Slides ensae 11bis

Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11
Actuariat de l’Assurance Non-Vie # 11
A. Charpentier (UQAM & Université de Rennes 1)
ENSAE ParisTech, Octobre 2015 - Janvier 2016.
http://freakonometrics.hypotheses.org
@freakonometrics 1

Regression Models in Claims Reserving
A natural idea is to assume that incremental payments Yi,j can be explained by
two factors: one related to occurrence year i, and one development factor, related
to j. Formally, we assume that
Yi,j ∼ L(θi,j), where θi,j = αi · βj
i.e. Yi,j is a random variable, with distribution L, where parameter(s) can be
related to the two factors.
@freakonometrics 2

Poisson regression in claims reserving
Renshaw & Verrall (1998) proposed to use a Poisson regression for incremental
payments to estimate claim reserve, i.e.
Yi,j ∼ P (exp [γ + αi + βj]) .
1 devF=as.factor(development); anF=as.factor(year)
2 REG=glm(vec.C~devF+anF , family = "Poisson")
Here,
1 > summary(REG)
2 Call:
3 glm(formula = vec.C ~ anF + devF , family = poisson(link = "log"),
4 data = triangle)
5
6 Deviance Residuals:
7 Min 1Q Median 3Q Max
8 -2.343e+00 -4.996e -01 9.978e -07 2.770e-01 3.936e+00
@freakonometrics 3

9
10 Coefficients :
11 Estimate Std. Error z value Pr(>|z|)
12 (Intercept) 8.05697 0.01551 519.426 < 2e-16 ***
13 anF1989 0.06440 0.02090 3.081 0.00206 **
14 anF1990 0.20242 0.02025 9.995 < 2e-16 ***
15 anF1991 0.31175 0.01980 15.744 < 2e-16 ***
16 anF1992 0.44407 0.01933 22.971 < 2e -16 ***
17 anF1993 0.50271 0.02079 24.179 < 2e -16 ***
18 devF1 -0.96513 0.01359 -70.994 < 2e -16 ***
19 devF2 -4.14853 0.06613 -62.729 < 2e-16 ***
20 devF3 -5.10499 0.12632 -40.413 < 2e -16 ***
21 devF4 -5.94962 0.24279 -24.505 < 2e-16 ***
22 devF5 -5.01244 0.21877 -22.912 < 2e -16 ***
23 ---
24
25 ( Dispersion parameter for poisson family taken to be 1)
26
@freakonometrics 4

27 Null deviance: 46695.269 on 20 degrees of freedom
28 Residual deviance: 30.214 on 10 degrees of freedom
29 AIC: 209.52
30
31 Number of Fisher Scoring iterations: 4
Again, it is possible to summarize this information in triangles....
Predictions can be used to complete the triangle.
1 ANew=rep(1 :Ntr),times=Ntr) ; DNew=rep (0 :(Ntr -1),each=Ntr)
2 P=predict(REG , newdata=data.frame(A=as.factor(ANew),D=as.factor(
DNew)))
3 payinc.pred= exp(matrix(as.numeric(P),nrow=n,ncol=n))
4 noise = payinc -payinc.pred
1 year development paycum payinc payinc.pred noise
2 1 1988 0 3209 3209 3155.699242 5.330076e+01
3 2 1989 0 3367 3367 3365.604828 1.395172e+00
4 3 1990 0 3871 3871 3863.737217 7.262783e+00
@freakonometrics 5

5 4 1991 0 4239 4239 4310.096418 -7.109642e+01
6 5 1992 0 4929 4929 4919.862296 9.137704e+00
7 6 1993 0 5217 5217 5217.000000 1.818989e-12
8 7 1988 1 4372 1163 1202.109851 -3.910985e+01
9 8 1989 1 4659 1292 1282.069808 9.930192e+00
10 9 1990 1 5345 1474 1471.824853 2.175147e+00
11 10 1991 1 5917 1678 1641.857784 3.614222e+01
12 11 1992 1 6794 1865 1874.137704 -9.137704e+00
13 12 1988 2 4411 39 49.820712 -1.082071e+01
14 13 1989 2 4696 37 53.134604 -1.613460e+01
15 14 1990 2 5398 53 60.998886 -7.998886e+00
16 15 1991 2 6020 103 68.045798 3.495420e+01
17 16 1988 3 4428 17 19.143790 -2.143790e+00
18 17 1989 3 4720 24 20.417165 3.582835e+00
19 18 1990 3 5420 22 23.439044 -1.439044e+00
20 19 1988 4 4435 7 8.226405 -1.226405e+00
21 20 1989 4 4730 10 8.773595 1.226405e+00
22 21 1988 5 4456 21 21.000000 -2.842171e-14
@freakonometrics 6

The pearson residuals are
εP
i,j =
Xi,j − µi,j
µi,j
,
The deviance residuals are
εD
i,j =
Xi,j − µi,j
di,j
,
Pearson’s error can be obtained from function resid=residuals(REG,"pearson"), and
summarized in a triangle
1 > PEARSON
2 [,1] [,2] [,3] [,4] [,5] [,6]
3 [1,] 9.4882e-01 -1.128012 -1.5330 -0.48996 -0.42759 -6.2021e-15
4 [2,] 2.4048e-02 0.277333 -2.2134 0.79291 0.41404 NA
5 [3,] 1.1684e-01 0.056697 -1.0241 -0.29723 NA NA
6 [4,] -1.0829e+00 0.891963 4.2373 NA NA NA
7 [5,] 1.3027e-01 -0.211074 NA NA NA NA
8 [6,] 2.5183e-14 NA NA NA NA NA
@freakonometrics 7

Errors in GLMs
@freakonometrics 8

log-Poisson regression and Chain-Ladder
The log-Poisson is interesting since it (usually) provides the same amount of
reserves as Chain Ladder.
1 > library( ChainLadder )
2 > an <- 10; ligne = rep (1:an , each=an); colonne = rep (1:an , an)
3 > passe = (ligne + colonne - 1) <=an; n = sum(passe)
4 > PAID=GenIns; INC=PAID
5 > INC [,2:an]= PAID [,2:an]-PAID [ ,1:(an -1)]
6 > Y = as.vector(INC)
7 > lig = as.factor(ligne)
8 > col = as.factor(colonne)
9 > base = data.frame(Y,col ,lig)
10 > reg=glm(Y~col+lig ,data=base ,family="poisson")
11 > sum(exp(predict(reg ,newdata=base))[passe!=TRUE ])
12 [1] 18680856
@freakonometrics 9

log-Poisson regression and Chain-Ladder
1 > MackChainLadder (GenIns)
2 MackChainLadder (Triangle = GenIns)
3 Latest Dev.To.Date Ultimate IBNR Mack.S.E CV(IBNR)
4 1 3 ,901 ,463 1.0000 3 ,901 ,463 0 0 NaN
5 2 5 ,339 ,085 0.9826 5 ,433 ,719 94 ,634 71 ,835 0.759
6 3 4 ,909 ,315 0.9127 5 ,378 ,826 469 ,511 119 ,474 0.254
7 4 4 ,588 ,268 0.8661 5 ,297 ,906 709 ,638 131 ,573 0.185
8 5 3 ,873 ,311 0.7973 4 ,858 ,200 984 ,889 260 ,530 0.265
9 6 3 ,691 ,712 0.7223 5 ,111 ,171 1 ,419 ,459 410 ,407 0.289
10 7 3 ,483 ,130 0.6153 5 ,660 ,771 2 ,177 ,641 557 ,796 0.256
11 8 2 ,864 ,498 0.4222 6 ,784 ,799 3 ,920 ,301 874 ,882 0.223
12 9 1 ,363 ,294 0.2416 5 ,642 ,266 4 ,278 ,972 970 ,960 0.227
13 10 344 ,014 0.0692 4 ,969 ,825 4 ,625 ,811 1 ,362 ,981 0.295
14 Totals
15 Latest: 34 ,358 ,090.00
16 Ultimate: 53 ,038 ,945.61
17 IBNR: 18 ,680 ,855.61
@freakonometrics 10

18 Mack S.E.: 2 ,441 ,364.13
19 CV(IBNR): 0.13
@freakonometrics 11

An explicit expression to quantify uncertainty
Recall that we while to estimate
E([R − R]2
) = E(R) − E(R)
2
+ Var(R − R)
≈ Var(R) + Var(R)
Classically, consider a log-Poisson model, were incremental payments satisfy
Yi,j ∼ P(µi,j) where µi,j = exp[ηi,j] = exp[γ + αi + βj]
Using the delta method, we get that asymptotically
Var(Yi,j) = Var(µi,j) ≈
∂µi,j
∂ηi,j
2
Var(ηi,j)
where, since we consider a log link,
∂µi,j
∂ηi,j
= µi,j
@freakonometrics 12

i.e., with an ODP distribution (i.e. Var(Yi,j) = ϕE(Yi,j),
E [Yi,j − Yi,j]2
≈ ϕ · µi,j + µ2
i,j · Var(ηi,j)
and
Cov(Yi,j, Yk,l) ≈ µi,j · µk,l · Cov (ηi,j, ηk,l)
Thus, since the overall amount of reserves satisﬁes
E [R − R]2
≈
i+j−1>n
ϕ · µi,j + µ Var(η)µ.
2 > passe = (ligne + colonne - 1) <=an; np = sum(passe)
3 > futur = (ligne + colonne - 1)> an; nf = sum(passe)
4 > INC=PAID
5 > INC [ ,2:6]= PAID [,2:6]- PAID [ ,1:5]
@freakonometrics 13

7 > lig = as.factor(ligne); col = as.factor(colonne)
8 >
9 > CL <- glm(Y~lig+col , family= quasipoisson )
10 > Y2=Y; Y2[is.na(Y)]=.001
11 > CL2 <- glm(Y2~lig+col , family= quasipoisson )
12 > YP = predict(CL)
13 > p = 2*6-1;
14 > phi.P = sum(residuals(CL ,"pearson")^2)/(np -p)
15 > Sig = vcov(CL)
16 > X = model.matrix(CL2)
17 > Cov.eta = X%*%Sig%*%t(X)
18 > mu.hat = exp(predict(CL ,newdata=data.frame(lig ,col)))*futur
19 > pe2 = phi.P * sum(mu.hat) + t(mu.hat) %*% Cov.eta %*% mu.hat
20 > cat("Total reserve =", sum(mu.hat), "prediction error =", sqrt(pe2
),"n")
21 Total reserve = 2426.985 prediction error = 131.7726
i.e. E [R − R]2
= 131.77.
@freakonometrics 14

Uncertainty and bootstrap simulations
Based on that theoretical triangle, it is possible to generate residuals to obtain a
simulated triangle. Since the size of the sample is small (here 21 observed
values), assuming normality for Pearson’s residuals can be too restrictive.
Resampling bootstrap procedure can then be more robust.
In order to get the loss distribution, it is possible to use bootstrap techniques to
generate a matrix of errors, see Renshaw & Verrall (1994). They suggest to
boostrap 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
@freakonometrics 15

• 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
Let resid.sim be resampled residuals. Note that REG$fitted.values (called here
payinc.pred) is the vector containing the µi,j’s. And further V (µi,j) is here simply
REG$fitted.values since the variance function for the Poisson regression is the
identity function. Hence, here
Y ∗
i,j = µi,j + r
(b)
i,j × µi,j
and thus, set
1 resid.sim = sample(resid ,Ntr*(Ntr +1)/2,replace=TRUE)
2 payinc.sim = resid.sim*sqrt(payinc.pred)+payinc.pred
3
4 [,1] [,2] [,3] [,4] [,5] [,6]
@freakonometrics 16

5 [1,] 3155.699 1216.465 42.17691 18.22026 9.021844 22.89738
6 [2,] 3381.694 1245.399 84.02244 18.20322 11.122243 NA
7 [3,] 3726.151 1432.534 61.44170 23.43904 NA NA
8 [4,] 4337.279 1642.832 74.58658 NA NA NA
9 [5,] 4929.000 1879.777 NA NA NA NA
10 [6,] 5186.116 NA NA NA NA NA
For this simulated triangle, we can use Chain-Ladder estimate to derive a
simulated reserve amount (here 2448.175). Figure below shows the empirical
distribution of those amounts based on 10, 000 random simulations.
@freakonometrics 17

@freakonometrics 18

Parametric or nonparametric Monte Carlo ?
A natural idea would be to assume that Pearson residual have a Gaussian
distribution, qqnorm(R); qqline(R)
The graph on the right draw point with a size proportional to its Cook’s distance.
@freakonometrics 19

Instead of resampling in the sample obtained, we can also directly draw from a
normal distribution, i.e.
1 > rnorm(length(R),mean=mean(R),sd=sd(R))
@freakonometrics 20

The second triangle is obtained using a Student t distribution (the blue line
being the bootstrap estimate).
0.80 0.85 0.90 0.95 1.00
2400245025002550260026502700
VaR for total reserves
probability level
quantilelevel
Student
Normal
bootstrap
Note that the bootstrap technique is valid only in the case were the residuals are
perfectly independent.
@freakonometrics 21

In R, it is also possible to use the BootChainLadder(Triangle , R = 999, process.distr
= "od.pois") function.
@freakonometrics 22

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 & Verrall (1999)
suggested
φ =
ε2
i,j
n − p
where the summation is over all past observations.
@freakonometrics 23

Bootstrap Chain-Ladder
1 > I=as.matrix(read.table("D: triangleC.csv",sep=";",header=FALSE))
2 > BCL <- BootChainLadder (Triangle = I, R = 999, process.distr = "od.
pois")
3 > BCL
4 BootChainLadder (Triangle = I, R = 999, process.distr = "od.pois")
5
6 Latest Mean Ultimate Mean IBNR SD IBNR IBNR 75% IBNR 95%
7 1 4 ,456 4 ,456 0.0 0.0 0 0
8 2 4 ,730 4 ,752 22.0 11.8 28 45
9 3 5 ,420 5 ,455 35.3 14.6 44 61
10 4 6 ,020 6 ,086 66.2 20.8 78 102
11 5 6 ,794 6 ,947 152.7 29.1 170 205
12 6 5 ,217 7 ,364 2 ,146.9 112.5 2 ,214 2 ,327
13
14 Totals
15 Latest: 32 ,637
16 Mean Ultimate: 35 ,060
@freakonometrics 24

17 Mean IBNR: 2 ,423
18 SD IBNR: 131
19 Total IBNR 75%: 2 ,501
20 Total IBNR 95%: 2 ,653
Histogram of Total.IBNR
Total IBNR
Frequency
2000 2200 2400 2600 2800 3000
0100200300
2000 2200 2400 2600 2800 3000
0.00.40.8
ecdf(Total.IBNR)
Total IBNR
Fn(x)
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Simulated ultimate claims cost
origin period
ultimateclaimscosts
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020004000
Latest actual incremental claims
against simulated values
origin period
latestincrementalclaims
qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq
qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq Latest actual
@freakonometrics 25

From Poisson to Over-Dispersed Poisson
Classical, in GLMs we consider distributions with density
f(z|θ, φ) = exp
zθ − b(θ)
φ
+ c(z, φ) ,
e.g. for the Poisson distribution P(λ) then
f(z|λ) = exp(−λ)
λz
z!
= exp z log λ − λ − log z! , z ∈ N,
with θ = log λ, φ = 1, b(θ) = exp θ = λ and c(z, φ) = − log z!.
Assume that φ = 1 becomes an additional parameter (that should be estimated).
Note that in that case f(z|λ) is not any more a density, but it is a quasidensity.
Further, note that
V ar(Z) = φE(Z).
Thus, if φ > 1 there is overdispersion.
@freakonometrics 26

On quasiPoisson regression
In order to understand the rule of the additional parameter, recall that for the
Gaussien linear model, N(µ, σ2
) it is an exponential distribution with θ = µ,
b(θ) = θ2
/2, φ = σ2
and
c(z, φ) = −
1
2
y2
σ2
+ log(2πσ2
) .
Thus, φ is the variance parameter
Y |X ∼ N(Xβ, σ2
)
In that linear model, estimation is done based on the following process,
• estimate β as β = (X X)−1
X Y
• derive the implied residuals, ε = Y − Xβ
• estimate σ as the variance of the implied residuals
@freakonometrics 27

Thus, φ does not impact the estimation of the coeﬃcient, but it will impact their
signiﬁcativity.
4 > INC[,2:an]= PAID [,2:an]-PAID [ ,1:(an -1)]
9 > reg1=glm(Y~col+lig ,data=base ,family="poisson")
10 > reg2=glm(Y~col+lig ,data=base ,family=" quasipoisson ")
11 > summary(reg1)
12 Call:
13 glm(formula = Y ~ col + lig , family = "poisson", data = base)
14 Coefficients :
15 Estimate Std. Error z value Pr(>|z|)
16 (Intercept) 12.5064047 0.0007540 16587.372 < 2e-16 ***
@freakonometrics 28

17 col2 0.3312722 0.0006694 494.848 < 2e-16 ***
18 col3 0.3211186 0.0006877 466.963 < 2e -16 ***
19 col4 0.3059600 0.0007008 436.570 < 2e -16 ***
20 col5 0.2193163 0.0007324 299.461 < 2e-16 ***
21 col6 0.2700770 0.0007445 362.755 < 2e -16 ***
22 col7 0.3722084 0.0007606 489.344 < 2e -16 ***
23 col8 0.5533331 0.0008133 680.377 < 2e-16 ***
24 col9 0.3689342 0.0010429 353.772 < 2e-16 ***
25 col10 0.2420330 0.0018642 129.830 < 2e -16 ***
26 lig2 0.9125263 0.0006490 1406.042 < 2e-16 ***
27 lig3 0.9588306 0.0006652 1441.374 < 2e -16 ***
28 lig4 1.0259970 0.0006840 1499.927 < 2e-16 ***
29 lig5 0.4352762 0.0008019 542.814 < 2e -16 ***
30 lig6 0.0800565 0.0009364 85.492 < 2e -16 ***
31 lig7 -0.0063815 0.0010390 -6.142 8.14e-10 ***
32 lig8 -0.3944522 0.0013529 -291.560 < 2e -16 ***
33 lig9 0.0093782 0.0013963 6.716 1.86e -11 ***
34 lig10 -1.3799067 0.0039097 -352.946 < 2e -16 ***
@freakonometrics 29

35 ---
36 ( Dispersion parameter for poisson family taken to be 1)
37 Null deviance: 10699464 on 54 degrees of freedom
38 Residual deviance: 1903014 on 36 degrees of freedom
39 (45 observations deleted due to missingness )
40 AIC: 1903877
1 > summary(reg2)
2 Call:
3 glm(formula = Y ~ col + lig , family = " quasipoisson ", data = base)
4 Coefficients :
5 Estimate Std. Error t value Pr(>|t|)
6 (Intercept) 12.506405 0.172924 72.323 < 2e -16 ***
7 col2 0.331272 0.153537 2.158 0.03771 *
8 col3 0.321119 0.157719 2.036 0.04916 *
9 col4 0.305960 0.160736 1.903 0.06499 .
10 col5 0.219316 0.167970 1.306 0.19994
11 col6 0.270077 0.170756 1.582 0.12247
@freakonometrics 30

12 col7 0.372208 0.174451 2.134 0.03976 *
13 col8 0.553333 0.186525 2.967 0.00532 **
14 col9 0.368934 0.239181 1.542 0.13170
15 col10 0.242033 0.427562 0.566 0.57485
16 lig2 0.912526 0.148850 6.131 4.65e-07 ***
17 lig3 0.958831 0.152569 6.285 2.90e-07 ***
18 lig4 1.025997 0.156883 6.540 1.33e-07 ***
19 lig5 0.435276 0.183914 2.367 0.02344 *
20 lig6 0.080057 0.214770 0.373 0.71152
21 lig7 -0.006381 0.238290 -0.027 0.97878
22 lig8 -0.394452 0.310289 -1.271 0.21180
23 lig9 0.009378 0.320249 0.029 0.97680
24 lig10 -1.379907 0.896690 -1.539 0.13258
25 ---
26 ( Dispersion parameter for quasipoisson family taken to be 52601.93)
@freakonometrics 31

30 AIC: NA
Thus, coefficients are identical so it not affect the best estimate of claims
reserves.... unless we take into account the fact that some variates are no longer
significant.....
2 > base$lig[base$lig=="7"]="1"
5 > base$col[base$col=="5"]="1"
@freakonometrics 32

14 > summary(glm(Y~col+lig ,data=base ,family=" quasipoisson "))
15 Call:
16 glm(formula = Y ~ col + lig , family = " quasipoisson ", data = base)
17 Coefficients :
18 Estimate Std. Error t value Pr(>|t|)
19 (Intercept) 12.73401 0.07764 164.022 < 2e-16 ***
20 col8 0.28877 0.14109 2.047 0.04618 *
21 lig2 0.96246 0.10984 8.763 1.59e -11 ***
22 lig3 0.99721 0.11232 8.878 1.07e -11 ***
23 lig4 1.06465 0.11481 9.273 2.82e -12 ***
24 lig5 0.45513 0.14622 3.113 0.00312 **
25 lig10 -1.60752 0.85482 -1.881 0.06611 .
26 ---
27 ( Dispersion parameter for quasipoisson family taken to be 49241.53)
@freakonometrics 33

31 AIC: NA
Thus,
1 > M= cbind(Y,predict(reg1 ,newdata=base0 ,type="response"),
2 + predict(reg2 ,newdata=base0 ,type="response"),
3 + predict(reg3 ,newdata=base , type="response"))
4 > sum(M[is.na(Y)==TRUE ,2])
5 [1] 18680856
7 [1] 18680856
9 [1] 18226919
Including an overdispersion parameter φ might impact the estimation of the
overall reserves.
@freakonometrics 34

Testing for overdispersion
In order to test for overdispersion in an econometric model, we need to specify
how overdispersion appears. A classical test is to assume that
V ar(Y |X) = E(Y |X) + τE(Y |X)2
which is a standard econometric model with random eﬀect. We want to test
H0 : τ = 0 against H1 : τ > 0
A standard test statistics is
T =
n
i=1[Yi − λi]2
− Yi
2
n
i=1 λ2
i
which has a N(0, 1) distribution under H0. An alternative is to consider
T =
n
i=1[Yi − λi]2
− Yi
n
i=1[[Yi − λi]2 − Yi]2
@freakonometrics 35

Those test can be found in R, respectively
1 > library(AER)
2 > dispersiontest (reglmp)
3 > dispersiontest (reglmp ,trafo = 2)
An alternative is simply the following
1 > library(ChainLadder)
5 > INC [,2:an]= PAID [,2:an]-PAID [ ,1:(an -1)]
10 > reg1=glm(Y~col+lig ,data=base ,family="poisson")
11 > reg2=glm(Y~col+lig ,data=base ,family=" quasipoisson ")
12 > dispersiontest (reg1)
@freakonometrics 36

13 Overdispersion test
14 data: reg1
15 z = 4.3942 , p-value = 5.558e -06
16 alternative hypothesis : true dispersion is greater than 1
@freakonometrics 37

Alternative models for overdispersion
There is overdispersion if Var(Y ) > E(Y ), which can be obtained with a negative
binomial distribution (with belongs to the exponential family)
1 > library(MASS)
2 > reg3=glm.nb(Y~col+lig ,data=base)
3 > summary(reg3)
4 ( Dispersion parameter for Negative Binomial (13.8349) family taken to
be 1)
5 Theta: 13.83
6 Std. Err.: 2.61
7 2 x log -likelihood: -1460.766
8 > sum(exp(predict(reg3 ,newdata=base))[passe!=TRUE ])
9 [1] 18085795
@freakonometrics 38

Uncertainty and overdispersion
Based on the explicit expression for the prediction error, it is possible to obtain
prediction error for those three models,
1 > predCL=function(reg=reg1 ,regb=reg1b){
2 + p = 2*6-1;
3 + phi.P = sum(residuals(reg ,"pearson")^2)/(np -p)
4 + Sig = vcov(reg)
5 + X = model.matrix(regb)
6 + Cov.eta = X%*%Sig%*%t(X)
7 + mu.hat = exp(predict(reg ,newdata=data.frame(lig ,col)))*futur
8 + pe2 = phi.P * sum(mu.hat) + t(mu.hat) %*% Cov.eta %*% mu.hat
9 + cat("Total reserve =", sum(mu.hat), " prediction error =", sqrt(pe2
),sqrt(pe2)/sum(mu.hat),"n")
10 + }
Avec nos trois modèles, Poisson, ODP et binomiale négative, on obtient,
1 > predCL(reg1 ,reg1b)
@freakonometrics 39

2 Total reserve = 18680856 prediction error = 896876.9 0.04801048
4 Total reserve = 18680856 prediction error = 4736425 0.2535443
6 Total reserve = 18085795 prediction error = 2058134 0.1137984
@freakonometrics 40

On the prediction error
In order to derive an estimation of the prediction error using bootstrap
techniques, we have not only to generate randomly possible triangles, but also to
add uncertainty in the developpement, using e.g. the fact that
Ci,j+1 = λjCi,j + σj Ci,j + εi,j
where the noise can be assume to be Gaussian, N(0, 1).
The statistical interpretation is that
Ci,j+1|Ci,j ∼ N(λjCi,j + σ2
j Ci,j)
Classically we use
1 > CL=function(triangle){
2 + n=nrow(triangle)
3 + LAMBDA=rep(NA ,n -1)
4 + for(i in 1:(n-1)){
@freakonometrics 41

5 + LAMBDA[i]= sum(triangle [1:(n-i),i+1])/
6 + sum(triangle [1:(n-i),i]) }
7 + DIAG=diag(triangle[,n:1])
8 + TOTO=c(1,rev(LAMBDA))
9 + return(sum(cumprod(TOTO)*DIAG -DIAG)) }
a natural idea is to consider
1 > CLboot=function(triangle ,l,s){
2 + m=nrow(triangle)
3 + for(i in 2:m){
4 + triangle [(m-i+2):m,i]= rnorm(i-1,
5 + mean=triangle [(m-i+2):m,i-1]*l[i-1],
6 + sd=sqrt(triangle [(m-i+2):m,i -1])*s[i -1])
7 + }
8 + ULT=triangle[,m]
9 + DIAG=diag(triangle[,m:1])
10 + return(sum(ULT -DIAG)) }
Then, we can run boostrap simulations,
@freakonometrics 42

1 > base=data.frame(Y,lig ,col)
2 > REG=glm(Y~lig+col ,family=poisson)
3 > YP=predict(REG ,newdata=base)
4 > E=residuals(REG ,"pearson")
5 > PROV.BE=rep(NA ,5000)
6 > PROVISION=rep(NA ,5000)
7 > for(k in 1:50000){
8 + simE=sample(E,size =36, replace=TRUE)
9 + bruit=simE*sqrt(exp(YP))
10 + INCsim=exp(YP)+bruit
11 + INCM=matrix(INCsim ,6 ,6)
12 + CUMM=INCM
13 + for(j in 2:6){CUMM[,j]= CUMM[,j -1]+ INCM[,j]}
14 + PROV.BE[k]=CL(CUMM)
15 + PROVISION[k]= CLboot(CUMM ,lambda ,sigma)}
@freakonometrics 43

Random Generation of a Quasi-Distribution
It is also possible to generate Poisson, or quasi-Poisson random variables.
Recall that the negative binomial distribution has probability function
P[N = k] =
Γ(k + r)
k!Γ(r)
· [1 − p]r
pk
where the expected value and the variance are
µ = r ·
p
1 − p
and σ2
= µ = r ·
p
(1 − p)2
Assume that σ2
= ϕ · µ, then
r =
µ
ϕ − 1
and p =
1
ϕ
1 > rqpois = function(n, lambda , phi) {
2 + return( rnbinom(n, size = lambda/(1-phi), prob = 1/phi) }
@freakonometrics 44

Using GAM for claims reserving
In the case of GAM’s, assume that
Yi,j ∼ L(θçi, j), where θi,j = ϕ(u(i) + v(j)),
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.
1 > library(gam)
2 > GAM=gam(payinc~s(year ,5)+s(development ,3),data=D,familly="Poisson")
3 > plot.gam(GAM ,se=T,col="red",ask=TRUE ,main="GAM model , df=5, df=3")
@freakonometrics 45

@freakonometrics 46

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 the
presence of negative incremental values. Unfortunately, one can hardly assume
that data are Gaussian because of the skewness. Renshaw & Verral (1994)
suggested to add a “small constant” to the past data and to substract this
constant from forecasts at the end.
@freakonometrics 47

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.
@freakonometrics 48

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.
q
q
q
0 1 2 3 4
012345
q
q
q
@freakonometrics 49

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 Chain
Ladder estimate.
Distribution of the reserves, GAM model with Chain Ladder
Total amount of reserves
Density
2000 4000 6000 8000
0e+001e−042e−043e−044e−045e−04
@freakonometrics 50

Tweedie ?
The density of a tweedie model with power function p would be
1 > ftweedie = function(y,p,mu ,psi){
2 + if(p==2){f = dgamma(y, 1/psi , 1/(psi*mu))} else
3 + if(p==1){f = dpois(y/psi , mu/psi)} else
4 + {lambda = mu^(2-p)/psi /(2-p)
5 + if(y==0){ f = exp(-lambda)} else
6 + { alpha = (2-p)/(p -1)
7 + beta = 1 / (psi * (p -1) * mu^(p -1))
8 + k = max (10, ceiling(lambda + 7*sqrt(lambda)))
9 + f = sum(dpois (1:k,lambda) * dgamma(y,alpha*(1:k),beta))
10 + }}
11 + return(f)
12 + }
A numerical problem is that we should have no missing values in the regression,
so artiﬁcially, consider
1 > source("http:// freakonometrics .free.fr/bases.R")
@freakonometrics 51

2 > library(statmod)
5 > INC=PAID
6 > INC [ ,2:6]= PAID [,2:6]- PAID [ ,1:5]
10 > y = Y[passe]
11 > Y[is.na(Y)]=.01
Then, we can run an econometric regression
1 > pltweedie <- function(pow){
2 + regt = glm(Y~lig+col , tweedie(pow ,0))
3 + reserve = sum(fitted.values(regt)[!passe ])
4 + dev = deviance(regt)
5 + phi.hat = dev/n
6 + mu = fitted.values(regt)[passe]
@freakonometrics 52

7 + hat.logL = 0
8 + for (k in 1: length(y)){
9 + hat.logL <- hat.logL + log(ftweedie(y[k], pow , mu[k], phi.hat))
}
10 + cat("Puissance =", round(pow ,3) , "phi =", round(phi.hat ,2),
11 + "Reserve (tot) =", round(reserve), "logL =", round(hat.logL ,3))
12 + hat.logL}
13 > for(pow in c(1 ,1.25 ,1.5 ,1.75 ,2)){pltweedie(pow)}
14 Puissance = 1 phi = 166.95 Reserve (tot) = 1345 logL = -Inf
15 Puissance = 1.25 phi = 42.92 Reserve (tot) = 1216 logL = -151.72
18 Puissance = 2 phi = 6.78 Reserve (tot) = 125 logL = -170.614
It is also possible to run a optimization routine,
1 > optimize(pltweedie , c(1.01 ,1.99) , tol =1e-4, maximum = TRUE)
2 -144.624
3 $maximum
@freakonometrics 53

4 [1] 1.427873
5
6 $objective
7 [1] -144.6237
Thus, here the Poisson model might not be the appropriate one,
1.0 1.2 1.4 1.6 1.8 2.0
−2000−1500−1000−500
power
Loglikelihood
@freakonometrics 54

Bayesian Models in Claims Reserving
The ﬁrst idea is to consider some credibility based model, with
Ci,n = Z · CMack
i,n + [1 − Z] · µi
given some a priori µi.
For instance Benkhtander (1976) and Hovinen (1981) suggested
Z = 1 − [1 − βi]2
where βi =
n−1
k=n−i
1
λk
Note that
Ci,n = Ci,n−i + [1 − βi] βi · ·CMack
i,n + [1 − βi] · µi
More generally, consider the Cape-Code technique,
Ci,n = Ci,n−i + 1 −
Ci,n−i
Ci,n
Ci,n
@freakonometrics 55

sous la forme
Ci,n = Ci,n−i + 1 −
Ci,n−i
Ci,n
LRi · Pi,
où LRi correspond au loss ratio pour l’année i, i.e. LRi = Ci,n/Pi. L’idée de la
méthode dite Cape-Code est d’écrire une forme plus générale,
Ci,n = Ci,n−i + (1 − πn−i) LRiPi
où πn−i correspond à une cadence de paiement, et peut être estimé par la
méthode Chain Ladder. Quant aux LRi il s’agit des loss ratio cibles,
correspondant à un avis d’expert. On peut aussi proposer un même ratio cible
pour plusieurs années de survenance. On posera alors
Ri = Ci,n − Ci,n−i = (1 − πn−i)LRAPi.
pour i ∈ A, où
LRA = k∈A Cn,n−k
k∈A πn−kPk
.
@freakonometrics 56

Dans un premier temps, on peut calculer les πi à partir de la méthode Chain
Ladder, i.e.
πn−i =
Ci,n−i
Ci,n
où la charge ultime est celle prédite pas la méthode Chain-Ladder.
1 > Cultime = MackChainLadder (PAID)$ FullTriangle [,6]
2 > (PI <- (1- Cdiag/Cultime))
3 1 2 3 4 5 6
4 0.00000 0.00471 0.00656 0.01086 0.02204 0.29181
5 > LR <- TRIANGLE [,6]/PREMIUM
6 > Cdiag <- diag(PAID [ ,6:1])
7 > (Cultime -Cdiag)/(LR*PREMIUM)
8 1 2 3 4 5 6
9 0.00000 0.00471 0.00656 0.01086 0.02204 0.29181
Si on suppose ensuite que A = {1, 2, · · · , n}, alors
1 > LR=sum(TRIANGLE [ ,6])/sum(PREMIUM)
@freakonometrics 57

2 > PI*LR*PREMIUM
3 1 2 3 4 5 6
4 0.0 24.6 35.6 62.7 139.6 2095.3
5 > sum(PI*LR*PREMIUM)
6 [1] 2358
On obtient ici un montant de provision total inférieur à celui obtenu par la
méthode Chain Ladder puisque le montant de provisions vaut ici 2357.756.
@freakonometrics 58

Modèles bayésiens et Chain Ladder
De manière générale, un méthode bayésienne repose sur deux hypothèses
• une loi a priori pour les paramètres du modèle (Xi,j, Ci,j, λi,j,
LRi,j = Ci,j/Pj, etc)
• une technique pour calculer les lois a posteriori, qui sont en général assez
complexes.
@freakonometrics 59

Modèles bayésiens pour les nombres de sinistres
Soit Ni,j l’incrément du nombre de sinistres, i.e. le nombre de sinistres survenus
l’année i, déclarés l’année i + j.
On note Mi le nombre total de sinistres par année de survenance, i.e.
Mi = Ni,0 + Ni,1 + · · · . Supposons que Mi ∼ P(λi), et que p = (p0, p1, · · · , pn)
désigne les proprotions des paiments par année de déroulé.
Conditionnellement à Mi = mi, les années de survenance sont indépenantes, et le
vecteur du nombre de sinistres survenus année l’année i suit une loi multinomiale
M(mi, p).
@freakonometrics 60

Modèles bayésiens pour les nombres de sinistres
La vraisemblance L(M0, M1, · · · , Mn, p|Ni,j) est alors
n
i=0
Mi!
(Mi − Nn−i)!Ni,0!Ni,1! · · · Ni,n−i!
[1 − pn−i]Mi−Nn−i p
Ni,0
0 p
Ni,1
1 · · · p
Ni,n−i
n−i
où Nn−i = N0 + N1 + · · · + Nn−i et pn−i = p0 + p1 + · · · + pn−i.
Il faut ensuite de donner une loi a priori pour les paramètres. La loi a posteriori
sera alors proportionnelle produit entre la vraisemblance et cette loi a priori.
@freakonometrics 61

Modèles bayésiens pour les montants agrégés
On pose Yi,j = log(Ci,j), et on suppose que Yi,j = µ + αi + βj + εi,j, où
εi,j ∼ N(0, σ2
). Aussi, Yi,j suit une loi normale,
f(yi,j|µ, α, β, σ2
) ∝
1
σ
exp −
1
2σ2
[yi,j − µ − αi − βj]
2
,
et la vraisemblance est alors
L(θ, σ|Y ) ∝ σ−m
exp


i,j
[yi,j − µ − αi − βj]
2


où m = (n(n + 1)/2 désigne le nombre d’observations passées. La diﬃculté est
alors de spéciﬁer une loi a priori pour (θ, σ2
), i.e. (µ, α, β, σ2
).
@freakonometrics 62

MCMC and Bayesian Models
We have a sample x = {x1, · · · , xd) i.i.d. from distribution fθ(·).
In predictive modeling, we need E(g(X)|x) = g(x)fθ|x(x)dx where
fθ|x(x) = f(x|x) = f(x|θ) · π(θ|x)dθ
How can we derive π(θ|x) ? Can we sample from π(θ|x) (and use monte carlo
technique to approximate the integral) ?
@freakonometrics 63

Hastings-Metropolis
Back to our problem, we want to sample from π(θ|x)
i.e. generate θ1, · · · , θn, · · · from π(θ|x).
Hastings-Metropolis sampler will generate a Markov Chain (θt) as follows,
• generate θ1
• generate θ and U ∼ U([0, 1]),
compute R =
π(θ |x)
π(θt|x)
P(θt|θ )
P(θ |θt−1)
if U < R set θt+1 = θ
if U ≥ R set θt+1 = θt
R is the acceptance ratio, we accept the new state θ with probability min{1, R}.
@freakonometrics 64

Using MCMC to generate Gaussian values
1 > metrop1 <- function(n=1000 , eps =0.5){
2 + vec <- vector("numeric", n)
3 + x=0
4 + vec [1] <- x
5 + for (i in 2:n) {
6 + innov <- runif(1,-eps ,eps)
7 + mov <- x+innov
8 + aprob <- min(1, dnorm(mov)/dnorm(x))
9 + u <- runif (1)
10 + if (u < aprob)
11 + x <- mov
12 + vec[i] <- x
13 + }
14 + return(vec)}
@freakonometrics 66

Using MCMC to generate Gaussian values
1 > plot.mcmc <- function(mcmc.out){
2 + op <- par(mfrow=c(2 ,2))
3 + plot(ts(mcmc.out),col="red")
4 + hist(mcmc.out ,30, probability =TRUE ,
5 + col="light blue")
6 + lines(seq(-4,4,by =.01) ,dnorm(seq
(-4,4,
7 + by =.01)),col="red")
8 + qqnorm(mcmc.out)
9 + abline(a=mean(mcmc.out),b=sd(mcmc.
out))
10 + acf(mcmc.out ,col="blue",lag.max
=100)
11 + par(op)}
12 > metrop.out <-metrop1 (10000 ,1)
13 > plot.mcmc(metrop.out)
@freakonometrics 67

Heuristics on Hastings-Metropolis
In standard Monte Carlo, generate θi’s i.i.d., then
1
n
n
i=1
g(θi) → E[g(θ)] = g(θ)π(θ)dθ
(strong law of large numbers).
Well-behaved Markov Chains (P aperiodic, irreducible, positive recurrent) can
satisfy some ergodic property, similar to that LLN. More precisely,
• P has a unique stationary distribution λ, i.e. λ = λ × P
• ergodic theorem
1
n
n
i=1
g(θi) → g(θ)λ(θ)dθ
even if θi’s are not independent.
@freakonometrics 68

Heuristics on Hastings-Metropolis
Remark The conditions mentioned above are
• aperiodic, the chain does not regularly return to any state in multiples of
some k.
• irreducible, the state can go from any state to any other state in some ﬁnite
number of steps
• positively recurrent, the chain will return to any particular state with
probability 1, and ﬁnite expected return time
@freakonometrics 69

MCMC and Loss Models
Example A Tweedie model, E(X) = µ and Var(X) = ϕ · µp
. Here assume that ϕ
and p are given, and µ is the unknown parameter.
→ need a predictive distribution for µ given x.
Consider the following transition kernel (a Gamma distribution)
µ|µt ∼ G
µt
α
, α
with E(µ|µt) = µt and CV(µ) =
1
√
α
.
Use some a priori distribution, e.g. G (α0, β0).
@freakonometrics 70

• generate µ1
• at step t : generate µ ∼ G α−1
µt, α and U ∼ U([0, 1]),
compute R =
π(µ ) · f(x|µ )
π(µt) · f(x|θt)
Pα(µt|θ )
Pα(θ |θt−1)
if U < R set θt+1 = θ
if U ≥ R set θt+1 = θt
where
f(x|µ) = L(µ) =
n
i=1
f(xi|µ, p, ϕ),
f(x · |µ, p, ϕ) being the density of the Tweedie distribution, dtweedie function
(x, p, mu, phi) from library(tweedie).
@freakonometrics 71

1 > p=2 ; phi=2/5
2 > set.seed (1) ; X <- rtweedie (50,p,10,phi)
3 > metrop2 <- function(n=10000 , a0=10,
4 + b0=1, alpha =1){
5 + vec <- vector("numeric", n)
6 + vec [1] <- rgamma (1,a0 ,b0)
7 + for (i in 2:n){
8 + mustar <- rgamma (1,vec[i -1]/alpha ,alpha)
9 + R=prod(dtweedie(X,p,mustar ,phi)/dtweedie
10 + (X,p,vec[i-1],phi))*dgamma(mustar ,a0 ,b0)
11 + dgamma(vec[i-1],a0 ,b0)* dgamma(vec[i-1],
12 + mustar/alpha ,alpha)/dgamma(mustar ,
13 + vec[i-1]/alpha ,alpha)
14 + aprob <- min(1,R)
15 + ifelse(runif (1) < aprob ,vec[i]<-mustar ,
16 + vec[i]<-vec[i -1])}
17 + return(vec)}
18 > metrop.output <-metrop2 (10000 , alpha =1)
@freakonometrics 72

Gibbs Sampler For a multivariate problem, it is possible to use Gibbs sampler.
Example Assume that the loss ratio of a company has a lognormal distribution,
LN(µ, σ2
), .e.g
1 > LR <- c(0.958 , 0.614 , 0.977 , 0.921 , 0.756)
Example Assume that we have a sample x from a N(µ, σ2
). We want the
posterior distribution of θ = (µ, σ2
) given x . Observe here that if priors are
Gaussian N µ0, τ2
and the inverse Gamma distribution IG(a, b), then



µ|σ2
, x ∼ N
σ2
σ2 + nτ2
µ0 +
nτ2
σ2 + nτ2
x,
σ2
τ2
σ2 + nτ2
2
i=1
σ2
|µ, x ∼ IG
n
2
+ a,
1
2
n
i=1
[xi − µ]2
+ b
More generally, we need the conditional distribution of θk|θ−k, x, for all k.
1 > x <- log(LR)
@freakonometrics 73

Gibbs Sampler
1 > xbar <- mean(x)
2 > mu <- sigma2=rep (0 ,10000)
3 > sigma2 [1] <- 1/rgamma (1, shape =1,rate =1)
4 > Z <- sigma2 [1]/(sigma2 [1]+n*1)
5 > mu [1] <- rnorm (1,m=Z*0+(1 -Z)*xbar ,
6 + sd=sqrt (1*Z))
7 > for (i in 2:10000){
8 + Z <- sigma2[i-1]/(sigma2[i -1]+n*1)
9 + mu[i] <- rnorm (1,m=Z*0+(1 -Z)*xbar ,
10 + sd=sqrt (1*Z))
11 + sigma2[i] <- 1/rgamma (1, shape=n/2+1,
12 + rate <- (1/2)*(sum((x-mu[i])${ }^{
wedge}$2))+1)
13 + }
@freakonometrics 74

Gibbs Sampler
Example Consider some vector X = (X1, · · · , Xd) with indépendent
components, Xi ∼ E(λi). We sample to sample from X given XT
1 > s for some
threshold s > 0.
• start with some starting point x0 such that xT
0 1 > s
• pick up (randomly) i ∈ {1, · · · , d}
Xi given Xi > s − xT
(−i)1 has an Exponential distribution E(λi)
draw Y ∼ E(λi) and set xi = y + (s − xT
(−i)1)+ until xT
(−i)1 + xi > s
E.g. losses and allocated expenses
@freakonometrics 75

Gibbs Sampler
1 > sim <- NULL
2 > lambda <- c(1 ,2)
3 > X <- c(3 ,3)
4 > s <- 5
5 > for(k in 1:1000){
6 + i <- sample (1:2 ,1)
7 + X[i] <- rexp(1, lambda[i])+
8 + max(0,s-sum(X[-i]))
9 + while(sum(X)<s){
10 + X[i] <- rexp(1, lambda[i])+
11 + max(0,s-sum(X[-i])) }
12 + sim <- rbind(sim ,X) }
@freakonometrics 76

JAGS and STAN
Martyn Plummer developed JAGS Just another Gibbs sampler in 2007 (stable
since 2013) in library(runjags). It is an open-source, enhanced, cross-platform
version of an earlier engine BUGS (Bayesian inference Using Gibbs Sampling).
STAN library(Rstan) is a newer tool that uses the Hamiltonian Monte Carlo
(HMC) sampler.
HMC uses information about the derivative of the posterior probability density
to improve the algorithm. These derivatives are supplied by algorithm
diﬀerentiation in C/C++ codes.
@freakonometrics 77

JAGS on the N(µ, σ2
) distribution
1 library(runjags)
2 jags.model <- "
3 + model {
4 + mu ~ dnorm(mu0 , 1/(sigma0 ^2))
5 + g ~ dgamma(k0 , theta0)
6 + sigma <- 1 / g
7 + for (i in 1:n) {
8 + logLR[i] ~ dnorm(mu , g^2)
9 + }
10 + }"
1 > jags.data <- list(n=length(LR),
2 + logLR=log(LR), mu0=-.2, sigma0 =0.02 ,
3 + k0=1, theta0 =1)
4
5 > jags.init <- list(list(mu=log (1.2) ,
6 + g=1/0.5^2) ,
7 + list(mu=log (.8) ,
8 + g=1/.2^2))
9
10 > model.out <- autorun.jags(jags.model ,
11 + data=jags.data , inits=jags.init ,
12 + monitor=c("mu", "sigma"), n.chains =2)
13 traceplot(model.out$mcmc)
14 summary(model.out)
@freakonometrics 78

STAN on the N(µ, σ2
) distribution
1 > library(rstan)
2 > stan.model <- "
3 + data {
4 + int <lower =0> n;
5 + vector[n] LR;
6 + real mu0;
7 + real <lower =0> sigma0;
8 + real <lower =0> k0;
9 + real <lower =0> theta0;
10 + }
11 + parameters {
12 + real mu;
13 + real <lower =0> sigma;
14 + }
1 + model }
2 + mu ~ normal(mu0 , sigma0);
3 + sigma ~ inv_gamma(k0 , theta0);
4 + for (i in 1:n)
5 + log(LR[i]) ~ normal(mu , sigma);
6 + }
7
8 > stan.data <- list(n=length(LR), r=LR , mu0=
mu0 ,
9 + sigma0=sigma0 , k0=k0 , theta0=theta0)
10 > stan.out <- stan(model_code=stan.model ,
11 + data=stan.data , seed =2)
12 > traceplot(stan.out)
13 > print(stan.out , digits_summary =2)
@freakonometrics 79

Example Consider some simple time series of Loss Ratios,
LRt ∼ N(µt, σ2
) where µt = φµt−1 + εt
E.g. in JAGS we can deﬁne the vector µ = (µ1, · · · , µT ) recursively
1 + model {
2 + mu [1] ~ dnorm(mu0 , 1/(sigma0 ^2))
3 + for (t in 2:T) { mu[t] ~ dnorm(mu[t-1], 1/(sigma0 ^2)) }
4 + }
@freakonometrics 80

A Bayesian version of Chain Ladder
Assume that λi,j ∼ N µj,
τj
Ci,j
.
We can use Gibbs sampler to get the distribution of the transition factors, as well
as a distribution for the reserves,
1 > source("http:// freakonometrics .free.fr/triangleCL.R")
2 > source("http:// freakonometrics .free.fr/bayesCL.R")
3 > mcmcCL <-bayesian.triangle(PAID)
@freakonometrics 81

1 > plot.mcmc(mcmcCL$Lambda [ ,1])
2 > plot.mcmc(mcmcCL$Lambda [ ,2])
3 > plot.mcmc(mcmcCL$reserves [ ,6])
4 > plot.mcmc(mcmcCL$reserves [ ,7])
5 > library( ChainLadder )
6 > MCL <-MackChainLadder (PAID)
7 > m<-sum(MCL$ FullTriangle [,6]-
8 + diag(MCL$ FullTriangle [ ,6:1]))
9 > stdev <-MCL$Total.Mack.S.E
10 > hist(mcmcCL$reserves [,7], probability=
TRUE ,
11 > breaks =20, col="light blue")
12 > x=seq (2000 ,3000 , by =10)
13 > y=dnorm(x,m,stdev)
14 > lines(x,y,col="red")
@freakonometrics 82

Autres modèles bayésiens
Dans le cadre des modèles de provisionnement, on suppose
λi,j|λj, σ2
j , Ci,j ∼ N λj,
σ2
j
Ci,j
Notons γj = log(λj). λ désigne l’ensemble des observations, i.e. λi,j, et le
paramètre que l’on cherche à estimer est γ. La log-vraisemblance est alors
log L(λ|γ, C, σ2
) =
i,j
log
Ci,j
σ2
j
−
Ci,j
σ2
j
[λi,j − exp(γj)]
2
En utilisant le théorème de Bayes
log L(λ|γ, C, σ2
)
a posteriori
= log π(γ)
a priori
+ log L(γ|λ, C, σ2
)
log vraisemblance
+constante
Si on utilise une loi uniforme comme loi a priori, on obtient
log L(λ|γ, C, σ2
) = log L(γ|λ, C, σ2
) + constante
@freakonometrics 83

Les calculs de lois conditionnelles peuvent être simples dans certains cas (très
limités). De manière gérérale, on utilise des méthodes de simulation pour
approcher les lois. En particulier, on peut utiliser les algorithmes de Gibbs ou
d’Hastings-Metropolis.
On part d’un vecteur initial γ(0)
= (γ
(0)
1 , · · · , γ
(0)
m ), puis



γ
(k+1)
1 ∼ f(·|γ
(k)
2 , · · · , γ
(k)
m , λ, C, σ)
γ
(k+1)
2 ∼ f(·|γ
(k+1)
1 , γ
(k)
3 , · · · , γ
(k)
m , λ, C, σ)
γ
(k+1)
3 ∼ f(·|γ
(k+1)
1 , γ
(k+1)
2 , γ
(k)
4 , · · · , γ
(k)
m , λ, C, σ)
...
γ
(k+1)
m−1 ∼ f(·|γ
(k+1)
1 , γ
(k+1)
2 , · · · , γ
(k+1)
m−2 , γ
(k)
m , λ, C, σ)
γ
(k+1)
m ∼ f(·|γ
(k+1)
1 , γ
(k+1)
2 , · · · , γ
(k+1)
m−1 , λ, C, σ)
A l’aide de cet algorithme, on simule alors de triangles C, puis on estime la
process error.
@freakonometrics 84

L’algorithme d’adaptative rejection metropolis sampling peut alors être utiliser
pour simuler ces diﬀérentes lois conditionnelle (cf Balson (2008)).
La méthode de rejet est basé sur l’idée suivante
• on souhaite tirer (indépendemment) suivant une loi f, qu’on ne sait pas
simuler
• on sait simuler suivant une loi g qui vériﬁe f(x) ≤ Mg(x), pour tout x, où M
peut être calculée.
L’algorithme pour tirer suivant f est alors le suivant
• faire une boucle
◦ tirer Y selon la loi g
◦ tirer U selon la loi uniforme sur [0, 1], indépendamment de Y ,
• tant que U >
f(Y )
Mg(Y )
.
@freakonometrics 85

• poser X = Y .
On peut utiliser cette technique pour simuler une loi normale à partir d’une loi
de Laplace, de densité g(x) = 0.5 · exp(−|x|), avec M =
√
2eπ−1. Mais cet
algorithme est très coûteux en temps s’il y a beaucoup de rejets,
L’adaptative rejection sampling est une extension de cet algorithme, à condition
d’avoir une densité log-concave. On parle aussi de méthode des cordes.
@freakonometrics 86

On majore localement la fonction log f par des fonctions linéaires. On construit
alors une enveloppe à log f.
On majore alors f par une fonction gn qui va dépendre du pas.
Formellement, on construit Li,j(x) la droite reliant les points (xi, log(f(xi))) et
@freakonometrics 87

(xj, log(f(xj))). On pose alors
hn(x) = min {Li−1,i(x), Li+1,i+2(x)} ,
qui déﬁnie alors une enveloppe de log(f) (par concavité de log(f). On utilise
alors un algorithme de rejet avec comme fonction de référence
gn(x) =
exp(hn(x))
exp(hn(t))dt
normalisée pour déﬁnir une densité.
• faire une boucle
◦ tirer Y selon la loi gn
◦ tirer U selon la loi uniforme sur [0, 1], indépendamment de Y ,
• tant que U >
f(Y )
exp(hn(Y ))
.
• poser X = Y .
@freakonometrics 88

Enﬁn, l’adaptative rejection metropolis sampling rajoute une étape
suppl ´mentaire, dans le cas des densité non log-concave. L’idée est d’utiliser la
technique préc´dante, même si hn n’est plus forcément une enveloppe de log(f),
puis de rajouter une étape de rejet supplémenataire. Rappelons que l’on cherche
à implénter un algorithme de Gibbs, c’est à dire créér une suite de variables
X1, X2, · · · .
Supposons que l’on dispose de Xk−1. Pour tirer Xk, on utilise l’algorithme
précédant, et la nouvelle étape de rejet est la suivante
• tirer U selon la loi uniforme sur [0, 1], indépendamment de X et de Xk−1,
◦ si U > min 1,
f(X) min{f(Xk−1), exp(hn(Xk−1))}
f(Xk−1) min{f(X), exp(hn(X))}
alors garder
Xk = Xk−1
◦ sinon poser Xk = X
@freakonometrics 89

Code R pour l’algorithme ARMS
Ces fonctions exponentielles par morceaux sont inéressantes car elles sont faciles
à simuler. La fonction hn est linéaires par morceaux, avec comme noeuds Nk, de
telle sorte que
hn(x) = akx + bk pour tout x ∈ [Nk, Nk+1].
Alors gn(x) =
exp(hn(x))
In
où
In = exp(hn(t))dt =
exp[hn(Nk+1)] − exp[hn(Nk)]
ak
. On calcule alors Gn, la
fonction de répartition associée à gn, et on fait utilise une méthode d’inversion
pour tirer suivant Gn.
@freakonometrics 90

Bayesian estimation for reserves
0 200 400 600 800 1000
220023002400250026002700
iteration
reserves(total)
@freakonometrics 91

@freakonometrics 92

@freakonometrics 93

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