第四章净均衡保费与毛保费netpremiunandgrosspremium第一节保费简介introductiontopremium1、保费的构成benefitofthelifeinsurance(netpremium)initiatingexpensesgrosspremiumexpensesmaintainingexpensesprofitmarginsmarginsadversedeviation1、保费的构成纯保费(将来保单受益的精算现值)附加费用(与保单相关的费用的精算现值)毛保费(购买费用)2、保费的分类按保费缴纳的方式分(thewaythepremiumispaid)一次性缴纳:趸缴(纯/毛)保费single(gross/net)premium以年金的方式缴纳:期缴(纯/毛)保费(gross/net)premiumpaidbyinstallments按保险的种类分:只覆盖死亡的保险:纯寿险保费lifeinsurancepremium只覆盖生存的保险:生存险保费pureendowmentpremium既覆盖死亡又覆盖生存的保险:两全险保费endowmentpremium3、常见险种的趸缴纯保费纯寿险趸缴纯保费(死亡受益死亡即刻支付)生存险趸缴纯保费(一次性生存受益期末支付,生存年金受益期初支付)两全保险趸缴纯保费(死亡受益死亡即刻支付,生存受益期末支付)1:,,,xxxmmnxnAAAA1::,,,,xxxxmmnnxnAaaaa:xnA第二节净均衡保费netlevelpremium1、净均衡保费与趸缴纯保费的关系纯保费厘定原则——平衡原则:E(给付金现值)=E(纯保费现值)保险人的潜在亏损均值为零。假设保单签发时,被保险人的年龄为x,保险人的未来损失现值变量为LL=给付金现值-纯保费现值E(L)=0净均衡保费与趸缴纯保费的关系E(趸缴纯保费现值)=E(净均衡保费现值)2、净均衡保费的种类完全连续净均衡保费fullycontinuousnetlevelpremiums死亡即刻给付lifeinsurancepayableimmediatelyonthedeathof(x)连续缴费premiumpaidcontinuously完全离散净均衡保费fullydiscretenetlevelpremiums死亡年末给付lifeinsurancepayableattheendofthepolicyyearinwhichdeathoccurs离散缴费premiumpaidonanniversariesofthepolicyissuedatewhiletheinsuredsurvives半连续净均衡保费semi-continuousnetlevelpremiums死亡即刻给付lifeinsurancepayableimmediatelyonthedeathof(x)离散缴费premiumpaidonanniversariesofthepolicyissuedatewhiletheinsuredsurvives3、完全连续年缴净均衡保费的厘定(以终身人寿保险为例)条件:(x)死亡即刻给付1单位的终身人寿保险,被保险人从保单生效起按年连续交付保费。(给付连续,缴费也连续)厘定过程:假设在全连续模型中,单位保额的以平衡原理计算的终身寿险纯保费为,在保单签发时被保险人年龄为(x),被保险人未来剩余寿命变量为t,保险人的未来损失现值变量为,其中损失现值=未来死亡给付现值-未来缴纳保费现值,1()()(2)()0()0()txxtxxxxxxLlTvPAaAELAPAaPAa()()xpAxL3、完全连续年缴净均衡保费的厘定(一般情形)在一般情形下-()()xLEZZpYpEY未来死亡给付现值未来缴纳保费现值,根据平衡原理,我们有4、常见险种的完全连续净均衡保费总结险种保费公式终身人寿保险n年定期寿险n年两全保险h年缴费终身人寿保险h年缴费n年两全保险n年生存保险m年递延终身生存保险()xxxPAAa11()xnxnxnPAAa:::()xnxnxnPAAa:::()xxhxhPAAa:11::()xnxnxnPAAa:1::()xxxmmmxmPaAaa()hxnxnxhPAAa:::5、完全连续L方差的计算2222202222()()[()]()()()()(())()()()1L()[1]()()()[1]()[1]()()txxttxtxxtttttxxxtxxxxxVarLELELELLvPAaVarLELvpAapdtpApAvLvpAvpApAVarLVarvAApA对于终身寿险,计算过程复杂,换一个思路作将的表达式进行化简,将2222222222222()()()(1)()2()112()(1)2()()1()()1xxxxxxxxxxxxxxxxxxAAAAAVarLaAaaAaaaaVarLaaAa带入,例2.1已知利息力为0.06,死亡力为0.04,求)()2()()1(LVarAPx例2.1答案00222222()0.41110,()()0.092(1)()0.04()(2)()0.25()2tttttxxxxxxxxxxxAEvvedteedtAaAAAPAaAAVarLa所以例2.2设L1是对(x)的保额为1的完全连续终身寿险,在签单时的损失现值随机变量。其均衡纯保费由平衡原理决定,已知,设L2是对该保险交纳4/3倍的均衡纯保费时,在签单时的损失现值随机变量,计算L2的期望与标准差之和。150.08()0.5625xaVarL,,例2.2答案2222112112122222221()(1)()4133110.085()00.125115()()()0.120.233343()(1)()43(1)(()()(1)txxttttxxxxxtxxpLvpaVarLAALvpaLpaAaELpaaELELpEaappVarLAApVarLpVarL222224310.12)9360.080.122525(1)0.08436()0.56250.810.925()0.90.7VarLLEL的标准差=6、完全离散纯净均衡保费厘定(终身寿险为例)条件:(x)死亡年末给付1单位终身人寿保险,被保险人从保单生效起按年期初缴费。厘定过程:111111121222222(1),K0,1,2,(2)()001(3)()()()[(1)](1)()()(1)[()]()KxKxxxxxxkkkxxkkkxxxxxxxxxLvPaAELAPaPavVarLVarvPaVarvPdPPPVarvVarvdddPAAAAdda7、常见险种的完全离散净均衡保费总结险种保费公式终身人寿保险n年定期寿险n年两全保险h年缴费终身人寿保险h年缴费n年两全保险n年生存保险m年递延终身生存保险xxxxxNMaAP)()(11nxxnxxnxnxnxNNMMaAP:::)()(nxxnxnxxnxnxnxNNDMMaAP:::)(hxxxhxxxhNNMaAP:)(11nxxnxnxnxnxNNDaAP:::)()(:1:mxxmxmxmxmxmxxmNNNDaaAaP)()(hxxnxnxxhxnxnxhNNDMMaAP:::例2.3设一个0岁生命的整值剩余寿命服从概率函数为在其死亡年末赔付1单位的保单,每年年初缴付保费P。当保费按平衡原理决定时,计算保险人亏损现值的期望值与方差(i=6%)。3,2,1,0410kqk例2.3答案0.17788)()1(412)1(41])1[()((2)3667.0478.60]4)1[(410])1[(0)()1()1(2%64%36.12422106.043001111dPadPdPadPdPvdPELVarPdPdPadPqdPvdPLEdPvdPaPvLkkkkkkkexample2.435:20203555135:200.0420.02990.6099calculateifppApsolutions1135:2035:2035:20135:20203535:2035:2035:2035:20120|3520|35113535:20552035203535:2035:2035:2035:2035:2035:201203535:2035:20,0.0420.6ApaAEppaaAAAAAEpppaaaaEpxyaxyxy设,135:200.0110.0110990.02990.031xpy即example2.5Considera10000fullydiscretewholelifeinsurance.letdenoteanannualpremiumforthispolicyandL()denotetheloss-at-issuerandomvariableforonesuchpolicyonthebasisoftheillustrativelifetable,6%interestandissueage35DeterminethepremiumsuchthatthedistributionofL()hasamean0.calculatethevarianceofL()ApproximatethelowestpremiumsuchthattheprobabilitythelossL()ispositiveislessthan0.5.calculatethevarianceofL()Determinethepremiumsuchthattheprobabilityofapositivetotallosson100suchindependentpoliciesis0.05bythenormalapproximationsolutions1111111111111111111(2)()10000[()0]0.5[100000]0.510000[100000][0][(1)]0.5[][(1)lnln][kkkrrkkkrrkkkrkrrrLvapLpvapvapvapvddpvpkvpdd为单位保额对应的保费,则351111111450.06450.06222235ln1]0.5lnln1[]0.5lnln11441,10000ln1var(())(10000)[()][1]10000rdkvdhpkhhvdvSSLAAd令,为满足这一条件的最大整数查表得h=44,solutions10012100110010011100100113()()()100()()()()100()()0()Pr[0]0.05Pr[]0.05()()()0()(01)1.()()iiiiiiiiiisLLLLEsELELELVarsVarLVarLVarLsEsEssVarsVarssEsEsNVarsVars