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### Slides delta-2

1. 1. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesQuelques problèmes liés aux tables demortalité, et quelques réponsesactuariellesArthur CharpentierENSAE/ENSAI-CRESTGroupe de travail Protection Sociale, Assurance, AnnuitésParis Jourdain - Sciences EconomiquesSeptembre 20061
2. 2. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesStandard notations in life insuranceLet T denote the random variable representing the lifelength for a givenindividual. The residual lifetime, or time-until-death Tx denotes theresidual lifetime for an individual with age x,TxL= (T − x|T > x).Denote by tqx = P(Tx ≤ t) = P(T ≤ x + t|T > x) the probability thatindividual with age x dies within t years (the so-called mortality ratio), andtpx = 1 −t qx the associated survival probability. Denote ﬁnally qx =1 qxand px =1 px.In a dynamic approach, deﬁne Tx(t) the remaining life time of an x-agedindividual in calendar year t, qx(t) the probability that an x-agedindividual dies in calendar year t, and px(t) = 1 − qx(t).2
3. 3. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesStandard notations in life insuranceLet Lx,t denote the number of individuals aged x alive on year t. Dx,tdenotes the number of deaths.The force of mortality (or instantaneous morality rate) at age x duringcalendar year t isµx(t) = − log px(t), or tpx = exp (−µx(t)) .The expected remaining lifetime of an individual aged x in year t is thenex(t) deﬁned as ex(t) = E(Tx(t)), and ﬁnally, recall that the whole lifeannuity-due isax(t) =∞k=1νk· P(Tx(t) > k).3
4. 4. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesJoint life insurance“The lifetimes of married couples are parallel data of the type referred to asseveral individuals. A married couple does not have common risks owing togenetics, but they might have them owing to selection of the partner; forexample, a non-smoker might prefer a non-smoker, leading to smokingconcordance within pairs, but it might also be due to shared risks orlifestyle. The latter could be the case even though a non-smoker chooses asmoker, as they will both have the risk from the smoke, one as an activesmoker, one as a passive smoker. Furthermore, they share diet and thelocal environment. We have previously mentioned the event-relateddependence seen after the death of one partner. Probably short-termdependence is more important than long-term dependence in this case. If awidow survives a couple of years after the loss of her husband, her risk isapproximately back to normal risk. There is also a risk of common events,as couples are physically together and can die simultaneously in accidents.However, this accounts for only a small part of the events”.4
5. 5. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesHouggard (1999)5
6. 6. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesAgeObservedmortalityrate25 30 35 40 45 50 55 60 65 70 75 80 85 900.00.030.060.090.120.150.18FemalesWidowsMarried WomenAgeObservedmortalityrate25 30 35 40 45 50 55 60 65 70 75 80 85 90-0.010.040.080.120.160.200.24MalesWidowersMarried MenFigure 1: Mortality rates, Denuit & al. (1999).6
7. 7. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesIn life insurance, analogous of those ﬁnancial derivatives can be considered.Consider a husband and his wife, and denote by Tx and Ty the survival lifelengths, assuming that the man has age x and his wife y when they buy alife-insurance contract. Several contracts can be considered, where capitalCk is due each year k,• as long as the spouses are both still alive,g(Tx, Ty) =∞k=1vkCk1(Tx > k and Ty > k),• as long as there is a survivor, g(Tx, Ty) =∞k=1vkCk1(Tx > k or Ty > k).Note that Ck can be stochastic if the capital is indexed on a ﬁnancial asset,or if the income is indexed by some stochastic interest rate. The associatedpure premium, called annuities when Ck = 1, can be written respectively7
8. 8. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles(with standard actuarial notations)axy:n =nk=1vkP(Tx > k, Ty > k) and axy:n =nk=1vkP(Tx > k or Ty > k).(annuités vie-jointes et annuités au dernier survivant). Those contracts areusually built for an husband and his wife, i.e. contracts with more risks canbe considered if children are involved, or even higher when dealing withcollective insurance contracts. Deﬁne similarly widow’s pension annuity asax|y = ay − axy =∞k=1vkP(Ty > k|Tx > k)(called annuités de veuvage). This can also be writtenax|y = ay − axy =∞k=1vkkpy −∞k=1vkkpxywhere kpxy = P(Tx > k, Ty > k).8
9. 9. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesUnder the assumption of independence,ax|y = ay − axy =∞k=1vkkpy −∞k=1vkkpx ·k py.9
10. 10. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesBounds for annuities for dependent lifetimesFrom Tchen (1980), recall that if φ : R2→ R is supermodular, i.e.φ(x2, y2) − φ(x1, y2) − φ(x2, y1) + φ(x1, y1) ≥ 0,for any x1 ≤ x2 and y1 ≤ y2, then for any (X, Y ),E g(X−, Y −) ≤ E (g(X, Y )) ≤ E g(X+, Y +) ,where (X−, Y −) and (X+, Y +) are respectively contercomonotonic andcomonotonic versions of (X, Y ), i.e.P(X−≤ x, Y −≤ y) = max{P(X ≤ x) + P(Y ≤ y) − 1, 0},P(X+≤ x, Y +≤ y) = min{P(X ≤ x), P(Y ≤ y)},(the lower and upper Fréchet-Hoeﬀding bounds).Since those annuities satisfy supermodular conditions.10
11. 11. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFor the n-year joint-life annuity,axy:n =nk=1vkP(Tx > k, Ty > k) =nk=1vkkpxy.Thena−xy:n ≤ axy:n ≤ a+xy:n, wherea−xy:n =nk=1vkmax{kpx +k py − 1, 0}( lower Fréchet bound ),a+xy:n =nk=1vkmin{kpx,k py}( upper Fréchet bound ).11
12. 12. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesx = y = 30nn-yearjoint-lifeannuities10 20 30 40 50 60891011121314151617181920x = y = 40nn-yearjoint-lifeannuities10 20 30 40 50 60891011121314151617x = y = 50nn-yearjoint-lifeannuities10 20 30 40 50 607.58.59.510.512.013.515.0x = y = 60nn-yearjoint-lifeannuities10 20 30 40 50 607.58.08.59.09.510.511.5Figure 2: Bounds for axy:n , Denuit & al. (1999).12
13. 13. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFor the n-year last-survivor annuity,axy:n =nk=1vkP(Tx > k or Ty > k) =nk=1vkkpxy,where kpxy = P(Tx > k or Ty > k) =k px +k py −k pxy.Thena−xy:n ≤ axy:n ≤ a+xy:n, wherea−xy:n =nk=1vk(1 − min{kqx,k qy}) ( upper Fréchet bound ),a+xy:n =nk=1vk(1 − max{kqx +k qy − 1, 0}) ( lower Fréchet bound ).13
14. 14. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesx = y = 30nn-yearlast-survivorannuities10 20 30 40 50 608910121416182022x = y = 40nn-yearlast-survivorannuities10 20 30 40 50 6089101214161820x = y = 50nn-yearlast-survivorannuities10 20 30 40 50 608910111213141516171819x = y = 60nn-yearlast-survivorannuities10 20 30 40 50 608910111213141516Figure 3: Bounds for axy:n , Denuit & al. (1999).14
15. 15. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFor the widow’s pension annuity,ax|y = ay − axy =∞k=1vkkpy −∞k=1vkkpxy.Thena−x|y ≤ ax|y ≤ a+x|y, wherea−x|y = ay − axy =∞k=1vkkpy −∞k=1vkmin{kpx,k py}.( upper Fréchet bound ),a+x|y = ay−axy =∞k=1vkkpy−∞k=1vkmax{kpx+kpy−1, 0}.( lower Fréchet bound ).15
16. 16. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesAGE x=yWIDOW’SPENSION20 25 30 35 40 45 50 55 60 65 70 75 80 85 900.01.22.43.64.86.0AGE x=y+5WIDOW’SPENSION20 25 30 35 40 45 50 55 60 65 70 75 80 85 900.01.22.43.64.86.0AGE x=y-5WIDOW’SPENSION20 25 30 35 40 45 50 55 60 65 70 75 80 85 900.01.22.43.64.86.0Figure 4: Bounds for axy:n , Denuit & al. (1999).16
17. 17. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesA random vector (X, Y ) is positively dependent by quadrant ifP(X ≤ x, Y ≤ y) ≥ P(X ≤ x) × P(Y ≤ y) = P(X⊥≤ x, Y ⊥≤ y),or, equivalently,P(X > x, Y > y) ≥ P(X > x) × P(Y > y) == P(X⊥> x, Y ⊥> y)where (X⊥, Y ⊥) is an independent version of (X, Y ).In this case, the lower bound in Tchen’s theorem can be improved,E g(X⊥, Y ⊥) ≤ E (g(X, Y )) ≤ E g(X+, Y +) ,17
18. 18. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesNorberg (1989) & Wolthius (1994)Norberg (1989) proposed a 4 states Markovian process,STATE 1STATE 2STATE 3STATE 4• both are still alive,• the husband is dead, the wife is alive,• the wife is dead, the husband is alive,• both are dead.18
19. 19. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles• µ12(t) mortality rate of a married man, aged x + t,• µ13(t) mortality rate of a married woman, aged y + t,• µ24(t) mortality rate of a widowed man, aged x + t,• µ34(t) mortality rate of a widowed man, aged x + t.Wolthius (1994) proposed the following model,• µ12(t) = (1 − α12)µM (x + t)• µ13(t) = (1 − α13)µW (y + t)• µ24(t) = (1 + α24)µW (y + t)• µ34(t) = (1 + α34)µM (x + t)where µM and µW are respectively men and women mortality rates,modeled using Makeham’s formula,µ·(z) = A· + B·Cz· , where A·, B· > 0 and C· > 1.19
20. 20. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesUsing this model, Wolthius (1994) obtained any joint distribution, e.g.P(Tx > s, Ty > t) =p11(0, t) + p11(0, s)p12(s, t) if 0 ≤ s ≤ tp11(0, s) + p11(0, t)p12(t, s) if 0 ≤ t ≤ swhere pi,j(s, t) is the probability to be at state j at date t given at time s,we where at state i. Hence, for all 0 < s < tp11(s, t) = exp −ts[µ12(ω) + µ13(ω)]dωp22(s, t) = exp −tsµ34(ω)dω and p33(s, t) = exp −tsµ24(ω)dωso that p1i(·, ·) can be writtenp1i(s, t) = exp −ts[p11(s, ωµ1i(ω)pii(ω, t)]dω .Based on Belgian dataset, Denuit et al. (1999) obtained the following,20
21. 21. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesfor widow’s pension annuity ax|y,21
22. 22. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesAGE x=yWIDOW’SPENSION20 40 60 80 1000123456AGE y=x+5WIDOW’SPENSION20 40 60 80 1000123456independencemaximumminimumMarkovAGE y=x-5WIDOW’SPENSION20 40 60 80 1000123456Figure 5: ax|y with the Markovian model, Denuit & al. (1999).22
23. 23. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesThe copula approach Frees, Carrière & Valdez(1996)Copula models were considered for coupling remaining lifetimes,P(Tx ≤ s, Ty ≤ t) = C(P(Tx ≤ s), P(Ty ≤ t)).• Shemyakin & Youn (1999, 2001) considered Gumbel copula,C(u, v) = exp − {(− log u)α+ (− log v)α}1α, α ≥ 1,• Denuit et al. (1999) considered Mardia copula,C(u, v) =α22C−(u, v) + (1 − α2)C⊥(u, v) +α22C+(u, v), α ∈ [0, 1],• Frees, Carrière & Valdez (1996) considered Frank copulaC(u, v) = −1αlog 1 +(e−αu− 1)(e−αv− 1)e−α − 1, α ≥ 0.23
24. 24. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFor instance,ax|y = ay − axy =∞k=1vkkpy −∞k=1vkkpxy,wherekpxy = P(min(Tx, Ty} > k) =k px + −kpy − 1 + C(1 −k px, 1 − −kpy).Remark Shemyakin & Youn (2001) proposed to consider a copula whichdepends on the age diﬀerence x − y, i.e. C = Cx−y.24
25. 25. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesn x = y = 40n-yearjoint-lifeannuities10 15 20 25 30 35 40 45 50 55 608.09.511.012.514.015.517.0independenceminimummaximumMardian x = y = 50n-yearjoint-lifeannuities10 15 20 25 30 35 40 45 50 55 607.58.59.510.512.013.515.0independenceminimummaximumMardian x = y = 40n-yearjoint-lifeannuities10 15 20 25 30 35 40 45 50 55 608.09.511.012.514.015.517.0independenceminimummaximumGumbeln x = y = 50n-yearjoint-lifeannuities10 15 20 25 30 35 40 45 50 55 607.58.59.510.512.013.515.0independenceminimummaximumGumbelFigure 6: axy:n with copulas, Denuit & al. (1999).25
26. 26. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesn x = y = 40n-yearlast-survivorannuities10 15 20 25 30 35 40 45 50 55 6089101214161820independencemaximumminimumMardian x = y = 50n-yearlast-survivorannuities10 15 20 25 30 35 40 45 50 55 608910111213141516171819independencemaximumminimumMardian x = y = 40n-yearlast-survivorannuities10 20 30 40 50 608101214161820independencemaximumminimumGumbeln x = y = 50n-yearlast-survivorannuities10 20 30 40 50 6081012141618independencemaximumminimumGumbelFigure 7: axy:n with copulas, Denuit & al. (1999).26
27. 27. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesn x = y = 40n-yearlast-survivorannuities10 15 20 25 30 35 40 45 50 55 6089101214161820independencemaximumminimumMardian x = y = 50n-yearlast-survivorannuities10 15 20 25 30 35 40 45 50 55 608910111213141516171819independencemaximumminimumMardian x = y = 40n-yearlast-survivorannuities10 20 30 40 50 608101214161820independencemaximumminimumGumbeln x = y = 50n-yearlast-survivorannuities10 20 30 40 50 6081012141618independencemaximumminimumGumbelFigure 8: ax|y with copulas, Denuit & al. (1999).27
28. 28. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesMotivation: demographic evolutionFigures 9: graph of x → Lx,t/L0,t, i.e. survival function of T (remaininglifetime at birth), the so-called "rectangularization eﬀect",Figures 10-11: graph of x → log(mx) where mx(t) = −∂ log(Lx,t)/∂x, theinstantaneous mortality rate,Figures 12-13: graph of x → ex(t), expected remaining lifetime at age x,Life expectancyat birth (e0) at 40 (e40) at 65 (e65)Period Male Female Male Female Male Female1910 49.52 53.37 26.92 30.06 10.65 11.951930 54.35 59.34 27.88 31.88 11.24 13.131950 63.45 69.21 30.68 35.21 12.18 14.601970 68.38 75.84 32.25 38.46 13.02 16.781990 72.76 80.96 35.52 42.47 15.56 19.9328
29. 29. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles0 20 40 60 80 100Age (male)0.00.20.40.60.81.01870191019501990Figure 9: Evolution of x → Lx,t/L0,t in France, at diﬀerent periods t.29
30. 30. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFigure 10: Evolution of x → log(mx(t)) in France, at diﬀerent periods t.30
31. 31. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles0 20 40 60 80 100Age (male) - MORTALITY RATIO10-4.000010-3.000010-2.000010-1.0000100.00004579234579234568234579234579Q.1870Q.1880Q.1890Q.1900Q.1910Q.1920Q.1930Q.1940Q.1950Q.1960Q.1970Q.1980Q.1990Q.2000Q.2010Q.2020Q.2030Figure 11: Evolution of x → log(mx(t)) in France, at diﬀerent periods t.31
32. 32. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFigure 12: Evolution of ex(t) in France, including some projections.32
33. 33. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles60 70 80 90AGE (Male) - REMAINING LIFE EXPECTANCY0102030E.1870E.1880E.1890E.1900E.1910E.1920E.1930E.1940E.1950E.1960E.1970E.1980E.1990E.2000E.2010E.2020E.2030Figure 13: Evolution of ex(t) in France, including some projections.33
34. 34. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFigure 14: Expected remaining lifetimes (Ulpien (170-228), in Rome andSimpson (1710-1761), in the United Kingdom).34
35. 35. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesWhat is a mortality table ?“Au XVIIème siècle, une autre impulsion au Calcul des probabilités et à laStatistique vint d’Angleterre et de Hollande par l’étude de problèmesd’assurance qui se ramenait essentiellement à trouver la probabilité pourqu’une personne d’âge x vive encore n années” (J. Neveu).Mortality table were introduced in 1662 by Graunt (Natural and politicalobservations upon bills of mortality), and studied into details in 1693 byHalley.Note that the assumption of stationarity of mortality has been ﬁrstintroduced in 1756 by A. Deparcieux (Essai sur les probabilités de ladurée de vie humaine).A mortality table (see next slide, e.g. TD88-90) is simply a (normalized)number of persons still in live at age x. It has been build using INSEEdata, collected from 1988 and 1990 on the French male population. FromApril, 27th1993 it has to be used for death insurance pricing.35
36. 36. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesAge Lx Age Lx Age Lx Age Lx Age Lx0 100000 22 97987 44 93515 66 73075 88 141331 99129 23 97830 45 93133 67 71366 89 116252 99057 24 97677 46 92727 68 69559 90 93893 99010 25 97524 47 92295 69 67655 91 74384 98977 26 97373 48 91833 70 65649 92 57635 98948 27 97222 49 91332 71 63543 93 43506 98921 28 97070 50 90778 72 61285 94 32117 98897 29 96916 51 90171 73 58911 95 23158 98876 30 96759 52 89511 74 56416 96 16359 98855 31 96597 53 88791 75 53818 97 111510 98835 32 96429 54 88011 76 51086 98 74011 98814 33 96255 55 87165 77 48251 99 45312 98793 34 96071 56 86241 78 45284 100 26313 98771 35 95878 57 85256 79 42203 101 14514 98745 36 95676 58 84211 80 39041 102 7615 98712 37 95463 59 83083 81 35824 103 3716 98667 38 95237 60 81884 82 32518 104 1717 98606 39 94997 61 80602 83 29220 105 718 98520 40 94746 62 79243 84 25962 106 219 98406 41 94476 63 77807 85 22780 107 020 98277 42 94182 64 76295 86 19725 108 021 98137 43 93868 65 74720 87 16843 109 036
37. 37. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesDecrement analysis and stochastic modelsHistorically, starting from life tables, actuaries obtained probabilities,Life tables =⇒ survival probabilitiespx.In decrement analysis, the aim is to build life tables,Survival probabilities px =⇒ life tables .With n independent lives, the number of deaths within a year is B(n, qx)distributed.Over much of the life span, the mortality rate qx is small, and the numberof deaths observed - at a particular age - can be accurately approximatedby the Poisson distribution, P(???).37
38. 38. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesLexis diagramOn the so-called Lexis diagram, time is plotted on the x-axis, and the ageon the y-axis. Each individual is represented by a life line, parallel with theﬁrst bissectrice, i.e. from (t, 0) to (t + x, x) where t is the date at birth andx the age at death (see Figure 15).Longitudinal mortality tables allow to estimate the remaining life length,for a given individual, not based on present information (as in transversetables), but for future tendencies. Time is here the central notion, andappears here through three variables,• the age of individuals, denoted x,• the date of observation, denoted t,• the generation of individuals, denoted g, where g = t − x.The link between those notions can be visualized using the Lewis diagram.38
39. 39. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles1860 1900 1940 1980 2020Time (period of observation)020406080AgetTRANSVERSALt+1x+1xLONGITUDINALFigure 15: Lexis diagram, static vs. dynamic lifetables.39
40. 40. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesStandard notations and graduation modelsRecall that for static mortality (Bowers et al. (1997)),Gompertz (1825) suggested to model instantaneous mortality rates aslog(µx) = B · Cx.Makeham (1860) suggested to model instantaneous mortality rates aslog(µx) = A + B · Cx.The survival probability is then deﬁned aspx = exp −10µx+tdt = P(T > x + 1|T > x)40
41. 41. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesThe Lee-Carter modelµx(t) denotes the death rates at age x in calendar year t,κt denotes the index of mortality change,ax and bx denote some age speciﬁc constants,ax denotes some general pattern,bx denotes the relative speed of change at each age,εx,t denotes the residual (with mean 0 and variance σ2ε ),log µx,t = ax + bxκt + εx,t. (1)assumed to be i.i.d. and N(0, σ2).41
42. 42. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesOver-parameterization of the modelThe model is here over-parameterized since the structure is invariant underthe following transformations,(ax, bx, κt) → (ax, bx/c, cκt)(ax, bx, κt) → (ax − cbx, bx, κt + c)for any constant c.κt is determined up to a linear transformation,bx is determined up to a multiplicative constant,ax is determined up to a linear adjustmentFor normalization purpose, assume that the sums of the (bx)x and the (κt)tare respectively 1 and 0,tnt=t1κt = 0,Tx=1bx = 1.42
43. 43. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesEstimation of the Lee-Carter modelThe parameters are obtained by Ordinary Least Squares techniques,(ax, bx, κt) = argmin(ax,bx,κt) x,t(εx,t)2.In order to forecast future mortality, typically assume that (κt) is a randomwalk with a negative drift,κt = κt−1 + c + ut. (2)43
44. 44. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesA space-time model for mortalityThe Lee-Carter model can also estimated as a space-state model. Ifunobserved variable κt is a random walk with drift, the two group ofequations can be considered as a equations of a space-state model.The model contains n observation equations (one for each age) and oneunique state equation that explains the dynamics of unobserved variable:log µx(t) = ax + bxκt + εx,tκt = κt−1 + c + ut(see Gouriéroux and Monfort (1990)). Therefore, those models can beestimated and used in prediction by a strong recursive algorithm (e.g.Kalman ﬁlter), and they allow some non-identical variance-covariancematrix of forecasting errors.Remark: We expect that the age-speciﬁc mortality rates share a commonstochastic trend (i.e. are cointegrated).44
45. 45. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesEstimation of the Lee-Carter modelLet (Dx,t) denote the number of death at age x and time t, and (Ex,t)denote the matching person-years of exposure risk of death (seeErlandt-Jonhson and Johnson (1980)).Empirical mortality rates are themx,t =Dx,tEx,tHence,ax =1tn − t1 + 1tnt=t1log mx,t = logtnt=t1m1/hx,t (3)where h = tn − t1 + 1, is a least square error estimator.In Equation (1) , there is no observable variable on the right hand side, andtherefore ordinary regression cannot be performed.45
46. 46. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesEstimation of the Lee-Carter modelLee and Carter (1992) present an approximate method using regressiontechniques:1. Estimate ax as in Equation (3)2. Compute the matrix Z of the (log mx,t − ax)x,t and estimate (κt) and(bx) as the ﬁrst right and ﬁrst left singular vectors in the SingularVector Decomposition of Z, subject to constraint the κ 1 = 0 andb 1 = 1.3. Adjust the estimated κt so that, for each year tTx=1ex,t exp ax + bxκt =Tx=1dx,t.46
47. 47. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesEstimation of the Lee-Carter modelThe κt’s are re-estimated so that the resulting death rates (with ax andbx), applied to the actual risk exposure, produce the total number of deathsactually observed for year t.This avoids sizable discrepancies between predicted and actual death (seeLee (2000) for a discussion).In order to build up dynamic lifetables, estimations for (κt) are needed.Note that (κt) is clearly nonstationnary, hence, some ARIMA(p, 1, d)models can be ﬁtted, i.eκt = κt−1 − 0.498 + εt − 0.488εt−1 (male)κt = κt−1 − 0.791 + εt − 0.495εt−1 (female)where (εt) is a Gaussian white noise, εt ∼ N (0, σ = 1.538) andεt ∼ N (0, σ = 1.177) respectively. Therefore, forecast of (κt) can be done(see Figure 16 to 18).47
48. 48. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles10 30 50 70 90Age (x)-8-6-4-2a(LeeCarter)Figure 16: Evolution of the ax’s, for French males.48
49. 49. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles10 30 50 70 90Age (x)0.000.010.020.03b(LeeCarter)Figure 17: Evolution of the bx’s, for French males.49
50. 50. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles1950 1960 1970 1980 1990 2000Calendar Year (t)-50-30-101030Kappa(LeeCarter)Figure 18: Evolution of the κt’s, for French males.50
51. 51. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles1940 1960 1980 2000 2020 2040Annee calendaire (t)-5.5-5.0-4.5-4.0-3.5-3.0Makeham modelLinear forecastQuadratic forecastFigure 19: Forecasting log µx(t), linear or quadratic trend, at age 65.51
52. 52. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesFrom Lee Carter to a Poisson log-bilinear modelNote that Lee Carter method models the logarithm of the force ofmortality, but not the number of deaths, which might be the variable ofinterest in actuarial applications.Let Ex,t denote the exposure-at-risk at age x during calendar year t, andrecall that Dx,t denotes the number of death.The idea is to assume that Dx,t follows some distribution L, with expectedvalueE(Dx,t) = Ex,t × µx(t) = ax + bxκt.52
53. 53. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesModelling the number of deaths using a Poisson modelThe Poisson assumption is plausible to model the number of death per year(see Brillinger (1986) and Brouhns, Denuit and Vermunt (2002)).Assume thatDx,t ∼ P (Ex,t exp (ax + bxκt)) ,where parameters are still subject to the previous constraints.Parameters ax, bx and κt are estimated here maximizing the log-likelihood,i.e.L (a, b, κ) =x,t[Dx,t (ax + βxκt) − Ex,t exp (ax + βxκt)] + constant.Basic regression programs can not be used since the model is not linear(due to the bilinear term βxκt).Note that Goodman (1979) proposed an iterative method for estimatinglog-linear models with bilinear terms.53
54. 54. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesModelling the number of deaths using a Poisson modelThe idea is the following: at step k + 1, update parameters θ asθk+1 = θk −∂Lk∂θ θ=θk∂2Lk∂θ2θ=θk−1(starting here with values (a0, b0, κ0) = (0, 1, 0)). It comes that, ifDx,t,k = Ex,t exp ax,k + bx,kκt,kat step k + 1ax,k+1 = ax,k −Dx,t − Dx,t,k− Dx,t,k, bx,k+1 = bx,k and κt,k+1 = κt,k54
55. 55. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesModelling the number of deaths using a Poisson modelat step k + 2ax,k+2 = ax,k+1, bx,k+2 = bx,k+1 and κt,k+2 = κt,k+1−Dx,t − Dx,t,k+1 bx,k+1− Dx,t,k+1 bx,k+12at step k + 3ax,k+3 = ax,k+2, bx,k+3 = bx,k+2−Dx,t − Dx,t,k+2 κt,k+2− Dx,t,k+2 (κt,k+2)2and κt,k+3 = κt,k+255
56. 56. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesModelling the number of deaths using a Poisson modelRemark: In the Lee-Carter approach, the (κt)’s where ﬁrst estimated, andthen re-estimated to ﬁt with the total number of deaths observed per year.In the Poisson log-bilinear models, it is not the case since likelihoodequations ensure to obtain exactly the observed number of deaths, since∂L∂ax= 0 if and only iftD (x, t) =tL (x, t) exp (αx + βxκt)56
57. 57. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles1950 1960 1970 1980 1990−40−2002040Comparison of mortality indices (K)YearFigure 20: Comparing mortality indices kt’s, Poisson v.s. Lee-Carter.57
58. 58. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles1950 2000 2050 2100−200−150−100−50050Comparing forecasts of mortality indices (K)YearFigure 21: Comparing forecasts of mortality indices kt’s, Poisson v.s. Lee-Carter.58
59. 59. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesA short additional word on very old agesClassical mortality ratios are quite instable for very old ages, mainly due tothe lack of reliable data. It could then be all the more interesting tosmooth them.Note also that the maximal age is already a diﬃcult issue: 99 years old isundoubtedly enough (recall that Jeanne Calment survived up to 122 yearsold).Practitioners and biologists usually think that the exists an upper boundfor human lives. Could it be 125 years old ?Vaupel (1997) considered 70 millions individuals, from 14 countries, olderthan 80 (including more than 200, 000 older than 100). It pointed out thatprobability of death increase with an increasing rate.Coale & Kisker (1990) proposed the following extrapolation technique:µx = µ65 · exp(γx(x − 65)),59
60. 60. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielleswhere γx = γ80 + s(x − 80) for all x ≥ 80, where γ80 = log(µ80/µ65)/15.60
61. 61. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesWhat is the impact on life expectancies ?Recall that life expectancy, at time t for individuals of age x isex(t) =1 − exp (−µx(t))µx(t)+k≥1k−1j=0exp (−µx+j(t + j)) 1 − exp (−µx+k(t + k))µx+k(t + k)61
62. 62. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesYear 2000 (t)Prospective TPRV TVAge (x) Male Female50 32.54 39.63 37.74 32.9165 18.58 23.91 22.46 19.7580 8.01 10.51 9.39 8.61Year 2005 (t)Prospective TPRV TVAge (x) Male Female50 33.51 40.48 38.66 32.9165 19.35 24.71 23.35 19.7580 8.43 11.03 10.00 8.6162
63. 63. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuariellesWhat is the impact on annuities ?Consider an insurance contract, where 1 is due between the age of 50 and63
64. 64. Arthur CHARPENTIER - Quelques problèmes liés aux tables de mortalité, et quelques réponses actuarielles70 (rate = 3.5%):age TD TV Poisson Lee Carter age TD TV Poisson Lee C20 4.07 4.74 5.23 5.15 +28.5% +10.4% 33 6.6421 4.22 4.91 5.42 5.33 +28.3% +10.4% 34 6.8922 4.39 5.08 5.61 5.52 +27.9 % +10.3% 35 7.1623 4.55 5.27 5.80 5.71 +27.4 % +10.1% 36 7.4524 4.73 5.46 6.00 5.91 +26.9 % +10.0% 37 7.7425 4.91 5.65 6.21 6.11 +26.5 % +9.8% 38 8.0526 5.10 5.86 6.44 6.33 +26.4 % +10.0% 39 8.3727 5.29 6.07 6.67 6.56 +26.1 % +9.9% 40 8.7128 5.49 6.29 6.91 6.80 +25.8 % +9.9% 41 9.0729 5.70 6.52 7.15 7.04 +25.4 % +9.7% 42 9.4530 5.92 6.75 7.40 7.28 +24.9 % +9.6% 43 9.8431 6.15 7.00 7.65 7.54 +24.4 % +9.3% 44 10.2632 6.39 7.25 7.92 7.80 +24.0 % +9.2% 45 10.7164
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