R code for Chapter 1 of Non-Life Insurance Pricing with GLM

On , wrote in CYBAEA Data:

Insurance pricing is backwards and primitive, harking back to an era before computers. One standard (and good) textbook on the topic is Non-Life Insurance Pricing with Generalized Linear Models by Esbjorn Ohlsson and Born Johansson (Amazon UK | US). We have been doing some work in this area recently. Needing a robust internal training course and documented methodology, we have been working our way through the book again and converting the examples and exercises to R, the statistical computing and analysis platform. This is part of a series of posts containing elements of the R code.

#!/usr/bin/Rscript
## PricingGLM-1.r - Code for Chapter 1 of "Non-Life Insurance Pricing with GLM"
## Copyright © 2012 CYBAEA Limited (http://www.cybaea.net/)

## @book{ohlsson2010non,
##   title={Non-Life Insurance Pricing with Generalized Linear Models},
##   author={Ohlsson, E. and Johansson, B.},
##   isbn={9783642107900},
##   series={Eaa Series: Textbook},
##   url={http://books.google.com/books?id=l4rjeflJ\_bIC},
##   year={2010},
##   publisher={Springer Verlag}
## }

With the preliminaries out of the way, let us get started.

Example 1.2

We grab the data for Table 1.2 from the book's web site and store it as an R object with lots of good meta information.

################
### Example 1.2
con <- url("http://www2.math.su.se/~esbj/GLMbook/moppe.sas")
data <- readLines(con, n = 200L, warn = FALSE, encoding = "unknown")
close(con)
## Find the data range
data.start <- grep("^cards;", data) + 1L
data.end   <- grep("^;", data[data.start:999L]) + data.start - 2L
table.1.2  <- read.table(text = data[data.start:data.end],
                         header = FALSE, sep = "", quote = "",
                         col.names = c("premiekl", "moptva", "zon", "dur",
                             "medskad", "antskad", "riskpre", "helpre", "cell"),
                         na.strings = NULL,
                         colClasses = c(rep("factor", 3), "numeric",
                             rep("integer", 4), "NULL"),
                         comment.char = "")
rm(con, data, data.start, data.end)     # Cleanup
comment(table.1.2) <-
    c("Title: Partial casco moped insurance from Wasa insurance, 1994--1999",
      "Source: http://www2.math.su.se/~esbj/GLMbook/moppe.sas",
      "Copyright: http://www2.math.su.se/~esbj/GLMbook/")
## See the SAS code for this derived field
table.1.2$skadfre = with(table.1.2, antskad / dur)
## English language column names as comments:
comment(table.1.2$premiekl) <-
    c("Name: Class",
      "Code: 1=Weight over 60kg and more than 2 gears",
      "Code: 2=Other")
comment(table.1.2$moptva)   <-
    c("Name: Age",
      "Code: 1=At most 1 year",
      "Code: 2=2 years or more")
comment(table.1.2$zon)      <-
    c("Name: Zone",
      "Code: 1=Central and semi-central parts of Sweden's three largest cities",
      "Code: 2=suburbs and middle-sized towns",
      "Code: 3=Lesser towns, except those in 5 or 7",
      "Code: 4=Small towns and countryside, except 5--7",
      "Code: 5=Northern towns",
      "Code: 6=Northern countryside",
      "Code: 7=Gotland (Sweden's largest island)")
comment(table.1.2$dur)      <-
    c("Name: Duration",
      "Unit: year")
comment(table.1.2$medskad)  <-
    c("Name: Claim severity",
      "Unit: SEK")
comment(table.1.2$antskad)  <- "Name: No. claims"
comment(table.1.2$riskpre)  <-
    c("Name: Pure premium",
      "Unit: SEK")
comment(table.1.2$helpre)   <-
    c("Name: Actual premium",
      "Note: The premium for one year according to the tariff in force 1999",
      "Unit: SEK")
comment(table.1.2$skadfre)  <-
    c("Name: Claim frequency",
      "Unit: /year")
## Save results for later
save(table.1.2, file = "table.1.2.RData")
## Print the table (not as pretty as the book)
print(table.1.2)
################
premiekl moptva zon dur medskad antskad riskpre helpre skadfre
1 1 1 1 62.9000 18256 17 4936 2049 0.2703
2 1 1 2 112.9000 13632 7 845 1230 0.0620
3 1 1 3 133.1000 20877 9 1411 762 0.0676
4 1 1 4 376.6000 13045 7 242 396 0.0186
5 1 1 5 9.4000 0 0 0 990 0.0000
6 1 1 6 70.8000 15000 1 212 594 0.0141
7 1 1 7 4.4000 8018 1 1829 396 0.2273
8 1 2 1 352.1000 8232 52 1216 1229 0.1477
9 1 2 2 840.1000 7418 69 609 738 0.0821
10 1 2 3 1378.3000 7318 75 398 457 0.0544
11 1 2 4 5505.3000 6922 136 171 238 0.0247
12 1 2 5 114.1000 11131 2 195 594 0.0175
13 1 2 6 810.9000 5970 14 103 356 0.0173
14 1 2 7 62.3000 6500 1 104 238 0.0161
15 2 1 1 191.6000 7754 43 1740 1024 0.2244
16 2 1 2 237.3000 6933 34 993 615 0.1433
17 2 1 3 162.4000 4402 11 298 381 0.0677
18 2 1 4 446.5000 8214 8 147 198 0.0179
19 2 1 5 13.2000 0 0 0 495 0.0000
20 2 1 6 82.8000 5830 3 211 297 0.0362
21 2 1 7 14.5000 0 0 0 198 0.0000
22 2 2 1 844.8000 4728 94 526 614 0.1113
23 2 2 2 1296.0000 4252 99 325 369 0.0764
24 2 2 3 1214.9000 4212 37 128 229 0.0305
25 2 2 4 3740.7000 3846 56 58 119 0.0150
26 2 2 5 109.4000 3925 4 144 297 0.0366
27 2 2 6 404.7000 5280 5 65 178 0.0124
28 2 2 7 66.3000 7795 1 118 119 0.0151

That was easy. Now for something a little harder.

Example 1.3

Here we are concerned with replicating Table 1.4. We do it slowly, step-by-step, for pedagogical reasons.

################
### Example 1.3
if (!exists("table.1.2"))
    load("table.1.2.RData")
## We calculate each of the columns individually and slowly here
## to show each step

## First we have simply the labels of the table
rating.factor <-
    with(table.1.2,
         c(rep("Vehicle class", nlevels(premiekl)),
           rep("Vehicle age", nlevels(moptva)),
           rep("Zone", nlevels(zon))))

## The Class column
class.num <- with(table.1.2, c(levels(premiekl), levels(moptva), levels(zon)))

## The Duration is the sum of durations within each class
duration.total <-
    c(with(table.1.2, tapply(dur, premiekl, sum)),
      with(table.1.2, tapply(dur, moptva, sum)),
      with(table.1.2, tapply(dur, zon, sum)))

## Calculate relativities in the tariff
## The denominator of the fraction is the class with the highest exposure
## (i.e. the maximum total duration): we make that explicit with the
## which.max() construct.  We also set the contrasts to use this as the base,
## which will be useful for the glm() model later.
class.base <- which.max(duration.total[1:2])
age.base   <- which.max(duration.total[3:4])
zone.base  <- which.max(duration.total[5:11])

rt.class <- with(table.1.2, tapply(helpre, premiekl, sum))
rt.class <- rt.class / rt.class[class.base]
rt.age   <- with(table.1.2, tapply(helpre, moptva, sum))
rt.age   <- rt.age / rt.age[age.base]
rt.zone  <- with(table.1.2, tapply(helpre, zon, sum))
rt.zone  <- rt.zone / rt.zone[zone.base]

contrasts(table.1.2$premiekl) <-
    contr.treatment(nlevels(table.1.2$premiekl))[rank(-duration.total[1:2],
                                                      ties.method = "first"), ]
contrasts(table.1.2$moptva) <-
    contr.treatment(nlevels(table.1.2$moptva))[rank(-duration.total[3:4],
                                                    ties.method = "first"), ]
contrasts(table.1.2$zon) <-
    contr.treatment(nlevels(table.1.2$zon))[rank(-duration.total[5:11],
                                                 ties.method = "first"), ]

The contrasts could also have been set with the base= argument, e.g. contrasts(table.1.2$zon) <- contr.treatment(nlevels(table.1.2$zon), base = zone.base), which would be closer in spirit to the SAS code. But I like the idiom presented here where we follow the duration order; it also extends well to other (i.e. not treatment) contrasts. I just wish rank() had an decreasing= argument like order() which I think would be clearer than using rank(-x) to get a decreasing sort order.

That was the easy part. At this stage in the book you are not really expected to understand the next step so do not despair! We just show how easy it is to replicate the SAS code in R. An alternative approach using direct optimization is outlined in Exercise 1.3 below.

## Relativities of MMT; we use the glm approach here as per the book's
## SAS code at http://www2.math.su.se/~esbj/GLMbook/moppe.sas
m <- glm(riskpre ~ premiekl + moptva + zon, data = table.1.2,
         family = poisson("log"), weights = dur)

## If the next line is a mystery then you need to
## (1) read up on contrasts or
## (2) remember that the link function is log() which is why we use exp here
rels <- exp( coef(m)[1] + coef(m)[-1] ) / exp(coef(m)[1])

rm.class <- c(1, rels[1])               # See rm.zone below for the
rm.age   <- c(rels[2], 1)               # general approach
rm.zone  <- c(1, rels[3:8])[rank(-duration.total[5:11], ties.method = "first")]

## Create and save the data frame
table.1.4 <-
    data.frame(Rating.factor = rating.factor, Class = class.num,
               Duration = duration.total,
               Rel.tariff = c(rt.class, rt.age, rt.zone),
               Rel.MMT    = c(rm.class, rm.age, rm.zone))
save(table.1.4, file = "table.1.4.RData")
print(table.1.4, digits = 3)
rm(rating.factor, class.num, duration.total, class.base, age.base, zone.base,
   rt.class, rt.age, rt.zone, rm.class, rm.age, rm.zone, m, rels)
################

The result is something like this:

Rating.factor Class Duration Rel.tariff Rel.MMT
1 Vehicle class 1 9833.20 1.00 1.00
2 Vehicle class 2 8825.10 0.50 0.43
3 Vehicle age 1 1918.40 1.67 2.73
4 Vehicle age 2 16739.90 1.00 1.00
5 Zone 1 1451.40 5.17 8.97
6 Zone 2 2486.30 3.10 4.19
7 Zone 3 2888.70 1.92 2.52
8 Zone 4 10069.10 1.00 1.00
9 Zone 5 246.10 2.50 1.24
10 Zone 6 1369.20 1.50 0.74
11 Zone 7 147.50 1.00 1.23

Note the rather unusual and apparently inconsistent rounding in the book: 147, 1.66, and 5.16 would be better as 148 (the value is 147.5), 1.67, and 5.17.

Exercise 1.3

Here it gets interesting as we get a different value from the authors. Possibly a small bug on our part but at least we provide the code for you to check. So if you spot a problem let us know in the comments.

################
## Exercise 1.3

## The values from the book
g0  <- 0.03305
g12 <- 2.01231
g22 <- 0.74288
dim.names <- list(Milage = c("Low", "High"),
                  Age = c("New", "Old"))
pyears <- matrix(c(47039, 56455, 190513, 28612), nrow = 2,
                 dimnames = dim.names)
claims <- matrix(c(0.033, 0.067, 0.025, 0.049), nrow = 2,
                 dimnames = dim.names)

## Function to calculate the error of the estimate
GvalsError <- function (gvals) {
    ## The current estimates
    g0  <- gvals[1]
    g12 <- gvals[2]
    g22 <- gvals[3]
    ## The current estimates in convenient matrix form
    G  <- matrix(c(1, 1, g12, g22), nrow = 2)
    G1 <- matrix(c(1, g12), nrow = 2, ncol = 2)
    G2 <- matrix(c(1, g22), nrow = 2, ncol = 2, byrow = TRUE)
    ## The calculated values
    G0  <- addmargins(claims * pyears)["Sum", "Sum"] / ( sum(pyears * G1 * G2) )
    G12 <- addmargins(claims * pyears)["High", "Sum"] /
        ( g0 * addmargins(pyears * G2)["High", "Sum"] )
    G22 <- addmargins(claims * pyears)["Sum", "Old"] /
        ( g0 * addmargins(pyears * G1)["Sum", "Old"] )
    ## The sum of squared errors
    error <- (g0 - G0)^2 + (g12 - G12)^2 + (g22 - G22)^2
    return(error)
}

## Minimize the error function to obtain our estimate
gamma <- optim(c(g0, g12, g22), GvalsError)
stopifnot(gamma$convergence == 0)
gamma <- gamma$par

values <- data.frame(legend = c("Our calculation", "Book value"),
                     g0  = c(gamma[1], g0),
                     g12 = c(gamma[2], g12),
                     g22 = c(gamma[3], g22),
                     row.names = "legend")
print(values, digits = 4)

## Close, but not the same.

rm(g0, g12, g22, dim.names, pyears, claims, gamma, values)
################

The resulting table is something like:

g0 g12 g22
Our calculation 0.0334 1.9951 0.7452
Book value 0.0331 2.0123 0.7429

Close, but not the same. Perhaps they used a different error function.

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