We continue working our way through the examples, case studies, and exercises of what is affectionately known here as “the two bears book” (Swedish björn = bear) and more formally as Non-Life Insurance Pricing with Generalized Linear Models by Esbjörn Ohlsson and Börn Johansson (Amazon UK | US).

At this stage, our purpose is to reproduce the analysis from the book using the R statistical computing and analysis platform, and to answer the data analysis elements of the exercises and case studies. Any critique of the approach and of pricing and modeling in the Insurance industry in general will wait for a later article.

In the following, we will assume that the reader has a copy of the book and a working installation of R. We will be using the `data.table`

, `foreach`

, and `ggplot2`

packages which are not part of the standard distribution, so the reader should install them first (e.g. by executing the following line from within an R session:

**Note that Hadley broke the API to package ggplot2 so the code as it stands here will not work with the latest versions.** This is one reason we normally use the

`lattice`

package instead and will take a little while to fix.The beginning of the code is not so important, but here goes:

## Example 2.5: Moped insurance continued

We continue the moped insurance example, and we use the data that we saved in our Chapter 1 session. The goal is to reproduce Table 2.7 so we start building that as a data frame after loading the data.

We next calculate the duration and number of claims for each level of each rating factor. We also set the contrasts for the levels, using the same idiom as in our Chapter 1 session. The foreach package is convenient to use here, but you can of course do it with a normal loop and a couple of variables.

Next, we need to build the frequency and severity models separately, as in the discussion at the beginning of section 2.3.4 on page 34.

### Frequency model

Note here the use of the `offset()`

term as opposed to the `weights=`

argument we have used before. The offset is `log(dur)`

because our link function is `log`

.

See `help("Insurance", package = "MASS")`

for a similar example and sections 7.1 (p. 189–190) and 7.3 of Modern Applied Statistics with S (Amazon UK | US ) for a (very brief) discussion.

### Severity model

There are a couple of points to note here:

- We will model using a Gamma distribution for the errors, following the discussion on page 20 and also pages 33-34. Note the point that this is only one of several plausible candidate distributions.
- Because we are using the Gamma distribution we need to remove the zero values from the data; we do this using the
`table.1.2[table.1.2$medskad > 0, ]`

construct. - To reproduce the values from the book, we use the non-canonical “
`log`

” link function even though the canonical function (“`inverse`

”) gives a slightly better fit (residual deviance 5.9 versus 8.0 on 16 degrees of freedom). This follows the approach discussed in Example 2.3 on page 30.

### Combining the models

Now it is trivial to combine and display the results.

rating.factor | class | duration | n.claims | rels.frequency | rels.severity | rels.pure.premium | |
---|---|---|---|---|---|---|---|

1 | Vehicle class | 1 | 9833.20 | 391 | 1.00 | 1.00 | 1.00 |

2 | Vehicle class | 2 | 8825.10 | 395 | 0.78 | 0.55 | 0.42 |

11 | Vehicle age | 1 | 1918.40 | 141 | 1.55 | 1.79 | 2.78 |

21 | Vehicle age | 2 | 16739.90 | 645 | 1.00 | 1.00 | 1.00 |

12 | Zone | 1 | 1451.40 | 206 | 7.10 | 1.21 | 8.62 |

22 | Zone | 2 | 2486.30 | 209 | 4.17 | 1.07 | 4.48 |

3 | Zone | 3 | 2888.70 | 132 | 2.23 | 1.07 | 2.38 |

4 | Zone | 4 | 10069.10 | 207 | 1.00 | 1.00 | 1.00 |

5 | Zone | 5 | 246.10 | 6 | 1.20 | 1.21 | 1.46 |

6 | Zone | 6 | 1369.20 | 23 | 0.79 | 0.98 | 0.78 |

7 | Zone | 7 | 147.50 | 3 | 1.00 | 1.20 | 1.20 |

## Case Study: Motorcycle Insurance

The authors use the term Case Study for larger exercises. There is quite a lot to cover here, and some of the questions are more about discussing the results. We will focus on getting the results.

### The data

#### Obtaining the data

First we read the (fixed width) data. The column names are the same (based on Swedish language) as in the book.

#### Adding meta-data information

We are more than a little obsessed with documenting our data, so here goes. See the book for the details.

#### Rating factors

We finally add the rating factors from the book.

#### Save the data

Never forget to save.

Now we can start tackling the exercises.

### Problem 1: Aggregate to cells of current tariff

Aggregate the data to the cells of the current tariff. Compute the empirical claim frequency and severity at this level.

We really like the `data.table`

package. The syntax is much easier on the eye (and the hand). As a bonus, it scales well to much larger data sets than `data.frame`

. If you are not already using it, now is the time to start.

### Problem 2: Determine how the duration and number of claims is distributed

Determine how the duration and number of claims is distributed for each of the rating factor classes, as an indication of the accuracy of the statistical analysis.

This is one of those ‘there is no right answer’ questions, but let us have a look at the data. First load it if needed and also load the libraries we will be using.

#### 1. Number of claims (antskad)

Let us start with the number of claims for the undelying data. First we plot the number of policies for each number of claims (range of number of claims is {0, 1, 2}); this is one way to do it:

There is certainly some variation there, especially on the first rating factor.

Recall from the discussion on page 18 the assumption that the number of claims for a single policy is a Poisson process and the distribution of the number of claims in a tariff cell is Poisson distributed. Let us first look at the whole data set (and note that the `data.table`

notation is used; if using `data.frame`

be sure to have `drop=FALSE`

):

antskad | n | predicted | |
---|---|---|---|

1 | 0 | 63878 | 63878.0 |

2 | 1 | 643 | 646.0 |

3 | 2 | 27 | 7.0 |

Not too bad, but then it can’t really be with that heavy weighting to the first value of antskad. We can show the same split by the levels of the first rating factor as an example:

rating.1 | antskad | n | predicted | |
---|---|---|---|---|

1 | 1 | 0 | 8409 | 8409.0 |

2 | 1 | 1 | 163 | 164.0 |

3 | 1 | 2 | 10 | 3.0 |

4 | 2 | 0 | 11632 | 11632.0 |

5 | 2 | 1 | 157 | 157.0 |

6 | 2 | 2 | 5 | 2.0 |

7 | 3 | 0 | 12604 | 12604.0 |

8 | 3 | 1 | 113 | 113.0 |

9 | 3 | 2 | 5 | 1.0 |

10 | 4 | 0 | 24626 | 24626.0 |

11 | 4 | 1 | 184 | 185.0 |

12 | 4 | 2 | 6 | 1.0 |

13 | 5 | 0 | 2368 | 2368.0 |

14 | 5 | 1 | 9 | 9.0 |

15 | 6 | 0 | 3867 | 3867.0 |

16 | 6 | 1 | 16 | 16.0 |

17 | 6 | 2 | 1 | 0.0 |

18 | 7 | 0 | 372 | 372.0 |

19 | 7 | 1 | 1 | 1.0 |

You get the idea….

If we can’t really see very much at the individual claims level, how does it appear when we look at the levels aggregated to the current tariff cells? Here, of course, we have much more data with up to 33 claims in one cell, but we will limit the display to the first few number of claims.

There is certainly something going on for low bonus classes and high vehicle ages that is not straightforward Poisson distribution. Looking at the two in combination we see the old motorcycle - low bonus class (rating.3 = 3 and rating.4 = 1) is the main culpit, but also note that there is relatively little data in these cells.

We can fit the poisson distribution using the same predictionPoisson() function as before. We also look at the fit in a simple way (the - much! - better approach is to use the residuals from the fitted model)

We will not show the tables here (as we said: there is a better way). Instead, we can show the fitted Poisson distribution:

#### 2. Duration

We can plot the distribution of duration as before. Here 440 is the 90% quantile for the duration (412) rounded up to the next multiplier of binwidth.

However, in this case it is much more illuminating to look at the overall distribution:

Big spikes near 1/2 and 1 year (and in general spikes around 1/2 year multiples) suggests that either (1) this is not a random sample of policies or (2) there is a strong seasonal effect in the time of year people initially sign up for the policy.

### Problem 3: Determine the relativities for claim frequency and severity

Determine the relativities for claim frequency and severity separately, by using GLMs; use the results to get relativities for the pure premium.

We first load the data (if needed) and create a data frame to hold the output, following the structure of Table 2.8 in the book.

rating.factor | class | relativity | |
---|---|---|---|

1 | zone | 1 | 7.678 |

2 | zone | 2 | 4.227 |

3 | zone | 3 | 1.336 |

4 | zone | 4 | 1.000 |

5 | zone | 5 | 1.734 |

6 | zone | 6 | 1.402 |

7 | zone | 7 | 1.402 |

8 | mc class | 1 | 0.625 |

9 | mc class | 2 | 0.769 |

10 | mc class | 3 | 1.000 |

11 | mc class | 4 | 1.406 |

12 | mc class | 5 | 1.875 |

13 | mc class | 6 | 4.062 |

14 | mc class | 7 | 6.873 |

15 | vehicle age | 1 | 2.000 |

16 | vehicle age | 2 | 1.200 |

17 | vehicle age | 3 | 1.000 |

18 | bonus class | 1 | 1.250 |

19 | bonus class | 2 | 1.125 |

20 | bonus class | 3 | 1.000 |

Close enough to the book.

First we set the contrasts so the baseline for the models is the level with the highest duration. This is the same approach we used before.

We next fit the models and combine the results as in Example 2.5 above. Here we also make a note of a simple measure of the goodness of fit: the residual deviance and the degrees of freedom. The frequency model is a good fit, the severity model is not.

Finally we convert, save, and compare with the current values.

rating.factor | class | duration | n.claims | skadkostK | relativity | rels.pure.prem | |
---|---|---|---|---|---|---|---|

1 | Zone | 1 | 6205.310 | 183 | 5540.000 | 7.678 | 6.239 |

2 | Zone | 2 | 10103.090 | 167 | 4811.000 | 4.227 | 3.179 |

3 | Zone | 3 | 11676.573 | 123 | 2523.000 | 1.336 | 0.993 |

4 | Zone | 4 | 32628.493 | 196 | 3775.000 | 1.000 | 1.000 |

5 | Zone | 5 | 1582.112 | 9 | 105.000 | 1.734 | 0.155 |

6 | Zone | 6 | 2799.945 | 18 | 288.000 | 1.402 | 0.223 |

7 | Zone | 7 | 241.288 | 1 | 1.000 | 1.402 | 0.002 |

8 | MC class | 1 | 5190.351 | 46 | 993.000 | 0.625 | 0.395 |

9 | MC class | 2 | 3990.115 | 57 | 883.000 | 0.769 | 0.536 |

10 | MC class | 3 | 21665.679 | 166 | 5372.000 | 1.000 | 1.000 |

11 | MC class | 4 | 11739.882 | 98 | 2192.000 | 1.406 | 0.574 |

12 | MC class | 5 | 13439.926 | 149 | 3297.000 | 1.875 | 1.413 |

13 | MC class | 6 | 8880.134 | 175 | 4161.000 | 4.062 | 4.876 |

14 | MC class | 7 | 330.723 | 6 | 145.000 | 6.873 | 0.602 |

15 | Vehicle age | 1 | 4955.403 | 126 | 4964.000 | 2.000 | 4.715 |

16 | Vehicle age | 2 | 9753.811 | 145 | 5507.000 | 1.200 | 2.021 |

17 | Vehicle age | 3 | 50527.597 | 426 | 6570.000 | 1.000 | 1.000 |

18 | Bonus class | 1 | 19893.370 | 207 | 4558.000 | 1.250 | 0.797 |

19 | Bonus class | 2 | 9615.764 | 121 | 3627.000 | 1.125 | 0.604 |

20 | Bonus class | 3 | 35727.677 | 369 | 8857.000 | 1.000 | 1.000 |

It is really rather different from the current relativity, as we will also see below. But remember that the severity model was a poor fit: that would be my first place to start looking if staying within the insurance pricing framework outlined in this book.

### Problem 4: Discussions

Just note here that the ratio for the relativity we calculated and for the existing one can be found as

Note the huge range of relativities in our new model (3552 versus 12).