A.5. NCCI’s 2007 Hazard Group Mapping

Question 1

Why the NCCI moved from 17 limits to 5 for ELFs

Answer 1

1. ELFs at any pair of excess limits are highly correlated across classes.
2. Limits below $100,000 were heavily represented in the prior
   17 limits.
3. They wanted to cover the range of limits commonly used for retrospective rating.

Question 2

Other credibility options considered

Answer 2

1. Using the median instead of the average for k.
2. Excluding Medical Only claims from the analysis.
3. Including only Serious claims in the analysis.
4. Requiring a minimum # of claims for classes used in the calculation of k.
5. Various square root rules, such as z =sqrt( n/384 ), which
   corresponds to a 95% chance of n being within 10% of its
   expected value.

Question 3

Measuring distance between vectors

Answer 3

The NCCI used the usual L2 or Euclidean distance, measured
as:

||x − y||2 =sqrt(∑i=1 (xi − yi)^2)

The NCCI also considered using the L1 distance of ||x − y||1 =∑i=1 |xi − yi|. This would have had the advantage in that it
minimizes the relative error in estimating excess premium. The relative error in estimating excess premium for class c
with limit L is PLR × |RHG(L) − Rc(L)|. Since the analysis
was not sensitive to the distance measure, they chose to use the traditional squared distance measure.

Question 4

Why the NCCI didn’t use standardization

Answer 4

1. The resultant hazard groups using standardization didn’t differ much from not using it.
2. Excess ratios at different limits have a similar unit of
   measure, which is dollars of excess loss per dollar of total loss. Standardization would have eliminated this common denominator.
3. All excess ratios are between 0 and 1, while standardization could have led to results outside this range.
4. There is a greater range of excess ratios at lower limits, and this is a good thing since it is based on actual data (compared to excess ratios at higher limits being based more on fitted loss distributions). Standardization would have given this real data less weight.

Question 5

Weighted k-means algorithm

Answer 5

1. Decide on the number of clusters (potential hazard groups) k to target.
2. Start with an abritrary initial assignment of classes into k clusters (denoted as HG below).
3. Compute the centroid of each cluster i for the k clusters as:

Ri =∑c∈HGi (wcRc)/∑c∈HGi (wc)

4. For each class, find the closest centroid using the L2
distance, and assign the class to that cluster. If any classes have been re-assigned in this step, go back to step 3. Continue this process until no classes are re-assigned.1. Decide on the number of clusters (potential hazard groups) k to target.

Question 6

Hierarchical clustering

Answer 6

Hierarchical clustering means any additional cluster would be a subset of a single existing cluster. Non-hierarchical clustering methods seek the best partition for any given number of clusters

Question 7

Desirable optimality properties from k-means

Answer 7

K-means maximizes the equivalent of R2 from linear regression, which means maximizing the percentage of total variation explained by the hazard groups. This is equivalent to stating that k-means minimizes the within variance and maximizes the between variance, which means the hazard groups will be homogeneous and well separated.

Question 8

Test statistics used to decide # of hazard groups

Answer 8

1. The Calinski and Harabasz statistic of Trace(B)/(k−1)/Trace(W)/(n−k), where n is the # of classes, k is the # of hazard groups. Higher values of this statistic indicate a better # of clusters. The test is also known as the Pseudo-F test since it resembles the F-test of regression analysis.
2. The Cubic Clustering Criterion (CCC) statistic, which
   compares the amount of variance explained by a given
   set of clusters to that expected when clusters are formed at random based on the multi-dimensional uniform distribution. Again, a high value of this statistic indicates better performance.

Question 9

Why the NCCI didn’t rely on the CCC test statistic

showing 9 groups

Answer 9

1. Milligan and Cooper found the Calinski and Harabasz
   statistic outperformed the CCC statistic.
2. The CCC statistic deserves less weight when correlation is present, which was the case.
3. The selection of # of hazard groups ought to be driven by the large classes where most of the experience was concentrated. Using these highly or fully credible classes showed 7 as the optimal number.
4. There was crossover in the excess ratios between hazard groups when using 9 groups, which is something that isn’t appealing in practice.

Question 10

Underwriter considerations

Answer 10

1. Similarity between class codes that were in different groups.
2. Degree of exposure to automobile accidents in a given class.
3. Extent heavy machinery is used in a given class.

Question 11

3 key ideas of hazard group remapping

Answer 11

1. Computing excess ratios by class.
2. Sorting classes based on excess ratios.
3. Cluster analysis.