How Fixed Deductibles Change LER Under Inflation

Why a Fixed Deductible Changes the LER

When claim severities trend upward but a policy deductible stays fixed in nominal dollars, the deductible represents a smaller share of a typical loss. More of each trended loss can therefore exceed the deductible. The loss elimination ratio (LER) measures the expected portion of ground-up losses eliminated by the deductible, so this change in relative size affects the LER.

The calculation depends on keeping two things distinct: the trended loss distribution and the deductible stated in dollars. A fixed deductible is not multiplied by the trend factor. Instead, proportional trend scales the losses relative to that unchanged deductible.

Definition: Let $X$ be a nonnegative ground-up loss with $0<E[X]<\infty$. Under a per-loss deductible $d\geq 0$, the amount eliminated from the loss is $\min(X,d)$, and the insurer payment is $(X-d)_+=\max(X-d,0)$. The loss elimination ratio is $$ \operatorname{LER}(d;X)=\frac{E[\min(X,d)]}{E[X]}. $$ It is the expected eliminated loss divided by expected ground-up loss, not the expected eliminated loss divided by expected insurer payment.

Suppose a positive proportional trend factor $c$ scales every loss, so that the post-trend loss is $X_c=cX$. For example, a single year of 4% proportional trend has $c=1.04$; after $t$ years at a constant annual rate $r$, the factor is $c=(1+r)^t$. Assume the same loss population is being compared, the trend is deterministic and proportional, and the nominal deductible remains $d$. This setup does not model changes in claim frequency, coverage terms, or the shape of severity beyond the proportional scaling.

The Effective Deductible on the Original Scale

The post-trend LER can be calculated using the trended loss distribution directly. It can also be calculated using the original distribution with a rescaled deductible. These are equivalent methods, provided the deductible is rescaled in the correct direction.

\operatorname{LER}(d;X_c) =\frac{E[\min(cX,d)]}{E[cX]} =\frac{E[\min(X,d/c)]}{E[X]} =\operatorname{LER}(d/c;X).

The notation $\operatorname{LER}(d;X_c)$ means the LER at deductible $d$ for the post-trend severity $X_c$. The effective deductible on the original scale is $d/c$: a fixed nominal deductible $d$ buys less protection relative to a loss that has been multiplied by $c$.

Write the post-trend loss.
Set $X_c=cX$, where $c>0$, and leave the fixed nominal deductible at $d$.

Express the eliminated amount on the original scale.
For each loss value $X$, $\min(cX,d)=c\min(X,d/c)$.

Cancel the common scale factor.
Since $E[cX]=cE[X]$, the factor $c$ in the numerator and denominator cancels, leaving $\operatorname{LER}(d/c;X)$.

This derivation also explains the direction of the effect. As $c$ increases while $d$ stays fixed, $d/c$ decreases. For a fixed nonnegative severity distribution, $\operatorname{LER}(a;X)$ is nondecreasing in deductible $a$, because $\min(X,a)$ cannot decrease when $a$ increases. Consequently, the LER under a fixed deductible cannot increase solely because of positive proportional severity trend. It can remain unchanged in special cases, such as when the deductible already exceeds all possible losses both before and after the effective rescaling.

A useful identity for calculations is $E[\min(X,d)]=\int_0^d S_X(x)\,dx$ for nonnegative $X$, where $S_X(x)=P(X>x)$. It follows by writing $\min(X,d)$ as the length of the interval from 0 to $d$ on which $X>x$, then taking expectations. For discrete severities, the same expected limited loss can instead be found by summing $\min(x_j,d)$ times each outcome probability.

Calculations for Common Severity Models

The scaling relationship applies to continuous, discrete, and mixed severities. The model-specific work is to calculate the expected eliminated amount and expected ground-up loss, or to use the equivalent effective deductible on the original scale. Each example below treats all losses as hypothetical.

Worked Example: Exponential Severity With a Fixed Deductible

A hypothetical book has original ground-up claim severity $X$ exponentially distributed with mean $\$10{,}000$. Every loss is increased proportionally by 25%, so $X_c=1.25X$, while the per-loss deductible remains fixed at $\$2{,}500$. Calculate the post-trend LER and the expected insurer payment per loss.

Method and model. For an exponential loss with mean $\theta$, the survival function is $S_X(x)=e^{-x/\theta}$ for $x\geq0$. Thus $E[X]=\theta$, and the expected limited loss at deductible $d$ is $$ E[\min(X,d)]=\int_0^d e^{-x/\theta}\,dx =\theta(1-e^{-d/\theta}). $$ After scaling by 1.25, the severity is still exponential, now with mean $1.25(\$10{,}000)=\$12{,}500$. The relevant parameters are therefore post-trend mean $\$12{,}500$ and fixed deductible $\$2{,}500$.

Expected eliminated loss. The exponent is $-\$2{,}500/\$12{,}500=-0.2$, so $$ E[\min(X_c,\$2{,}500)] =\$12{,}500(1-e^{-0.2}) =\$12{,}500(0.181269246) =\$2{,}265.87. $$

LER and insurer payment. The expected post-trend ground-up loss is $\$12{,}500$. Therefore $$ \operatorname{LER}=\frac{\$2{,}265.8656\ldots}{\$12{,}500} =0.181269246\ldots \approx 0.1813. $$ The expected insurer payment is the ground-up mean less the expected eliminated loss: $$ E[(X_c-\$2{,}500)_+]=\$12{,}500-\$2{,}265.87=\$10{,}234.13. $$

Check. The eliminated and insurer-paid expected amounts sum to $\$2{,}265.87+\$10{,}234.13=\$12{,}500.00$, the expected ground-up loss. The result also equals $1-e^{-(\$2{,}500/1.25)/\$10{,}000}=1-e^{-0.2}$, using the effective deductible $\$2{,}000$ on the original scale. The method avoids applying the unchanged deductible to the original mean after trending only the losses.

Worked Example: Uniform Severity Before and After Trend

A hypothetical portfolio has original severity uniformly distributed from $\$0$ to $\$20{,}000$, written $X\sim\operatorname{Uniform}(0,\$20{,}000)$. Losses are trended by a factor of 1.5 and the nominal deductible remains $\$6{,}000$. Calculate the LER after trend and compare it with the LER before trend at the same nominal deductible.

Method. For $X\sim\operatorname{Uniform}(0,b)$, the mean is $b/2$. When $0\leq d\leq b$, integration over the uniform density gives $$ E[\min(X,d)] =\frac{1}{b}\left(\int_0^d x\,dx+\int_d^b d\,dx\right) =d-\frac{d^2}{2b}. $$ After the trend, $X_c$ is uniform on $[0,\$30{,}000]$, since both endpoints are multiplied by 1.5. Its mean is $\$15{,}000$.

Post-trend eliminated loss. Use $b=\$30{,}000$ and $d=\$6{,}000$: $$ E[\min(X_c,\$6{,}000)] =\$6{,}000-\frac{(\$6{,}000)^2}{2(\$30{,}000)} =\$6{,}000-\$600 =\$5{,}400. $$ The post-trend LER is $\$5{,}400/\$15{,}000=0.36$.

Pre-trend comparison. Before trend, the mean is $\$10{,}000$, and $$ E[\min(X,\$6{,}000)] =\$6{,}000-\frac{(\$6{,}000)^2}{2(\$20{,}000)} =\$6{,}000-\$900 =\$5{,}100. $$ The pre-trend LER is $\$5{,}100/\$10{,}000=0.51$. Thus the LER decreases from 0.51 to 0.36 while the nominal deductible is held fixed.

Check using the original scale. The effective deductible is $\$6{,}000/1.5=\$4{,}000$. On the original severity, the eliminated mean at that deductible is $\$4{,}000-(\$4{,}000)^2/[2(\$20{,}000)]=\$3{,}600$; dividing by the original mean $\$10{,}000$ again gives 0.36. This comparison avoids treating the LER as unchanged just because the deductible itself has not been rewritten.

Worked Example: Discrete Severity With Several Loss Sizes

A hypothetical repair-cost model has original losses of $\$2{,}000$, $\$8{,}000$, and $\$20{,}000$, with respective probabilities 0.50, 0.30, and 0.20. Severity is trended by a factor of 1.5, and the fixed nominal deductible is $\$6{,}000$. Calculate the post-trend LER and verify it using the effective deductible.

Method. For a discrete severity, calculate each post-trend loss, apply the deductible to that loss, and probability-weight the results. The post-trend losses are $1.5(\$2{,}000)=\$3{,}000$, $1.5(\$8{,}000)=\$12{,}000$, and $1.5(\$20{,}000)=\$30{,}000$. Their eliminated amounts are $\$3{,}000$, $\$6{,}000$, and $\$6{,}000$, respectively.

Expected eliminated loss. $$ E[\min(X_c,\$6{,}000)] =0.50(\$3{,}000)+0.30(\$6{,}000)+0.20(\$6{,}000) =\$1{,}500+\$1{,}800+\$1{,}200 =\$4{,}500. $$ Expected ground-up loss. $$ E[X_c] =0.50(\$3{,}000)+0.30(\$12{,}000)+0.20(\$30{,}000) =\$1{,}500+\$3{,}600+\$6{,}000 =\$11{,}100. $$ Therefore the post-trend LER is $\$4{,}500/\$11{,}100=0.405405\ldots$, or 0.4054 to four decimal places.

Verification on the original scale. The effective deductible is $\$6{,}000/1.5=\$4{,}000$. The original expected loss is $$ E[X]=0.50(\$2{,}000)+0.30(\$8{,}000)+0.20(\$20{,}000) =\$1{,}000+\$2{,}400+\$4{,}000 =\$7{,}400. $$ At deductible $\$4{,}000$, the eliminated amounts are $\$2{,}000$, $\$4{,}000$, and $\$4{,}000$. Their expected value is $0.50(\$2{,}000)+0.30(\$4{,}000)+0.20(\$4{,}000)=\$3{,}000$. The ratio $\$3{,}000/\$7{,}400=0.405405\ldots$ agrees. The calculation avoids capping the original losses at $\$6{,}000$, which would use the wrong deductible on the original scale.

Common Exam Trap: Trending a Fixed Deductible

A fixed nominal deductible stays at the same dollar amount after severity trend. It is the loss distribution that changes. A common incorrect approach is to multiply both the loss and the deductible by the trend factor, which silently changes the contract to one with a trended deductible.

For the exponential example, the original mean is $\$10{,}000$, the trend factor is 1.25, and the stated deductible is $\$2{,}500$. The incorrect approach would multiply the deductible by 1.25 and calculate an original-scale LER of $$ 1-e^{-(\$3{,}125/\$10{,}000)} =1-e^{-0.3125} =0.2684. $$ That is not the LER for the fixed $\$2{,}500$ deductible. The correct post-trend LER is $1-e^{-(\$2{,}500/\$12{,}500)}=0.1813$, equivalently the original-scale LER at effective deductible $\$2{,}500/1.25=\$2{,}000$. Multiplying the deductible by the trend factor rather than dividing it for the original-scale calculation reverses the adjustment.

A second check is to keep the denominator consistent with the loss scale used in the numerator. If calculating directly after trend, divide expected eliminated loss by $E[X_c]$. If calculating on the original scale using $d/c$, divide by $E[X]$. Mixing a post-trend numerator with an original-scale denominator does not produce an LER.

Key takeaway: Under deterministic proportional severity trend $X_c=cX$, a fixed nominal deductible $d$ gives $\operatorname{LER}(d;X_c)=\operatorname{LER}(d/c;X)$. The trend factor scales losses, not the fixed contract deductible. As the effective deductible falls on the original scale, the LER is nonincreasing under the assumptions stated.

Check Your Understanding

For each hypothetical model, calculate the requested post-trend LER. Give ratios to four decimal places where needed.

Original severity is exponential with mean $\$12{,}000$. Losses are multiplied by 1.2, and the fixed deductible is $\$3{,}600$. Find the post-trend LER. (Answer: 0.2212)
Original severity is uniform on $[0,\$24{,}000]$. The trend factor is 1.5 and the fixed deductible is $\$6{,}000$. Find the post-trend LER. (Answer: 0.3056)
Original losses are $\$1{,}000$, $\$5{,}000$, and $\$12{,}000$, with probabilities 0.40, 0.40, and 0.20. The trend factor is 2 and the fixed deductible is $\$8{,}000$. Find the post-trend LER. (Answer: 0.5833)
Original severity is uniform on $[0,\$20{,}000]$. Losses are multiplied by 1.25, the deductible stays at $\$5{,}000$, and the question asks for both the LER and expected insurer payment per loss. (Answer: LER $=0.3600$; expected insurer payment $=\$8{,}000.00$)
Original losses are $\$4{,}000$ and $\$10{,}000$, each with probability 0.50. Losses are multiplied by 1.25, and the fixed deductible is $\$5{,}000$. Find the post-trend LER. (Answer: 0.5714)

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