风险模型与非寿险精算学 (33).pdf

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1、1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions Casualty Actuarial Science CS2 Actuarial Statistics 2 Ch.4 RISK MODELS 2 Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggre

2、gate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions Index 11 Aggregate claim distributions under proportional and excess of loss reinsurance 1.1 Proportional reinsurance 1.2 Excess of loss rein

3、surance 22 The individual risk model 33 Parameter variability/uncertainty 3.1 Introduction 3.2 Variability in a heterogeneous portfolio 3.2.1 Example 3.2.2 Example 3.3 Variability in a homogeneous portfolio 3.3.1 Example 3.4 Variability in claim numbers and claim amounts and parameter uncertainty 3.

4、4.1 Example 3.4.2 Example 44 Exam-style questions Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions Reference Institute and Facult

5、y of Actuaries. Subject CT6: Statistical Methods Course Notes (for the 2014 exams). Actuarial Education Company (ActEd), 2013. Institute and Faculty of Actuaries. Subject CT6: Statistical Methods Core Technical Core Reading (for the 2014 exams). Institute and Faculty of Actuaries, 2013. Stuart A. Kl

6、ugman, Harry H. Panjer, Gordon E. Willmot. Loss Models: From Data to Decisions (Second Edition). John Wiley & Sons, Inc, 2004, ISBN: 0471215775. Stuart A. Klugman, Harry H. Panjer, Gordon E. Willmot. Loss Models: From Data to Decisions (Forth Edition). John Wiley & Sons, Inc, 2012, ISBN: 1118315324.

7、 Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions Syllabus objectives I 1 Construct models appropriate for short term insurance c

8、ontracts in terms of the numbers of claims and the amounts of individual claims. 2 Describe the major simplifying assumptions underlying the models in (iii) 3 Derive the mean, variance and coeffi cient of skewness for compound binomial, compound Poisson and compound negative binomial random variable

9、s. 4 Repeat 3. for both the insurer and the reinsurer after the operation of simple forms of proportional and excess of loss reinsurance. Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model

10、3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance 1 Aggregate claim distributions under proportional and excess of loss reinsurance1.1 Proportional reinsurance The distribution of the number of claims involving the reinsurer is the sa

11、me as the distribution of the number of claims involving the insurer, as each pays a defi ned proportion of every claim. For a retention level (0 6 6 1), the ithindividual claim amount for the insurer is Xiand for the reinsurer is (1 - )Xi. The aggregate claims amounts are S and (1 - )S respectively

12、. Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance 1.2 Excess of loss r

13、einsurance The amount that an insurer pays on the i-th claim under individual excess of loss reinsurance with retention level M is Yi= min(Xi, M). The amount that the reinsurer pays is Zi= max(0, Xi M). Thus, the insurers aggregate claims net of reinsurance can be represented as SI= Y1+ Y2+ + YN and

14、 the reinsurers aggregate claims as SR= Z1+ Z2+ . + ZN Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsuran

15、ce1.2 Excess of loss reinsurance If, for example, N Poi(), SIhas a compound Poisson distribution with Poisson parameter , and the ithindividual claim amount is Yi. Similarly, SRhas a compound Poisson distribution with Poisson parameter , and the ithindividual claim amount is Zi. Note, however that i

16、f F(M) 0, as will usually be the case, then Zitakes the value 0. In other words, 0 is counted as a possible claim amount for the reinsurer. From a practical point of view, this defi nition of SR is rather artifi cial. The insurer will know the observed value of N, but the reinsurer will probably kno

17、w only the number of claims above the retention level M since the insurer may notify the reinsurer only of claims above the retention level. Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk mo

18、del3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance Example Example 1.1 The annual aggregate claim amount from a risk has a compound Poisson distribution with Poisson parameter 10. Individual claim amounts are uniformly distributed o

19、n (0,2000). The insurer of this risk has eff ected excess of loss reinsurance with retention level 1,600. Calculate the mean, variance and coeffi cient of skewness of both the insurers and reinsurers aggregate claims under this reinsurance arrangement. Casualty Actuarial Science CS2 Actuarial Statis

20、tics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance Solution Let SIand SR be as above. To fi nd ESI calculate EYi. Now EY

21、i = M 0 xf(x)dx + MP(Xi M) where f(x) = 0.0005 is the U(0,2000) density function and M = 1,600. This gives EYi = 0.0005x2 2 |M 0 + 0.2M = 960 so that ESI = 10EYi = 9,600 . Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss rein

22、surance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance To fi nd varSI calculate EY 2 i from EY 2 i = M 0 x2f(x)dx + M2P(Xi M) = 0.0005x3 3 |M 0 + 0.2M2 = 1,194,666.7 so that varSI = 10EY 2 i = 11,946,667.

23、Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance To fi nd the coeffi ci

24、ent of skewness of the insurers claims, calculate EY 3 i from EY 3 i = M 0 x3f(x)dx + M3P(Xi M) = 0.0005x4 4 |M 0 + 0.2M3 = 1,638,400,000 so that E(SI E(SI)3 = 10EY 3 i = 16,384,000,000 and the coeffi cient of skewness of SIis 16,384,000,000/(11,946,667)3/2= 0.397. Casualty Actuarial Science CS2 Act

25、uarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance To fi nd ESR note that the expected annual aggregate cla

26、im amount from the risk is ES = EX = 10 100 = 10,000 so that ESR = 10,000 - ESI = 400. To fi nd varSR calculate EZ2 i from EZ2 i = 2000 M (x M)2f(x)dx = 2000M 0 0.0005y2dy where y = x M = 0.0005y3 3 |2000M 0 = 10,666.7 so that varSR = 10EZ2 i = 106,667 . Casualty Actuarial Science CS2 Actuarial Stat

27、istics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance To fi nd the coeffi cient of skewness of the reinsurers claims, cal

28、culate EZ3 i from EZ3 i = 2000 M (x M)3f(x)dx = 2000M 0 0.0005y3dy where y = x M = 3,200,000 so that E(SR ESR)3 = 10EZ3 i = 32,000,000 and the coeffi cient of skewness of SRis 32,000,000/(106,667)3/2= 0.92. Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under p

29、roportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance The reinsurers aggregate claims can also be represented by SR= W1+ W2+ . + WNR where the random variable NRdenotes

30、the actual number of (non-zero) payments made by the reinsurer. For example, suppose that the risk above gave rise to the following eight claim amounts in a particular year: 403 1490 1948 443 1866 1704 1221 823. Then in formula (1.1) the observed value of N is 8, and the third, fi fth and sixth clai

31、ms require payments from the reinsurer of 348, 266 and 104 respectively. The reinsurer makes a “payment” of 0 on the other fi ve claims. In formula (1.2), the observed value of NRis 3 and the observed values of W1, W2and W3are 348, 266 and 104 respectively. Note that the observed value of SRis the s

32、ame (i.e. 718) under each defi nition. Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of

33、loss reinsurance Chapter 4 Section 1.2 shows that Wihas density function g(w) = f(w+M) 1F(M), w 0. To specify the distribution for SRas given in formula (1.2) the distribution of NR is needed. This is found as follows. Defi ne NR= I1+ I2+ . + IN where N denotes the number of claims from the risk (as

34、 usual). Ij is an indicator random variable which takes the value 1 if the reinsurer makes a (non-zero) payment on the j-th claim, and takes the value 0 otherwise. Thus NRgives the number of payments made by the reinsurer. Since Ijtakes the value 1 only if Xj M, Casualty Actuarial Science CS2 Actuar

35、ial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance P(Ij= 1) = P(Xj M) = , say, and P(Ij= 0) = 1 . Further, Ijh

36、as MGF MI(t) = expt + 1 - and by formula (3.7) NRhas MGF MNR(t) = MN(logMI(t). Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.

37、1 Proportional reinsurance1.2 Excess of loss reinsurance Example Example 1.2 Continuing the above example and using formula (1.2) as the model for SR, it can be seen that SRhas a compound Poisson distribution with Poisson parameter 0.2 10 = 2. Individual claims, Wi, have density function g(w) = f(w+

38、M) 1F(M) = 0.0005/0.2 = 0.0025, for 0 w 400 i.e. Wiis uniformly distributed on (0,400). EWi = 200 and EW2 i = 53,333.33 and EW3 i = 16,000,000, giving the same result as before. Thus, there are two ways to specify and evaluate the distribution of SR. Casualty Actuarial Science CS2 Actuarial Statisti

39、cs 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance Question 8.5 The aggregate claims from a risk have a compound Poisson d

40、istribution with parameter . Individual claim amounts(in pounds) have a Pareto distribution with parameters =3 and =1,000 . The insurer of the risk calculates the premium using a premium loading factor of 0.2( ie it charges 20% in excess of the risk premium).The insurer is considering eff ecting exc

41、ess of loss reinsurance with retention limit ?1,000. The reinsurance premium would be calculated using a premium loading factor of 0.3. The insurances profi t is defi ned to be the premium charged by the insurer less the reinsurance premium and less the claims paid by the insurer, net of reinsurance

42、. 1 Show that the insurers expected profi t before reinsurance is 100 . 2 Calculate the insurers expected profi t after eff ecting the reinsurance, and hence fi nd the percentage reduction in the insurers expected profi t. of the insurers profi t as a result of eff ecting the reinsurance. Casualty A

43、ctuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance Solution Let S be the aggregate

44、claim amount before reinsurance. Then: S = X1+ X2+ .XN Where X is Pareto(),and =3 and =1000. So: E(X) = 1 = 500and var(X) = 2 (1)2(2) = 750,000 So if the Poisson parameter (whose value is unknown) is , say ,we have: E(S) = 500 and : var(S) = E(X2) = (750,000 + 5002) = 1,000,000 The insurers expected

45、 profi t without reinsurance is equal to premiums minus expected claims. But if a loading factor of 500=600, and the expected profi t is 600-500=100. Casualty Actuarial Science CS2 Actuarial Statistics 2 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individua

46、l risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance Now consider the eff ect of the reinsurance . For each Xiwe have Xi= Yi+ Zi, where Yi= Xiif Xi 1000 and : Zi= 0if Xi 1000 So for the reinsurer, total aggregate claims are g

47、iven by SR= Z1+ Z2+ . + ZN, a compound Poisson distribution, where each Zihas the distribution given above. The reinsurance premium is given by 1.3E(SR),where: E(SR) = E(Z)E(N) = E(Z) Now: E(Z) = Z 1000 (x 1,000) 3 1,0003 (1,000 + x)4 dx Putting = x 1,000 in this integral: E(Z) = Z 0 3 1,0003 (2,000

48、 + u)4 du = ?1,000 2,000 ?3 Z 0 u 3 2,0003 (2,000 + u)4 du Recognising this integral as the mean of the Pareto(3,2,000) distribution, we see that: E(Z) = ?1 2 ?3 2000 3 1 = 125 and: E(SR) = 125 which gives a reinsurance premium of 1.3 125 = 162.5 Casualty Actuarial Science CS2 Actuarial Statistics 2

49、 1 Aggregate claim distributions under proportional and excess of loss reinsurance2 The individual risk model3 Parameter variability/uncertainty4 Exam-style questions1.1 Proportional reinsurance1.2 Excess of loss reinsurance Alternatively you could evaluate this integral using the substitution t = 1000 + x or using integration by parts. So the expected profi t with reinsurance is 600 162.5 E(S SR), since S SRis what the insurer pays net of reinsurance. But: E(S SR) = E(S) E(SR) = 375 So the expected profi t