ACTSC 331 Note : Life Contingency

Transcription

1 ACTSC 331 Note : Life Contingency Johnew Zhang December 3, 212 Contents 1 Review Survival Moel Insurance Annuities - same iea as insurance Variance Increasing contract Premium Policy values Retrospective Policy Value Thiele Differential Equation Avance topic: Asset shares an analysis of surplus Avance topics: Contracts where benefit is a% of t V Avance topics: policy alternations Review for Test Multiple state moels Notations Kolmogorov forwar equations Benefits in MSM Premiums a Policy Values in MSM Thiele DE for MSM Multiple ecrement moel, multiple life moel Multiple Decrement Moels (MDM) KFEs for a MDM Premiums an Policy Value in MDM Eample

2 4.3 Depenent an Inepenent Probabilities Eample Builing a MDM from SDMs Eample Multiple Life Functions Benefits for Multiple Lives Gompertz an Makeham Mortality Common Shock Moel Eample Avance Topics Interest Rate Risk Profit Testing Final Review 45 2

3 1 Review 1.1 Survival Moel This part of notes is in the stuy note for mlc. 1.2 Insurance b T = benefit payable if the life ies at time T Z = present value ranom var = PV of any payments on a contract Z epens on T or K or K (m) where K = T an K (m) = 1 m mt For insurance, we always have a ma of one payment, such as, for benefits payable on eath 1. Z = v T = e δt for whole life { v T if T < n 2. Z = for term if T n { if T < n 3. Z = v n for pure enowment if T n { v T if T < n 4. Z = v n for enowment insurance if T n { if T < n 5. Z = for a eferre contract something epens on the contract if T n If we annual benefits (payable at en of the year), use K + 1. For benefits pai of the en of the 1 m year of eath, use K(m) + 1 m. For any Z, we can always fin E[Z] = EP V = AV by first principle (E[Z] = zp (Z = z) = v t tp µ t t) - the sum or integral over all ates of the amount pai iscount factor probability of part. If we have a iscrete survival moel given by a life table, it is teious to calculate EPVs of whole life or long term contracts, so the values for whole life insurance are often inclue in the table. E[Z] = 1 v k+1 k q = A k= Also, the relationship between A s in the table is a recursion: Trivial relationships coul be observe A = A 1 :n + n E A +n = vq + E A +1 3

4 term + pure enowment = Enowment term + eferre whole life = whole life eferre = pure enowment the contract The relationships all hol within payment timing options. Relationship between ifferent timing (UDD) (pretty intuitive) 1.3 Annuities - same iea as insurance Y = PVRV as before, a function of T, K or K (m). We can also have annuities payable at the en (ue) or start (immeiate) of each perio. We can evaluate the EPVs with ä (m) first principles, amount iscount probability over all ates use the relationship to A s recall ä n = 1 vn EPV ä = 1 A. Similarly, ä :n = 1 A :n. Relationships:. Then Y = ä = 1 vk+1 K+1. Hence 1. annuity ue = 1 + annuity immeiate. In other wors, ä = 1 + a, ä (m) 2. Term: ä :n = 1 + a :n v n np 3. whole life = term + eferre, ä = ä :n + n ä 4. eferre = n E any contract for age + n = 1 m + a(m) To get relationship between annual, an mthly cases, we nee UDD. Iea: convert to A (m), use UDD on that, convert back to ä. Result: ä (m) = α(m)ä β(m), ä (m) :n = α(m)ä :n β(1 n E ) where α(m) = 1.4 Variance i i (m) (m), β = i i(m) i (m) (m) For insurance, it s easy to fin the secon moment of Z. E[Z 2 ] = same calculation as E[Z] but with v 2 instea of v = 2 A something. Then V ar(z) = 2 A A 2 for annuities, it s not easy to o this way because payments are not inepenent relationship. Instea, we use Y = 1 Z V ar(y ) = 1 2 V ar(z) = 2 A A 2 2 4

5 1.5 Increasing contract (IA) pays $k if they ie in the kth year. (Iä) pays $k + 1 at the kth year. 1.6 Premium Loss-at-issue RV L = PV future benefits PV future premiums To fin P by the equivalence principle, set E[L ] =, i.e. set P so that EPV premiums is equal to the EPV benefits. We can also inclue epense an profit margin. L g = PV future benefits + epenses PV future premiums Then set P g such that the EPV premium is equal to EPV benefits plus EPV epenses. Apparently, the gross premium is always higher than the net premium. Epenses can be fie or as a percentage of the premiums 2 Policy values Definition. The time t future loss RV is L t = PV at t of future benefits PV at t of future premiums conitional on the contract being in force at time t. If the contact has annual payments an t is an integer, then there may be a payment at t. The convention is to consier premium payment at time t to be in the future an benefits in the past. (i.e. P at t is in future premiums. S at t is not in future benefits.) For annuities (where benefit is a series of payments) it can be either way. For enowment insurance or pure enowment, the payment is at time n. But L n oes not inclue the time n payment as future benefit so L t =. But what we o have is lim t n L t = S = S for enowment insurance an pure enowment insurance. For term Eample lim L t = t n 5-year enowment insurance with annual premiums P, sum insure S payable at the en of year issue to (). { Sv K+1 P ä L = K+1 if K < 5 Sv 5 P ä 5 if K 5 5

6 We also know L 5 =. What about L 1? If the policy is in force at that point, the person is alive an age + 1. They have 4 years of benefit an 4 premiums left to pay. { Sv K+1+1 P ä L 1 = K+1 +1 if K +1 3 Sv 4 P ä 4 if K +1 4 Similarly we can efine L 2, L 3 an L 4. Now let s put in some number Makeham rule ω = 12, A =.1, B =.35, c = 1.75, i = 6%, = 5 With these parameters, we get P 5 = , A 5 = , A 51 =.34723, A 55 =.39449, 5 E 5 =.69562, 4 E 51 = To fin P, set E[L ] = = S(A 5 5 E 5 A E 5 ) P (ä 5 5 E 5 ä 55 ). Solving P Then E[L 1 K 5 1] = SA 51:4 P ä 51:4 = S(A 51 4 E 51 A E 51 ) P (ä 51 4 E 51 ä 55 ) = If + 1 is alive, their remaining premiums are not sufficient to cover their remaining benefits. The insurance company shoul hol capital in reserve to make up the short fall. Each year the epecte value of L t goes up for an enowment insurance. Insurer can buil up capital from early premium payments to cover the later benefit payments. Logically, the policy value at time t is the amount which, when combine with future premiums, will eactly cover the future benefits. In other wors, t V + EP V at t future premiums = EP V at t future benefits. Mathematically, t V = E[L t T > t] It s the amount that will, along with future premiums, cover future benefits. Where oes the $ come from? From other policies. say we have N ientical policies (from last eample - 5 year insurance). We collect N at. It earns 6% N at 1, but some people ie in age 5 51 an each get 1, at 1. The number who ie on average is Nq 5 =.1357N so we have N 135.7N = N. There are n policyholers still in force at time 1, so each has N N = which is eactly what we ha for E[L 1 T 5 > 1]. Fomr this to work at, we neee the same interest rate earne as assume mortality eperience to be the same as epecte. In reality, there are two versions of the policy value Net Premium Policy Value (NPPV) - EPV of the future benefits minus the premiums on the policy value basis with an artificial premium recalculate using the equivalence principle (no ep) an the policy value basis Gross Premium Policy Value (GPPV) - EPV of future benefits minus premiums on the policy value basis with actual gross premiums an incluing epenses 6

7 Two ifferences for these two versions coul be epenses vs non-epenses an actual vs artificial premiums Basis : the set of interest, mortality, an epense assumptions use in an actuarial calculation. Premium Basis - use to calculate P Policy Value basis - use to calculate t V If they are, an they assume no epenses, then GP P V = NP P V. In general, the policy value basis is more conservative than the premium basis. Premium basis nees to be realistic but competitive. Policy values are about ensuring solvency so the basis is more pessimistic. (worse mortality, lower interest rates, higher epenses) Eample Whole life $1, issue to (5). Premiums payable for 15 years ma. Basis: Markham rule, ω = 12, A =.1, B =.35, C = 1.75, 6% interest rate, 1% of premium plus $1 initial. P: 1A P ä 5:15 = P ä 5:15 then For gross, L 1 = P = 1A ä 5:15 = { 1, v K 6+1 P ä K6 +1 K 6 4 1, v K 6+1 P ä 5 L 6 5 L 2 = 1, v K 7+1 since no more premiums are ue If policy value basis is the same L g 1 = { 1, v K P ä K6 +1 K 6 4 1, v K P ä 5 K 6 5 1V g = E[L g 1 T 5 > 1] = 1, A 6.99P ä 6:5 = = The gross for L g 2 is the same as the net. 2V g = E[L g 2 T 5 > 2] = 1, A 7 = Now instea, assume the policy value basis is: same mortality, same epense, 5% interest. Then 1 V g = 1, A 6:5% ä 6:5 =

8 = > 1 V g 5%. More cautious assumption is going to result in higher policy values. Similarly, 2 V g = 1, A 7:5% = Back to Net, if we use policy value basis of 5%, we nee to calculate P (artificial premium).p is the theoretical premium that woul have been charge at time if we ha use the policy value basis, an no epenses. We nee 1, A 5 = P ä 5:15 5%. Then P = = 4.26 Then 1 V n = E[L 1 T 5 > 1] = 1, A 6 P ä 6:5 = ; 2 V n = same = Again if policy value basis is premium basis, where 2V n = V n = 1, A 6 P ä 6:5 P = 1, A 5 ä 5:15 Why artificial premiums? Consier E[L 1 T 5 > 1] where there were never any epenses is equal to SA 6 P ä 6:5 = SA 6:5% SA 6:6% ä 5:5 6% ä 6:5 5% There is no way to simplify this because the interest rates on t match. Theoretically, it s nicer if that the basis use to calculate P in the policy value is the same. i.e E[L 1 T 6 > 1] = SA 6 SA 5 ä 5:15 a 6:5 Take a general case of $1 whole life issue at (X). t V = E[L t T X > t] = A X+t A ä ä +t If everything is on the same basis Similarly for enowment insurance. = (1 ä +t ) ( 1 ä )ä +t = 1 ä+t ä ä tv = A +t:n t A :n ä :n ä +t:n t = 1 ä+t:n t ä :n We can only use these simplifications if with the same basis an premium assumption with no epenses. 8

9 2.1 Retrospective Policy Value If the policy value basis is equal to the premium basis an the eperience matches the assumptions (actual interest, mortality, an epenses are as preicte) then we can also epress t V in terms of cash flows from time to time t. tv = V + EP V at of premiums in (, t) EP V at of benefits in (, t) te Compare to prospective policy value tv = EP V at t of future benefits EP V at t of future premiums We on t actually use retrospective policy values much. We use asset shares instea which are base on the actual eperience in time to t. Proof. V = EP V B EP V P = EP V at of (, t) benefits + EP V at of (t, ) benefits EP V at of (, t) premiums EP V at of (t, ) premiums = EPV at of (, t) benefits (, t) premiums + (policy value at t brought back to time for interest an survival = t E t V ) Solving for t V gives the result. Recursion: we know A = vq + vp A +1 an ä = 1 + vp ä +1 tv = EP V at t of benefits Premiums = EP V at t of (t, t + s) benefits premiums + EP V at t of (t + s, ) benefits premiums = EP V at t of (t, t + s) benefit Premiums + E +tt+s V In particular, let s = 1, tv = EPV at t of net year of benefits premiums + vp +tt+1 V In general terms, let P t = Premium at t, e t = premium-relate epenses at t, S t+1 = sum insure payable at if K = t at t + 1, E t+1 = benefit-relate epenses at t + 1, i t = interest rate earne in (t, t + 1). then we can obtain the recursion ( t V + P t e t )(1 + i t ) = q +t (S t+1 + E t+1 ) + P +tt+1 V Policy value at t plus net income at t accumulate for 1 year is eactly enough to provie the eath benefit for those who ie an the t + 1 policy value for survivors. 9

10 Proof. Assume whole life, no epense, constant interest rate i. tv = SA +t P ä +t but we also know t+1v = SA +t+1 P ä +t+1 A +t = vq +t + vp +t A +t+1 so ä +t = 1 + v p +t ä +t+1 tv = S(vq +t + vp +t A +t+1 ) P (1 + vp +t ä +t+1 ) tv + P = Sq +t V + vp +t (SA +t+1 P ä +t+1 ) ivie both sies by v. Hence we get the recursion formula. In general for any contract, we can always erive the recursive relationship by splitting out the cash flow in the net year (or 1 m year) from the cash-flow from the onwars. What we have at t + what we get at t, accumulate for one perio must equal what we nee to provie then. At time t + 1 we must provie Policy value for t + 1 ( t+1 V ) enough etra to increase that to the benefit payable if the life ies The etra amount is S t+1 + E t+1 t+1 V which is calle NAAR (net amount at risk) or DSAR (eath strain at risk). The NAAP t+1 can be thought at as the sensitivity to mortality in the year t t + 1. Eample: 5-year iscrete enowment insurance to (5) usual mortality an 6% interest, no epenses. P was an 1 V was (Calculate between 1A 51:4 P ä 51:4 but we coul also get 1 V using recursion ( V )(1.6) = q 5 (1) + p 51 V so 1 V = Also then NAAR 1 = 1 1 V = Then for time 2, ( 1 V )(1.6) = q 51 (1) + p 512 V Hence 2 V = an NAAP 2 = Policy value is higher than NAAR is lower in the secon year. Similarly NAAR 3 = , 3 V = , 4 V = , NAAR 4 = , 5 V = but 5 V = 1, so over the entre contract, reserve goes up an NAAR goes own. For a 5-year term insurance instea, P = , 1 V = 2.14, 2 V = 31.69, 3 V = 33.27, 4 V = 23.31, 5 V = = 5 V ; NAAR s are huge compare enowment insurance. 1

11 Term insurance is much more sensitive to mortality risk then enowment insurance. Cash flow more frequent than usual. We coul calculate (or approimate using UDD), the premium an the policy value at any payment ate. At time t + s (t Z, s < 1), t+sv = EP V at t + sof future benefits premiums (e.g.) = SA (m) +t+s P ä(m) +t+s but we on t usually have A s an ä for non-integer ages. This is where recursions are useful. ( t V + P t e t )(1 + i t ) 1/m = 1 q +t (S m t+ 1 + E m t+ 1 ) + 1 p +t ( m m t+ 1 V ) m so that gives us t+ 1 V then m ( t+ 1 V + P m t+ 1 e m t+ 1 )(1 + i) 1 m = 1 q m m +t+ 1 )(S m t+ 2 + E m t+ 2 ) + 1 p m m +t+ 1 ( m t+ 2 V ) m For enowment insurance, NAAR ecrease over time (same for whole life). For term, NAAR is always large. Enowment insurance is less sensitive to mortality risk than term insurance is. On the other han, enowment insurance is more sensitive to interest rate risk than term insurance is, because the reserves hol are much larger. In any year of a contract at time t, suppose there are N policies in force. The insurer has N t V in reserves (they get N(P t e t )) The epecte etra amount neee at time t+1 to pay eath benefits is Nq +t NAAR t+1 The actual amount neee is the actual number of ollars NAAR t+1.the ifference Nq +t NAARt + 1(actual number of ollars Nq +t ) is the mortality loss (gain) in the year t t + 1. We looke at 1 mly payment contracts an everything is the same as annual. Recursion was ( t+ 1 V + P m + 1 e m + k )(1 + i) 1 m = 1 q m m +t+ k (S m t+ k+1 + E m t+ k P m m +t+ k ( m t+ k+1 V ) m But if benefit an premium payment frequencies are ifferent, we nee to leave out the corresponing term from the recursion on ates where no payment is mae. e.g. premiums semiannual, benefits monthly ( t V + P t )(1 + i) 1 12 = 1 q +t S 12 t p +t ( t+ 1 V ) 12 ( t+ 1 V + )(1 + i) 1 12 = 1 q t+ 1 S 12 t p t+ 1 ( 12 t+ 2 V ) 12 What if m an we have continuous payment. No new principles! At any time t, tv = EPV at time t of future benefits premiums given in force at time t We on t nee to worry about when benefits or premiums are payable since it s all continuous T is continuous = P (T = t) =. 11

12 Eample Whole life, no epenses, continuous tv = SĀ+t P ā +t L t T t = Sv T +t P ā T+t Hence we can fin P (L t > l T t) = Sv T +t P ( 1 vt ) = (S + P δ δ )vt +t P δ P ((S + P δ )vt +t P δ > l) = P (vt +t > l + P δ S + P ) δ = P ( δt +t > ln ( l + P δ S + P )) δ = P (T +t < ln (l + P δ ) ln (S + P δ ) ) = k q +t δ We coul o a similar proceure for a iscrete contract but it s not as nice. Also V ar(l t T t) = (S + P δ )2 ( 2 Ā +t Ā2 +t) This approach also works for enowment insurance. But not for term, eferre, or other cases where the RV governing the premiums is ifferent from the RV for benefits. Similarly, for n-year enowment insurance with premiums for n. Let H +t = min{t +t, n t} an then L t T t = Sv H +t P ā H+t so V ar(l t T t) = (S+ P δ )2 ( 2 Ā +t:n t Ā2 +t:n t ) an P (L t > l T t) = P (H +t k ) = P (T +t k an n t k ) { k q +t if k n t Hence it is ene up with if k > n t But if benefits an premiums have ifferent RVs, it s more complicate. For eample, term insurance { Sv T +t P ā L t T t = T+t T +t n P ā n T +t > n eferre L t T t = { P ā T+t T +t < n t v T +t P ā n t T +t n t 12

13 Then for V ar(l t T t) we coul nee V ar(p V benefits), V ar( PV Premiums) an their covariance L t = SZ t P Y t T t Then V ar(l t T t) = S 2 V ar(z t T t) + P 2 V ar(y t T t) 2SP Cov(Z t, Y t T t) Cov(Z t, Y t ) = E[Z t Y t T t] E[Z t T t]e[y t t] We can now fin the policy value (mean of L t ) or any payment ate (integer t for annual, multiple of 1 m for 1 mly, an any real t for continuous contract. But what about in between payment ate? No new principles! Still PV at time t of future benefits premiums. But A :t or a :t on t eist if t / Z nor o A :t (m) or a (m) for t not a multiple of 1 :t m. We can use the policy value at the nearest payment ate. Two approaches 1. start with t+1 V, iscount back (1 s) of a year for interest an survival; ajust for any income or outgo ue to events in (t + s, t + 1]. t+sv = t+1 V v 1 s 1 sp +t+s + Sv 1 s 1 sq +t+s 2. start with t V an accumulate for S of a year for interest an survival an ajust ue to events in (t, t + s] equivalently t+sv = ( tv + P t ) se +t t+sv = ( tv + P t )(1 + i) s sp +t] Sv1 s s q +t v 1 s sp +t S sq +t V 1 s sp +t In both cases we use an ajacent policy value, bring it to the correct time, an ajust for what i/i not happen in the time between. Eample Use illustrative life table 6%. Whole life insurance for (4). Fully iscrete S = 1. P = 1A 4 ä 4, Fin 2.25 V. 21V = 1A 61 P ä 61 = , 2 V = Uner UDD.75 p 6.25 = p 6 or 2.25V = v.75.75p p v.75 = = an.75q 6.25 = p 6 = V = ( 2V + P ).25 q 6 1v v.25.25p 6 =

14 Uner UDD.25 p 6 = = an.25 q 6 =.344 2V assumes that the P at time 2 is in the future 2+ɛ V assumes it is in the past. Over an entire whole life contract, the overall tren is increasing but there are iscontinuities at every payment ate. We can approimate t+s V by using interpolation between t V an t+1 V. But we nee to take into account the iscontinuities cause by the premiums In our eample, ( t V + P )(1 s) + ( t+1 V )s = t+s V 2.25V.75( 2 V + P ) +.25( 21 V ) = For more accuracy, we can also incorporate interest in our interpolation. In our eample t+sv = ( t V + P )(1 + i) s (1 s) + ( t+1 V v 1 s )s 2.25V =.75( 2 V + P )(1.6) ( 21 V )( ).75 = Thiele Differential Equation We can calculate t V for any t in continuous integer t in annual t + k m in 1 mly. Any time t in annual or 1 mly, we also have 2 approimations. Discrete contracts have iscontinuities ue to payments, but not for the continuous case. We can erive a ifferential equation to fin the rate of change of t V. We can then use the DE to approimate the change in t V for a small interval (t, t + t). Use the principle that t V = EPV of benefits premiums tv = (S t+n + E t+n )e t+u t δ ss u p +t µ +t+u u (P t+u e t+u )e t+u t δ ss u p +t u Note this allows δ to be a function of time if δ s = δ, s then it s e δu. But if not, e t+u t δ ss = v(t+u) v(t). Substitute r = t + u an u = r, t = r u. u = r = t, u = r =. tv = t up +t : t p u p +t = t+u p = r p so u p +t = r p tp ((S r +E r )µ +r (P r e r )) v(r) v(t) rp tp r = 1 v(t) t p ( Differentiate t V v(t) t p with respect to t an it will be equal to v(t) t p (p t e t (S t + E t )µ +t ) t ((S r +E r )µ +r (P r e r ))v(r) r p r 14

15 If we assume compoun interest, v(t) = e δt = v t equate, cancel out e δt tp t e δt tp t V = ( e δt tp (δ + µ +t )) t V + e δt tp ( t t V ) t t V = δ t V + (P t e t ) (S t + E t t V ) u +t Above represent the rat eof change of policy value at t. δ t V is the amount hel instant interest rate an (P t e t ) is the premium income instant rate minus epenses. For last part, if eath occurs, provie NAAR to pay benefits. If we also have bounary conitions for the DE ( V =?) then we can ientify all the contract etails. There is a one to one relationship between contracts an Thiele s DE s. Eample: 1. t t V = (1 t V )µ +t + δ t V with V = P. This is a whole life insurance with single premium of P at time. 2. t t V = (1 t V )µ +t +δ t V +P if t < n. t V = for t n an lim t n tv = n V = 2. continuously pai P throughout n years an benefit of 1, if ies before n, benefit of 2, if they survive. Ientify DE from contact etails an ientify contract from DE an bounary conitions. If the De was ientical ecept n V =, it woul be a term insurance contract with 1, on eath within n. 3. t t V = (S t V )µ +t = δ t V X with V = P. There is a single premium pai at time an eath benefit of S pai on eath. X is an annuity payment. Using Thiele to approimate t V, since a erivative is t+hv t V t t V = lim h h We can choose a small h an then use the right sie of the DE to approimate the change in policy value over any interval of length h for a continuous contract. This gives us: (1 + δh) t V + (P t e t )h t+h V + NAAR t hµ +t (more accurate if h is smaller) It is essentially the continuous recursion for a small step size h. We can isolate t V or t+hv (whichever we on t know) an solve. We nee a starting point (usually either or time n in a term/enowment insurance contract) an then we can work iteratively to fin the t V at any time that is a multiple of h. 15

16 Eample: $1 1 enowment insurance contract to (4), 6% interest Usual Makeham mortality µ +t = A + Bc +t where A =.1, B =.35, c = 1.75 We know the DE is t t V = δ t V + P (1, t V )µ 4+t 1 V = 1, where δ = log(1.6). However for P, we will set P so that V =. Each step, Solving for t V, t+hv t V h δ t V + P (1, t V ) µ 4+t tv = t+h V P h + 1, h µ 4+t hδ hµ 4+t 2.3 Avance topic: Asset shares an analysis of surplus Retrospective policy value(base on an assumptions) is tv R = V + EP V at of premiums in (, t) EP V at of benefits in (, t) te If policy value basis = premium basis an equivalence P is use, then it equals t V P = EPV of future benefits EPV of future premium. It represents the amount the insure nees ti have per policy to cover future obligations. Asset share is similar but base on actual eperience in time (, t) an represents the amount the insurer actually has per policy. If AS t < t V, we have a loss. If AS t > t V, we have a profit. AS t = t V is possible but it woul only happen if eperience eactly matche assumptions (unlikely). Eample: Usual Makeham moel, 6% interest 15-year eferre iscrete whole life insurance of $1, is issue to (5). Death benefit in eferre perio is return of premiums without interest. Epense, 15% of first P an 2% of other premiums an $1 on payment of benefit. First, calculate P using equivalence principle EPV premiums = EPV benefits + Epense P ä 5:15 = 1, 1 15 A %P + 2%P ä 5:15 + P (IA) 1 5:15 + 1A 5 Hence P = Net fin 5 V (some basis) 16

17 retrospective prospective recursion Retrospective: Prospective: 5V = 11, P ä 5:5.13P P (IA) 1 5:5 1A1 5:5 5E 5 = 11, P (IA) 1 55:1 + SP A1 55:1 + 1, 1E 55 A P a : 55:1 +1A 55 P ä 55:1 Recursively: ( V + P.13P )(1 +.6) = q 5 (P + 1) + P 51 V ( 1 V + P.2P )(1.6) = q 51 (2P + 1) + p 512 V Keep oing the same process. We will get the value for the year-5 reserve. Now suppose actual interest was 6%, 5.5%, 6.5%, 6%, 7% per year. Actual epense were 1% of first P, 1% of rest, an $5 on eath. Actual mortality was q =.14 for = 5, 51,, 54. Then we can calculate AS 5 = 11, In total, profit of per policy. t amount policy epenses at t 1 value at at at t 1 t 1 1 P 1 1%P 1 1 acc at t value (a 1 + P 1 e 1 )(1 + 6%) 2 rem 1 P 1%P 2 (a 2 + P 2 surv 1 e 2 )(15.5%)..... amount pai at t remaining survivors AS t (P + 5).14 accv1 b (2P + 5)(surv 1 ) (.14). accv 2 b 2 surv 1 (1.14).. rem 1 surv 1 rem 2 surv 2. We have a surplus of AS 5 5 V = Interest was higher, epense were lower, mortality was close. Analysis of surplus can tell us how much of the surplus (or loss) is cause by each of: epenses, mortality, interest. The iea is to change one factor at a time from assume to actual. Orer matters!. 17

18 tv (all assume ) = AS epense 5 (actuall epense assume, mort, int) mortality epense = AS5 (actuall epense, mort assume int) = AS 5 AS epense mortality epense 5 = 5 V = effect of epense an ASt AS epense 5 = effect of mortality. mortality epense Lastly AS 5 AS5 = effect of interest. By calculation, 365 surplus = 25 epenses 1 mortality interest. 2.4 Avance topics: Contracts where benefit is a% of t V Sometimes the benefit in an insurance contract be a percentage of the policy value at the time of eath. We can t calculate P irectly so we have to use recursion (or Thiele). Eample: fully iscrete, eath benefit is c t+1 V for t =, 1,, n 1, premium P t payable at times t =, 1,, n 1. Benefit in the very last year is n V = S. Recursion: where p +t = p +t + cq +t ( t V + P )(1 + i) = q +t (c t+1 V ) + p +t ( t+1 V ) = p +tt+1v multiply by t E = v t p p +1 p +t 1 p +tv t+1 V t V = P v t+1 t+1p t+1v t EtV = P t E Sum from t = to n 1, mile terms cancel. n 1 nenv EV = P v t tp t= then If n V = known number, then If c = 1, tv = V + P ä :n te P = n V n E ä :n tv = P ä t v t (p = 1) 18

19 2.5 Avance topics: policy alternations Sometime ofter inception, the policyholer may request to cancel (lapse) the policy. change premium terms Change the sum insure change the benefit type (from whole life/enowment insurance to something else) By law, if the policy has been in force for 2 + years, the insurer must provie some surrener value if the policy is lapse. The surrener value (or cash value) can be agree upon ahea of time (at t = ) some calculation base on the policy value at time t For eample some% of t V (< 1%) possibly fie epense subtracte. Some % of AS t. Let the cash value at time t be C t. If the policy lapses, policyholer receives C t at t. Otherwise, we ca easily calculate the new (moifie) terms of the contract using: C t +EPV t of future (moifie) premiums = EPV t of future (moifie) benefit+epense Eample Whole life to (4) S = 1,, interest rate in the long term is 6%. No epenses (fully iscrete). From the ILT, A 4 = , A 6 = , A 65 = , l 6 = , l 65 = Fin P = 1,A 4 ä 4 = Suppose at time 2, policyholer wants to moify the policy by 1. Surrener 2. stop policy premium but keep whole life 3. turn it into a 5 eferre annuity of X per year with a 1, eath benefit in eferre perio, an pay 5 more premium. The insurer calculates surrener value as C t = 9% of t V $1 epense. 2V = 1, A ä 6 = C 2 = 9% = They get $213.2 (not great for policyholer, less than contributors) 19

20 2. C 2 + = S A 6 + (have now + future premiums = future benefit + future epense) S = C 2 A 6 = C ä 6:5 = 1, A 1 6:5 + X 5 ä 6 + C 2 = 18.88(ä 6 5 E 6 ä 65 ) = 1, (A 6 5 E 6 A 65 ) + X( 5 E 6 ä 65 ) Solving for X, we get Review for Test 1 time t loss rv L t = PV at t of future benefit premiums E[L t T > t] = t V = EPV at t of benefits premiums gross v.s. net policy value. If the policy value basis = Premium basis an P is equivalence principle premiums, then net policy value has simplifie calculation t V = 1 ä+t ä t Prospective v.s. Retrospective EPV of premium (, t) EPV of benefit (, t) tv = te equal to prospective if same basis an equivalence principle recursive equation ( t V + P t e t )(1 + i) = q +t (S t+1 + E t+1 ) + p +t ( t+1 V ) or ( t V + P t e t )(1 + i) = t+1 V + NAAR t+1 q +t 1 where NAAR t+1 = S t+1 + E t+1 t+1 V is mortality risk. Some principles if m ly contract. Values between payment ates. accurate metho linear interpolation with or without interest Continuous contracts: same principles for everything. easier to fin the istance of L t ifferential equation instea of recursion t t V = δ t V + (P t e t ) NAAR t µ +t use as approimation for a small h: t+hv t V h 2 = same RS

21 3 Multiple state moels We are use to the moel Alive Dea 1 We use { the RV T = time until eath of (). But really, we have a continuous process if () is alive at time t Y (t) = 1 if () is ea at time t. Then T = ma{t : Y (t) = } (i.e. latest time t where Y (t) is still in the alive state). In general, there are policies that cannot be moelle with 2 states. Eamples: 1. Insurance which pays a ifferent amount for acciental eath than other eath. We nee Alive Acciental Death 1 Other Death 2 2. Disability (permanent)/ Critical illness insurance pays a lump sum benefit on becoming permanently isable or iagnosis. We nee Alive Disable/Ill1 Dea 2 3. Disability income policy may pay (non-permanent) or annuity while policyholer is isable. We nee Healthy Sick 1 Dea 2 21

22 4. Joint life insurance that epens on the survival or eath of more than one policyholers. We nee &y alive y alive 1 alive 2 both eath 3 5. Pension plan - employees may entitle to ifferent benefits base on how they eit the plan. We nee Active... 5 Disability Retirement 4 Retire 1 Dea 2 Withrawn 3 Assumptions 1. Always in star when the policy is purchase. All possibilities are counte in the sates of the moel. (i.e. it s not possible to leave the moel entirely). Can t be in 2 states at once. That means at any time t, Y (t) = {, 1, }. 2. Markov property (future movements of Y (t) epen only on the present state, not on the history before the presents). In some cases, this assumption may not be realistic. 3. P (2 or more transitions in (t, t + h)) = o(h) (a function is o(h) if lim h o(h) h = ). can t have 2 transitions at eactly the same time. 4. time until transition for all states is ifferentiable functions. 22

23 3.1 Notations tp ij = P (Y ( + t) = j Y () = i) = P (policyholers in state j at age + t given that they were in state i at ) i can be = j or j µ 1 tp ii = P (Y ( + s) = i, s t Y () = i) = P (policyholer stays in state i throughout age + t given in i at ) epening on the moel an the nature of state i, t p ii may or may not = t p ii They woul be equal if It is impossible to leave state i (i is an absorbing state e.g. eath), then t p ii an tp ii = 1. It is impossible to re-enter state i after leaving it. Hazar Rate: tp ij h + µ ij = lim h for i j = force of transition from state i to state j at age In the alive-ea moel, t p = t p = t p ; t p 1 = t q ; t p 11 = 1 = t p 11 ; t p 1 = µ. At any time t, Y (t) must equal one of its states, so n j tp ij Lemma 1. n p ij = hµ ij + o(h) Proof. µ ij = npij h = h p ij o(h) h since o(h) h, so hµ ij probability of going from i to j in a time interval of length h is hµ ij Lemma 2. t p ii = t p ii + o(t). Proof. = ; = 1 for any i. = n p ij o(h) so for a small h, the plus some error. tp ii = P ( in i at age + t in i at ) = P (stay in i in i at ) + P (leave an not come back in i at ) = t p ii + P (2 + transitions in (, t)) = t p ii + o(t) If we cannot leave i or cannot reenter i, then we on t nee the o(t) since t p ii = t p ii 23

24 Lemma 3. h p ii = 1 h j i µij + o(h) where j i µij is the total force of eit from i. Proof. 1 h p ii = 1 P (o not leave i in i at ) = P (o leave i by + t in i at ) = j i = j i = h j i hp ij + P (leave an come back in i at ) [hµ ij + o(h)] + P (2 + transitions in (, h)) µ ij + o(h) so h p ii = 1 h j i µij + o(h). j i µ represents the total force of transition at of state i at age. Recall in the 2-state moel that Similarly for MSM, to start, Notice t+hp ii t p ii h Let h t t p ii using the chain rule tp = e t µ +rr t+np ii = t p ii hp ii +t t+hp ii = t p ii (1 hµ i +t + o(h) t t p ii tp ii integrate from to t an eponentiate = t p ii µ i = t p ii µ i +t + t p ii o(h) h +t + = t (log( tp ii )) = µ i +t tp ii = e t µi +r r+c but since p ii = 1, so c =. Make sure when compare to the 2-state moel because with t p ii, we only care about i an not i. 24

25 3.2 Kolmogorov forwar equations In a general MSM for any two states i an j (where i can equal to j) t+hp ij = t p ij hp ij +t + tp ik hp kj +t k j Apply those lemmas t+hp ij t p ij h t+hp ij = t p ij ( h p jj + o(h)) + tp ik (hµ kj +t + o(h)) k j = t p ij (1 hµ j +t + o(h)) + h ( t p ik µ kj +t + o(h)) k j = t p ij h t p ij µ j +t + h tp ik µ kj +t + o(h) k j = h tp ij µ j +t h t t p ij = t p ij µ j +t + tp ik µ kj +t k j + h k j tpik µ kj +t + o(h) h h but µ j +t = k j µjk +t by efinition so we can write t t p ij = ( t p ik µ kj +t tp ij µ jk +t ) k j where this measures the change in probability i to j; for t p ik µ kj +t, it measures that go somewhere other than j an then transition to j; for t p ij µ jk +t, it goes to j but transition elsewhere at last secon. Let s consier the healthy-isable-an-eath eample. Since an 1 cannot be re-entere once left, t+hp 1 = t p 1 hp 11 +t + t p hp 1 +t tp = t p an so tp 11 = t p 11 t+hp 1 = t p 1 hp 11 +t + t p (hµ 1 +t + o(h)) 25

26 where µ i = µ 12 an t+hp 1 = t p 1 (e h µ12 +t+r r ) + (e t µ +r r )hµ 1 +t + o(h) t t p 2 = 1 ( t p k k= µ k2 +t t p 2 = t p µ 2 +t + t p 1 µ 12 +t µ 2k +t) t t p 11 = 2 k=,k 1 ( t p 1k µ k1 +t t p 11 µ 1k +t) = t p 11 µ 12 +t but t p 1 = µ 21 =, µ 1 =. Similarly, we can fin the KFEs for any probability an solve the DEs simultaneously for the t p ij s. This is ouble but teious. So we can use another approach = conition on the time of first transition (say r) an then integrate r from to t. Let to 1 be σ =.5, 1 to 2 be ν =.1 an to 2 be µ =.2. Constant force o not epen on age. tp = t p = e t (σ+µ)r = e t(σ+µ) = e.7t tp 11 = t p 11 = e tν = e.1t Obviously, t p 22 = 1 an t p 2, t p 21, an t p 1 =. Since at time t, the policyholer must be in some state {, 1, 2}, t p 12 = 1 e.1t t t t tp 1 = rp σ t r p 11 +rr = e.7r.5e.1(t r) r =.5e.1t e.6r r = 5 6 (e.1t e.7t ) so again since we must be in some state at time t, t p + t p 1 + t p 2 = 1 so t p 2 = 1 e.7t 5 6 (e.1t e.7t ) or we coul calculate t p 2 irectly t p 2 = P ( to 2 irectly)+ P ( to 1 to 1) = t r p µ 1r + P ( 1 2). P ( 1 2) = t rp σ t r p 12 +rr or t rp 1 ν 1r both integrals must evaluate to the same thing.then P ( 2 irectly ) + P ( 1 2) as up to the same thing we ha before for t p 2. With some moels, we can get epressions for t p 1j by conitioning on the time of first transition. We can fin an analytical epression in terms of t if the transition forces are constant, but with more comple moels an or transition forces that epens on + t, the KFEs give us valuable info. Eample: 26

27 Healthy Disable 1 Dea 2 because we can renter state an 1. then t p 1 = t s p µ 1 Instea we use KFEs t t p 1 = +st sp 11 2 k=, 1 +ss = t ( t p k µ k1 tp t p tp 11 t p 11 r p µ 1 +rt rp 11 +rr +t t p 1 µ 1k +t) = t p µ 1 +t t p 1 (µ 1 +t + µ 12 Similarly, we can obtain the KFEs for other t p ij s an solve. We can use the KFs an a small time step h to approimate t p ij s. (Why? If mortality was Makeham or other moels that epen on age, we coul nee to use numerical integration to evaluate the integrals) We have t t p ij = ( t p ik µ k+j +t tp ij µ jk +t ) k j +t) approimate with t+h p ij tp ij h for some h so t+hp ij = t p ij + h ( t p ik µ kj +t tp ij µ jk +t ) k j Start at t =, p ii = 1 an p ij = for j i. then we can fin h p ij Hence we can get 2h p ij from those h p ij s an the µ ij +h Smaller h = more accurate, well h = Benefits in MSM if we have the µ ij s. s, an then for 3h, 4h,. gives ecent accuracy. ā ij = EPV of an annuity that pays 1 per year payable continuously whenever in state j, given that they start at age in state i. Ā ij = EPV of 1 pai at the instant of each transition into state j, given that they start at age in state i. 27

28 We coul moify these benefits by making them iscrete (annual or 1 mly) or a a term of n to either benefit (no payments after time n) or a require payment only on the first transition to j for A or only on the first visit to j (for a). To calculate the EPV of any benefit in a MSM, we use the same principle ( )amt pai at t iscount factor prob of payment at t ā ij = A 1 :n ij = Ā 1 :n ij = n n 1 k= v k i j e δt tp ij t e δt l j kp il p lj +k ( t p il µ lj +t )t In both cases for A, we start in i but the transition into j (that triggers payment) can be from any state l j. 3.4 Premiums a Policy Values in MSM Some principles set P so EPV premiums = EPV benefits + epenses (at inception) t V is EPV of future benefits + epense EPV of future premiums (at time t). In alive an ea, t V was conitional on the policy being in force at t. Which is just being alive at age + t, but with MSM, there may be more than one way for a policy to be in force at time t. so t V (1) = EPV benefits premiums at t, given policyholer is in state i at t. tv (i) = if i is an absorbing state such as ea. Eample consier the health, isable an ea moel again. From to 1, σ =.5; from 1 to 2, ν =.1; from to 2, µ =.2 for age to + 2. We ha epressions for all t p ij s. Contract: premiums P payable continuously while healthy; annuity of B payable while isable. eath benefit S = 1, payable on eath from either state or 1. All 2 year term. δ = 6% no epense. Calculate P: P ā :2 = Bā1 :2 + SĀ1 2 :2 28

29 so ā ā :2 = :2 = 2 2 Ā 1 :2 2 = e δt tp t = e δt tp 1 t = at time 1, calculate t V (i) for all i. 2 2 e.6t e.7t t = e.6t (e.7t e.1t )t = e δt ( t p µ 2 +t + t p 1 µ 12 +t)t =.3342 P = Bā1 :2 + SĀ1 2 :2 ā :2 1V (2) = = V () = Bā 1 +1:1 + SĀ1 2 +1:1 P ā +1:1 ā 11 Ā 1 +1:1 12 = 1V (1) = Bā 11 +1:1 + SĀ :1 +1:1 = 1 1 e δt tp 11 +1t = e δt tp 11 +1µ tt = V () = V (1) = Huge ifference in 1 V epening on the state of the policyholer healthy lives are subsiizing isable lives. 3.5 Thiele DE for MSM t t V (i) = δ t V (i) + (P (i) t In our eample we have from last time, e (i) t ) B (i) t µ ij +t (S(ij) + t V (j) t V (i) ) j i t t V () = δ t V () µ 1 +t( + t V (1) t V () ) µ 2 +t(1, + t V () ) t t V (1) = δ t V (1) + 1, µ 12 +t(1, + t V (1) ) Just like before, we might be able to solve analytically for t V (i) s, or we can use a small time interval h an use the DEs to approimate. Bounary conitions usually at the en of contract. 29

30 4 Multiple ecrement moel, multiple life moel 4.1 Multiple Decrement Moels (MDM) Special case of MSM, one active state (state ) an n absorbing states (1,, n). So only one transition (ma) can occur! 1 Active n At time t, policyholer must still be in or have mae one transition to j (1, 2,, n) an then still be in. tp = t p an as always, p ii = 1, p ij = KFEs for a MDM (we can solve this DE to get t p = e t µ tp j = t t t p = +r r = e t sp µ j +s s n i=1 tp µ i +t = e t n j=1 µj +r r ). t t p j = t p µ j +t n j=1 µj +t r We efine t p = 1 t p = n j=1 tp j (not in at t) (similar to µ +t = n j=1 µj +t ) tp is the total probability of leaving active state by t. if the transition forces are simple enough, we can evaluate the integrals an get eact epressions t p j. But if we want accuracy, we can use a life table just like in alive ea. Recall: = starting age(integer), l = number of starting lives, l = number of still alive at age = l p an = l l +1 From the MDM case, we nee l = avg number of still in state at age = l ( p ) 3

31 (j) = l p j = number who go to j in age an + 1 = l ( p j ) Then = n j=1 (j) = l l +1. We nee a fractional age assumption to use the table. 1. UDD: Assume that for each ecrement in the table, t p j j = 1, 2,, n. 2. CFT: Assume that each force of transition within each year is constant. for j = 1, 2,, n. With CFT, µ +t = n j=1 µ j +t = µj, t < 1 µ j +t = n j=1 Thus, t p = 1 t p = 1 e t µ +r r = 1 e tµ In particular, p = 1 e µ. Consier s 1, j = 1,, n s sp j = rp µ j +r r = s µ j, t < 1 = µ e rµ µj r = µj µ But if we let e sµ = s p from before, so we have s p j it makes sense to be proper trivial to the force from to j) If we let s = 1, we get So finally, tp j = µj µ sp j p j = µj µ = pj p (1 p ) (p ) (1 (p ) s ) = µj µ = tp j, t 1 for (1 e sµ ) We can obtain p j, p an p from the MD table since they are annual. (1 ( 1 p ) s ) (logically, p = l +1 l, p = 1 p = l l +1 l p j 31 = j l = l

32 Awesomely, we get the eact same result for s p j if we assume UDD in the MDM (proof is in the supplementary notes). Compare to CFM result in Alive to Dea. Eample: Suppose we have n = 2 sp = s p = (p ) s = (p ) s sp 1 = s q = 1 (p ) s = (1 (p ) s ) In force Dea 1 Lapes 2 Fin 1. probability (65) is still in force at 67 l probability (65) lapses between 66 an 67 2p 65 = l 67 l 65 = contract over by 68 1 p 2 65 = (p 65)(p 2 66) = (65) ies before age p 1 65 = 3 p p 2 65 = 1 3 p 65 = 1 l 68 l 65 = p 1 65 = p (p 65)( 2 p 1 66) = 1 65 l 65 + l 66 l 65 From UDD,.2 p 1 66 =.2(p1 66 ) an From CFT,.2p 1 66 = p1 66 p 66 (1 (p 66 ).2 ) 32

33 4.2 Premiums an Policy Value in MDM Some principles, policy values are the same as in the alive-ea moel since there is only one active state Eample 3 year fully iscrete contract for (65), 1, at the en of year of ea. If they lapse or survive 3 years, return of 1 2 of total premiums pai. Calculate the annual premium P. No epenses, multiple ecrement table given, i = 8% EPV of premiums = P ( l 65 + l 66 v + l 67 v 2 l 65 ) = P EPV of benefits = 1, ( (1) 65 v + (1) 66 v2 + (1) 67 v3 l 65 = EPV ROP benefit = so using the equivalent principle, Hence P = P 2 (2) 65 v + P (2) 66 v2 + 3P 2 ((2) 67 + l 68)v 3 l 65 = P P = P 4.3 Depenent an Inepenent Probabilities tp j s are known as epenent probability since they assume that the other ecrements are present in the MDM Define the inepenent probability tp j = e t µj +r r (the eact result you shoul get in a moel with only ecrement j) Also efine t p j = 1 t p j Comparing t q j an t p j ; both are the probability of going from to j in t years but t p j epens on the other ecrements being available whereas tq j as sues j is the only possible ecrement. If we have epressions for µ j s, we can obtain eact inepenent probabilities. But to go from a iscrete MD table to associate Single Decrement Tables, we nee a fractional age assumption. Assume CFT, we know sp j sp = µj µ 33

34 where s 1 an with respect to SDMs tp = n tp =1 (Another justification ) so Let t = 1, p = e µ tp = e t µ +r r = an p j = e µj, thus µ j µ = log pj log p n tp =1 = s p j sp p j = (p ) pj /p Eample q65 1 = 1 p 1 65 = 1 ( l 66 ) ( (1) 65 )/(1 l 66 ) l 65 l 65 = l Builing a MDM from SDMs We can get eact epressions if we know µ j +t s for all j, but if we have iscrete tables, we nee a fractional age assumption. Recall tp j = e t µj +r r tp j = tq j = 1 t p j t rp µ j +r r tp = 1 t p = 1 e t µ +r r If we assume UDD in the multiple secrete table or CFT, we got p j = (p ) pj /p so we can take apart a MDM. To assemble a MDM from SDMs, we can just reverse the relationship p j = log pj log p 34 p

35 but remember t p = n i=1 tp j so in particular p = n j=1 pj an p = 1 n i=1 pi so t p j = log p j log n (1 i=1 pi This relationship hols if we assume CFT or UDD in the MDM. On the other han, we can assume each SDM has the UDD property. (i.e. t q j = tq j = t p j µ j +t = qj ) Then t t t n tp j = tp µ j +r r = rp 1 rp 2 rp n µ j +r r = qj tp j = q j (1 r qr i = q j i j t n i=1 p i (1 rq)r i i j rp i r i=1 = epens on n, q j egree n polynomial with coefficients from q i s Eample Take apart the MDM into two SDMs an increase lapses by 2%. Lastly, reassemble into a MDM. We have been assuming that the time until transition was ifferentiable (hence continuous), but it s possible in real life to have transitions occur iscretely. For eample, retirement at age 65 eactly lapses at policy anniversary etc. How o we incorporate iscrete transitions into a MDM? Easy! The iscrete ecrement(s) is not competing with other ecrements for lives uring each year. So if we have the inepenent probabilities for the iscrete ecrement, we inclue them at the en of the year base on the lives remaining from continuous ecrements. Use the same eample an the same ecrement table, but we are aing a thir ecrement retirement with the following probabilities: 5% of people reaching age 66 retire eactly then. 7% of people reaching 67 retire eactly then; the rest efinitely will retire reaching age of 68. l eact eact eactly

36 4.5 Multiple Life Functions &y alive y alive 1 alive 2 both eath 3 We have t p 11 y = t p11 y for all state i because no state can be reentere but it is possible to have, 1, or 2 transitions in a time perio of length t. Future lifetime RV framework, T = time until eath of () T y = time until eath of (y) Define T y = min{t, T y } (the joint life status (earliest eath of () an (y))) S y (t) = P r(t y > t) = t p y = P (min{t, T y } > t) = P (T > t, T y > t) If we assume T an T y are inepenent, then t p y = t p t p y We have t p y (both survive t years) an t q y = 1 t p y (at least one ies within t years) u tq y = ( u p y )( t q +u:y+u ) (both survive to u, at least one ies by u + t) Note that y is a status. Define T y = ma{t, T y } (the last survivor status (latest eath of () an (y))) Notice that T + T y = T y + T y so t p y = t p + t p y t p y an t q y = t q + t q y t q y tp y = t p y tq y = t p 1 y + t p 2 y + t p 3 y = t p y tp y = t p y + t p 1 y + t p 3 y tq y = 1 t p 3 y Also some new probabilities base on the orer in which the eaths occur tq 1y = ies first within t = tq 2y = ies secon within t = 36 t t tp yµ 2 +r:y+rr tp 1 yµ 13 +r:y+rr

37 tq 1y + t q 2y = t q tq 1 + t q 2 = t q y y y tq 1y + t q 2 = t q y y tq 2y + t q 2 = t q y y for p s, q s, A s, a s, etc. tp y = t p + t p y t p y 4.6 Benefits for Multiple Lives Insurance: Joint lit insurance (pays an first eath); last survivor insurance (pays on secon eath); contingent insurance (epens on orer). (all can be whole life, term, enowment insurance, an continuous, annual, or 1 m ly) Annuities: joint life annuity (pay until the first eath); last survivor annuity (pays until secon eath); reversionary annuity (pays to while alive, starting after the eath of y) In the continuous case, we can evelop epression for the EPVs of any of these benefits. joint whole life annuity ā y = e δt tp y t = e δt tp yt last survivor whole life annuity ā y = e δt tp y t = e δt ( t p e δt tp yt + e δt tp 1 yt + e δt tp 2 yt = ā y + ā + ā y y+ t p 1 y+ t p 2 y)t = Logical Eplanation ā y pays until the last eath, i.e. pays while either is alive. ā + ā y pays to each while alive (2 while both are alive) so subtract the etra ā y Reversionary annuity ā y (means to pay to y after eath of ) = e δt tp 2 yt ( must be ea an y alive to get pai) = e δt ( t p 2 y + t p yt e δt tp yt. Hence ā y = ā y ā y Logical Eplanation ā y pays y after is ea. ā y pays while y is alive (1 while both alive) so subtract the etra ā y. Āy = e δt tp y(µ 1 +t:y+t + µ 2 +t:y+t)t (either or y ying trig- Joint life insurance gers payment) Last survivor insurance triggers the payment) Āy = e δt ( t p 1 yµ 13 +t + t p 2 yµ 23 y+t)t (either or y ying last But if we have the 2-life moel with a common shock (the force of transition from 3 irectly). We woul just a t p yµ 3 +t:y+t to the probability of payment at t insie both the integrals. Ā y = Ā + Āy Āy 37

38 Contingent Insurance We can see Ā1 y = e δt tp yµ 2 +t:y+tt. Similarly, Ā 1 y = tp yµ 1 +t:y+tt. Ā 1y + Ā 1 y = Āy an Ā2 y = e δt tp 1 yµ 13 +tt an Ā 2 y = e δt tp 2 yµ 23 y+tt. We can see Ā 2y + Ā 2 y = Āy We can a a term to any of these benefits t integrate up to n instea of. We can have annual or 1 mly payments to sum instea of integrals. Special Case inepenent lives; not necessarily realistic because spouses usually have similar health an activity habits; also they spen more time together than any 2 ranom people, so they can be subject to accients. If lives are inepenent, that means µ 3 y = (no common shock). µ 1 y = µ y = µ 23 y µ 2 y = µ = µ 13 Then µ y = µ y = µ 1 y + µ 2 y = µ + µ y so t p y = t p y e t (µ +r+µ y+r )r = t p t p y = e t µ +r:y+r r = Eample: q =.2, q +1 =.25, q +2 =.3, q y =.3.q y+1 =.35, q y+2 =.4. Assuming () an (y) are inepenent, calculate 1. 2 p y = 2 p 2 p y = (1.2)(1.25)(1.3)(1.35) = p y = 2 p + 2 p y 2 p y = q y = p y q +1:y+1 = q y = Gompertz an Makeham Mortality Gompetz µ = Bc so if () an (y) are inepenent with same mortality, then µ y = µ + µ y = Bc + Bc y. Define w such that c w = c + c y (i.e. w = log(c +c y ) log c then µ y = Bc w an the joint life status y can be replace by the single life status w. w is the equivalent single age) Makeham µ = A + Bc so µ y = A + Bc + A + Bc y = 2A + B(c + c y ). Define v such that c + c y = 2c v. Hence µ y = 2(A + Bc v ) = µ v + µ v so v is the equivalent equal age y can be replace with vv. 38