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在精算科学和保险费率制定中，考虑到风险敞口可能是一场噩梦。不知何故，简单的结果是因为计算起来更加复杂，只是因为我们必须考虑到暴露是一个异构变量这一事实。
保险费率制定中的风险敞口可以看作是审查数据的问题（在我的数据集中，风险敞口始终小于1，因为观察结果是合同，而不是保单持有人），利息变量是未观察到的变量，因为我们必须为保险合同定价一年（整年）的保险期。因此，我们必须对保险索赔的年度频率进行建模。
在我们的数据集中，我们考虑索赔总数与总风险承担比率。例如，如果我们考虑泊松过程
考虑以下数据集，
> nombre=rbind(nombre1,nombre2)> baseFREQ = merge(contrat,nombre)
在这里，我们确实有两个感兴趣的变量，即每张合约的敞口，
> E <- baseFREQ$exposition
和（观察到的）索赔数量（在该时间段内）
> Y <- baseFREQ$nbre
无需协变量，可以计算每个合同的平均（每年）索赔数量以及相关的方差
> (mean=weighted.mean(Y/E,E))[1] 0.07279295> (variance=sum((Y-mean*E)^2)/sum(E))[1] 0.08778567
看起来方差（略）大于平均值（我们将在几周后看到如何更正式地对其进行测试）。
可以在保单持有人居住的地区添加协变量，例如人口密度
Density, zone 11 average = 0.07962411 variance = 0.08711477Density, zone 21 average = 0.05294927 variance = 0.07378567Density, zone 22 average = 0.09330982 variance = 0.09582698Density, zone 23 average = 0.06918033 variance = 0.07641805Density, zone 24 average = 0.06004009 variance = 0.06293811Density, zone 25 average = 0.06577788 variance = 0.06726093Density, zone 26 average = 0.0688496 variance = 0.07126078Density, zone 31 average = 0.07725273 variance = 0.09067Density, zone 41 average = 0.03649222 variance = 0.03914317Density, zone 42 average = 0.08333333 variance = 0.1004027Density, zone 43 average = 0.07304602 variance = 0.07209618Density, zone 52 average = 0.06893741 variance = 0.07178091Density, zone 53 average = 0.07725661 variance = 0.07811935Density, zone 54 average = 0.07816105 variance = 0.08947993Density, zone 72 average = 0.08579731 variance = 0.09693305Density, zone 73 average = 0.04943033 variance = 0.04835521Density, zone 74 average = 0.1188611 variance = 0.1221675Density, zone 82 average = 0.09345635 variance = 0.09917425Density, zone 83 average = 0.04299708 variance = 0.05259835Density, zone 91 average = 0.07468126 variance = 0.3045718Density, zone 93 average = 0.08197912 variance = 0.09350102Density, zone 94 average = 0.03140971 variance = 0.04672329
可以可视化该信息
> plot(meani,variancei,cex=sqrt(Ei),col="grey",pch=19, > points(meani,variancei,cex=sqrt(Ei))
圆圈的大小与组的大小有关（面积与组内的总暴露量成正比）。第一个对角线对应于泊松模型，即方差应等于均值。也可以考虑其他协变量
汽车品牌
也可以将驾驶员的年龄视为分类变量
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R语言贝叶斯广义线性混合（多层次/水平/嵌套）模型GLMM、逻辑回归分析教育留级影响因素数据
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让我们更仔细地看一下不同年龄段的人，
在右边，我们可以观察到年轻的（没有经验的）驾驶员。那是预料之中的。但是有些类别 低于 第一个对角线：期望的频率很大，但方差不大。也就是说，我们 可以肯定的 是，年轻的驾驶员会发生更多的车祸。相反，它不是一个异类：年轻的驾驶员可以看作是一个相对同质的类，发生车祸的频率很高。
使用原始数据集（在这里，我仅使用具有50,000个客户的子集），我们获得了以下图形：
由于圈正在从18岁下降到25岁，因此具有明显的经验影响。
同时我们可以发现有可能将曝光量视为标准变量，并查看系数实际上是否等于1。如果没有任何协变量，
Call:glm(formula = Y ~ log(E), family = poisson("log")) Deviance Residuals:Min 1Q Median 3Q Max-0.3988 -0.3388 -0.2786 -0.1981 12.9036 Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) -2.83045 0.02822 -100.31 <2e-16 ***log(E) 0.53950 0.02905 18.57 <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: 12931 on 49999 degrees of freedomResidual deviance: 12475 on 49998 degrees of freedomAIC: 16150 Number of Fisher Scoring iterations: 6
也就是说，该参数显然严格小于1。它与重要性均不相关，
Linear hypothesis test Hypothesis:log(E) = 1 Model 1: restricted modelModel 2: Y ~ log(E) Res.Df Df Chisq Pr(>Chisq)1 499992 49998 1 251.19 < 2.2e-16 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
我也没有考虑协变量，
Deviance Residuals:Min 1Q Median 3Q Max-0.7114 -0.3200 -0.2637 -0.1896 12.7104 Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) -14.07321 181.04892 -0.078 0.938042log(exposition) 0.56781 0.03029 18.744 < 2e-16 ***carburantE -0.17979 0.04630 -3.883 0.000103 ***as.factor(ageconducteur)19 12.18354 181.04915 0.067 0.946348as.factor(ageconducteur)20 12.48752 181.04902 0.069 0.945011
因此，假设暴露是此处的外生变量可能是一个过强的假设。
接下来我们开始讨论建模索赔频率时的过度分散。在前面，我讨论了具有不同暴露程度的经验方差的计算。但是我只使用一个因素来计算类。当然，可以使用更多的因素。例如，使用因子的笛卡尔积，
Class D A (17,24] average = 0.06274415 variance = 0.06174966Class D A (24,40] average = 0.07271905 variance = 0.07675049Class D A (40,65] average = 0.05432262 variance = 0.06556844Class D A (65,101] average = 0.03026999 variance = 0.02960885Class D B (17,24] average = 0.2383109 variance = 0.2442396Class D B (24,40] average = 0.06662015 variance = 0.07121064Class D B (40,65] average = 0.05551854 variance = 0.05543831Class D B (65,101] average = 0.0556386 variance = 0.0540786Class D C (17,24] average = 0.1524552 variance = 0.1592623Class D C (24,40] average = 0.0795852 variance = 0.09091435Class D C (40,65] average = 0.07554481 variance = 0.08263404Class D C (65,101] average = 0.06936605 variance = 0.06684982Class D D (17,24] average = 0.1584052 variance = 0.1552583Class D D (24,40] average = 0.1079038 variance = 0.121747Class D D (40,65] average = 0.06989518 variance = 0.07780811Class D D (65,101] average = 0.0470501 variance = 0.04575461Class D E (17,24] average = 0.2007164 variance = 0.2647663Class D E (24,40] average = 0.1121569 variance = 0.1172205Class D E (40,65] average = 0.106563 variance = 0.1068348Class D E (65,101] average = 0.1572701 variance = 0.2126338Class D F (17,24] average = 0.2314815 variance = 0.1616788Class D F (24,40] average = 0.1690485 variance = 0.1443094Class D F (40,65] average = 0.08496827 variance = 0.07914423Class D F (65,101] average = 0.1547769 variance = 0.1442915Class E A (17,24] average = 0.1275345 variance = 0.1171678Class E A (24,40] average = 0.04523504 variance = 0.04741449Class E A (40,65] average = 0.05402834 variance = 0.05427582Class E A (65,101] average = 0.04176129 variance = 0.04539265Class E B (17,24] average = 0.1114712 variance = 0.1059153Class E B (24,40] average = 0.04211314 variance = 0.04068724Class E B (40,65] average = 0.04987117 variance = 0.05096601Class E B (65,101] average = 0.03123003 variance = 0.03041192Class E C (17,24] average = 0.1256302 variance = 0.1310862Class E C (24,40] average = 0.05118006 variance = 0.05122782Class E C (40,65] average = 0.05394576 variance = 0.05594004Class E C (65,101] average = 0.04570239 variance = 0.04422991Class E D (17,24] average = 0.1777142 variance = 0.1917696Class E D (24,40] average = 0.06293331 variance = 0.06738658Class E D (40,65] average = 0.08532688 variance = 0.2378571Class E D (65,101] average = 0.05442916 variance = 0.05724951Class E E (17,24] average = 0.1826558 variance = 0.2085505Class E E (24,40] average = 0.07804062 variance = 0.09637156Class E E (40,65] average = 0.08191469 variance = 0.08791804Class E E (65,101] average = 0.1017367 variance = 0.1141004Class E F (17,24] average = 0 variance = 0Class E F (24,40] average = 0.07731177 variance = 0.07415932Class E F (40,65] average = 0.1081142 variance = 0.1074324Class E F (65,101] average = 0.09071118 variance = 0.1170159
同样，可以将方差与平均值作图，
> plot(vm,vv,cex=sqrt(ve),col="grey",pch=19,+ xlab="Empirical average",ylab="Empirical variance")> points(vm,vv,cex=sqrt(ve))> abline(a=0,b=1,lty=2)
一种替代方法是使用树。树可以从其他变量获得，但它应该是相当接近我们理想的模型。在这里，我确实使用了整个数据库（超过60万行）
树如下
> plot(T)> text(T)
现在，每个分支都定义了一个类，可以使用它来定义一个类。应该被认为是同质的。
Class 6 average = 0.04010406 variance = 0.04424163Class 8 average = 0.05191127 variance = 0.05948133Class 9 average = 0.07442635 variance = 0.08694552Class 10 average = 0.4143646 variance = 0.4494002Class 11 average = 0.1917445 variance = 0.1744355Class 15 average = 0.04754595 variance = 0.05389675Class 20 average = 0.08129577 variance = 0.0906322Class 22 average = 0.05813419 variance = 0.07089811Class 23 average = 0.06123807 variance = 0.07010473Class 24 average = 0.06707301 variance = 0.07270995Class 25 average = 0.3164557 variance = 0.2026906Class 26 average = 0.08705041 variance = 0.108456Class 27 average = 0.06705214 variance = 0.07174673Class 30 average = 0.05292652 variance = 0.06127301Class 31 average = 0.07195285 variance = 0.08620593Class 32 average = 0.08133722 variance = 0.08960552Class 34 average = 0.1831559 variance = 0.2010849Class 39 average = 0.06173885 variance = 0.06573939Class 41 average = 0.07089419 variance = 0.07102932Class 44 average = 0.09426152 variance = 0.1032255Class 47 average = 0.03641669 variance = 0.03869702Class 49 average = 0.0506601 variance = 0.05089276Class 50 average = 0.06373107 variance = 0.06536792Class 51 average = 0.06762947 variance = 0.06926191Class 56 average = 0.06771764 variance = 0.07122379Class 57 average = 0.04949142 variance = 0.05086885Class 58 average = 0.2459016 variance = 0.2451116Class 59 average = 0.05996851 variance = 0.0615773Class 61 average = 0.07458053 variance = 0.0818608Class 63 average = 0.06203737 variance = 0.06249892Class 64 average = 0.07321618 variance = 0.07603106Class 66 average = 0.07332127 variance = 0.07262425Class 68 average = 0.07478147 variance = 0.07884597Class 70 average = 0.06566728 variance = 0.06749411Class 71 average = 0.09159605 variance = 0.09434413Class 75 average = 0.03228927 variance = 0.03403198Class 76 average = 0.04630848 variance = 0.04861813Class 78 average = 0.05342351 variance = 0.05626653Class 79 average = 0.05778622 variance = 0.05987139Class 80 average = 0.0374993 variance = 0.0385351Class 83 average = 0.06721729 variance = 0.07295168Class 86 average = 0.09888492 variance = 0.1131409Class 87 average = 0.1019186 variance = 0.2051122Class 88 average = 0.05281703 variance = 0.0635244Class 91 average = 0.08332136 variance = 0.09067632Class 96 average = 0.07682093 variance = 0.08144446Class 97 average = 0.0792268 variance = 0.08092019Class 99 average = 0.1019089 variance = 0.1072126Class 100 average = 0.1018262 variance = 0.1081117Class 101 average = 0.1106647 variance = 0.1151819Class 103 average = 0.08147644 variance = 0.08411685Class 104 average = 0.06456508 variance = 0.06801061Class 107 average = 0.1197225 variance = 0.1250056Class 108 average = 0.0924619 variance = 0.09845582Class 109 average = 0.1198932 variance = 0.1209162
在这里，当根据索赔的经验平均值绘制经验方差时，我们得到
在这里，我们可以识别剩余异质性的类。
