All classes, especially Class 1, exhibit strong trends over time. It is thusnatural that the above premiums (the intercept values) be larger that thoseobtained with B¨uhlmann–Straub’s model. The choice of credibility model isthus very important. A good fit to data should always be seeked once theestimated parameter values have been obtained. Disregarding the trend herecould result in a serious under-rating for all classes.Here the collective parameter estimates are:(1) a collective intercept and slope of 1,885 and-32, respectively,(2) an expected variance of ˆs2= 5.0×107,(3) and variance of conditional means of the regression parametersˆA=bracketleftbigg145,359-6,623-6,623302bracketrightbiggThese compare favourably to the B¨uhlmann–Straub’s estimators ˆm=¯XZW=1,684, ˆs2= 1.39×108and ˆa= 89,639. Recall that in both modelsEbracketleftbigCov(Xj|Θj)bracketrightbig=Ebracketleftbigσ2(Θj)bracketrightbigVj=s2w-1j10. . .00w-1j2. . .0............00. . .w-1jn,where the estimator ˆs2ofs2is more than half smaller in the regressionmodel, so there is less unexplained residual within-class variance left whenfitting Hachemeister’s model; that is good.
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68CHAPTER 6.CREDIBILITY REGRESSION MODELSThe comparison of the other variance component, the scalara=V[μ(Θ)]in B¨uhlmann–Straub’s model, to the corresponding value in Hachemeister’smodel is more complicated; the latter is an×ndiagonal matrix, becausethe premium changes at each period:Covbracketleftbigμ(Θj)bracketrightbig= CovbracketleftbigYβ(Θj)bracketrightbig=Y A Yprime.Using the plug–inˆAestimators above for Hachemeister’s data set we get thefollowing estimated getˆCov[μ(Θj)] =YˆA Yprime=298953289435893388924189144890478895088853887568865988562884328943619539496427974609849399527005598159302626036590469205358933949643099467025035053908575716109464717683207192375526388924279746702506075451258417623226620770132740377794281847418914609850350545125871962926671337132075547797548396188168448904939953908584176292667435719447643380962854718998094489478895270057571623226713371944767558154686377911889599910081050888559816109466207713207643381546866599179296905102018107131538875930264717701327554780962863779179297207102622108037113452568866260368320740377975485471911889690510262210833911405611977359885659047192377942839618998095999102018108037114056120075126094628846920575526818478816894489100810107131113452119773126094132415.Compare the diagonal variance values to ˆa= 89,639 in B¨uhlmann–Straub’smodel. We see that because of the reverted time scale, the matrix values forthe older periods are smaller than ˆa, but larger than it for the last 4 periods.Remark 6.1.The method could easily be extended to the case when thedesign matricesYjchange from one class to another. For instance, if someclass does not exhibit any inflation, thenpcould be set to 1 for those classes.One can also model linear inflation (that isp= 2) for some classes andquadratic for others (p= 3).
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