Statistique de l’assurance STT6705V, partie 12 bis
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In the previous post (here) discussing forecasts of actuarial quantities, I did not mention much how to forecast the temporal component in the Lee-Carter model. Actually, many things can be done. Consider here some exponential smoothing techniques (see here for more details). For instance, with a simple model, with an additive error, no trend and no seasonal component,Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
or equivalently
Then
, and if we want a confidence interval > (ETS.ANN=ets(Y,model="ANN")) ETS(A,N,N) Call: ets(y = Y, model = "ANN") Smoothing parameters: alpha = 0.7039 Initial states: l = 75.0358 sigma: 11.4136 AIC AICc BIC 941.0089 941.1339 946.1991 > plot(forecast(ETS.ANN,h=100))
Here, a natural idea will probably be to take into account a linear trend on the series, i.e.
where
then the forecast we make is
i.e.ETS.AAN=ets(Y,model="AAN") plot(forecast(ETS.AAN,h=100))
It is also possible to ask for an “optimal” exponential smoothing model (without any specification)> ETS
ETS(A,A,N)
Call:
ets(y = Y)
Smoothing parameters:
alpha = 0.6107
beta = 1e-04
Initial states:
l = 74.5622
b = -1.9812
sigma: 11.0094
AIC AICc BIC
937.8695 938.2950 948.2500
> plot(forecast(ETS,h=100))
Another idea can be to consider some autoregressive integrated moving average models (ARIMA), here with a linear trend
For instance, if we want only the integrated component, i.e.
then the code is
> ARIMA010T=Arima(Y,order=c(0,1,0),include.drift=TRUE) Series: Y ARIMA(0,1,0) with drift Call: Arima(x = Y, order = c(0, 1, 0), include.drift = TRUE) Coefficients: drift -2.0337 s.e. 1.1904 sigma^2 estimated as 138.9: log likelihood = -380.8 AIC = 765.6 AICc = 765.73 BIC = 770.77 > plot(forecast(ARIMA010T,h=100))
But note that any kind of ARIMA model can be considered, e.g. integrated with an autoregressive component (here of order 1)
or with also a moving average component (again of order 1)
Actually, note that, once again, we can ask for an automatic selection of the model,> (autoARIMA=auto.arima(Y,allowdrift=TRUE))
Series: Y
ARIMA(0,1,1) with drift
Call: auto.arima(x = Y, allowdrift = TRUE)
Coefficients:
ma1 drift
-0.3868 -2.0139
s.e. 0.0970 0.6894
sigma^2 estimated as 122.3: log likelihood = -374.65
AIC = 755.29 AICc = 755.55 BIC = 763.05
> plot(forecast(autoARIMA,h=100))
Finally, it is possible to use also also some specific functions of the package, for instance to consider a random walk with a drift,
RWF=rwf(Y,h=100,drift=TRUE) plot(RWF)
or Holt model (with a trend)HOLT=holt(Y,h=100,damped=TRUE) plot(HOLT)
And if all that is not enough, it is also possible to go further by changing the size of the series we use to fit the model. A question that naturally arises is the treatment of wars in our model: shouldn’t we assume that the forecast should be based only on the last 50 years (and exclude wars from our analysis) ? In that case, for instance, the exponential smoothing technique gives
while the ARIMA procedure returns
And finally, with the Holt technique, we have
So, it looks like we have a lot of techniques that can be used to provide a forecast for the temporal component in the Lee-Carter model,
All those estimators can be used to estimate annuities of insurance contracts (as here),BASEB=BASEB[,1:99]
BASEC=BASEC[,1:99]
AGE=AGE[1:99]
library(gnm)
D=as.vector(BASEB)
E=as.vector(BASEC)
A=rep(AGE,each=length(ANNEE))
Y=rep(ANNEE,length(AGE))
base=data.frame(D,E,A,Y,a=as.factor(A),
y=as.factor(Y))
LC2=gnm(D~a+Mult(a,y),offset=log(E),
family=poisson,data=base)
A=LC2$coefficients[1]+LC2$coefficients[2:99]
B=LC2$coefficients[100:198]
K0=LC2$coefficients[199:297]
Y=as.numeric(K01)
K1=c(K0,forecast(ets(Y),h=120)$mean)
K2=c(K0,forecast(auto.arima(Y,allowdrift=TRUE),h=120)$mean)
K3=c(K0,rwf(Y,h=120,drift=TRUE)$mean)
K4=c(K0,holt(Y,h=120,drift=TRUE)$mean)
K5=c(K0,forecast(ets(Y[50:99]),h=120)$mean)
K6=c(K0,forecast(auto.arima(Y[50:99],allowdrift=TRUE),h=120)$mean)
K7=c(K0,rwf(Y[50:99],h=120,drift=TRUE)$mean)
K8=c(K0,holt(Y[50:99],h=120,drift=TRUE)$mean)
MU=matrix(NA,length(A),length(K1))
MU1=MU2=MU3=MU4=MU5=MU6=MU7=MU8=MU
for(i in 1:length(A)){
for(j in 1:length(K1)){
MU1[i,j]=exp(A[i]+B[i]*K1[j])
MU2[i,j]=exp(A[i]+B[i]*K5[j])
MU3[i,j]=exp(A[i]+B[i]*K3[j])
MU4[i,j]=exp(A[i]+B[i]*K4[j])
MU5[i,j]=exp(A[i]+B[i]*K5[j])
MU6[i,j]=exp(A[i]+B[i]*K6[j])
MU7[i,j]=exp(A[i]+B[i]*K7[j])
MU8[i,j]=exp(A[i]+B[i]*K8[j])
}}
t=2000
x=40
s=seq(0,98-x-1)
Pxt1=cumprod(exp(-diag(MU1[x+1+s,t+s-1898])))
Pxt2=cumprod(exp(-diag(MU2[x+1+s,t+s-1898])))
Pxt3=cumprod(exp(-diag(MU3[x+1+s,t+s-1898])))
Pxt4=cumprod(exp(-diag(MU4[x+1+s,t+s-1898])))
Pxt5=cumprod(exp(-diag(MU5[x+1+s,t+s-1898])))
Pxt6=cumprod(exp(-diag(MU6[x+1+s,t+s-1898])))
Pxt7=cumprod(exp(-diag(MU7[x+1+s,t+s-1898])))
Pxt8=cumprod(exp(-diag(MU8[x+1+s,t+s-1898])))
r=.035
m=70
h=seq(0,21)
V1=1/(1+r)^(m-x+h)*Pxt1[m-x+h]
V2=1/(1+r)^(m-x+h)*Pxt2[m-x+h]
V3=1/(1+r)^(m-x+h)*Pxt3[m-x+h]
V4=1/(1+r)^(m-x+h)*Pxt4[m-x+h]
V5=1/(1+r)^(m-x+h)*Pxt5[m-x+h]
V6=1/(1+r)^(m-x+h)*Pxt6[m-x+h]
V7=1/(1+r)^(m-x+h)*Pxt7[m-x+h]
V8=1/(1+r)^(m-x+h)*Pxt8[m-x+h]
Hence, here the difference is significant,> M=cbind(V1,V2,V3,V4,V5,V6,V7,V8) > apply(M,2,sum) V1 V2 V3 V4 V5 V6 V7 V8 4.389372 4.632793 4.406465 4.389372 4.632793 4.468934 4.482064 4.632793or graphically,
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