ge u$, the expectation depends only on the modeled tail. The tail-integral identity follows by writing a nonnegative payment $Z$ as the area under its exceedance indicators: $Z=\int_0^\infty I(Z>t)\,dt$. Taking expectations gives $E[Z]=\int_0^\infty P(Z>t)\,dt$. Applying this to $(X-d)_+$ yields

E[(X-d)_+]=\int_d^\infty S_X(x)\,dx.

For $d\ge u$ and $\alpha>1$, substitute the unconditional Pareto tail and integrate:

E[(X-d)_+] =q\int_d^\infty \left(\frac{u}{x}\right)^\alpha dx =\frac{q u^\alpha d^{1-\alpha}}{\alpha-1}.

The condition $\alpha>1$ is essential: if $\alpha\le 1$, the modeled tail has an infinite expected excess above any finite deductible. When the desired quantity is conditional on $X>u$, omit $q$; that result is not yet the expected payment per arbitrary claim.

Worked Example: Expected Payment With a Deductible and Policy Limit

Use the fitted tail from the previous example: $u=\$1{,}000$, $q=0.08$, and $\alpha=2$. A hypothetical policy has a $1,500 deductible and pays at most $2,500 per claim. Find the expected insurer payment per claim in the full claim population.

Method. The payment is $\min((X-d)_+,L)$, where $d=\$1{,}500$ and $L=\$2{,}500$. It equals the difference between the excess above $d$ and the excess above $d+L=\$4{,}000$. For $\alpha=2$, the unconditional excess above any $a\ge u$ is

E[(X-a)_+]=\frac{q u^2}{a}.

At the deductible, this is

E[(X-1{,}500)_+] =\frac{0.08(1{,}000)^2}{1{,}500} =\frac{80{,}000}{1{,}500} =\$53.3333\ldots.

At the point where the policy limit is exhausted, it is

E[(X-4{,}000)_+] =\frac{0.08(1{,}000)^2}{4{,}000} =\frac{80{,}000}{4{,}000} =\$20.

Subtract to obtain the expected capped payment per claim:

E[\min((X-1{,}500)_+,2{,}500)] =53.3333\ldots-20 =\$33.33.

The result is rounded to the nearest cent only at the final step. Check. It is positive and less than the uncapped expected payment of $53.3333. A common error is to treat the conditional tail expectation as the full-population expectation; the factor $q=0.08$ has already been included here.

Pricing a Per-Claim Excess-of-Loss Layer

A per-claim layer attaching at $a$ with limit $L$ pays $\min((X-a)_+,L)$ on each claim. Its upper exhaustion point is $b=a+L$. For $a\ge u$, its expected payment per claim is the area under the unconditional survival curve from $a$ to $b$:

E[\min((X-a)_+,L)] =\int_a^{a+L} S_X(x)\,dx.

For a Pareto tail with $\alpha>1$, this becomes

E[\min((X-a)_+,L)] =\frac{q u^\alpha}{\alpha-1} \left(a^{1-\alpha}-(a+L)^{1-\alpha}\right).

The formula is per claim. If the expected number of claims is $m$, and claim severities follow the modeled distribution independently of claim count, the expected total layer recovery is $m$ times the expected layer payment per claim. This multiplication uses linearity of expectation; it does not require claim severities to be mutually independent.

Worked Example: Expected Cost of a Per-Claim Reinsurance Layer

A hypothetical reinsurer covers the layer from $2,000 to $5,000 on each claim. Use $u=\$1{,}000$, $q=0.08$, and $\alpha=2$. The insurer expects 1,200 claims during the year. Calculate the expected recovery per claim in the full claim population and the expected total recovery for the year.

Method. The attachment is $a=\$2{,}000$, the limit is $L=\$3{,}000$, and the exhaustion point is $a+L=\$5{,}000$. For $\alpha=2$, the layer expectation per claim is

\int_{2{,}000}^{5{,}000}0.08\left(\frac{1{,}000}{x}\right)^2dx =0.08(1{,}000)^2 \left(\frac{1}{2{,}000}-\frac{1}{5{,}000}\right).

Evaluate the terms inside the parentheses:

\frac{1}{2{,}000}-\frac{1}{5{,}000} =0.0005-0.0002=0.0003.

Therefore, the expected recovery per claim in the full claim population is

0.08(1{,}000{,}000)(0.0003)=\$24.

Multiplying by the expected claim count gives

1{,}200(\$24)=\$28{,}800.

Check. Conditional on exceeding $1,000, the layer expectation is $\$24/0.08=\$300$; multiplying that conditional amount by the threshold-exceedance probability returns $24 per arbitrary claim. The common mistake is to report $300 as the per-claim cost for the full portfolio, overstating it by a factor of $1/q=12.5$.

Keep the Bases Aligned

A reliable solution keeps three quantities distinct: the probability of reaching the modeled threshold, the severity distribution conditional on reaching it, and the payment expectation for an arbitrary claim. For a deductible or layer attaching above $u$, either integrate the unconditional survival function directly or calculate conditionally and multiply by $q$ exactly once.

Identify the data denominator.
Use claims above the lower threshold as the denominator for a conditional tail ratio. Use all claims as the denominator when estimating the proportion $q$ that reaches the threshold.

Fit and check the tail.
Estimate $\alpha$ from the threshold ratio, then verify that the fitted conditional exceedance probability reproduces the observed count proportion.

Apply the contract once.
For an attachment $a$ and limit $L$, integrate from $a$ to $a+L$. Do not apply a deductible or attachment again after the payment has already been calculated.

Convert to the requested basis.
Use the unconditional per-claim expectation for a portfolio total, then multiply by expected claim count when the severity and count bases are consistent.

The Pareto mean is finite only when $\alpha>1$, so a fitted value at or below one is a warning that expected-cost pricing under this unbounded tail model is infinite. A policy limit can make the expected payment finite even when the uncapped mean is infinite, because its layer integral has a finite upper endpoint. That does not make the uncapped expectation finite.

Key takeaway: A conditional Pareto tail fit does not by itself determine an expected payment per claim. Carry the threshold probability $q$ into the unconditional survival function, integrate over the contract’s actual attachment and exhaustion points, and multiply by claim count only after obtaining the per-claim expectation.

Check Your Understanding

For each problem, state whether the requested amount is conditional on exceeding the tail threshold or is per claim in the full population.

A Pareto tail begins at $1,000. Among claims exceeding $1,000, 90 of 810 exceed $3,000. Estimate $\alpha$ by the grouped tail-ratio method. (Answer: $\widehat{\alpha}=2$.)
A fitted tail has $u=\$1{,}000$, $q=0.12$, and $\alpha=2$. Find the expected payment per claim above a $2,000 deductible, with no upper limit. (Answer: $60.00.)
A fitted tail has $u=\$500$, $q=0.04$, and $\alpha=3$. A per-claim layer attaches at $1,000 and has a $1,000 limit. Find its expected payment per claim in the full population, rounded to cents. (Answer: $1.88.)
Among 600 claims exceeding $2,000, 150 exceed $4,000. Estimate the Pareto shape above $2,000. If 6% of all claims exceed $2,000, find the expected payment per claim above a $3,000 deductible, with no limit. (Answer: $\widehat{\alpha}=2$; $80.00.)
A fitted tail has $u=\$1{,}000$, $q=0.10$, and $\alpha=2$. There are 2,500 expected claims, and a per-claim layer attaches at $2,500 with a $2,500 limit. Find the expected total recovery. (Answer: $50,000.)

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