Selecting a Rate Under Uncertainty

Introduction

A rate indication is an estimate, not a guarantee that one precise rate change will be sufficient. The indication can depend on uncertain assumptions about losses, trends, expenses, or other components of the analysis. If plausible assumptions produce a range of indications, choosing the midpoint without explanation may not reflect the insurer’s actual tolerance for inadequate rates or unnecessary premium increases.

A selection under uncertainty is a decision rule: it combines the plausible indications with a stated way of evaluating the consequences of choosing a rate change that differs from them. The selection is therefore not determined by the range alone. It depends on the scenarios considered, any probabilities assigned to them, and the relative cost assigned to shortfall and excess.

Definition: Let \(R\) denote the indicated rate change under a particular plausible set of assumptions, and let \(c\) denote the selected rate change. A shortfall occurs when \(c<R\); an excess occurs when \(c>R\). A selection rule specifies how these differences are evaluated and how \(c\) is chosen. All rate differences below are measured in percentage points.

Expected Loss With Different Costs for Shortfall and Excess

When a decision maker can assign probabilities to a finite set of plausible indications, one approach is to choose the rate change with the lowest expected loss. The probabilities represent the decision maker’s assessment of the scenarios, not a certainty that the scenarios exhaust every possible outcome. They must be nonnegative and sum to one.

Suppose the loss from a difference between the selected and indicated rate is squared. Let \(\lambda>0\) be the penalty weight for shortfall and \(\mu>0\) the penalty weight for excess. A higher \(\lambda\) expresses greater concern about selecting too little; a higher \(\mu\) expresses greater concern about selecting too much.

Model: For indication \(R_j\) with probability \(p_j\), define the loss at selection \(c\) as \(\lambda(R_j-c)^2\) when \(c<R_j\), and \(\mu(c-R_j)^2\) when \(c>R_j\). The loss is zero when \(c=R_j\). The expected loss is the probability-weighted average of these scenario losses. This model is appropriate when squared deviations and the chosen probability and penalty weights represent the decision maker’s objective; it does not establish that those weights or probabilities are objectively correct.

L(c)=\sum_{j:c<R_j}p_j\lambda(R_j-c)^2+ \sum_{j:c>R_j}p_j\mu(c-R_j)^2

For any interval between adjacent indicated values, the scenarios on either side of \(c\) are fixed. On that interval, the expected loss is a quadratic function of \(c\). Its minimum can be found by differentiating, checking that the resulting candidate is in the interval, and comparing it with any relevant boundary values. The loss is convex: each scenario’s squared loss is convex, and a probability-weighted sum of convex functions is convex. Thus a valid stationary point is a global minimum.

Why the Weighted Selection Formula Holds

For a candidate \(c\) between some indications, separate indications below \(c\), which contribute excess losses, from indications above \(c\), which contribute shortfall losses. Differentiating the expected loss on this interval gives

\frac{1}{2}L'(c) = \sum_{j:R_j<c}p_j\mu(c-R_j) + \sum_{j:R_j>c}p_j\lambda(c-R_j).

Setting the derivative to zero and collecting the terms in \(c\) gives the candidate

c= \frac{ \mu\sum_{j:R_j<c}p_jR_j+ \lambda\sum_{j:R_j>c}p_jR_j }{ \mu\sum_{j:R_j<c}p_j+ \lambda\sum_{j:R_j>c}p_j }.

This expression is conditional on which indications fall below and above \(c\). It is not safe to use it without checking that the candidate lies in the interval assumed when deriving it. If all scenario losses use the same squared-deviation weight, the minimizer reduces to the probability-weighted mean of the indications. With asymmetric penalties, indications on the shortfall side receive weight \(\lambda\), while those on the excess side receive weight \(\mu\).

Worked Example: Asymmetric Penalties Across Three Indications

A hypothetical insurer’s sensitivity analysis produces indicated rate changes of 5%, 9%, and 13%, with respective probabilities 0.25, 0.50, and 0.25. The decision maker assigns a shortfall penalty \(\lambda=2\) and an excess penalty \(\mu=1\), using squared deviations. Find the selection that minimizes expected loss.

Method and interval. First test the interval from 9% to 13%. In that interval, the 5% and 9% indications are below the selection, so they incur excess penalties; the 13% indication is above it, so it incurs a shortfall penalty. The derivative, divided by 2, is

$$ 0.25(c-5)+0.50(c-9)+2(0.25)(c-13) = 1.25c-12.25. $$

Setting this equal to zero gives \(c=12.25/1.25=9.8\%\). This is inside the tested interval, so it is a valid stationary point. Convexity makes it a global minimum.

Check the expected loss. At \(c=9.8\%\), the 5% scenario has an excess of 4.8 percentage points and contributes \(0.25(1)(4.8^2)=5.76\). The 9% scenario has an excess of 0.8 percentage points and contributes \(0.50(1)(0.8^2)=0.32\). The 13% scenario has a shortfall of 3.2 percentage points and contributes \(0.25(2)(3.2^2)=5.12\). Total expected loss is \(5.76+0.32+5.12=11.20\) squared percentage points.

For comparison, at \(c=10\%\), the contributions are \(0.25(5^2)=6.25\), \(0.50(1^2)=0.50\), and \(0.25(2)(3^2)=4.50\), for a total of 11.25. The lower loss at 9.8% confirms the selection. Common mistake avoided: rounding a candidate to 10% before checking the loss would not give the minimum.

Sensitivity to Scenario Probabilities and Risk Preferences

A calculated optimum should be interpreted together with the assumptions that produced it. A selection can move if the probability assigned to a high indication changes, if the shortfall penalty changes, or if the scenarios themselves change. These shifts do not automatically make the original calculation wrong; they show which judgments matter to the decision.

Worked Example: Testing the Effect of the Shortfall Penalty

For a hypothetical product, the indications are 4%, 8%, and 12%, with probabilities 0.25, 0.50, and 0.25. First use equal squared-loss penalties, \(\lambda=\mu=1\). Then repeat with \(\lambda=2\) and \(\mu=1\). Compare the selections and identify what the sensitivity says.

Equal penalties. With the same penalty on both sides, the minimizer is the probability-weighted mean:

0.25(4)+0.50(8)+0.25(12)=1+4+3=8\%.

Greater shortfall penalty. Test a selection between 8% and 12%. Indications of 4% and 8% incur excess penalties, while 12% incurs a shortfall penalty. The derivative divided by 2 is

$$ 0.25(c-4)+0.50(c-8)+2(0.25)(c-12) =1.25c-11. $$

The zero is \(c=11/1.25=8.8\%\), which lies between 8% and 12%. It is therefore the global minimum for this convex loss function.

As a direct check, the expected loss at 8.8% is \(0.25(4.8^2)+0.50(0.8^2)+0.25(2)(3.2^2)=5.76+0.32+5.12=11.20\). At 8%, it is \(0.25(4^2)+0+0.25(2)(4^2)=4+8=12\). The selection moves from 8% to 8.8% because the decision maker assigned greater cost to shortfall. This is a preference sensitivity, not evidence that the underlying indications changed.

Sensitivity analysis is most useful when it reports both the selected value and the assumptions that move it. For example, a recommendation might state that the central selection is 8.8% under the specified shortfall penalty, while equal penalties give 8.0%. If the two results differ materially, the decision maker should be explicit about which risk preference supports the recommendation rather than presenting one value as assumption-free.

When Scenario Probabilities Are Not Credible

A probability-weighted decision rule can create false precision if there is no defensible basis for assigning probabilities. In that situation, a robust rule can focus on the range of plausible indications instead. One simple option is to minimize the maximum absolute deviation from any indication in the range. This rule treats every indication in the specified range as possible and does not assign greater likelihood to one than another.

Minimax absolute-deviation rule: For plausible indications ranging from \(R_{\min}\) to \(R_{\max}\), choose \(c\) to minimize \(\max(|c-R_{\min}|,|c-R_{\max}|)\). When selections between the endpoints are allowed and both endpoints are included, the minimizing selection is the midpoint of the range. The rule is appropriate only when minimizing the worst absolute deviation is a suitable representation of risk tolerance; it does not distinguish shortfall from excess.

Worked Example: A Robust Selection From a Range

A hypothetical insurer has three plausible indications: 6%, 10%, and 14%. It cannot defend probabilities for these outcomes. Its stated risk tolerance is to minimize the largest absolute deviation from any of the three indicated changes. Find the selection and the resulting worst-case deviation.

Method. The range endpoints are 6% and 14%. The midpoint is \((6+14)/2=10\%\). At this selection, the deviations from 6%, 10%, and 14% are 4, 0, and 4 percentage points, respectively. The maximum is 4 percentage points.

Why this is optimal. For any selection \(c\), the triangle inequality gives

14-6 \leq |14-c|+|c-6|.

The right-hand side must therefore be at least 8 percentage points, so at least one of its two terms must be at least 4. No selection can have a maximum endpoint deviation below 4. The midpoint attains that bound, and the intermediate 10% indication also has zero deviation. Thus 10% is the minimax selection, with a worst-case deviation of 4 percentage points.

Common mistake avoided: using a probability-weighted mean when no probabilities have been justified would answer a different decision problem. Here, the midpoint follows from the stated worst-case criterion.

Turning a Calculation Into a Defensible Selection

A numerical optimum is only as useful as the decision statement supporting it. State the plausible indications, any assigned probabilities, the loss or risk measure, the decision-maker’s risk tolerance, and the selected rate change. If probabilities are used, confirm that they sum to one. If an asymmetric loss is used, identify which side carries the greater penalty. If a robust rule is used, identify the range over which the worst case is evaluated.

Keep the interpretation tied to the assumptions. An expected-loss minimum is optimal only relative to its specified scenario probabilities and loss function. A minimax selection is optimal only relative to its specified range and worst-case criterion. Neither method removes uncertainty; each makes the treatment of uncertainty explicit. The process is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Define the decision.
Let \(c\) be the proposed rate change and express the plausible indications on the same percentage-point scale.

State the uncertainty model.
Specify scenario probabilities when they are defensible, or define the plausible range when they are not.

Represent risk tolerance.
Choose a loss function or worst-case criterion that clearly distinguishes any concern about shortfall from concern about excess.

Calculate and verify.
Check interval assumptions for a piecewise optimum, evaluate the resulting loss, and test sensitivity to important assumptions.

Explain the recommendation.
Report the selected change together with the assumptions and risk tolerance that justify it.

Common Mistakes

Treating the midpoint as automatic. A midpoint is justified by a particular criterion, such as minimizing the maximum absolute deviation, not merely by having a range.
Calling a probability-weighted result assumption-free. Scenario probabilities and penalty weights affect the expected-loss minimum and should be stated.
Using a piecewise formula without checking the interval. The candidate formula assumes particular indications lie below or above \(c\); verify that the solution satisfies those assumptions.
Rounding before comparison. Retain sufficient precision through the calculation, then round the selected change as appropriate for the decision. A nearby rounded value need not be the exact loss minimum.
Confusing a selected rate with certainty about the indication. A selection is a decision under uncertainty; it does not make the scenarios or their probabilities certain.

Check Your Understanding

For each problem, identify the decision rule before calculating. Treat all selections and deviations as percentage points.

A sensitivity analysis gives indications of 2%, 6%, and 10% with probabilities 0.25, 0.50, and 0.25. Under symmetric squared loss, what selection minimizes expected loss? (Answer: 6%)
Indications are 4%, 8%, and 12%, with probabilities 0.20, 0.50, and 0.30. The shortfall penalty is \(\lambda=2\) and the excess penalty is \(\mu=1\), with squared deviations. Find the minimizing selection. (Answer: 9.2308%)
A range of plausible indications is 5% to 15%, and the decision maker wants to minimize the largest absolute deviation from the range. Find the selection and its worst-case deviation. (Answer: 10%; 5 percentage points)
Indications are 5%, 9%, and 13%, with probabilities 0.40, 0.35, and 0.25. The decision rule is to choose the smallest \(c\) for which the probability that the indication exceeds \(c\) is no more than 0.25. What is the selection? (Answer: 9%)
Indications are 3% and 7%, with probabilities 0.40 and 0.60. Under squared loss, let \(\lambda=1.5\) for shortfall and \(\mu=1\) for excess. Find the minimizing selection, assuming it lies between the two indications, and verify that interval condition. (Answer: 5.7692%; it lies between 3% and 7%)

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