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CAS Exam 5 · Territory Rating Methods · Tutorial 453 of 989

Smoothing Relativities Across Neighboring Territories

Apply and check territory-smoothing rules while distinguishing a plausible geographic relationship from an assumption based only on map proximity.

Intermediate 11 min read

What You'll Learn

  • Apply a one-pass neighbor-averaging rule to territory relativities.
  • Identify how adjacency choices and smoothing strength affect the result.
  • Rebalance smoothed factors to a required exposure-weighted average.
  • Formulate and solve a weighted penalized-smoothing problem.
  • Assess whether geographic proximity is a credible complement to sparse local experience.

Why Smooth Territory Relativities?

Territory indications can vary sharply from one area to the next, especially when local experience is limited. Some variation may reflect genuine differences in expected cost; some may be random fluctuation. Smoothing is a way to moderate variation between areas judged to be related. It does not establish that neighboring territories have the same risk, and adjacency alone is not evidence that they do.

The method must therefore make its assumptions visible. The analyst specifies which territories count as neighbors, how strongly each neighbor contributes, and how much smoothing to apply. The resulting factors should then be checked against the rating objective, the underlying experience, and plausible geographic or underwriting differences. This topic is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Definition: Let \(r_j\) be the preliminary relativity for territory \(j\), before the smoothing rule is applied, and let \(q_j\) be its smoothed relativity. A neighborhood rule specifies the set \(N(j)\) of territories used as neighbors of \(j\), along with any neighbor weights. A smoothing rule combines \(r_j\) with information from those selected neighbors. The rule is a modeling choice, not a universal actuarial formula.

The preliminary relativities may already reflect the experience measure, credibility approach, and complement selected for the analysis. Smoothing is a separate geographic adjustment to those inputs; it should not silently replace those choices or count the same information twice. In particular, territory factors are not smoothed merely because adjacent areas can be drawn on a map.

A One-Pass Neighbor-Averaging Rule

One transparent rule blends each preliminary relativity with a weighted average of its neighbors. For territory \(j\), let \(a_{jk}\) be the weight assigned to neighbor \(k\), where the weights are nonnegative and sum to one across \(N(j)\). Let \(0\leq\alpha\leq1\) be the smoothing strength. Then

$$ q_j=(1-\alpha)r_j+\alpha\sum_{k\in N(j)}a_{jk}r_k. $$

When \(\alpha=0\), there is no smoothing; when \(\alpha=1\), the preliminary factor is replaced by the weighted neighbor average. Values between zero and one produce a convex blend, so the result lies between the territory’s own indication and its neighbor average. This rule is a one-pass calculation: all neighbor averages use the preliminary \(r_k\), not factors updated earlier in the calculation. Repeatedly updating factors would be a different procedure and must be stated as such.

Worked Example: One-Pass Smoothing and Statewide Balance

A hypothetical insurer has three territories in a line. Their preliminary relativities are \(r_1=0.80\), \(r_2=1.40\), and \(r_3=1.00\). Territory 1 has only territory 2 as a neighbor, territory 3 has only territory 2, and territory 2 gives equal weight to territories 1 and 3. Use \(\alpha=0.25\). The exposure weights are 100, 50, and 150, respectively. Apply the one-pass rule, then restore an exposure-weighted average relativity of 1.00.

Method and smoothing step. For territory 1, the neighbor average is \(1.40\), so \(q_1=0.75(0.80)+0.25(1.40)=0.60+0.35=0.95\). For territory 2, the neighbor average is \((0.80+1.00)/2=0.90\), so \(q_2=0.75(1.40)+0.25(0.90)=1.05+0.225=1.275\). For territory 3, the neighbor average is \(1.40\), so \(q_3=0.75(1.00)+0.25(1.40)=0.75+0.35=1.10\).

Check the required balance. The preliminary exposure-weighted mean is \((100(0.80)+50(1.40)+150(1.00))/300=(80+70+150)/300=1.00\). The smoothed weighted mean is \((100(0.95)+50(1.275)+150(1.10))/300=(95+63.75+165)/300=1.0791667\). Thus smoothing alone does not preserve the required mean.

Rebalance. Multiply every smoothed relativity by \(1.00/1.0791667=240/259\). The balanced factors are \(0.95(240/259)=228/259=0.8803\), \(1.275(240/259)=306/259=1.1815\), and \(1.10(240/259)=264/259=1.0193\), rounded to four decimals. The weighted sum is \((100(228)+50(306)+150(264))/259=77700/259=300\); dividing by 300 exposures gives 1.00. The calculation is checked.

Common mistake avoided: Reporting the unbalanced \(q_j\) as final factors would change the exposure-weighted average from 1.00 to 1.0792. Rebalancing is necessary here because the indicated target is explicitly 1.00.

The rebalance in this example multiplies all smoothed factors by one constant. It preserves their ordering and their ratios to one another, but changes their absolute levels. The appropriate target depends on the rating plan’s balancing requirement; it should not be assumed to be 1.00 without checking that requirement.

Choosing the Neighborhood Is an Actuarial Judgment

A geographic neighborhood is defensible when location plausibly conveys information about expected cost after considering the other features already reflected in the indications. Potential reasons for similarity could include a continuous area with comparable exposure conditions. Potential reasons against it include a major geographic barrier, a sharp change in urban density, different weather or traffic patterns, or a boundary that separates substantially different risks. These are hypotheses to evaluate, not facts established by a shared boundary.

When local experience is sparse, nearby territories can be considered as a complement, but proximity should not be used as an automatic substitute for assessing the selected complement or credibility. A defensible analysis asks whether the relationship is plausible for the insured population and coverage, whether the chosen neighbors are consistent with that relationship, and whether the degree of smoothing is proportionate to the evidence. If the geographic relationship is weak or contradicted by relevant risk differences, the analyst should reduce the smoothing, change the neighbor set, or leave the factors unsmoothed rather than force similarity.

One practical way to represent unequal neighbor influence is to assign weights \(a_{jk}\), such as weights that give more influence to a genuinely similar adjacent territory than to a less similar one. The weights must be specified and normalized to sum to one for each territory with neighbors. A territory with no selected neighbors has no neighbor average under this rule; the analyst must not invent one by treating every territory as a neighbor.

Worked Example: Excluding an Unconvincing Geographic Link

Suppose preliminary relativities for four hypothetical territories are \(0.80, 1.20, 1.00,\) and \(1.40\), with exposure weights \(100, 50, 80,\) and \(70\). The proposed smoothing strength is \(\alpha=0.25\). A map shows a continuous boundary between territories 1 and 2 and between 3 and 4, but territories 2 and 3 are separated by a substantial water barrier and have different traffic patterns. Apply one-pass smoothing only within the two plausible pairs, then rebalance to the original exposure-weighted mean.

Method and adjacency decision. Use neighbor pairs \(1\leftrightarrow2\) and \(3\leftrightarrow4\), with no link between 2 and 3. The factors after smoothing are \(q_1=0.75(0.80)+0.25(1.20)=0.90\) and \(q_2=0.75(1.20)+0.25(0.80)=1.10\). For the second pair, \(q_3=0.75(1.00)+0.25(1.40)=1.10\) and \(q_4=0.75(1.40)+0.25(1.00)=1.30\).

Calculate the target mean. The preliminary weighted sum is \(100(0.80)+50(1.20)+80(1.00)+70(1.40)=80+60+80+98=318\). Total exposure is 300, so the target mean is \(318/300=1.06\). The smoothed weighted sum is \(100(0.90)+50(1.10)+80(1.10)+70(1.30)=90+55+88+91=324\), giving a smoothed mean of \(324/300=1.08\).

Rebalance and check. The multiplier is \(1.06/1.08=53/54\). The balanced factors are \(0.90(53/54)=0.8833\), \(1.10(53/54)=1.0796\), \(1.10(53/54)=1.0796\), and \(1.30(53/54)=1.2759\), rounded to four decimals. Their weighted sum is \(324(53/54)=318\), so their weighted mean is \(318/300=1.06\), as required.

Common mistake avoided: Treating all areas that touch on a map as equivalent neighbors would include the 2–3 link despite the stated barrier and risk differences. The chosen neighborhood is part of the actuarial assumption, not merely a geometric convention.

Penalized Smoothing: Balancing Fit and Neighbor Similarity

A second formulation makes the tradeoff explicit. Let \(w_j>0\) measure the relative precision assigned to preliminary indication \(r_j\), and let \(c_{jk}\geq0\) be the strength of an included neighbor link. A penalized method chooses factors \(q_j\) to minimize a weighted measure of departure from the indications plus a penalty for differences across linked territories:

$$ \sum_j w_j(q_j-r_j)^2+\lambda\sum_{\{j,k\}}c_{jk}(q_j-q_k)^2, \qquad \lambda\geq0. $$

Each pair \(\{j,k\}\) is included once in the second sum. The first term rewards fidelity to the preliminary indications; the second discourages differences across selected links. Larger \(\lambda\) imposes stronger similarity. This is a different rule from one-pass averaging: \(\lambda\) is a penalty strength relative to the \(w_j\) and \(c_{jk}\), not a direct percentage blend. The weights need an actuarial rationale and should reflect the reliability assigned to each indication, not be chosen just to obtain a desired result.

Derivation: For two linked territories with preliminary indications \(r_1,r_2\), weights \(w_1,w_2\), and link strength \(c\), minimize \(w_1(q_1-r_1)^2+w_2(q_2-r_2)^2+\lambda c(q_1-q_2)^2\). Setting each partial derivative to zero gives \(w_1(q_1-r_1)+\lambda c(q_1-q_2)=0\) and \(w_2(q_2-r_2)+\lambda c(q_2-q_1)=0\). Equivalently, \((w_1+\lambda c)q_1-\lambda c q_2=w_1r_1\) and \(-\lambda c q_1+(w_2+\lambda c)q_2=w_2r_2\). The equations show why the selected link pulls the two fitted factors toward each other, while each indication continues to exert influence through its weight.

Worked Example: Solving a Two-Territory Penalized Model

Two hypothetical adjacent territories have preliminary factors \(r_1=0.80\) and \(r_2=1.20\). Use precision weights \(w_1=4\) and \(w_2=1\), link strength \(c=1\), and penalty parameter \(\lambda=2\). Find the penalized factors and check their weighted mean.

Method. Substitution into the normal equations gives \(6q_1-2q_2=3.2\) and \(-2q_1+3q_2=1.2\). Multiply the first equation by 3 to obtain \(18q_1-6q_2=9.6\), and multiply the second by 2 to obtain \(-4q_1+6q_2=2.4\). Adding gives \(14q_1=12\), so \(q_1=12/14=6/7=0.8571\).

Substitute \(q_1=6/7\) into the first equation: \(6(6/7)-2q_2=3.2\), so \(36/7-2q_2=16/5\). Thus \(2q_2=36/7-16/5=(180-112)/35=68/35\), and \(q_2=34/35=0.9714\). Values are rounded to four decimals only at the end.

Check. The preliminary weighted mean is \((4(0.80)+1(1.20))/5=4.40/5=0.88\). The fitted weighted mean is \((4(6/7)+34/35)/5=(120/35+34/35)/5=(154/35)/5=0.88\). The mean is preserved because the pairwise penalty depends only on the difference between factors; summing the two normal equations cancels the link terms. The solution is checked.

Common mistake avoided: Applying \(\lambda=2\) as a 200% blend would not solve the stated optimization problem. Here \(\lambda\) affects the equations jointly with the precision and link weights; it is not the \(\alpha\) in the one-pass rule.

When Smoothing Helps—and When It Does Not

Smoothing can reduce unstable jumps when adjacent areas are plausibly related and local indications contain substantial random variation. It can also obscure a real difference if the neighbor set or penalty is poorly chosen. Neither a visually smooth map nor a single numerical fit demonstrates that the factors are actuarially appropriate.

Before applying a rule, make the following checks:

  • Define the input. State whether the starting factors are raw indications or already adjusted for credibility, and what they represent.
  • State the graph and rule. Identify included neighbors, weights, smoothing strength, and whether the calculation is one-pass or iterative.
  • Assess plausibility. Ask whether proximity is meaningful for this coverage and population, and whether barriers or sharp risk changes undermine the link.
  • Check the rating constraint. Recalculate the required weighted average after smoothing and rebalance if the rating objective requires it.
  • Retain judgment. A mathematical smoothing rule cannot resolve an unsupported geographic assumption. If the relationship is not defensible, do not impose it.

The central distinction is between using geography as evidence and using geography as a convenient label. Proximity may be a useful complement to sparse local experience when the underlying risk relationship is credible. It is not, by itself, a reason to move one territory’s factor toward another’s.

Check Your Understanding

For each problem, state the rule you use and check any requested weighted average.

  1. Three territories form a chain with preliminary factors \(0.70, 1.10, 1.50\). End territories each have one neighbor; the middle territory weights its two neighbors equally. With one-pass \(\alpha=0.20\), find the three smoothed factors. (Answer: \(0.78,\ 1.10,\ 1.42\).)
  2. Two adjacent territories have factors \(0.90\) and \(1.30\). Use a one-pass rule with \(\alpha=0.25\), with each territory’s only neighbor being the other. What are the smoothed factors? (Answer: \(1.00,\ 1.20\).)
  3. Preliminary factors for three territories are \(0.80, 1.20, 1.00\), with exposure weights \(50,100,50\). Apply one-pass smoothing at \(\alpha=0.20\) on a chain, then find the multiplier that restores the preliminary exposure-weighted mean. (Answer: Smoothed factors \(0.88,\ 1.04,\ 1.04\); target mean \(1.05\); multiplier \(1.05/1.01=105/101\), approximately \(1.0396\).)
  4. For two linked territories, \(r_1=0.70\), \(r_2=1.30\), \(w_1=w_2=1\), \(c=1\), and \(\lambda=1\). Solve the penalized model. (Answer: \(q_1=0.90,\ q_2=1.10\).)
  5. Two territories share a boundary, but a major geographic barrier separates their insured populations and the available information indicates different risk conditions. Is adjacency alone a sufficient reason to smooth their factors toward one another? (Answer: No. The analyst must establish a plausible relationship for the coverage and population; otherwise the link should not be imposed solely because the territories touch.)
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