Why Indexed Deductibles Preserve LER Under Proportional Trend

Why Common Scaling Matters

When claim severities trend upward, an insurer may also index a policy’s deductible. The key question is not whether the expected dollar amount eliminated stays the same: it generally does not. The question is whether the proportion of expected ground-up loss eliminated stays the same when every loss and the deductible increase by the same factor. For an ordinary deductible, the answer follows directly from how both the eliminated amount and the ground-up loss scale.

The result is useful for comparing coverage across periods when a severity model is subject to a common proportional trend. It requires a particular relationship between the trend and deductible: scaling only the loss distribution, or only the deductible, is not enough.

Definition: Let $X$ be a nonnegative ground-up loss with $0<E[X]<\infty$, and let $d\geq 0$ be an ordinary deductible. The insurer’s payment per loss is $(X-d)_+=\max(X-d,0)$. The amount eliminated by the deductible is $X-(X-d)_+=\min(X,d)$. The ordinary-deductible loss elimination ratio is $$ \operatorname{LER}(X,d)=\frac{E[\min(X,d)]}{E[X]}. $$ It is the expected amount eliminated divided by the expected ground-up loss.

The definition uses expected amounts, not the probability that a loss is below the deductible. Losses above $d$ contribute $d$ to the eliminated amount, while losses below $d$ contribute their full value. If $E[X]=0$, the ratio is undefined; throughout, assume the expected ground-up loss is positive and finite.

The Proportional-Scaling Result

Suppose the loss in a later period is $X_c=cX$, where $c>0$ is a deterministic scaling factor, and the ordinary deductible is indexed to $d_c=cd$. For example, $c=1.08$ represents an 8% proportional increase. The same factor must apply to the loss and to the deductible.

Theorem: If $X_c=cX$ and $d_c=cd$ for the same $c>0$, then $$ \operatorname{LER}(X_c,d_c)=\operatorname{LER}(X,d). $$ The expected eliminated amount and the expected ground-up loss are both multiplied by $c$, so their ratio is unchanged.

The justification is pointwise: for every possible loss value $x\geq 0$, multiplying both arguments of the minimum by the same positive factor multiplies the minimum by that factor. Taking expectations preserves this equality.

\min(cx,cd)=c\min(x,d) \quad\Longrightarrow\quad E[\min(cX,cd)]=cE[\min(X,d)]

Also, $E[cX]=cE[X]$. Substituting both equalities into the LER definition gives the result:

\operatorname{LER}(cX,cd) =\frac{E[\min(cX,cd)]}{E[cX]} =\frac{cE[\min(X,d)]}{cE[X]} =\operatorname{LER}(X,d).

This argument does not require a continuous severity distribution. It applies to discrete, continuous, and mixed distributions, provided the losses are nonnegative with a positive finite mean. An equivalent check uses the insurer payment: $(cX-cd)_+=c(X-d)_+$. Thus expected insurer payment and expected ground-up loss also scale by the same factor.

Specify the original coverage.
Identify the ground-up severity $X$, its mean, and the ordinary deductible $d$.

Apply one common trend factor.
For a factor $c>0$, use $cX$ for the trended loss and $cd$ for the indexed deductible.

Scale both expected amounts.
The expected eliminated amount and expected ground-up loss are each multiplied by $c$.

Take their ratio.
The common factor cancels, leaving the original LER.

Worked Applications

Worked Example: Exponential Losses With an Indexed Deductible

A hypothetical portfolio has ground-up loss $X$ exponentially distributed with mean $\$5{,}000$. Its ordinary deductible is $\$2{,}000$. Losses and the deductible are both increased by 8%. Find the LER before and after the increase, and check the expected eliminated amounts. Give the LER to four decimal places and dollar amounts to the nearest cent.

Method and assumption. For an exponential loss with mean $\theta$, the survival function is $S_X(x)=e^{-x/\theta}$ for $x\geq 0$. For a nonnegative loss, $E[\min(X,d)]=\int_0^d S_X(x)\,dx$. Here $\theta=5{,}000$ and $d=2{,}000$.

Before trend. Integrate the survival function to find the expected eliminated amount:

E[\min(X,2{,}000)] =\int_0^{2{,}000}e^{-x/5{,}000}\,dx =5{,}000(1-e^{-0.4}) =1{,}648.3998\ldots

The ground-up mean is $\$5{,}000$, so

\operatorname{LER}(X,2{,}000) =\frac{1{,}648.3998\ldots}{5{,}000} =0.3296799\ldots \approx 0.3297.

The expected insurer payment is $\$5{,}000-\$1{,}648.40=\$3{,}351.60$, which checks that the expected eliminated amount and expected insurer payment sum to the ground-up mean.

After trend. The factor is $c=1.08$. The trended exponential mean is $1.08(5{,}000)=\$5{,}400$, and the indexed deductible is $1.08(2{,}000)=\$2{,}160$. The expected eliminated amount is also multiplied by 1.08:

1.08(1{,}648.3998\ldots)=\$1{,}780.27, \qquad \operatorname{LER}=\frac{1{,}780.27\text{ (unrounded)}}{5{,}400} =0.3297.

The expected insurer payment after trend is $\$5{,}400-\$1{,}780.27=\$3{,}619.73$. The LER is unchanged, although the expected eliminated dollar amount has increased. Common mistake avoided: treating an unchanged LER as an unchanged expected eliminated amount.

Worked Example: A Mixed Severity With an Atom

In a hypothetical model, a loss is an equal-probability mixture of an exponential loss with mean $\$2{,}000$ and a deterministic loss of $\$6{,}000$. The ordinary deductible is $\$3{,}000$. Both losses and the deductible are scaled by 1.10. Find the LER before and after scaling, to four decimal places.

Method and assumption. Compute the expected ground-up loss and expected eliminated amount separately for the two mixture components, then weight each component by its probability of 0.5. This handles the deterministic component, or atom, directly.

The expected ground-up loss is

$$ E[X]=0.5(2{,}000)+0.5(6{,}000)=4{,}000. $$

For the exponential component, the expected amount eliminated is $2{,}000(1-e^{-3{,}000/2{,}000})=2{,}000(1-e^{-1.5})=\$1{,}553.74$ (rounded to cents). For the deterministic component, it is $\min(6{,}000,3{,}000)=\$3{,}000$. Therefore,

E[\min(X,3{,}000)] =0.5\bigl(2{,}000(1-e^{-1.5})\bigr)+0.5(3{,}000) =2{,}276.8698\ldots

The original ratio is $2{,}276.8698\ldots/4{,}000=0.5692174\ldots$, or 0.5692.

After scaling, the component means or values are $\$2{,}200$ and $\$6{,}600$, and the indexed deductible is $\$3{,}300$. The trended ground-up mean is $0.5(2{,}200)+0.5(6{,}600)=\$4{,}400$. The expected eliminated amount is

0.5\bigl(2{,}200(1-e^{-1.5})\bigr)+0.5(3{,}300) =2{,}504.5568\ldots

Thus the trended LER is $2{,}504.5568\ldots/4{,}400=0.5692174\ldots$, or 0.5692. As a check, the eliminated amount is 1.10 times its original value: $1.10(2{,}276.8698\ldots)=2{,}504.5568\ldots$. Common mistake avoided: dropping or mishandling a point mass when calculating the expected eliminated amount.

Worked Example: Uniform Severity and a Direct Check

A hypothetical loss is uniform on $[0,\$10{,}000]$, and the ordinary deductible is $\$2{,}500$. Both are scaled by 1.20. Find the original and trended LERs, and verify the expected eliminated amounts. Give the LER to four decimal places and dollar amounts to the nearest cent.

Method and assumption. For $X$ uniform on $[0,b]$, with $0\leq d\leq b$, its mean is $b/2$ and its density is $1/b$ on that interval. Integrating the eliminated amount gives

E[\min(X,d)] =\int_0^d \frac{x}{b}\,dx+\int_d^b\frac{d}{b}\,dx =\frac{d^2}{2b}+\frac{d(b-d)}{b} =d-\frac{d^2}{2b}.

For $b=\$10{,}000$ and $d=\$2{,}500$, the ground-up mean is $\$5{,}000$, and the expected eliminated amount is

2{,}500-\frac{2{,}500^2}{2(10{,}000)} =2{,}500-312.50 =\$2{,}187.50.

The original LER is $\$2{,}187.50/\$5{,}000=\mathbf{0.4375}$. After scaling, $b=\$12{,}000$, $d=\$3{,}000$, and $E[X]=\$6{,}000$. The expected eliminated amount is

3{,}000-\frac{3{,}000^2}{2(12{,}000)} =3{,}000-375 =\$2{,}625.00.

The trended ratio is $\$2{,}625/\$6{,}000=\mathbf{0.4375}$, and $\$2{,}625=1.20(\$2{,}187.50)$. The calculation confirms that the numerator and denominator scale together.

When the Invariance Does Not Apply

The theorem concerns common proportional scaling, not inflation in general. In particular, if the deductible stays fixed while losses scale, the two expected amounts need not share a common factor. For the exponential model in the first example, keeping the deductible at $\$2{,}000$ while increasing the mean from $\$5{,}000$ to $\$5{,}400$ gives

\operatorname{LER} =1-e^{-2{,}000/5{,}400} =0.3093\text{ (approximately),}

rather than the original 0.3297. The correct indexed deductible is $\$2{,}160$, which restores the common factor and the original ratio. Similarly, an additive change such as $X_c=cX+k$, with a deductible merely set to $cd$, is not the common scaling $X_c=cX$; the theorem alone cannot establish invariance in that case.

A practical check is to write down both transformations before calculating: is the new loss exactly $cX$, and is the new deductible exactly $cd$, with the same $c>0$? If not, calculate the new expected eliminated amount and ground-up mean under their actual definitions rather than assuming the ratio is preserved.

Key takeaway: Under an ordinary deductible, proportional scaling preserves LER because $\min(cX,cd)=c\min(X,d)$ and $E[cX]=cE[X]$. The expected amounts change in dollars; their ratio does not.

Check Your Understanding

For each problem, identify whether the loss and deductible share a common proportional scaling before using the invariance result.

A hypothetical exponential severity has mean $\$8{,}000$, an ordinary deductible of $\$1{,}600$, and a common trend factor of 1.05 for losses and deductible. Find the LER and indexed deductible. (Answer: LER $=1-e^{-0.2}=0.1813$; indexed deductible $=\$1{,}680$.)
A hypothetical loss is uniform on $[0,\$12{,}000]$, with deductible $\$3{,}000$. Both are scaled by 0.90. Find the LER before and after scaling. (Answer: $0.4375$ before and after.)
A hypothetical loss is $\$1{,}000$ with probability 0.4 and $\$5{,}000$ with probability 0.6. With deductible $\$2{,}000$, find the LER, then state it after both loss and deductible are scaled by 1.25. (Answer: $E[X]=\$3{,}400$, $E[\min(X,d)]=\$1{,}600$, and LER $=0.4706$ before and after.)
A hypothetical exponential severity has mean $\$4{,}000$ and deductible $\$1{,}000$. Losses increase by 20%, but the deductible remains fixed at $\$1{,}000$. Find the new LER and compare it with the LER if the deductible were indexed. (Answer: Fixed-deductible LER $=1-e^{-1{,}000/4{,}800}=0.1881$; indexed-deductible LER $=1-e^{-0.25}=0.2212$.)
With equal probabilities, a hypothetical loss is $\$2{,}000$ or $\$6{,}000$, and the deductible is $\$3{,}000$. Losses are instead changed to $1.10X+\$500$, while the deductible is set to $1.10d$. Calculate the new LER and explain whether common-scaling invariance applies. (Answer: New losses are $\$2{,}700$ and $\$7{,}100$, the new mean is $\$4{,}900$, and the expected eliminated amount is $\$3{,}000$; new LER $=0.6122$. The invariance result does not apply because the loss change includes an additive $\$500$.)

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