Condition Correctly on Deductible-Exceeding Claims

Why the Claim Basis Matters

A deductible changes which claims are observed in a data set. If a file contains only claims that exceeded the deductible, its severity distribution describes exceeding claims, not all losses. This distinction matters when estimating average payment per observed claim, pricing payments per original exposure, or comparing model results with data. The key question is whether an expectation is conditional on exceedance or averaged over every original loss.

Consider a hypothetical policy with loss $X$ and an ordinary per-claim deductible $d$, with no upper payment limit. The payment for one loss is $Y^L=(X-d)_+$, where $(z)_+=\max(z,0)$. For an original loss, this payment is zero when $X\le d$. For a claim known to exceed the deductible, the payment is $X-d$. Those are different random quantities on different bases.

Definition: For a nonnegative loss $X$, the severity conditional on deductible exceedance is the distribution of $X\mid X>d$. The associated payment per exceeding claim is $X-d\mid X>d$. The expected payment per original loss is $E[(X-d)_+]$, which includes zero payments for losses that do not exceed $d$.

Throughout the examples, the scenarios are hypothetical, payments are undiscounted (interest rate $i=0\%$), and mortality is not part of the model. Unless stated otherwise, there is no policy limit. The focus is solely on conditioning on deductible exceedance, rather than applying additional coverage changes.

Conditional Severity and Payment Formulas

Let $F_X$ be the distribution function of a nonnegative loss and $S_X(x)=P(X>x)=1-F_X(x)$. For a strict exceedance, the probability of selection is $P(X>d)=S_X(d)$. When this probability is positive, the conditional distribution function of the loss is

F_{X\mid X>d}(x) = \begin{cases} 0, & x\le d,\\[4pt] \dfrac{F_X(x)-F_X(d)}{S_X(d)}, & x>d. \end{cases}

The numerator for $x>d$ is $P(d<X\le x)$; dividing by $P(X>d)$ conditions that probability on selection. For the payment $Z=X-d\mid X>d$, where $Z>0$, the corresponding distribution is

P(Z\le y\mid X>d) = \frac{F_X(d+y)-F_X(d)}{S_X(d)}, \qquad y\ge 0.

These formulas use the strict condition $X>d$, including when $X$ has probability mass at the deductible. For continuous severities, $P(X=d)=0$, so strict exceedance and exceedance including equality have the same probability. For a discrete severity, they need not. If the data instead select losses with $X\ge d$, the conditioning event and denominator must be changed to $P(X\ge d)$; it is not safe to ignore a possible atom at $d$.

The expected payment per exceeding claim can be found from the survival function of the excess. For a nonnegative payment $Z$, its mean is the area under its survival function. Thus, for strict exceedance,

E[X-d\mid X>d] = \int_0^\infty \frac{S_X(d+y)}{S_X(d)}\,dy = \frac{\displaystyle\int_d^\infty S_X(x)\,dx}{S_X(d)}.

The expected payment per original loss is instead

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

The last equality also follows by conditioning on whether the loss exceeds $d$: payment is zero otherwise. For nonnegative losses, the tail-integral identity remains valid for discrete distributions; isolated points do not affect the integral.

Why the Tail-Integral Formula Holds

For every realized loss, its payment can be written as the length of an interval: it is the length of the set of $y\ge0$ for which $X>d+y$. Equivalently,

(X-d)_+ = \int_0^\infty \mathbf{1}_{\{X>d+y\}}\,dy.

Taking expectations and interchanging the nonnegative integral and expectation gives

E[(X-d)_+] = \int_0^\infty P(X>d+y)\,dy = \int_d^\infty S_X(x)\,dx.

On the event $X\le d$, the payment is zero; on $X>d$, it is $X-d$. Therefore conditioning on exceedance yields $E[(X-d)_+]=P(X>d)E[X-d\mid X>d]$, which proves the ratio formula when $P(X>d)>0$. The conditional mean is undefined if the deductible is never exceeded.

Worked Examples

Worked Example: Exponential Severity and Two Different Averages

A hypothetical insurer models individual losses as exponential with rate $\lambda=1/1000$ per dollar, so $E[X]=1000$. The deductible is $d=\$500$. Find (1) the probability of exceeding the deductible, (2) the mean payment per exceeding claim, and (3) the mean payment per original loss. Use no discounting ($i=0\%$); mortality is not modeled.

Method and model. For an exponential loss, $S_X(x)=e^{-\lambda x}$ for $x\ge0$. First calculate the selection probability:

P(X>500)=S_X(500)=e^{-(1/1000)(500)}=e^{-0.5}=0.6065.

The conditional survival function of the payment $Z=X-500\mid X>500$ is

P(Z>y\mid X>500) = \frac{e^{-\lambda(500+y)}}{e^{-\lambda 500}} = e^{-\lambda y}, \qquad y\ge0.

It is again exponential with rate $1/1000$, so the conditional mean payment is

E[X-500\mid X>500]=\frac{1}{\lambda}=\$1{,}000.00.

The expected payment per original loss includes the probability of no payment:

E[(X-500)_+] = 0.6065307(\$1{,}000.00) = \$606.53.

Check. The unconditional amount is below the conditional amount because $0.6065$ of original losses exceed the deductible and the others pay zero. The common error is reporting $\$606.53$ as the mean payment among observed exceeding claims; the correct selected-claim mean is $\$1{,}000.00$.

Worked Example: Conditional Excess for a Pareto Type II Loss

A hypothetical portfolio has losses with Pareto Type II (Lomax) survival function $S_X(x)=(\theta/(\theta+x))^\alpha$, where $\theta=\$1{,}000$ and $\alpha=3$. For deductible $d=\$1{,}000$, find the exceedance probability, conditional mean payment, and expected payment per original loss. There is no discounting ($i=0\%$); mortality is not modeled.

Method and model. The selection probability is

P(X>1{,}000) = \left(\frac{1{,}000}{1{,}000+1{,}000}\right)^3 = \left(\frac12\right)^3 = 0.1250.

For an excess payment $Z=X-d\mid X>d$, its conditional survival function is the ratio of the original survival probabilities:

P(Z>y\mid X>1{,}000) = \frac{S_X(1{,}000+y)}{S_X(1{,}000)} = \left(\frac{2{,}000}{2{,}000+y}\right)^3,\qquad y\ge0.

Integrating this survival function gives the conditional mean:

E[Z\mid X>1{,}000] = \int_0^\infty\left(\frac{2{,}000}{2{,}000+y}\right)^3dy = 2{,}000^3\int_{2{,}000}^\infty u^{-3}\,du = 2{,}000^3\left(\frac{1}{2(2{,}000)^2}\right) = \$1{,}000.00.

Therefore, the mean severity among exceeding claims is $d+E[Z\mid X>d]=\$2{,}000.00$. The mean payment per original loss is

E[(X-1{,}000)_+] = 0.1250(\$1{,}000.00) = \$125.00.

Check. The conditional payment is eight times the original-loss average, matching the fact that only one eighth of losses exceed the deductible. Using the unconditional mean severity $E[X]=\$500.00$ and subtracting the deductible would give a negative amount, which is not an estimate of either payment average. The correct method conditions the severity or uses the tail integral.

Worked Example: Discrete Severity With a Mass Below the Deductible

A hypothetical product has loss distribution $P(X=\$400)=0.2$, $P(X=\$800)=0.3$, and $P(X=\$1{,}600)=0.5$. With deductible $d=\$600$, find the probability of exceedance, the mean severity and payment among exceeding claims, and the expected payment per original loss. Use $i=0\%$; mortality is not modeled.

Method and model. Only losses of $\$800$ and $\$1{,}600$ exceed $\$600$. Hence

$$ P(X>600)=0.3+0.5=0.8. $$

The conditional probabilities for those two severities are $0.3/0.8=0.375$ and $0.5/0.8=0.625$. Thus

E[X\mid X>600] = (800)(0.375)+(1{,}600)(0.625) = 300+1{,}000 = \$1{,}300.00.

Subtracting the deductible from the selected severity gives the payment mean:

E[X-600\mid X>600] = \$1{,}300.00-\$600.00 = \$700.00.

The payment per original loss is zero at $X=\$400$, $\$200$ at $X=\$800$, and $\$1{,}000$ at $X=\$1{,}600$. Therefore,

E[(X-600)_+] = (0.2)(0)+(0.3)(200)+(0.5)(1{,}000) = \$560.00.

Check. $0.8(\$700.00)=\$560.00$, as required. A common incorrect calculation is to find $E[X]=0.2(400)+0.3(800)+0.5(1{,}600)=\$1{,}120$, then subtract $\$600$, obtaining $\$520$. That subtracts the deductible from the unconditional mean, not from the conditional severity, and does not give the expected payment on either basis.

Common Exam Traps and Practical Checks

Identify the population behind the expectation.
“Average payment per reported claim” generally means $E[X-d\mid X>d]$. “Expected payment per original loss” means $E[(X-d)_+]$.

Calculate the selection probability.
For strict exceedance, use $P(X>d)=S_X(d)$. Check whether the severity has an atom at $d$, and whether the stated selection rule is $X>d$ or $X\ge d$.

Condition before averaging selected claims.
Use the conditional distribution or divide the expected payment per original loss by the exceedance probability. Do not substitute an unconditional severity mean for a conditional one.

Check the relationship between the answers.
When $P(X>d)>0$, the expected payment per original loss must equal the exceedance probability multiplied by the expected payment per exceeding claim.

A particularly tempting shortcut is $E[X]-d$. It is generally neither $E[X-d\mid X>d]$ nor $E[(X-d)_+]$. It subtracts $d$ from a mean that includes losses below the deductible, even though those losses are excluded from the selected data and have zero payment. The shortcut only works under additional special conditions that must be demonstrated, not assumed.

Correct conditioning is important when a severity sample contains only deductible-exceeding claims. Fitting or summarizing that sample as if it came from the original severity distribution can distort its mean and tail. The same distinction is essential when a problem asks for an observed-claim average rather than a cost per exposure. Keeping the selection probability and the expectation basis explicit is a reliable way to avoid mixing the two.

Key takeaway: For a strict deductible exceedance, $E[X-d\mid X>d]=E[(X-d)_+]/S_X(d)$. The numerator is expected payment per original loss; dividing by the exceedance probability converts it to expected payment per exceeding claim. This conditional-severity distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical model, state the selection probability and keep conditional payments distinct from payments per original loss. Assume no discounting and no mortality unless the problem states otherwise.

Losses are exponential with mean $\$2{,}500$, and the deductible is $\$1{,}000$. Find the conditional mean payment per exceeding claim and expected payment per original loss. (Answer: conditional mean $\$2{,}500.00$; expected original-loss payment $\$1{,}675.80$; exceedance probability $0.6703$.)
A Lomax loss has $\theta=\$600$, $\alpha=4$, and deductible $d=\$600$. Find the exceedance probability and both payment means. (Answer: exceedance probability $0.0625$; conditional mean payment $\$400.00$; expected original-loss payment $\$25.00$.)
A discrete loss takes values $\$300,\$900,\$1{,}800$ with probabilities $0.25,0.40,0.35$, respectively. At deductible $\$700$, find the conditional mean payment and expected payment per original loss. (Answer: exceedance probability $0.7500$; conditional mean payment $\$620.00$; expected original-loss payment $\$465.00$.)
A Lomax loss has $\theta=\$800$, $\alpha=2$, and $d=\$400$. Find the probability that an exceeding claim’s payment is greater than $\$500$, and find the two payment means. (Answer: conditional tail probability $0.4983$; conditional mean payment $\$1{,}200.00$; expected original-loss payment $\$533.33$.)
A discrete loss is $\$500$ with probability $0.20$, $\$1{,}000$ with probability $0.30$, and $\$2{,}500$ with probability $0.50$. For strict exceedance of deductible $d=\$1{,}000$, find the conditional payment mean and the payment mean per original loss. (Answer: exceedance probability $0.5000$; conditional mean payment $\$1{,}500.00$; expected original-loss payment $\$750.00$.)

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