Exam FAM · Short-Term Insurance and Reinsurance Coverages · Free Lesson

Loss Elimination Ratio and the Effect of Inflation

Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) lesson in Short-Term Insurance and Reinsurance Coverages. 11 min read, ~1,614 words.

A health insurer's $2,000 deductible eliminates 45% of expected losses. Three years later, medical costs have inflated 30% but the deductible stays at $2,000. The LER drops to 37%. That 8-point shift means millions more in claims flowing through.

HIGH-FREQUENCY: The LER and its behavior under inflation is one of the most frequently tested concepts in Topic 1.

The loss elimination ratio measures the fraction of expected losses eliminated by a deductible:

LER(d)=E[X∧d]E[X]\text{LER}(d) = \frac{E[X \wedge d]}{E[X]}

KEY: The LER ranges from 0 to 1. At d=0 d = 0 , LER = 0. As d→∞ d \to \infty , LER approaches 1.

For exponential with mean θ \theta : LER(d)=1−e−d/θ \text{LER}(d) = 1 - e^{-d/\theta} .

For Pareto with α>1 \alpha > 1 : LER(d)=1−(θ/(θ+d))α−1 \text{LER}(d) = 1 - (\theta/(\theta + d))^{\alpha-1} .

For uniform on (0,b) (0, b) : LER(d)=2d/b−d2/b2 \text{LER}(d) = 2d/b - d^2/b^2 for d≤b d \leq b .

Suppose nonnegative losses with a finite positive mean inflate by factor 1+r 1 + r , where r>0 r>0 . The inflated loss is X∗=(1+r)X X^* = (1+r)X . With a fixed ordinary deductible d≥0 d\geq0 and no payment limit:

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When you see positive inflation and a fixed ordinary deductible, compute deff=d/(1+r) d_{\text{eff}} = d/(1+r) . Use the pre-inflation distribution's LEV formula at that value. The new LER cannot exceed the original. A percentage cost increase requires positive original expected payment. For a strict LER decrease, check that the deductible is positive and that some losses exceed its reduced real value. Otherwise equality is possible.

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