Exponential Severity Deductible Calculations

Introduction

An ordinary deductible makes the insurer’s payment depend on how much of a loss exceeds a specified threshold. When claim severity follows an exponential distribution, the survival function gives a particularly direct way to calculate the insurer’s expected payment. This tutorial develops that calculation, interprets the resulting deductible factor, and applies the formulas at multiple deductible amounts.

The calculations concern severity per claim. They do not include policy limits, expenses, or a change in claim frequency caused by the deductible. Unless an example explicitly says otherwise, payments are undiscounted: the effective annual interest rate is $i=0$, and mortality is not applicable to these property-and-casualty claim-severity calculations. The methods are aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Exponential Severity and an Ordinary Deductible

Definition: A ground-up loss $X$ has an exponential severity distribution with rate parameter $\lambda>0$ if, for $x\geq 0$, $$ F_X(x)=1-e^{-\lambda x}, \qquad S_X(x)=e^{-\lambda x}. $$ Its mean is $\theta=E[X]=1/\lambda$. Under an ordinary deductible $d\geq 0$, the insurer’s payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. The deductible factor relative to ground-up expected severity is $E[Y^L]/E[X]$.

The superscript $L$ indicates a payment measured per loss, including losses for which the insurer pays zero. That distinction matters: $E[Y^L]$ is not the expected payment conditional on the loss exceeding $d$. For an exponential severity, these two quantities are especially easy to distinguish because of the distribution’s memoryless property.

Deriving the Expected Payment from the Survival Function

For any nonnegative random variable $Z$, its expectation can be written as the area under its survival function: $E[Z]=\int_0^\infty P(Z>y)\,dy$, when the integral is finite. Apply this identity to the nonnegative insurer payment $Y^L=(X-d)_+$. For $y\geq 0$, the event $Y^L>y$ is equivalent to $X>d+y$. Therefore,

\begin{aligned} E[Y^L] &=\int_0^\infty P(Y^L>y)\,dy\\ &=\int_0^\infty P(X>d+y)\,dy\\ &=\int_0^\infty S_X(d+y)\,dy\\ &=\int_0^\infty e^{-\lambda(d+y)}\,dy\\ &=e^{-\lambda d}\int_0^\infty e^{-\lambda y}\,dy\\ &=\frac{e^{-\lambda d}}{\lambda} =\theta e^{-d/\theta}. \end{aligned}

The factor $e^{-\lambda d}$ is the probability that a loss exceeds the deductible. The remaining integral, $1/\lambda=\theta$, is the mean exponential severity. Equivalently, given that a loss exceeds $d$, the excess amount $X-d$ has mean $\theta$, so the expected payment per loss is the probability of payment multiplied by the mean payment conditional on payment.

Result: If $X$ is exponential with mean $\theta$ and an ordinary deductible $d$ applies, then $$ E[(X-d)_+]=\theta e^{-d/\theta}, \qquad \frac{E[(X-d)_+]}{E[X]}=e^{-d/\theta}. $$ The second expression is the deductible factor relative to ground-up expected severity. It is a dimensionless ratio; multiply it by ground-up expected severity to obtain expected insurer payment per loss.

The formula includes the boundary case $d=0$: the factor is $e^0=1$, and expected insurer payment is $\theta=E[X]$. For $d>0$, the factor is between zero and one and decreases as the deductible increases. The expression assumes an exponential ground-up severity and an ordinary per-loss deductible. It does not by itself account for a policy limit, an aggregate deductible, or expenses.

Worked Example: Payments at Two Deductible Levels

A hypothetical property policy has ground-up claim severity $X$ that is exponential with mean $\theta=\$2{,}000$, so $\lambda=1/2000$ per dollar. Calculate the insurer’s expected payment per loss and deductible factor at deductibles of $\$1{,}000$ and $\$3{,}000$. No discounting is applied ($i=0$); mortality is not applicable.

Method and setup. For each deductible, use $E[(X-d)_+]=\theta e^{-d/\theta}$ and divide by $\theta$ to obtain the factor. The severity distribution and its parameters are the same at both deductible levels.

Deductible of $\$1{,}000$. The exponent is $-d/\theta=-1000/2000=-0.5$. Thus the factor is $e^{-0.5}=0.6065306597\ldots$, and the expected payment is

\$2{,}000e^{-0.5}=\$1{,}213.061319\ldots

So the deductible factor is $0.6065$, and the expected payment is $\$1{,}213.06$ to the nearest cent.

Deductible of $\$3{,}000$. Now $-d/\theta=-3000/2000=-1.5$. The factor is $e^{-1.5}=0.2231301601\ldots$, and the expected payment is

\$2{,}000e^{-1.5}=\$446.260320\ldots

So the factor is $0.2231$, and the expected payment is $\$446.26$ to the nearest cent.

Check and common mistake. The higher deductible has the smaller factor and expected payment, as expected. The factors are relative to the same ground-up mean of $\$2{,}000$. Avoid treating $e^{-d/\theta}$ as the dollar payment: it is a ratio, so the mean must be multiplied by it.

Using the Factor in a Loss-Cost Calculation

When claim frequency is unchanged by the deductible, the factor also converts a ground-up expected loss cost per exposure to an expected insurer loss cost per exposure. To see why, let $N$ be the number of claims for one exposure, and suppose the claim severities are independent of $N$, each with the same distribution as $X$. The expected total insurer payment is the expected claim count multiplied by the expected insurer payment per claim:

E\left[\sum_{j=1}^{N}(X_j-d)_+\right] =E[N]E[(X-d)_+].

The comparable ground-up expected loss is $E[N]E[X]$. Their ratio is therefore $E[(X-d)_+]/E[X]=e^{-d/\theta}$. This calculation requires that the deductible does not alter the number of claims counted and that the severity distribution used is appropriate for the claims being modeled. If observed claim counts are defined only for claims above a reporting threshold, or if the deductible affects claim reporting, the simple unchanged-frequency interpretation needs adjustment.

Worked Example: Converting Ground-Up Loss Cost to Insurer Loss Cost

A hypothetical portfolio has a ground-up expected loss cost of $\$1{,}500$ per exposure. Its claim severity is exponential with mean $\theta=\$1{,}500$, and the policy deductible is $\$600$. Calculate the deductible factor and expected insurer loss cost per exposure. No discounting is applied ($i=0$); mortality is not applicable.

Method and setup. Use $X\sim\operatorname{Exp}(\lambda=1/1500)$, where the rate is per dollar and $E[X]=\$1{,}500$. The ground-up expected loss cost is assumed to use the same severity basis, and the deductible is assumed not to change claim frequency.

Step 1: Calculate the factor. The exponent is $-600/1500=-0.4$, so

e^{-0.4}=0.6703200460\ldots

The deductible factor is $0.6703$ to four decimal places.

Step 2: Calculate insurer expected payment per claim. Retain the unrounded factor in the multiplication:

E[(X-600)_+]=\$1{,}500e^{-0.4} =\$1{,}005.480069\ldots

The expected insurer payment per claim is $\$1{,}005.48$ to the nearest cent.

Step 3: Calculate the insurer loss cost per exposure. Because the ground-up expected loss cost per exposure is $\$1{,}500$, multiply it by the factor:

\$1{,}500e^{-0.4}=\$1{,}005.480069\ldots \approx \$1{,}005.48.

The per-exposure result agrees with the per-claim result here because the stated ground-up loss cost equals the severity mean, corresponding to one expected claim per exposure. In general, the ground-up loss cost already incorporates expected claim frequency; multiply that loss cost by the deductible factor once, rather than multiplying it again by a claim count.

Check and common mistake. The factor is less than one, so the insurer loss cost is below the ground-up loss cost. Rounding the factor to $0.6703$ before multiplying gives a slightly different intermediate amount; retain full precision and round the dollar result only at the end.

From Expected Payment per Claim to Aggregate Payment

If the expected number of claims is known, aggregate expected payment follows by multiplying that expected count by the per-claim expected payment, provided each claim follows the specified severity model. The claim-count distribution is not needed for the expectation once its mean is known, as long as the severity and count assumptions support this product. The example below states a Poisson model explicitly to make the frequency assumption clear.

Worked Example: Aggregate Payments With a Poisson Claim Count

For a hypothetical group of policies, the claim count $N$ is Poisson with mean $320$, and claim severities are independent exponential random variables with mean $\theta=\$1{,}250$, or rate $\lambda=1/1250$ per dollar. Each claim has an ordinary deductible of $\$500$. Calculate the expected aggregate insurer payment. No discounting is applied ($i=0$); mortality is not applicable.

Method and setup. Since $N\sim\operatorname{Poisson}(320)$, $E[N]=320$. The claim severities are assumed independent of the claim count, with the same exponential distribution for each claim.

Step 1: Find expected payment per claim. The exponent is $-500/1250=-0.4$, and hence

E[(X-500)_+] =\$1{,}250e^{-0.4} =\$837.9000575\ldots

Step 2: Multiply by the expected claim count. Keep the unrounded per-claim amount:

E[\text{aggregate payment}] =320(\$1{,}250e^{-0.4}) =\$268{,}128.0184\ldots \approx \$268{,}128.02.

The expected aggregate payment is $\$268{,}128.02$ to the nearest cent. As a check, the expected ground-up aggregate loss is $320(\$1{,}250)=\$400{,}000$; applying the deductible factor $e^{-0.4}$ gives $\$400{,}000e^{-0.4}=\$268{,}128.0184\ldots$, the same result.

Common mistake avoided. Multiplying the rounded per-claim amount $\$837.90$ by $320$ gives $\$268{,}128.00$. That is the product of a rounded intermediate value, not the correctly rounded aggregate expectation. Retain full precision until the final rounding.

Common Exam Traps and Interpretation

Using the probability of payment as the expected payment. The probability a loss exceeds $d$ is $e^{-d/\theta}$; the expected payment per loss is $\theta e^{-d/\theta}$. The former is dimensionless, while the latter is in dollars.
Confusing per-loss and conditional expectations. For exponential severity, $E[X-d\mid X>d]=\theta$, but this is conditional on a payment occurring. Per loss, including zero-payment claims, expected payment is $\theta e^{-d/\theta}$.
Subtracting the deductible from the mean. In general, $E[(X-d)_+]$ is not $\max(E[X]-d,0)$. The payment function is applied to each possible loss before taking its expectation.
Mixing units. The deductible and the severity mean must be in the same currency units. A deductible of $\$1{,}000$ and a mean stated as $\$2$ thousand cannot be placed in the exponent as 1000 and 2 without converting one of them.
Rounding too early. Use the exponential expression or a sufficiently precise factor through the final multiplication, especially when scaling a per-claim amount to an aggregate amount.
Applying the formula outside its assumptions. The result is for exponential ground-up severity and an ordinary deductible with no limit. A different severity distribution or a policy limit requires a different payment calculation.

Key takeaway: For exponential severity with mean $\theta$, an ordinary deductible $d$ leaves an expected insurer payment per loss of $\theta e^{-d/\theta}$. Relative to ground-up expected severity, the deductible factor is $e^{-d/\theta}$. Check units, distinguish conditional from per-loss payment, and defer rounding until the final result.

Check Your Understanding

For each problem, use an exponential ground-up severity model and an ordinary per-loss deductible. Unless otherwise stated, assume no discounting ($i=0$); mortality is not applicable. Round dollar answers to the nearest cent and factors to four decimal places.

A hypothetical equipment policy has $X$ exponential with mean $\$1{,}000$ and deductible $\$250$. Find the deductible factor and expected insurer payment per loss. (Answer: factor $0.7788$; payment $\$778.80$.)
A hypothetical liability portfolio has exponential severity with mean $\$2{,}400$ and deductible $\$1{,}200$. Find the probability a loss exceeds the deductible and the expected payment per loss. (Answer: probability $0.6065$; payment $\$1{,}455.67$.)
A hypothetical warranty claim has exponential severity with mean $\$800$ and deductible $\$1{,}600$. Calculate the deductible factor and expected insurer payment per loss. (Answer: factor $0.1353$; payment $\$108.27$.)
For exponential severity with mean $\$1{,}800$, what deductible produces a factor of $0.25$? (Answer: $d=-1{,}800\ln(0.25)=\$2{,}495.33$.)
A hypothetical book has an expected claim count of $75$ and exponential severity with mean $\$2{,}000$. With a $\$1{,}000$ deductible, calculate the expected aggregate insurer payment. (Answer: $75(2{,}000e^{-0.5})=\$90{,}979.60$.)

What You'll Learn