LER for Exponential Severities

Why the Exponential Model Gives a Closed-Form LER

A deductible removes some ground-up loss from the insurer’s payment. The loss elimination ratio (LER) measures the expected amount removed as a proportion of expected ground-up loss. For exponential severity, the survival function has a particularly simple form, so both the expected payment above a deductible and the LER can be calculated directly from the mean and deductible.

Throughout this tutorial, consider a single nonnegative ground-up loss $X$ and an ordinary per-loss deductible $d\geq 0$. Assume there is no policy limit, coinsurance, or other coverage modification: the insurer pays the part of the loss above the deductible. This isolates the deductible effect.

Definition: For a ground-up loss $X$ with finite positive mean and an ordinary per-loss deductible $d$, the insurer payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. The loss elimination ratio is $$ \operatorname{LER}(d)=\frac{E[X]-E[Y^L]}{E[X]}. $$ Equivalently, it is the expected amount eliminated by the deductible divided by expected ground-up loss.

For an individual loss $x\leq d$, the insurer pays zero and the entire loss $x$ is eliminated. For $x>d$, the insurer pays $x-d$, so the eliminated amount is $x-(x-d)=d$. Thus the eliminated amount is $\min(X,d)$, not $d$ plus the loss below the deductible. This distinction matters when translating a payment calculation into an LER.

Exponential Severity Formulas

Assume ground-up severity is exponentially distributed with rate parameter $\lambda>0$, measured in inverse dollars if losses are measured in dollars. Its mean is $\theta=1/\lambda$. The density and survival function are

f_X(x)=\lambda e^{-\lambda x},\quad S_X(x)=P(X>x)=e^{-\lambda x},\quad x\geq 0,\qquad E[X]=\theta=\frac{1}{\lambda}.

The ordinary deductible payment is zero unless $X>d$. Conditional on exceeding $d$, the memoryless property says that the excess $X-d$ has the same exponential distribution as $X$. Therefore, the expected payment is the probability of exceeding the deductible multiplied by the mean excess payment, $\theta$.

E[Y^L]=P(X>d)E[X-d\mid X>d] =e^{-\lambda d}\theta =\theta e^{-d/\theta}.

The expected eliminated amount is the ground-up mean less the expected insurer payment. Dividing that amount by $E[X]=\theta$ gives the LER:

E[\min(X,d)]=\theta-\theta e^{-d/\theta} =\theta(1-e^{-d/\theta}), \qquad \operatorname{LER}(d)=1-e^{-d/\theta}.

Result: If $X$ is exponential with mean $\theta$ and the contract has only an ordinary per-loss deductible $d$, then $$ E[Y^L]=\theta e^{-d/\theta},\qquad \operatorname{LER}(d)=1-e^{-d/\theta}. $$ The result assumes a ground-up exponential severity and an unlimited insurer payment above the deductible. It does not directly give expected payment or LER when a policy limit, coinsurance, or a different deductible structure is also present.

Why the Expected Payment Formula Holds

The formula can also be justified without conditioning. For any nonnegative random variable $Z$, its mean is the area under its survival function: $E[Z]=\int_0^\infty P(Z>y)\,dy$. For $Y^L=(X-d)_+$, the event $Y^L>y$ is the event $X>d+y$. Under the exponential model, that probability is $e^{-\lambda(d+y)}$.

Write the payment survival probability.
For $y\geq 0$, $P(Y^L>y)=P(X>d+y)=e^{-\lambda(d+y)}$.

Integrate to obtain its mean.
$E[Y^L]=\int_0^\infty e^{-\lambda(d+y)}\,dy=e^{-\lambda d}\int_0^\infty e^{-\lambda y}\,dy=e^{-\lambda d}/\lambda$.

Substitute the exponential mean.
Since $1/\lambda=\theta$, $E[Y^L]=\theta e^{-\lambda d}=\theta e^{-d/\theta}$. Subtracting this from $E[X]=\theta$ and dividing by $\theta$ yields $\operatorname{LER}(d)=1-e^{-d/\theta}$.

This derivation also shows why the survival probability at the deductible appears as a factor: only losses that exceed $d$ can generate a payment, and the expected excess conditional on exceeding $d$ is still $\theta$ for exponential severity.

Worked Example: Expected Payment and LER for a Single Deductible

A hypothetical policy covers losses with ground-up severity $X$ exponential with mean $\theta=\$1{,}000$. It has a per-loss deductible of $d=\$400$, with no limit or coinsurance. Find the probability a loss exceeds the deductible, expected insurer payment, expected amount eliminated, and LER.

Method. Use the exponential survival function for the exceedance probability, then apply the deductible formulas. The parameter is $\lambda=1/\theta=1/1000$ per dollar.

First, the exceedance probability is $P(X>400)=e^{-400/1000}=e^{-0.4}=0.670320\ldots$, or $0.6703$ to four decimals.

The expected insurer payment is $E[Y^L]=1000e^{-400/1000}=1000(0.670320\ldots)=\$670.320\ldots$, which rounds to \$670.32.

The expected amount eliminated is $E[X]-E[Y^L]=\$1{,}000-\$670.32=\$329.68$. Equivalently, it is $1000(1-e^{-0.4})=\$329.679\ldots$, which also rounds to $\$329.68$.

Finally, $\operatorname{LER}(400)=1-e^{-0.4}=0.329679\ldots$, or 0.3297 to four decimals (32.97%). Check: $\$329.68/\$1{,}000=0.32968$, consistent with the ratio after rounding.

Common mistake avoided: For losses above $\$400$, the eliminated amount is $\$400$, not $\$400$ plus the dollars of loss below $\$400$. Those dollars are already part of the deductible amount.

Calculating Payments at Several Deductibles

The same exponential mean can be used to compare deductible options. The ratio $d/\theta$ is the deductible measured in units of the mean severity. A larger ratio produces a smaller expected insurer payment and a larger LER. When reporting amounts, retain the unrounded exponential value during the calculation and round only the final dollar results to cents.

Worked Example: Comparing Two Deductibles

A hypothetical equipment-loss portfolio is modeled with exponential ground-up severity of mean $\theta=\$600$. Two otherwise identical contracts have deductibles of $\$300$ and $\$900$, respectively, and no limits or coinsurance. For each contract, calculate the expected payment, expected amount eliminated, and LER.

Method. Apply $E[Y^L]=\theta e^{-d/\theta}$ and $\operatorname{LER}(d)=1-e^{-d/\theta}$ separately. Here $\lambda=1/600$ per dollar.

For $d=\$300$, $d/\theta=300/600=0.5$ and $e^{-0.5}=0.60653066\ldots$. Thus $E[Y^L]=600(0.60653066\ldots)=\$363.9184\ldots=$\$363.92. The expected amount eliminated is $\$600-\$363.92=\$236.08$, and the LER is $1-e^{-0.5}=0.393469\ldots=$0.3935 to four decimals.

For $d=\$900$, $d/\theta=900/600=1.5$ and $e^{-1.5}=0.22313016\ldots$. Thus $E[Y^L]=600(0.22313016\ldots)=\$133.8781\ldots=$\$133.88. The expected amount eliminated is $\$600-\$133.88=\$466.12$, and the LER is $1-e^{-1.5}=0.776869\ldots=$0.7769 to four decimals.

As a check, for each option the expected payment plus expected eliminated amount equals the $\$600$ ground-up mean after rounding. The larger deductible has the lower expected payment and the higher LER.

Common mistake avoided: Do not treat the deductible amount as the expected payment. The payment is made only when a loss exceeds the deductible, and its expectation is reduced by the factor $e^{-d/\theta}$.

Applying Inflation Before a Fixed Deductible

If a severity distribution is multiplied by an inflation factor, its mean is multiplied by the same factor. The deductible must then be compared with the severity on the same price basis. A fixed dollar deductible is not multiplied by the severity trend unless the contract specifically indexes it. Once the post-trend mean is identified, the exponential formulas apply to the post-trend loss and the stated deductible.

Worked Example: Fixed Deductible After Severity Trend

A hypothetical claim model has pre-trend exponential ground-up severity with mean $\$800$. Loss amounts are trended by a factor of $1.25$, while the policy retains a fixed $\$250$ deductible in the trended price basis. Assume the multiplication preserves exponential severity and that there is no limit or coinsurance. Find the post-trend mean, expected payment, expected amount eliminated, and LER.

Method. Trend the severity first, then apply the deductible to the trended loss. If $X_{\text{new}}=1.25X_{\text{old}}$, then $E[X_{\text{new}}]=1.25E[X_{\text{old}}]$; hence the post-trend mean is $1.25(\$800)=\$1{,}000$. The exponential rate on this basis is $1/1000$ per dollar.

The relevant ratio is $d/\theta_{\text{new}}=250/1000=0.25$. Since $e^{-0.25}=0.77880078\ldots$, the expected insurer payment is $1000e^{-0.25}=\$778.80078\ldots=$\$778.80.

The expected amount eliminated is $\$1{,}000-\$778.80078\ldots=\$221.19922\ldots=$\$221.20. The LER is $1-e^{-0.25}=0.22119922\ldots=$0.2212 to four decimals. Check: $\$221.20/\$1{,}000$ is approximately $0.2212$.

Common mistake avoided: The deductible is specified as fixed at $\$250$ on the trended basis. Replacing it with $1.25(\$250)=\$312.50$ would incorrectly index the deductible and answer a different contract question.

Common Exam Traps and Interpretation

The LER is a ratio of expected losses, not the probability that a loss is eliminated. For an exponential loss, $P(X\leq d)=1-e^{-d/\theta}$, which happens to equal the LER. That equality follows from the exponential formulas; it is not the definition of LER and should not be assumed for other severity distributions. In general, the expected eliminated dollars and the probability of having a loss below the deductible are different quantities.

A second trap is switching between the exponential rate and mean. If the model is stated using rate $\lambda$, use $e^{-\lambda d}$; if stated using mean $\theta$, use $e^{-d/\theta}$. These are equivalent only when $\theta=1/\lambda$. The exponent must be dimensionless: a deductible in dollars divided by a mean in dollars, or multiplied by a rate per dollar.

Finally, distinguish the deductible payment from a limited payment. The formulas here assume $Y^L=(X-d)_+$ with no upper cap. If coverage also imposes a limit, the insurer payment is no longer simply the excess above the deductible for every loss, so the result must be adjusted for that coverage term.

Key takeaway: For exponential ground-up severity with mean $\theta$ and an ordinary per-loss deductible $d$, expected insurer payment is $\theta e^{-d/\theta}$, expected eliminated loss is $\theta(1-e^{-d/\theta})$, and LER is $1-e^{-d/\theta}$. Keep the severity and deductible on the same price basis, and use the assumed coverage terms exactly.

Check Your Understanding

For each problem, assume an ordinary per-loss deductible, no limit or coinsurance, and exponential ground-up severity unless the problem says otherwise. Round dollar answers to cents and probabilities or LERs to four decimals.

A hypothetical policy has exponential ground-up severity with mean $\$500$ and deductible $\$200$. Find the expected insurer payment and LER. (Answer: expected payment $=\$335.16$; LER $=0.3297$.)
Loss severity is exponential with mean $\$1{,}200$, and the deductible is $\$600$. Find the exceedance probability and expected eliminated amount. (Answer: exceedance probability $=0.6065$; expected eliminated amount $=\$472.16$.)
A hypothetical shipment model has exponential severity with mean $\$750$ and a deductible of $\$150$. Find expected payment and LER. (Answer: expected payment $=\$614.05$; LER $=0.1813$.)
Pre-trend severity is exponential with mean $\$900$. A multiplicative trend factor of $1.10$ is applied, and a fixed deductible of $\$495$ applies on the post-trend basis. Find the post-trend expected payment and LER. (Answer: expected payment $=\$600.47$; LER $=0.3935$.)
A hypothetical loss has exponential severity with mean $\$400$ and a deductible of $\$800$. Find the expected payment and expected amount eliminated, and verify that they sum to the ground-up mean. (Answer: expected payment $=\$54.13$; expected amount eliminated $=\$345.87$; sum $=\$400.00$.)

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