Slides ibnr-belo-horizonte-master

Arthur CHARPENTIER - IBNR: quantification of uncertainty

IBNR : quantification of uncertainty

Arthur Charpentier (Universite Rennes 1)

http ://blogperso.univ-rennes1.fr/arthur.charpentier/

Universidade Federal de Minas Gerais, March 2009

1


A short introduction

2



3



Claims reserving is an important and difficult issue. It might take years until aclaim is finally settled,• reporting delay, between accident date and reporting date (notification at

insurance company)• settlement delay, between reporting date and final settlement (recovery

process, court decisions)• possible reopenings, due to unexpected developments”It is hoped that more casualty actuaries will involve themselves in this importantarea. IBNR reserves deserve more than just a clerical or cursory treatment andwe believe, as did Mr. Tarbell Chat ”the problem of incurred but not reportedclaim reserves is essentially actuarial or statistical”. Perhaps in today’senvironment the quotation would be even more relevant if it stated that theproblem ”...is more actuarial than statistical”.” Bornhuetter & Ferguson

(1972)

=⇒ claims reserving is a prediction problem

4


On reserving risk (and uncertainty)

From July to November 2004, stock price of Converium Holding -90% afterreserve increase annoncements

510

1520

25

2002 2003 2004 2005 2006 2007 2008

5


Reserve cycle

Note that there is a reserving cycle, i.e. one should focus on boni-mali modeling.

6


A short overview on prediction uncertainty

Basically, we have to predict some (random) future cash flow, denoted X.

Let Ft denote the information available at some time t.

Let X denote the prediction made at time t.

The (conditional) mean square error of prediction (MSE) is simply

mset(X) = E(

[X − X]2|Ft

)= V ar(X|Ft)︸︷︷︸

process variance

+(E (X|Ft)− X

)2

︸︷︷︸parameter estimation error

i.e X is

a predictor for X

an estimator for E(X|Ft).

7


On claims reserving techniques

8



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10



11


Notations on triangles

Information on claims is usually summarizes in payment triangles, eitherincremental triangles, or cumulated payments.

Development year j Occurence year i Calendar year i+ j

? ? ? ?

?

?

?

? ? ? ?

?

?

?

?

12





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? ? ? ?

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?

? ? ? ?

?

?

?

?

15



• Xi,j denotes incremental payments, payments of year j, for claims occurredyear i

• Ci,j denotes cumulated payments Ci,j = Xi,0 +Xi,1 + · · ·+Xi,j , i.e. paymentsseen as at year i+ j.

0 1 2 3 4 5

0 3209 1163 39 17 7 21

1 3367 1292 37 24 10

2 3871 1474 53 22

3 4239 1678 103

4 4929 1865

5 5217

et et

0 1 2 3 4 5

0 3209 4372 4411 4428 4435 4456

1 3367 4659 4696 4720 4730

2 3871 5345 5398 5420

3 4239 5917 6020

4 4929 6794

5 5217

from Partrat et al. (2005)

16



• Pi,j denotes earned premium for year i• Ni,j denotes cumulated number of claims, accdient year i seen as at year i+ j

(in thousands).

0 1 2 3 4 5

0 4563 4589 4590 4591 4591 4591

1 4718 4674 4671 4672 4672

2 4836 4861 4861 4863

3 5140 5168 5173

4 5633 5668

5 6389

etet

0 1 2 3 4 5

0 1043.4 1045.5 1047.5 1047.7 1047.7 1047.7

1 1043.0 1027.1 1028.7 1028.9 1028.7

2 965.1 967.9 967.8 970.1

3 977.0 984.7 986.8

4 1099.0 1118.5

5 1076.3

from Partrat et al. (2005). Using a simple Chain Ladder algorithm, thefollowing earned premium can be considered

Year i 0 1 2 3 4 5

Pi 4591 4672 4863 5175 5673 6431

17



And finally,

• Ci,j denotes cumulated payments Ci,j = Xi,0 +Xi,1 + · · ·+Xi,j , i.e. paymentsseen as at year i+ j.

• Ei,j denotes cumulated estimated final charge, seen as as at.

0 1 2 3 4 5

0 3209 4372 4411 4428 4435 4456

1 3367 4659 4696 4720 4730

2 3871 5345 5398 5420

3 4239 5917 6020

4 4929 6794

5 5217

et et

0 1 2 3 4 5

0 4795 4629 4497 4470 4456 4456

1 5135 4949 4783 4760 4750

2 5681 5631 5492 5470

3 6272 6198 6131

4 7326 7087

5 7353

those triangles are sometimes denoted paid (P ) and incurred (I) losses triangles.

18


The Chain Ladder estimate

We assume here that

Ci,j+1 = λjCi,j for all i, j = 1, · · · , n.

A natural estimator for λj based on past history is

λj =∑n−j

i=1 Ci,j+1∑n−ji=1 Ci,j

for all j = 1, · · · , n− 1.

Hence, it becomes possible to estimate future payments using

Ci,j =[λn+1−i...λj−1

]Ci,n+1−i.

19


Practical calculation of the Chain Ladder estimate

1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735

2001 56,762

211,961

λ0 =440, 438211, 961

= 2.07792.

20



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735

2001 56,762 ?

211,961

λ0 =440, 438211, 961

= 2.07792

21



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735

2001 56,762 ?

211,961 440,438

λ0 =440, 438211, 961

= 2.07792

22



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735

2001 56,762 117,945

211,961 440,438

λ0 =440, 438211, 961

= 2.07792 and C2001,2 = 56, 762× 2.07792 = 117, 945

23



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735 ?

2001 56,762 117,945 ?

211,961

λ0 =440, 438211, 961

= 2.07792 and C2001,2 = 56, 762× 2.07792 = 117, 945

24



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735 ?

2001 56,762 117,945 ?

211,961

λ1 =469, 635337, 781

= 1.3091 and C2001,2 = 56, 762× 1.3091 = 117, 945

25



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735 ?

2001 56,762 117,945 ?

211,961 337,781 469,635

λ1 =469, 635337, 781

= 1.3091and C2001,2 = 56, 762× 1.3091 = 117, 945

26



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854

2000 50,420 102,735 142,870

2001 56,762 117,945 164,025

211,961 337,781 469,635

λ1 =469, 635337, 781

= 1.3091 and C2001,3 = 117, 945× 1.3091 = 164, 025

27



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562

1999 49,756 101,587 136,854 158,471

2000 50,420 102,735 142,870 165,438

2001 56,762 117,945 164,025 189,935

211,961 332,781 385,347

λ2 =385, 347332, 781

= 1.1579 and C2001,4 = 164, 025× 1.1579 = 189, 935

28



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604

1998 30,470 65,482 90,973 103,562 111,364

1999 49,756 101,587 136,854 158,471 170,411

2000 50,420 102,735 142,870 165,438 177,903

2001 56,762 117,945 164,025 189,935 204,245

211,961 281,785 303,016

λ3 =303, 016281, 785

= 1.0753 and C2001,5 = 189, 935× 1.0753 = 204, 245

29



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210

1997 26,312 57,779 82,451 95,506 101,604 106,422

1998 30,470 65,482 90,973 103,562 111,364 116,577

1999 49,756 101,587 136,854 158,471 170,411 178,387

2000 50,420 102,735 142,870 165,438 177,903 186,203

2001 56,762 117,945 164,025 189,935 204,245 213,805

211,961 201,352 210,776

λ4 =210, 776201, 352

= 1.0468 and C2001,6 = 204, 245× 1.0468 = 213, 805

30



1 2 3 4 5 6 7

1995 23,758 49,114 68,582 79,840 86,298 90,566 92,878

1996 31,245 63,741 90,775 106,439 115,054 120,210 123,278

1997 26,312 57,779 82,451 95,506 101,604 106,422 109,139

1998 30,470 65,482 90,973 103,562 111,364 116,577 119,553

1999 49,756 101,587 136,854 158,471 170,411 178,387 182,941

2000 50,420 102,735 142,870 165,438 177,903 186,203 190,984

2001 56,762 117,945 164,025 189,935 204,245 213,805 219,263

211,961 90,566 92,878

λ5 =92, 87890, 566

= 1.0255 and C2001,7 = 213, 805× 1.0255 = 219, 263

31



1 2 3 4 5 6 7

1995

1996 120,210 123,278

1997 101,604 109,139

1998 103,562 119,553

1999 136,854 182,941

2000 102,735 190,984

2001 56,762 219,263

211,961

32



0 1 2 3 4 5

0 3209 4372 4411 4428 4435 4456

1 3367 4659 4696 4720 4730

2 3871 5345 5398 5420

3 4239 5917 6020

4 4929 6794

5 5217

et et

0 1 2 3 4 5

0 3209 4372 4411 4428 4435 4456

1 3367 4659 4696 4720 4730 4752.4

2 3871 5345 5398 5420 5430.1 5455.8

3 4239 5917 6020 6046.15 6057.4 6086.1

4 4929 6794 6871.7 6901.5 6914.3 6947.1

5 5217 7204.3 7286.7 7318.3 7331.9 7366.7

One the triangle has been completed, we obtain the amount of reserves, withrespectively 22, 36, 66, 153 and 2150 per accident year, i.e. the total is 2427.

33


The Chain-Ladder estimate

The Chain-Ladder estimate is probably the most popular technique to estimateclaim reserves. Let Ft denote the information avalable at time t, or more formallythe filtration generated by {Ci,j , i+ j ≤ t} - or equivalently {Xi,j , i+ j ≤ t}

Assume that incremental payments are independent by occurence years, i.e. Ci1,·

Ci2,· are independent for any i1 and i2.

Further, assume that (Ci,j)j≥0 is Markov, and more precisely, there exist λj ’sand σ2

j ’s such that E(Ci,j+1|Fi+j) = E(Ci,j+1|Ci,j) = λj · Ci,j

Var(Ci,j+1|Fi+j) = Var(Ci,j+1|Ci,j) = σ2j · Ci,j

Under those assumption, one gets

E(Ci,j+k|Fi+j) = E(Ci,j+k|Ci,j) = λj · λj+1 · · ·λj+k−1Ci,j

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Underlying assumptions in the Chain-Ladder estimate

Recall, see Mack (1993), properties of the Chain-Ladder estimate rely on thefollowing assumptions

H1 E (Ci,j+1|Ci,1, ..., Ci,j) = λj .Cij for all i = 0, 1, .., n and j = 0, 1, ..., n− 1

H2 (Ci,j)j=1,...,n and (Ci′,j)j=1,...,n are independent for all i 6= i′.

H3 V ar (Ci,j+1|Ci,1, ..., Ci,j) = Ci,jσ2j for all i = 0, 1, ..., n and j = 0, 1, ..., n− 1

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Properties of the Chain-Ladder estimate

Further

λj =∑n−j−1

i=0 Ci,j+1∑n−j−1i=0 Ci,j

is an unbiased estimator for λj , given Gj , and λj and λj + h are non-correlated,given Fj . Hence, an unbiased estimator for E(Ci,j |Fi) is

Ci,j = λn−i · λn−i+1 · · · λj−2

(λj−1 − 1

)· Ci,n−i.

Recall that λj is the estimator with minimal variance among all linear estimatorsobtained from λi,j = Ci,j+1/Ci,j ’s. Finally, recall that

σ2j =

1n− j − 1

n−j−1∑i=0

(Ci,j+1

Ci,j− λj

)2

·Xi,j

is an unbiased estimator of σ2j , given Gj (see Mack (1993) or Denuit &

Charpentier (2005)).

36


Import/export of datasets with R

In the case of Excel files, library(RODBC)

library(RODBC)

file= odbcConnectExcel("D:\\triangle.xls")

BASE <- sqlQuery(file, "select * from [Feuil1$D9:I15]")

odbcCloseAll()

TRIANGLE.D=as.matrix(BASE)

base = read.table("D:\\triangle.csv",header=FALSE,sep=";")

A more convenient way is to use source(base.R).

37


Example of payment triangle

file=paste("D:/reserves/triangles/xls/","triangle.csv",sep="")

base = read.table(file,header=FALSE,sep=";")

Consider the following cumulated trianglebase = read.table("D:\\vect-triangle.csv",header=TRUE,sep=";")

year = base$year

development = base$development

paycum = base$paycum

TRIANGLE.C = tapply(paycum,list(year,development),sum)

> TRIANGLE.C

[,1] [,2] [,3] [,4] [,5] [,6]

[1,] 3209 4372 4411 4428 4435 4456

[2,] 3367 4659 4696 4720 4730 NA

[3,] 3871 5345 5398 5420 NA NA

[4,] 4239 5917 6020 NA NA NA

[5,] 4929 6794 NA NA NA NA

[6,] 5217 NA NA NA NA NA

38



From cumulated triangles, it is possible to obtain increments

TRIANGLE.X = TRIANGLE.D; Ntr=nrow(TRIANGLE.C)

for(i in 2:Ntr){

TRIANGLE.X[1:(Ntr+1-i),i]= TRIANGLE.C[1:(Ntr+1-i),i]- TRIANGLE.C[1:(Ntr+1-i),i-1]}

i.e.

> TRIANGLE.C > TRIANGLE.X

[,1] [,2] [,3] [,4] [,5] [,6] [,1] [,2] [,3] [,4] [,5] [,6]

[1,] 3209 4372 4411 4428 4435 4456 [1,] 3209 1163 39 17 7 21

[2,] 3367 4659 4696 4720 4730 NA [2,] 3367 1292 37 24 10 NA

[3,] 3871 5345 5398 5420 NA NA [3,] 3871 1474 53 22 NA NA

[4,] 4239 5917 6020 NA NA NA [4,] 4239 1678 103 NA NA NA

[5,] 4929 6794 NA NA NA NA [5,] 4929 1865 NA NA NA NA

[6,] 5217 NA NA NA NA NA [6,] 5217 NA NA NA NA NA

39



In regression models, it will be useful to have data in a dataset, i.e. we have totransform matrices in vectors.vec.D=as.vector(TRIANGLE.C)[is.na(as.vector(TRIANGLE.C))==FALSE]

vec.C=as.vector(TRIANGLE.X)[is.na(as.vector(TRIANGLE.X))==FALSE]

year=NA; dev=NA

for(i in 1:Ntr){

year=c(year,1:(Ntr-i+1)); dev=c(dev,rep(i,Ntr-i+1))}

year=year[is.na(year)==FALSE]; dev=dev[is.na(dev)==FALSE]

triangle= data.frame(year,dev,vec.D,vec.C)

> triangle

year dev vec.C vec.X year dev vec.C vec.X year dev vec.C vec.X

1 1 1 3209 3209 8 2 2 4659 1292 15 4 3 6020 103

2 2 1 3367 3367 9 3 2 5345 1474 16 1 4 4428 17

3 3 1 3871 3871 10 4 2 5917 1678 17 2 4 4720 24

4 4 1 4239 4239 11 5 2 6794 1865 18 3 4 5420 22

5 5 1 4929 4929 12 1 3 4411 39 19 1 5 4435 7

6 6 1 5217 5217 13 2 3 4696 37 20 2 5 4730 10

7 1 2 4372 1163 14 3 3 5398 53 21 1 6 4456 21

40


An alternative to obtain incremental triangles from cumulated ones is simply touse inc <- cbind(cum[,1], t(apply(cum,1,diff))), and dualy, to obtain cumulatedtriangles from incremental ones cum <- t(apply(inc,1, cumsum)).

41


With R, those two vectors can be obtained using functions lambda(triangle) andsigma(triangle), the algorithm being simply

LAMBDA = matrix(NA,1,Ntr-1)

for(i in 1:(Ntr-1)){

LAMBDA[i] = sum(TRIANGLE.C[1:(Ntr-i),i+1])/

sum(TRIANGLE.C[1:(Ntr-i),i])}

Hence,

> LAMBDA

[,1] [,2] [,3] [,4] [,5]

[1,] 1.380933 1.011433 1.004343 1.001858 1.004735

An alternative is to remember that the chain ladder estimate is obtained as thecoefficient of a weighted regression,

x <- TRIANGLE.C[,1]

y <- TRIANGLE.C[,2]

lm(y ~ x + 0, weights=1/x)

we obtain here

Call:

42


lm(formula = y ~ x + 0, weights = 1/x)

Coefficients:

x

1.381

which is the value of the first link ratio. Actually more details can be obtained

> summary(lm(y ~ x + 0, weights=1/x))

Call:

lm(formula = y ~ x + 0, weights = 1/x)

Residuals:

1988 1989 1990 1991 1992

-1.048825 0.161975 -0.009507 0.971088 -0.179734

Coefficients:

Estimate Std. Error t value Pr(>|t|)

x 1.380933 0.005176 266.8 1.18e-09 ***

---

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

43


Residual standard error: 0.7249 on 4 degrees of freedom

(1 observation deleted due to missingness)

Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999

F-statistic: 7.119e+04 on 1 and 4 DF, p-value: 1.184e-09

If f denotes the triangle of λi,j = Di,j+1/Di,j ,

f=TRIANGLE.D[,2:Ntr]/TRIANGLE.D[,1:(Ntr-1)]

SIGMA = matrix(NA,1,Ntr-1)


D=TRIANGLE.D[,i]*(f[,i]-t(rep(LAMBDA[i],Ntr)))^2

SIGMA[i]<-1/(Ntr-i-1)*sum(D[,1:(Ntr-1)])}

SIGMA[Ntr-1]<-min(SIGMA[(Ntr-3):(Ntr-2)])

44


Hence,

> SIGMA

[,1] [,2] [,3] [,4] [,5]

[1] 0.525418785 0.102633234 0.002104330 0.000660780 0.000660780

In order to complete the triangle, one can simply use


TRIANGLE.D[(Ntr-i+1):(Ntr),i+1]=LAMBDA[i]*TRIANGLE.D[(Ntr-i+1):(Ntr),i]}

Hence,

> TRIANGLE.D

0 1 2 3 4 5

1988 3209 4372.000 4411.000 4428.000 4435.000 4456.000

1989 3367 4659.000 4696.000 4720.000 4730.000 4752.397

1990 3871 5345.000 5398.000 5420.000 5430.072 5455.784

1991 4239 5917.000 6020.000 6046.147 6057.383 6086.065

1992 4929 6794.000 6871.672 6901.518 6914.344 6947.084

1993 5217 7204.327 7286.691 7318.339 7331.939 7366.656

45


The total amount of reserve is the obtained comparing the last column (estimatedultimate loss amont) and the second diagonal (total payments as at now).

ultimate = TRIANGLE.D[,6]*(1+0.00)

payment.as.at = diag(TRIANGLE.D[,6:1])

RESERVES = ultimate-payment.as.at

> RESERVES

[1] 0.00000 22.39103 35.79342 65.67668 153.36790 2149.65640

Hence, here sum(RESERVES) is equal to 2426.885.

46


Mack’s approach with R

A ChainLadder-package grew out of presentations the author gave at the StochasticReserving Seminar at the Institute of Actuaries in November 2007. This packageimplements the Mack and Munich Chain Ladder model using weighted linearregression.

A link with Excel (through the RExcel-Addin) can be used.

MackChainLadder can be used to obtain λj ’s and σj ’s.

MunichChainLadder

47



> MackChainLadder(TRIANGLE.D)

Latest Dev.To.Date Ultimate IBNR Mack.S.E CoV

1 4,456 1.000 4,456 0.0 0.000 NaN

2 4,730 0.995 4,752 22.4 0.639 0.0285

3 5,420 0.993 5,456 35.8 2.503 0.0699

4 6,020 0.989 6,086 66.1 5.046 0.0764

5 6,794 0.978 6,947 153.1 31.332 0.2047

6 5,217 0.708 7,367 2,149.7 68.449 0.0318

Totals:

Sum of Latest: 32,637

Sum of Ultimate: 35,064

Sum of IBNR: 2,427

Total Mack S.E.: 79

Total CoV: 3

The total amount of reserves is here 2,427.

48



> MackChainLadder(TRIANGLE.D)

Latest Dev.To.Date Ultimate IBNR Mack.S.E CoV

1 4,456 1.000 4,456 0.0 0.000 NaN

2 4,730 0.995 4,752 22.4 0.639 0.0285

3 5,420 0.993 5,456 35.8 2.503 0.0699

4 6,020 0.989 6,086 66.1 5.046 0.0764

5 6,794 0.978 6,947 153.1 31.332 0.2047

6 5,217 0.708 7,367 2,149.7 68.449 0.0318

Totals:

Sum of Latest: 32,637

Sum of Ultimate: 35,064

Sum of IBNR: 2,427

Total Mack S.E.: 79

Total CoV: 3

The link-ratios σj ’s.

49



> MackChainLadder(TRIANGLE.D)$FullTriangle

F1 F2 F3 F4 F5 F6

1 3209 4372.000 4411.000 4428.000 4435.000 4456.000

2 3367 4659.000 4696.000 4720.000 4730.000 4752.397

3 3871 5345.000 5398.000 5420.000 5430.072 5455.784

4 4239 5917.000 6020.000 6046.147 6057.383 6086.065

5 4929 6794.000 6871.672 6901.518 6914.344 6947.084

6 5217 7204.327 7286.691 7318.339 7331.939 7366.656

> MackChainLadder(TRIANGLE.D)$f

[1] 1.380933 1.011433 1.004343 1.001858 1.004735 1.000000

> TRIANGLE=MackChainLadder(TRIANGLE.D)$FullTriangle

> sum(TRIANGLE[,Ntr]-rev(diag(TRIANGLE[Ntr:1,])))

[1] 2426.985

It is also possible to use plot(MackChainLadder(TRIANGLE.D))

50



1 2 3 4 5 6

IBNRLatest

Mack Chain Ladder Results

Origin year

Am

ount

s

030

0070

00

● ●●

●

●●

1 2 3 4 5 6

4000

6000

Chain ladder developments by origin year

Development year

Am

ount

s

1

1 1 1 1 1

2

2 2 2 23

3 3 3

4

4 4

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6

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4500 5000 5500 6000 6500

−1.

50.

01.

5

Fitted

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ndar

dise

d re

sidu

als

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Origin year

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50.

01.

5

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ndar

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1.0 1.5 2.0 2.5 3.0 3.5 4.0

−1.

50.

01.

5

Development year

Sta

ndar

dise

d re

sidu

als

51


Munich Chain-Ladder

> P=as.matrix(read.table("D:\\triangleE.csv",sep=";",header=FALSE))

> I=as.matrix(read.table("D:\\triangleC.csv",sep=";",header=FALSE))

> MCL=MunichChainLadder(Paid=P, Incurred=I)

> MCL

LatestPaid LatestIncurred Latest.P.I.Ratio UltimatePaid UltimateIncurred Ultimate.P.I.Ratio

1 4,456 4,456 1.00 4,456 4,456 1

2 4,750 4,730 1.00 4,750 4,753 1

3 5,470 5,420 1.01 5,454 5,455 1

4 6,131 6,020 1.02 6,085 6,086 1

5 7,087 6,794 1.04 6,980 6,983 1

6 7,353 5,217 1.41 7,537 7,544 1

> plot(MCL)

52


1 2 3 4 5 6

MCL PaidMCL Incurred

Munich Chain Ladder Results

origin year

Am

ount

s

020

0050

00

1 2 3 4 5 6

SCL P/IMCL P/I

Munich Chain Ladder vs. Standard Chain Ladder

origin year

%

020

4060

80

●

●

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●

●

●

●

●

●

●

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●

−2 −1 0 1 2

−2

−1

01

2

Paid residual plot

Paid residuals

Incu

rred

/Pai

d re

sidu

als

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−2 −1 0 1 2−

2−

10

12

Incurred residual plot

Incurred residuals

Pai

d/In

curr

ed r

esid

uals

53


Basics on GLM with R

X=c(1,2,3); Y=c(1,2,4); D=data.frame(X,Y)

REG=lm(Y~X,data=D); summary(REG)

x0=seq(0.1,3.9,by=0.05); D0=data.frame(X=x0)

y0=predict(REG,newdata=D0)

●

●

●

0 1 2 3 4

01

23

45

Figure 1 – Gaussian LM model.

54



REG=glm(Y~X,data=D,family=gaussian(link = "identity"))



●

●

●

0 1 2 3 4

01

23

45

Figure 2 – Gaussian GLM model, Identity link function (canonical).

55



REG=glm(Y~X,data=D,family=poisson(link = "log"))


y0=exp(predict(REG,newdata=D0))

●

●

●

0 1 2 3 4

01

23

45

Figure 3 – Poisson GLM model, log link function (canonical).

56



REG=glm(Y~X,data=D,family=Gamma(link = "inverse"))


y0=1/predict(REG,newdata=D0)

●

●

●

0 1 2 3 4

01

23

45

Figure 4 – Gamma GLM model, Identity link function (canonical).

57



REG=glm(Y~X,data=D,family=poisson(link = "identity"))



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●

0 1 2 3 4

01

23

45

Figure 5 – Poisson GLM model, Identity link function (noncanonical).

58



REG=glm(Y~X,data=D,family=poisson(link = "inverse"))


y0=1/predict(REG,newdata=D0)

●

●

●

0 1 2 3 4

01

23

45

Figure 6 – Poisson GLM model, inverse link function (noncanonical).

59


A factor model in claims reserving

A natural idea is to assume that incremental payments Yi,j can be explained bytwo factors : one related to occurrence year i, and one development factor,related to j. Formally, we assume that

Yi,j ∼ L(ϕ(1(occurrence year = i),1(development year = j))),

i.e. Yi,j is a random variable, with distribution L, where parameter(s) can berelated to the two factors, and where ϕ is a given function, called link function.

60


Poisson regression in claims reserving

Renshaw & Verrall (1998) proposed to use a Poisson regression for incrementalpayments to estimate claim reserve, i.e.

Yi,j ∼ P

(exp

[α+

∑u

βu1(occurrence year u = i) +∑

v

γv1(development year v = j)

]).

devF=as.factor(development); anF=as.factor(year)

REG=glm(vec.C~devF+anF, family = "Poisson")

Here,

> summary(REG)

Call:

glm(formula = vec.C ~ anF + devF, family = poisson(link = "log"),

data = triangle)

Deviance Residuals:

Min 1Q Median 3Q Max

-2.343e+00 -4.996e-01 9.978e-07 2.770e-01 3.936e+00

61


Coefficients:

Estimate Std. Error z value Pr(>|z|)

(Intercept) 8.05697 0.01551 519.426 < 2e-16 ***

anF1989 0.06440 0.02090 3.081 0.00206 **

anF1990 0.20242 0.02025 9.995 < 2e-16 ***

anF1991 0.31175 0.01980 15.744 < 2e-16 ***

anF1992 0.44407 0.01933 22.971 < 2e-16 ***

anF1993 0.50271 0.02079 24.179 < 2e-16 ***

devF1 -0.96513 0.01359 -70.994 < 2e-16 ***

devF2 -4.14853 0.06613 -62.729 < 2e-16 ***

devF3 -5.10499 0.12632 -40.413 < 2e-16 ***

devF4 -5.94962 0.24279 -24.505 < 2e-16 ***

devF5 -5.01244 0.21877 -22.912 < 2e-16 ***

---

Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1

(Dispersion parameter for poisson family taken to be 1)

Null deviance: 46695.269 on 20 degrees of freedom

Residual deviance: 30.214 on 10 degrees of freedom

AIC: 209.52

62


Number of Fisher Scoring iterations: 4

Again, it is possible to summarize this information in triangles....

Predictions can be used to complete the triangle.

on recupere alors les predictions pour completer le triangle.

ANew=rep(1 :Ntr),times=Ntr) ; DNew=rep(0 :(Ntr-1),each=Ntr)

P=predict(REG, newdata=data.frame(A=as.factor(ANew),D=as.factor(DNew)))

payinc.pred= exp(matrix(as.numeric(P),nrow=n,ncol=n))

noise = payinc-payinc.pred

year development paycum payinc payinc.pred noise

1 1988 0 3209 3209 3155.699242 5.330076e+01

2 1989 0 3367 3367 3365.604828 1.395172e+00

3 1990 0 3871 3871 3863.737217 7.262783e+00

4 1991 0 4239 4239 4310.096418 -7.109642e+01

5 1992 0 4929 4929 4919.862296 9.137704e+00

6 1993 0 5217 5217 5217.000000 1.818989e-12

7 1988 1 4372 1163 1202.109851 -3.910985e+01

8 1989 1 4659 1292 1282.069808 9.930192e+00

9 1990 1 5345 1474 1471.824853 2.175147e+00

63


10 1991 1 5917 1678 1641.857784 3.614222e+01

11 1992 1 6794 1865 1874.137704 -9.137704e+00

12 1988 2 4411 39 49.820712 -1.082071e+01

13 1989 2 4696 37 53.134604 -1.613460e+01

14 1990 2 5398 53 60.998886 -7.998886e+00

15 1991 2 6020 103 68.045798 3.495420e+01

16 1988 3 4428 17 19.143790 -2.143790e+00

17 1989 3 4720 24 20.417165 3.582835e+00

18 1990 3 5420 22 23.439044 -1.439044e+00

19 1988 4 4435 7 8.226405 -1.226405e+00

20 1989 4 4730 10 8.773595 1.226405e+00

21 1988 5 4456 21 21.000000 -2.842171e-14

Residuals are obtained using the residual function, with one of the followingoptions deviance, pearson, working, response or partial.

The pearson residuals are

εPi,j =

Xi,j − µi,j√µi,j

,

64


The deviance residuals are

εDi,j =

Xi,j − µi,j√di,j

,

Pearson’s error can be obtained from function resid=residuals(REG,"pearson"), andsummarized in a triangle

> PEARSON

[,1] [,2] [,3] [,4] [,5] [,6]

[1,] 9.488238e-01 -1.12801295 -1.533031 -0.4899687 -0.4275912 -6.202125e-15

[2,] 2.404895e-02 0.27733318 -2.213449 0.7929194 0.4140426 NA

[3,] 1.168421e-01 0.05669707 -1.024162 -0.2972380 NA NA

[4,] -1.082940e+00 0.89196334 4.237393 NA NA NA

[5,] 1.302749e-01 -0.21107479 NA NA NA NA

[6,] 2.518371e-14 NA NA NA NA NA

65


Errors in GLMs

●

●●

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●

●

●

●

●

●

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●

●

1988 1989 1990 1991 1992 1993

−2

−1

01

23

4

Pearson's error

Year of occurence

Err

or ●

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●

●

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●

●

●

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●

●

●

●●

●

●

1988 1989 1990 1991 1992 1993−

2−

10

12

34

Deviance's error

Year of occurence

Err

or

66


Errors in GLMs

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●

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●

●

●

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0 1 2 3 4 5

−2

−1

01

23

4

Pearson's error

Delai

Err

or ●

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0 1 2 3 4 5−

2−

10

12

34

Deviance's error

Delai

Err

or

67


Finally, the theoretical triangles of Yi,j ’s, defined as

> resid*sqrt(payinc.pred)+payinc.pred

[,1] [,2] [,3] [,4] [,5] [,6]

[1,] 3209 1163 39 17 7 21

[2,] 3367 1292 37 24 10 NA

[3,] 3871 1474 53 22 NA NA

[4,] 4239 1678 103 NA NA NA

[5,] 4929 1865 NA NA NA NA

[6,] 5217 NA NA NA NA NA

68


Uncertainty and bootstrap simulations

Based on that theoretical triangle, it is possible to generate residuals to obtain asimulated triangle. Since the size of the sample is small (here 21 observed values),assuming normality for Pearson’s residuals can be too restrictive. Resamplingbootstrap procedure can then be more robust.

In order to get the loss distribution, it is possible to use bootstrap techniques togenerate a matrix of errors, see Renshaw & Verrall (1994). They suggest toboostrap Pearson’s residuals, and the simulation procedure is the following• estimate the model parameter (GLM), β,

• calculate fitted values µi,j , and the residuals ri,j =Yi,j − µi,j√V (µi,j)

,

• forecast with original data µi,j for i+ j > n.Then can start the bootstrap loops, repeating B times• resample the residuals with resample, and get a new sample r(b)i,j ,

• create a pseudo sample solving Y ∗i,j = µi,j + r(b)i,j ×

√V (µi,j),

• estimate the model using GLM procedure and derive boostrap forecast

69


Let resid.sim be resampled residuals. Note that REG$fitted.values (called herepayinc.pred) is the vector containing the µi,j ’s. And further V (µi,j) is here simplyREG$fitted.values since the variance function for the Poisson regression is theidentity function. Hence, here

Y ∗i,j = µi,j + r(b)i,j ×

√µi,j

and thus, set

resid.sim = sample(resid,Ntr*(Ntr+1)/2,replace=TRUE)

payinc.sim = resid.sim*sqrt(payinc.pred)+payinc.pred

[,1] [,2] [,3] [,4] [,5] [,6]

[1,] 3155.699 1216.465 42.17691 18.22026 9.021844 22.89738

[2,] 3381.694 1245.399 84.02244 18.20322 11.122243 NA

[3,] 3726.151 1432.534 61.44170 23.43904 NA NA

[4,] 4337.279 1642.832 74.58658 NA NA NA

[5,] 4929.000 1879.777 NA NA NA NA

[6,] 5186.116 NA NA NA NA NA

For this simulated triangle, we can use Chain-Ladder estimate to derive a

70


simulated reserve amount (here 2448.175). Figure 7 shows the empiricaldistribution of those amounts based on 10, 000 random simulations.

2200 2300 2400 2500 2600 2700

0.0000.0010.0020.0030.0040.0050.0060.007

Estimated density of total reserves

(with Gaussian fitted distribution)

0.0 0.2 0.4 0.6 0.8 1.0

2300

2400

2500

2600

Estimated quantile of total reserves

(with Gaussian fitted distribution)

0.95 0.96 0.97 0.98 0.99 1.00

2550

2650

Figure 7 – Distribution of claim reserves, using bootstrap techniques.

71


Parametric or nonparametric Monte Carlo ?

A natural idea would be to assume that Pearson residual have a Gaussiandistribution, qqnorm(R) ; qqline(R)

●

● ●

●

●●

●

●●

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●

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●

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●

●

−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5

−2

−1

01

23

4

QQ plot of Pearson residuals

Theoritical quantiles

Em

piric

al q

uant

iles

●

●● ●

●

●

●

●

●

●

●

●

●

●

●

●●

●

−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5−

2−

10

12

34

QQ plot of Pearson residuals (Cook's distance)


Em

piric

al q

uant

iles

●

●● ●

●

●

●

●

●

●

●

●

●

●

●

●●

●

The graph on the right draw point with a size proportional to its Cook’s distance.

72


Instead of resampling in the sample obtained, we can also directly draw from anormal distribution, i.e.

rnorm(length(R),mean=mean(R),sd=sd(R))

●

● ●

●

●●

●

●●

●

●

●

●

●

●

●

●

●●

●

●

−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5

−2

−1

01

23

4

QQ plot of Pearson residuals


Em

piric

al q

uant

iles

Distribution of the reserves, B=10,000

Total amount of reserves

Den

sity

2100 2200 2300 2400 2500 2600 2700

0.00

00.

001

0.00

20.

003

0.00

40.

005

73


●

● ●

●

●●

●

●●

●

●

●

●

●

●

●

●

●●

●

●

−3 −2 −1 0 1 2 3

−2

−1

01

23

4QQ plot of Pearson residual (Student)


Em

piric

al q

uant

iles

Distribution of the reserves, B=10,000


Den

sity

2100 2200 2300 2400 2500 2600 2700

0.00

00.

001

0.00

20.

003

0.00

40.

005

The second triangle is obtained using a Student t distribution (the blue linebeing the bootstrap estimate).

74


0.80 0.85 0.90 0.95 1.00

2400

2450

2500

2550

2600

2650

2700

VaR for total reserves

probability level

quan

tile

leve

l

StudentNormalbootstrap

Note that the bootstrap technique is valid only in the case were the residuals areperfectly independent.

In R, it is also possible to use the BootChainLadder(Triangle , R = 999, process.distr

= "od.pois") function.

75


Going further

So far, we have derived a ditrisbution for the best estimate of total reserves.

Note tat it is possible to estimate a scale parameter φ. England and Verrall(1999) suggested

φ =

∑ε2i,j

n− pwhere the summation is over all past observations.

It is possible to sample from the estimated process distribution, i.e. generate asingle Poisson P(

∑αiβj), where the sum is over the future.

P286

The overall algorithm is simply

76


Analytical estimate of the prediction error

So far, we have been using bootstrap technique to derive a confidence interval offuture payments. But it is possible to use outputs of a GLM function.

Recall that

E(

[Xi,j − Xi,j ]2)

=(E[Xi,j ]− E[Xi,j ]

)2

+V ar(Xi,j−Xi,j) ≈ V ar(Xi,j)+V ar(Xi,j),

since• the squared bias is small and can be neglected,• the future loss and its forecast (computed from past losses) are independent.In the case of a log-Poisson model, E

(Xi,j

)= µi,j = exp(ηi,j) and

V ar(Xi,j = ϕ · µi,j , hence,

V ar(Xi,j) =≈∣∣∣∣∂µi,j

∂ηi,j

∣∣∣∣2 V ar(ηi,j).

Thus,E(

[Xi,j − Xi,j ]2)≈ ϕµi,j + µ2

i,j V ar(ηi,j).

77


So finally, E(R− R)2 can be computed and

E(R− R)2 ≈∑i,j

ϕµi,j + µ′V ar(η)µ.

78


Bootstrap Chain-Ladder

> I=as.matrix(read.table("D:\\triangleC.csv",sep=";",header=FALSE))

> BCL <- BootChainLadder(Triangle = I, R = 999, process.distr = "od.pois")

> BCL

BootChainLadder(Triangle = I, R = 999, process.distr = "od.pois")

Latest Mean Ultimate Mean IBNR SD IBNR IBNR 75% IBNR 95%

1 4,456 4,456 0.0 0.0 0 0

2 4,730 4,752 22.0 11.8 28 45

3 5,420 5,455 35.3 14.6 44 61

4 6,020 6,086 66.2 20.8 78 102

5 6,794 6,947 152.7 29.1 170 205

6 5,217 7,364 2,146.9 112.5 2,214 2,327

Totals

Latest: 32,637

Mean Ultimate: 35,060

Mean IBNR: 2,423

SD IBNR: 131

Total IBNR 75%: 2,501

Total IBNR 95%: 2,653

79


Histogram of Total.IBNR

Total IBNR

Fre

quen

cy

2000 2200 2400 2600 2800 3000

010

020

030

0

2000 2200 2400 2600 2800 3000

0.0

0.4

0.8

ecdf(Total.IBNR)

Total IBNR

Fn(

x)

● ● ●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●● ●● ●●●●●●● ● ●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●

●●●●●●●

●

●●●●●●●

1 2 3 4 5 6

4500

6000

7500

Simulated ultimate claims cost

origin period

ultim

ate

clai

ms

cost

s

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80


Using GAM for claims reserving

In the case of GAM’s, assume that

Yi,j ∼ L(ϕ(u(occurrence year) + v(development year))),

where here u and v are two unknown functions. We still have an additive form,but on unknown transformations of explanatory variates.

Spline functions are considered to estimation functions u and v.

library(gam)

GAM=gam(payinc~s(year,5)+s(development,3),data=D,familly="Poisson")

plot.gam(GAM,se=T,col="red",ask=TRUE,main="GAM model, df=5, df=3")

81


1988 1989 1990 1991 1992 1993

−500

0500

1000

1500

2000

GAM model, df=5, df=3

year

s(year, 5)

0 1 2 3 4 5

−1000

01000

2000

GAM model, df=5, df=3

development

s(development, 3)

82


Dealing with negative increments

Negative incremental values can arise due to timing of reinsurance, recoveries,cancellation of outstanding claims.

One might argue that the problem is more with the data than with the methods.

England & Verall (2002) mention that the Gaussian model is less affected by thepresence of negative incremental values. Unfortunately, one can hardly assumethat data are Gaussian because of the skewness.

Renshaw & Verral (1998) use a quasi-likelihood approach.

Renshaw & Verral (1994) suggest to add a “small constant” to the past data andto substract this constant from forecasts at the end.

83



A classical technique to avoid negative payments is to

consider a translation of the incremental triangle,

i.e. Y +i,j = Yi,j + κ such that Y +

i,j > 0 for all i, j.

●

●

●

0 1 2 3 4

01

23

45

●

●

●

Figure 8 – Translation in GLMs

84



A classical technique to avoid negative payments is to

consider a translation of the incremental triangle,

i.e. Y +i,j = Yi,j + κ such that Y +

i,j > 0 for all i, j.

●

●

●

0 1 2 3 4

01

23

45

●

●

●

Figure 9 – Translation in GLMs

85


Why a Poisson regression model ?

The is no reason to assume that incremental payments are Poisson distribution.The only motivation here is that the expected value is the same as the ChainLadder estimate.

Distribution of the reserves, GAM model with Chain Ladder


Den

sity

2000 4000 6000 8000

0e+

001e

−04

2e−

043e

−04

4e−

045e

−04

86


References

Stanard, J. N. (1985). A simulation test of prediction errors of loss reserveestimation techniques. Proceedings of the Casualty Actuarial Society, LXXII.124-148. http ://www.casact.org/pubs/proceed/proceed85/85124.pdf

Taylor, G. (2003). Chain ladder bias. ASTIN Bulletin, 33 No. 2, 313-330.

http ://www.casact.org/library/astin/vol33no2/313.pdf

Mack, T. (1993). Distribution free calculation of the standard error of chainladder reserve estimates. ASTIN Bulletin, 23, 213-225.


Mack, T. (1994). Which stochastic model is underlying the chain ladder method ?Insurance : Mathematics and Economics, 15, 133-138.

Mack T. (1994). Measuring the variability of chain ladder reserve estimates.Casualty Actuarial Society, Spring Forum, 101-182.

http ://www.casact.org/pubs/forum/94spforum/94spf101.pdf

87


References

Mack, T. (1997). Measuring the variability of chain ladder reserve estimates.Claims Reserving Manual volume 2. London : Institute of Actuaries.

http ://www.actuaries.org.uk/files/pdf/library/crm2-D6.pdf

Mack, T. (1999). The standard error of chain ladder reserve estimates : Recursivecalculation and inclusion of a tail factor. ASTIN Bulletin, 29 No. 2, 361-366.


Renshaw, A. E. Verral, R. J. (1998). A stochastic model underlying the chainladder technique. British Actuarial Journal, 4, 903-923.

England, P. D. Verrall, R. J. (1999). Analytic and bootstrap estimates ofprediction errors in claims reserving. Insurance : Mathematics and Economics,25, 281-293.

England, P. D. Verrall, R. J. (2002). Stochastic claims reserving in generalinsurance. British Actuarial Journal, 8, 443-544.

88


References

Benjamin, S. Eagles, L. (1997). A curve fitting method and a regression method.Claims Reserving Manual, volume 2. London : Institute of Actuaries.

http ://www.actuaries.org.uk/files/pdf/library/crm2-D3.pdf

England, P. D. Verrall, R. J. (2002). Stochastic claims reserving in generalinsurance. British Actuarial Journal, 8, 443-544.

England, P. D. (2002). A beginners guide to Bayesian modelling. GeneralInsurance Convention.

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