Turning a Target LER into a Deductible
A deductible determines how much of each ground-up loss is not paid by the insurer. When a coverage design specifies a target loss elimination ratio, the practical question is the reverse of an ordinary LER calculation: what deductible produces that target? The answer may be an algebraic inverse, a solution on one interval of a piecewise formula, or a numerical root.
Throughout, let \(X\) be a nonnegative ground-up loss with finite, positive mean, and suppose the policy has an ordinary per-loss deductible \(d\) and no limit or coinsurance. The insurer payment per loss is \((X-d)_+=\max(X-d,0)\). The loss eliminated by the deductible is \(\min(X,d)\). Losses and deductibles must be expressed in the same monetary units and on the same severity basis.
The two forms are equivalent because \(X=\min(X,d)+(X-d)_+\) for every nonnegative loss. Thus, the LER is the expected amount eliminated divided by expected ground-up loss; it is not the expected insurer payment divided by expected ground-up loss. This distinction determines which expected amount the target equation must match.
Why the Target Equation Holds
The expected eliminated amount can also be written using the survival function \(S_X(t)=P(X>t)\). For each fixed loss \(x\geq 0\), the area under the indicator \(1_{\{x>t\}}\) from zero to \(d\) is exactly \(\min(x,d)\). Taking expectations gives the survival-integral expression:
The expectation and integral can be interchanged because the integrand is nonnegative. Consequently, solving for a target \(r\) is equivalent to solving \( \int_0^d S_X(t)\,dt=rE[X] \). This equation applies to continuous, discrete, and mixed severities. It is also useful when a closed-form expression for the integral is unavailable.
For nonnegative losses, the LER starts at zero when \(d=0\) and approaches one as the deductible increases without bound. A bounded severity can attain one at a finite deductible: any \(d\) at least as large as the maximum possible loss eliminates the entire loss. A target outside the attainable range has no solution. If the survival function is zero over an interval, increasing \(d\) across that interval does not change the LER, so uniqueness also requires attention to the severity’s support.
Closed-Form Inverses
When the expected eliminated amount has a simple formula, invert that formula directly. For example, if \(X\) is exponential with rate \(\lambda>0\), then \(E[X]=1/\lambda\), and the survival-integral equation gives \(E[\min(X,d)]=(1-e^{-\lambda d})/\lambda\). The target equation reduces to \(1-e^{-\lambda d}=r\).
Worked Example: Exponential Severity and a Target LER
A hypothetical portfolio has exponential ground-up losses with mean $4,000, so \(\lambda=1/4000\) per dollar. Find the ordinary deductible that produces an LER of \(0.6000\). Assume there is no policy limit.
Method. Use the exponential closed-form inverse. The target eliminated amount is \(0.6000(4000)=\$2,400\), or 60% of the mean loss.
For an exponential loss, \(\operatorname{LER}(d)=1-e^{-\lambda d}\). Set this equal to the target and solve:
Check. At this deductible, \(e^{-d/4000}=0.4000\), so the insurer’s expected payment is 40% of the mean loss. The eliminated share is \(1-0.4000=0.6000\), as required. The deductible is rounded to the nearest cent.
Common mistake avoided. Setting the insurer-paid share equal to \(0.6000\) would instead give \(e^{-d/4000}=0.6000\), or \(d=-4000\ln(0.6000)=\$2{,}043.30\). That deductible produces an LER of \(0.4000\), not \(0.6000\): it matches the wrong share.
A uniform severity provides another direct inverse. If \(X\) is uniform on \([0,b]\) and \(0\leq d\leq b\), then \(E[X]=b/2\) and \(E[\min(X,d)]=d-d^2/(2b)\). With \(z=d/b\), the target equation becomes \(2z-z^2=r\). The deductible in this range is \(d=b(1-\sqrt{1-r})\). A candidate inverse must still be checked against the range used to derive it; if \(d\geq b\), the LER is already one.
Discrete Severities and Interval Checks
For a discrete severity, the expected eliminated amount is a sum of probability-weighted payments. Between successive possible loss amounts, each loss above the deductible contributes the deductible itself, while each loss below it contributes its full value. The resulting function is piecewise linear. Solve on a proposed interval, then confirm that the solution actually lies in that interval.
Worked Example: A Discrete Severity with an Interior Solution
A hypothetical equipment-loss model assigns probabilities \(0.25\), \(0.50\), and \(0.25\) to losses of $1,000, $3,000, and $6,000, respectively. Find a deductible that gives an LER of \(0.5000\).
Method. Calculate the mean, choose the interval containing the deductible, and solve the corresponding piecewise-linear equation. The mean ground-up loss is \(0.25(1000)+0.50(3000)+0.25(6000)=250+1500+1500=\$3,250\). The target eliminated amount is \(0.5000(3250)=\$1,625\).
Suppose first that \(1000\leq d\leq 3000\). The $1,000 loss is fully eliminated, while the $3,000 and $6,000 losses each contribute \(d\). Thus,
Equating this to the target gives \(250+0.75d=1625\), so \(0.75d=1375\) and \(d=1375/0.75=\$1,833.33\). This value lies between $1,000 and $3,000, confirming the assumed interval.
Check. The expected eliminated amount is \(250+0.75(1833.33)=\$1,625.00\), to the nearest cent. Dividing by the mean gives \(1625/3250=0.5000\). The deductible is rounded to the nearest cent.
Common mistake avoided. Using the interval formula outside its stated range can produce a plausible-looking but invalid answer. The interval check confirms that the $1,000 loss is fully eliminated and that both larger losses are only partially eliminated.
Numerical Solutions When There Is No Useful Inverse
For some mixtures or other severity models, the target equation cannot be rearranged into a convenient closed form. A numerical root-finding method is then appropriate. Bisection is reliable when an interval \([a,b]\) brackets a solution: evaluate the target difference at both endpoints, retain the half-interval where the sign changes, and repeat. For an ordinary deductible, monotonicity of \(E[\min(X,d)]\) makes this method particularly transparent.
Calculate \(T=rE[X]\), using consistent loss units.
Use \(g(d)=E[\min(X,d)]-T\); a solution satisfies \(g(d)=0\).
Find endpoints where \(g\) has opposite signs, then bisect or use another suitable numerical root finder.
Substitute the reported deductible into the expected eliminated amount and recompute the LER.
Worked Example: Bisection for a Mixture of Exponentials
A hypothetical severity is an equal-probability mixture of two exponential distributions: conditional on the first component, the mean loss is $2,000; conditional on the second, it is $8,000. Find a deductible giving an LER of \(0.5000\). Work in thousands of dollars and assume no policy limit.
Method. For an exponential loss with mean \(m\), the expected amount eliminated is \(m(1-e^{-d/m})\). Averaging over the two equally likely components gives \(E[X]=0.5(2)+0.5(8)=5\) thousand dollars and \(E[\min(X,d)]=1(1-e^{-d/2})+4(1-e^{-d/8})\). The target eliminated amount is \(0.5000(5)=2.5\) thousand dollars.
Define \(g(d)=1-e^{-d/2}+4(1-e^{-d/8})-2.5\). At \(d=4.0000\), the two eliminated components are \(1-e^{-2}=0.8647\) and \(4(1-e^{-0.5})=1.5739\), giving \(2.4385\) thousand dollars, below target. At \(d=4.5000\), they are approximately \(0.8946\) and \(1.7209\), giving \(0.8946+1.7209=2.6155\) thousand dollars, above target. Thus a solution is bracketed between 4.0000 and 4.5000.
Bisection narrows this bracket by repeatedly testing its midpoint. For example, at \(d=4.2500\), the eliminated amount is approximately \(0.8806+1.6485=2.5291\), above target, so the bracket becomes \([4.0000,4.2500]\). At \(d=4.1250\), it is approximately \(0.8729+1.6115=2.4844\), below target, so the bracket becomes \([4.1250,4.2500]\). Continuing bisection gives \(d\approx4.1684\) thousand dollars.
Check. Substitution at \(d=4.1684\) gives \(1-e^{-2.0842}+4(1-e^{-0.52105})\), approximately \(0.8756+1.6244=2.5000\) thousand dollars. Dividing by the mean of 5 thousand dollars gives an LER of \(2.5000/5=0.5000\). The deductible is approximately $4,168.40.
Common mistake avoided. The mixture weights must be applied to each component’s expected eliminated amount. The terms \(1\) and \(4\) above already include the component probabilities: they are \(0.5(2)\) and \(0.5(8)\), not the component means alone.
Practical Checks Before Reporting a Deductible
A computed root is only meaningful if the model and target equation match the coverage being priced or analyzed. The formulas here concern an ordinary deductible applied to a nonnegative ground-up loss, without a policy limit or coinsurance. If the loss has first been trended, use the severity after that trend and express the deductible in the corresponding monetary units, consistent with the coverage terms.
- Check feasibility. For a nonnegative loss and ordinary deductible, an LER target must lie between zero and one. Consider whether a finite deductible can attain the target, particularly for bounded severity.
- Check the denominator. The LER denominator is \(E[X]\), the expected ground-up loss, not the expected insurer payment.
- Check the units. If \(X\) is measured in thousands of dollars, report \(d\) in thousands or convert it to dollars before rounding.
- Check the formula’s range. A piecewise solution must satisfy the interval assumed in its derivation, and a closed-form inverse must respect its parameter restrictions.
- Check the result by substitution. Recalculate the eliminated amount and its ratio to \(E[X]\); do not rely only on the output of a root finder.
The central workflow is simple: translate the target ratio into an expected eliminated amount, select the appropriate formula for the severity, solve within the formula’s valid range, and verify the resulting ratio. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, find the deductible that produces the stated target LER. Round money to the nearest cent unless the problem asks for thousands of dollars.
- A hypothetical exponential loss has mean $2,500. What deductible produces an LER of \(0.2000\)? (Answer: $557.86)
- A hypothetical loss is uniform on \([0,\$12{,}000]\). What deductible produces an LER of \(0.4375\)? (Answer: $3,000.00)
- A discrete loss takes values of $2,000, $5,000, and $9,000 with probabilities \(0.20\), \(0.50\), and \(0.30\), respectively. Find a deductible for an LER of \(0.4000\). (Answer: $2,300.00)
- In thousands of dollars, a hypothetical severity is an equal-probability mixture of exponential distributions with means 1 and 3. For target LER \(0.5000\), set \(y=e^{-d/3}\), solve \(y^3+3y-2=0\), and report the deductible to four decimals in thousands. (Answer: approximately 1.5522 thousand dollars)