Loss Elimination Ratios and Deductible Savings

What a Loss Elimination Ratio Measures

An ordinary deductible shifts part of each covered loss from the insurer to the insured. The loss elimination ratio measures the expected amount shifted, relative to the expected ground-up loss before applying the deductible. It therefore describes a change in expected loss cost, not a change in claim frequency or a percentage of claims eliminated.

Let $X$ be a nonnegative ground-up loss with finite, positive expected value. For an ordinary per-loss deductible $d\geq 0$, the insurer payment is $(X-d)_+=\max(X-d,0)$. The portion of the ground-up loss not paid by the insurer is $\min(X,d)$. These amounts add to the original loss for every possible value of $X$.

Definition: The loss elimination ratio at deductible $d$ is $$ \operatorname{LER}(d)=\frac{E[\min(X,d)]}{E[X]}. $$ It is the expected loss amount removed by the deductible divided by the ground-up expected loss. The deductible factor relative to ground-up loss is $$ DF(d)=\frac{E[(X-d)_+]}{E[X]}=1-\operatorname{LER}(d). $$ These definitions require $0<E[X]<\infty$. They apply to the ordinary per-loss deductible considered here, not automatically to other deductible structures.

The ratio is based on dollars of expected loss. For a portfolio with a fixed expected number of claims, the same ratio applies to aggregate expected loss if the same severity distribution and deductible apply to each claim. This tutorial focuses on severity per loss; it does not model how a deductible changes claim frequency.

Why the Complement Relationship Holds

For every realized loss $X\geq 0$, the loss is divided into the amount below the deductible and the amount paid above it:

X=\min(X,d)+(X-d)_+.

If $X\leq d$, the two terms on the right are $X$ and 0. If $X>d$, they are $d$ and $X-d$. Thus the identity holds in both cases. Taking expectations gives

E[X]=E[\min(X,d)]+E[(X-d)_+].

Dividing by the positive quantity $E[X]$ shows that the shares of ground-up expected loss eliminated and paid by the insurer sum to one:

\operatorname{LER}(d)+DF(d)=1.

This derivation also explains a useful consistency check: an LER must be between 0 and 1, and its complement must equal the ground-up deductible factor. At $d=0$, the LER is 0 and the factor is 1. As $d$ increases, the amount eliminated cannot decrease, so the LER cannot decrease and the ground-up factor cannot increase.

Calculating the LER from a Severity Distribution

For a discrete severity distribution, calculate $\min(x,d)$ at each possible loss value $x$, weight each amount by its probability, and add. For a continuous nonnegative loss, the same expected amount can be expressed using the survival function $S_X(x)=P(X>x)$:

E[\min(X,d)]=\int_0^d S_X(x)\,dx, \qquad \operatorname{LER}(d)=\frac{\int_0^d S_X(x)\,dx}{E[X]}.

The integral identity follows by representing $\min(X,d)$ as the accumulated indicator that the loss exceeds each threshold from 0 to $d$: $\min(X,d)=\int_0^d \mathbf{1}_{\{X>x\}}\,dx$. Taking expectations gives the integral of $P(X>x)=S_X(x)$. The formula is useful when a survival function is known; the direct expectation definition remains valid for discrete, continuous, or mixed severities.

Worked Example: LER from a Discrete Severity Distribution

A hypothetical equipment policy has a ground-up loss $X$ with probabilities $P(X=0)=0.20$, $P(X=400)=0.50$, and $P(X=1{,}200)=0.30$, with amounts in dollars. Find the LER and ground-up deductible factor for an ordinary deductible of $d=\$500$. This is a discrete severity model; there is no interest or mortality assumption.

Step 1: Calculate ground-up expected loss.

E[X]=0.20(0)+0.50(400)+0.30(1{,}200) =0+200+360=\$560.

Step 2: Calculate the amount eliminated at each possible loss. The eliminated amount is $\min(X,500)$. It is \$0 when $X=0$, \$400 when $X=400$, and \$500 when $X=1{,}200$.

E[\min(X,500)] =0.20(0)+0.50(400)+0.30(500) =0+200+150=\$350.

Step 3: Calculate the ratio and factor.

\operatorname{LER}(500)=\frac{350}{560}=0.6250, \qquad DF(500)=1-0.6250=0.3750.

Check. The insurer pays \$0 for the first two outcomes and $\$1{,}200-\$500=\$700$ for the third. Its expected payment is $0.30(700)=\$210$, and $\$210/\$560=0.3750$, matching the deductible factor. The common mistake avoided here is to subtract the deductible from every loss, including losses below it.

Exponential Severity: A Closed-Form Calculation

For an exponential severity distribution with mean $\theta>0$, the survival function is $S_X(x)=e^{-x/\theta}$ for $x\geq0$, and $E[X]=\theta$. Applying the survival-integral formula yields

E[\min(X,d)] =\int_0^d e^{-x/\theta}\,dx =\theta\left(1-e^{-d/\theta}\right).

Consequently, for this model,

\operatorname{LER}(d)=1-e^{-d/\theta}, \qquad DF(d)=e^{-d/\theta}.

These expressions rely on the exponential severity assumption. They should not be applied to a different severity distribution merely because its mean is also $\theta$.

Worked Example: Exponential LER and Expected Savings

A hypothetical property portfolio has exponential ground-up severity with mean $\theta=\$1{,}200$ per loss. Calculate the expected amount eliminated, the LER, and the ground-up deductible factor at $d=\$600$. There is no interest or mortality assumption; the stated exponential distribution and mean fully specify the severity model.

Step 1: Calculate the exponent.

\frac{d}{\theta}=\frac{600}{1{,}200}=0.5, \qquad e^{-0.5}\approx 0.6065307.

Step 2: Calculate the LER and factor.

\operatorname{LER}(600)=1-e^{-0.5} \approx 1-0.6065307 =0.3934693,

DF(600)=e^{-0.5}\approx 0.6065307.

Step 3: Convert the ratio to expected dollars. The expected amount eliminated is the LER times the ground-up mean:

E[\min(X,600)] =1{,}200(0.3934693) \approx \$472.16.

The insurer's expected payment is $1{,}200(0.6065307)\approx \$727.84$. Rounded to four decimal places for the ratios and to the nearest cent for money, the LER is 0.3935, the factor is 0.6065, and the expected amount eliminated is \$472.16. The check is $\$472.16+\$727.84=\$1{,}200.00$, the ground-up expected loss.

Converting LERs to Factors Relative to a Base Deductible

A ground-up deductible factor compares expected insurer payment at deductible $d$ with expected ground-up loss. A rating plan may instead compare two deductible options, using a base deductible $b$. Define the factor for deductible $d$ relative to base deductible $b$ as the ratio of their expected insurer payments:

DF(d;b)=\frac{E[(X-d)_+]}{E[(X-b)_+]}.

This relative factor is not generally equal to $1-\operatorname{LER}(d)$, because its denominator is the payment at the base deductible, not ground-up loss. Using the ground-up factors in numerator and denominator gives the conversion:

DF(d;b) =\frac{DF(d)}{DF(b)} =\frac{1-\operatorname{LER}(d)}{1-\operatorname{LER}(b)}.

The denominator must be positive, meaning the base deductible leaves positive expected insurer payment. A relative factor below 1 indicates that the selected deductible produces lower expected insurer loss cost than the base deductible. Reversing the numerator and denominator would instead calculate the factor for the base relative to the selected deductible.

Worked Example: Convert LERs to a Base-Deductible Factor

A hypothetical severity model assigns probabilities 0.20, 0.50, and 0.30 to losses of \$100, \$500, and \$1,500, respectively. Compare a selected deductible of $d=\$600$ with a base deductible of $b=\$200$. The model is discrete, with no interest or mortality assumption.

Step 1: Find the ground-up expected loss.

E[X]=0.20(100)+0.50(500)+0.30(1{,}500) =20+250+450=\$720.

Step 2: Calculate the base-deductible eliminated amount and LER. At $b=200$, the eliminated amounts at losses \$100, \$500, and \$1,500 are \$100, \$200, and \$200.

E[\min(X,200)] =0.20(100)+0.50(200)+0.30(200) =20+100+60=\$180,

\operatorname{LER}(200)=\frac{180}{720}=0.2500, \qquad DF(200)=1-0.2500=0.7500.

Step 3: Calculate the selected-deductible eliminated amount and LER. At $d=600$, the eliminated amounts are \$100, \$500, and \$600.

E[\min(X,600)] =0.20(100)+0.50(500)+0.30(600) =20+250+180=\$450,

\operatorname{LER}(600)=\frac{450}{720}=0.6250, \qquad DF(600)=1-0.6250=0.3750.

Step 4: Convert to the factor relative to the base deductible.

DF(600;200) =\frac{1-\operatorname{LER}(600)}{1-\operatorname{LER}(200)} =\frac{0.3750}{0.7500}=0.5000.

Check. The expected payment at $b=200$ is $\$720-\$180=\$540$; at $d=600$, it is $\$720-\$450=\$270$. Their ratio is $\$270/\$540=0.5000$, as calculated. The selected deductible produces half the expected insurer loss cost of the base deductible under this severity model.

Interpretation and Common Traps

The LER is a proportion of expected ground-up dollars, not a claim-count statistic. In general, $P(X\leq d)$ is not the LER: it counts how often losses lie below the deductible, while the LER also reflects the size of the losses and the amount removed from losses above the deductible. A large loss above $d$ still has exactly $d$ dollars eliminated, not the entire loss.

A frequent incorrect calculation is to treat the deductible as though it were removed from every claim in full. For the discrete model in the first worked example, that error would calculate expected eliminated loss as $d=\$500$ and report an LER of $\$500/\$560=0.8929$. The correct amount is \$350: losses below the deductible contribute only their actual amount, and the probability-weighted calculation gives $0.20(0)+0.50(400)+0.30(500)=350$. The correct LER is therefore 0.6250.

Another exam trap is confusing the ground-up factor with a factor relative to a base deductible. The ground-up factor is $1-\operatorname{LER}(d)$; the relative factor divides that quantity by $1-\operatorname{LER}(b)$. Finally, these are loss-only comparisons. If a premium also includes expenses or other provisions that do not vary with expected loss, the loss factor alone does not determine the change in total premium.

Key takeaway: For an ordinary deductible, calculate the expected amount removed as $E[\min(X,d)]$, then divide by $E[X]$ to obtain the LER. Its complement is the factor relative to ground-up loss; divide two such complements to obtain a factor relative to a base deductible.

Check Your Understanding

For each problem, calculate the requested ratio or factor and check that the LER and ground-up factor sum to one.

A hypothetical severity distribution has probabilities 0.20, 0.50, and 0.30 at losses \$0, \$300, and \$900. For an ordinary deductible of \$400, find $E[X]$, the expected amount eliminated, the LER, and the ground-up deductible factor. (Answer: $E[X]=\$420$; expected amount eliminated $=\$270$; LER $=0.6429$; factor $=0.3571$.)
Ground-up severity is exponential with mean \$1,500. For a \$750 deductible, find the LER, ground-up factor, and expected amount eliminated. (Answer: LER $=0.3935$; factor $=0.6065$; expected amount eliminated $=\$590.20$.)
A deductible has LER 0.20, and a selected deductible has LER 0.44 under the same severity model. Find the selected deductible factor relative to the base deductible. (Answer: $(1-0.44)/(1-0.20)=0.70$.)
A discrete severity distribution assigns probabilities 0.20, 0.50, and 0.30 to losses of \$100, \$500, and \$1,000. Calculate the factor for a \$700 deductible relative to a \$200 base deductible. (Answer: $E[X]=\$570$; LERs are 0.3158 at \$200 and 0.8421 at \$700; relative factor $=0.2308$.)
For a severity model, ground-up expected loss is \$2,400 and expected insurer payment after a deductible is \$840. Find the expected amount eliminated, LER, and ground-up deductible factor. (Answer: expected amount eliminated $=\$1,560$; LER $=0.6500$; factor $=0.3500$.)

What You'll Learn