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Department
Actuarial Science
Course
ACTSC 331
Professor
Zhenyu Cui
Semester
Winter

Description
ACTSC 331 Note : Life Contingency Johnew Zhang December 3, 2012 Contents 1 Review 3 1.1 Survival Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3. . 1.2 Insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 . . 1.3 Annuities - same idea as insurance . . . . . . . . . . . . . . . . . . . . . .4. 1.4 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4. . . 1.5 Increasing contract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 . . 1.6 Premium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5. . . 2 Policy values 5 2.1 Retrospective Policy Value . . . . . . . . . . . . . . . . . . . . . . . . . .9. . 2.2 Thiele Di▯erential Equation . . . . . . . . . . . . . . . . . . . . . . . . .14 . 2.3 Advanced topic: Asset shares and analysis of surplus . . . . . . . . . . . . 16 2.4 Advanced topics: Contracts where bene▯t is a% of Vt . . . . . . . . . . . .18 2.5 Advanced topics: policy alternations . . . . . . . . . . . . . . . . . . . . 19. 2.6 Review for Test 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20 . . 3 Multiple state models 21 3.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23 . . 3.2 Kolmogorov forward equations . . . . . . . . . . . . . . . . . . . . . . . . 25 3.3 Bene▯ts in MSM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27. 3.4 Premiums a Policy Values in MSM . . . . . . . . . . . . . . . . . . . . . . .28 3.5 Thiele DE for MSM . . . . . . . . . . . . . . . . . . . . . . . . . . . .29 . . 4 Multiple decrement model, multiple life model 30 4.1 Multiple Decrement Models (MDM) . . . . . . . . . . . . . . . . . . . . . . 30 4.1.1 KFEs for a MDM . . . . . . . . . . . . . . . . . . . . . . . . . 30. . 4.2 Premiums and Policy Value in MDM . . . . . . . . . . . . . . . . . . . . . . 33 4.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33. . 1 4.3 Dependent and Independent Probabilities . . . . . . . . . . . . . . . . . . .33 4.3.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34. . 4.4 Building a MDM from SDMs . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.4.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35. . 4.5 Multiple Life Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 36. . 4.6 Bene▯ts for Multiple Lives . . . . . . . . . . . . . . . . . . . . . . . . . 37. . 4.7 Gompertz and Makeham Mortality . . . . . . . . . . . . . . . . . . . . . . . 38 4.8 Common Shock Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.8.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40. . 5 Advanced Topics 41 5.1 Interest Rate Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41. . 5.2 Pro▯t Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42 . . 6 Final Review 45 2 1 Review 1.1 Survival Model This part of notes is in the study note for mlc. 1.2 Insurance bTx= bene▯t payable if the life dies at tixe T Z = present value random var = PV of any payments on a contract (m) (m) 1 Z depends on Txor K xr K x where K x bT cxand K x = m bmT x For insurance, we always have a max of one payment, such as, for bene▯ts payable on death 1. Z = vTx= e▯▯Tx for whole life ( Tx 2. Z = v if x < n for term 0 if T ▯ n x ( 0 if T < n 3. Z = x for pure endowment vn if x ▯ n ( vTx if x < n 4. Z = for endowment insurance vn if x ▯ n ( 0 if Tx< n 5. Z = for a deferred contract something depends on the contract if Tx▯ n If we annual bene▯ts (payable at end of the year), use K + 1. For bene▯ts paid of the (m) x end of the1 year of death, usexK + . For any Z, we can always ▯nd E[Z] = EPV = AV m P m R t by ▯rst principle (E[Z] = zP(Z = z) = v t x tt) - the sum or integral over all dates of the amount paid x discount factor x probability of part. If we have a discrete survival model given by a life table, it is tedious to calculate EPVs of whole life or long term contracts , so the values for whole life insurance are often included in the table. X1 E[Z] = 1 ▯ v+1kjx= A x k=0 Also, the relationship between A ’s in the table is a recursion: x A x A x:n+ En▯xA x+n = vqx+ E Ax x+1 Trivial relationships could be observed 3 ▯ term + pure endowment = Endowment ▯ term + deferred whole life = whole life ▯ deferred = pure endowment ▯ the contract The relationships all hold within payment timing options. Relationship between di▯er- ent timing (UDD) (pretty intuitive) 1.3 Annuities - same idea as insurance (m) Y = PVRV as before , a function of T ;x orxK x . We can also have annuities payable at the end (due) or start (immediate) of each period. We can evaluate the EPVs with ▯ ▯rst principles, ▯ amount ▯ discount▯ probability over all dates n Kx+1 ▯ use the relationship to A’s recall n = 1▯v . Then Y = a  = 1▯v . Hence 1▯A d K x1 d EPV ax= dx . 1▯A xn Similarly,x:n= d . Relationships: 1. annuity due = 1 + annuity immediate. In other words, a x= 1 + a x  xm)= 1 + a xm) m n 2. Term: ax: = 1 + a x: ▯ v n x 3. whole life = term + deferred,x= a x: + njx 4. deferred = E ▯ any contract for age x + n n x To get relationship between annual, and mthly cases, we need UDD. Idea: convert (m) (m) (m) x to A x , use UDD on that, convert back to a x. Result: ax = ▯(m)a  x ▯(m), (m) id i▯i(m) x:n = ▯(m)a x:n▯ ▯(1 ▯ En) xhere ▯(m) = i(m)d(m);▯ = i(m)d(m) 1.4 Variance ▯ For insurance, it’s easy to ▯nd the second moment of Z. 2 2 2 E[Z ] = same calculation as E[Z] but with v instead of v = A something Then V ar(Z) = 2A ▯ A 2 ▯ for annuities, it’s not easy to do this way because payments are not independent 1▯Z relationship. Instead, we use Y = d 2 2 V ar(Y ) = 12V ar(Z) = A▯2 d d 4 1.5 Increasing contract ▯ (IA) xays $k if they die in the kth year. ▯ (Iax pays $k + 1 at the kth year. 1.6 Premium Loss-at-issue RV L = 0V future bene▯ts ▯ PV future premiums To ▯nd P by the equivalence principle, set E[0 ] = 0, i.e. set P so that EPV premiums is equal to the EPV bene▯ts. We can also include expense and pro▯t margin. g L 0 PV future bene▯ts + expenses ▯ PV future premiums Then set P such that the EPV premium is equal to EPV bene▯ts plus EPV expenses. Apparently, the gross premium is always higher than the net premium. Expenses can be ▯xed or as a percentage of the premiums 2 Policy values De▯nition. The time t future loss RV is t = PV at t of future bene▯ts▯PV at t of future premiums conditional on the contract being in force at time t. If the contact has annual payments and t is an integer, then there may be a payment at t. The convention is to consider premium payment at time t to be in the future and bene▯ts in the past. (i.e. P at t is in future premiums. S at t is not in future bene▯ts.) For annuities (where bene▯t is a series of payments) it can be either way. For endowment insurance or pure endowment, the payment is at time n. But Lndoes not include the time n payment as future bene▯t so L t 0. But what we do have is lim L t S ▯ 0 = S t!n ▯ for endowment insurance and pure endowment insurance. For term t!nm▯L t 0 Example 5-year endowment insurance with annual premiums P, sum insured S payable at the end of year issued to (x). ( K +1 Sv x ▯ PaKx+1 if Kx< 5 L0= 5 Sv ▯ Pa 5 if Kx▯ 5 5 We also know L = 0. What about L ? If the policy is in forced at that point, the 5 1 person is alive and age x + 1. They have 4 years of bene▯t and 4 premiums left to pay. ( K +1 Sv x+1 ▯ Pa Kx+1+1 if Kx+1 ▯ 3 L1= 4 Sv ▯ Pa 4 if Kx+1 ▯ 4 Similarly we can de▯ne L 2L a3d L . N4w let’s put in some number Makeham rule ! = 120;A = 0:0001;B = 0:00035;c = 1:075;i = 6%;x = 50 With these parameters, we get P 50 = 0:986493;A 50= 0:335868;A 51 = 0:347203;A 55 = 0:394409; 5 50 = 0:690562; 4 51= 0:742018. To ▯nd P, set E[L ] = 0 = S(A ▯ E A + E ) ▯ P(a  ▯ E a  ). 0 50 5 50 55 5 50 50 5 50 55 Solving P + 1735:55 Then E[L 1K 50▯ 1] = SA 514 ▯Pa51:4 = S(A 51E 4 +51 55P(4 51 51 E4a51 55 = 1727:95 If x + 1 is alive, their remaining premiums are not su▯cient to cover their remaining bene▯ts. The insurance company should hold 1727.95 capital in reserve to make up the short fall. Each year the expected value of Ltgoes up for an endowment insurance. Insurer can build up capital from early premium payments to cover the later ben- e▯t payments. Logically, the policy value at time t is the amount which, when com- bined with future premiums, will exactly cover the future bene▯ts. In other words, V t EPV at tture premiums = EPV at tuture bene▯ts. Mathematically, t = E[L jTt> x] It’s the amount that will, along with future premiums, cover future bene▯ts. Where does the $ come from? From other policies. say we have N identical policies (from last example - 5 year insurance). We collect 1735.55N at 0. It earns 6% ! 1839.68N at 1, but some people die in age 50 ▯ 51 and each get 10, 000 at 1. The number who die on average is Nq50 = 0:013507N so we have 1839:68N ▯ 135:07N = 1704:61N. There are 0.986493N 1704:61N policyholders still in force at time 1, so each 0:986493N = 1727:95 which is exactly what we had for E[L jT > 1]. Fomr this to work at, we needed 1 50 ▯ the same interest rate earned as assumed ▯ mortality experience to be the same as expected. In reality, there are two versions of the policy value ▯ Net Premium Policy Value (NPPV) - EPV of the future bene▯ts minus the premiums on the policy value basis with an arti▯cial premium recalculated using the equivalence principle (no exp) and the policy value basis ▯ Gross Premium Policy Value (GPPV) - EPV of future bene▯ts minus premiums on the policy value basis with actual gross premiums and including expenses 6 Two di▯erences for these two versions could be expenses vs non-expenses and actual vs arti▯cial premiums Basis : the set of interest, mortality, and expense assumptions used in an actuarial calcu- lation. Premium Basis - used to calculate P Policy Value basis - used to calculate V t If they are, and they assume no expenses, then GPPV = NPPV . In general, the policy value basis is more conservative than the premium basis. Premium basis needs to be realistic but competitive. Policy values are about ensuring solvency so the basis is more pessimistic. (worse mortality, lower interest rates, higher expenses) Example Whole life $10,000 issued to (50). Premiums payable for 15 years max. Basis: Markham rule, ! = 120;A = 0:001;B = 0:00035;C = 1:075, 6% interest rate, 1% of premium plus $100 initial. P: 10000A 50+ 100 + 0:01Pa = P then 50:15 50:15 10000A 50+ 100 P = = 377:41 0:9950:15 ( K +1 10;000v 60 ▯ PaK 601 K 60▯ 4 L 10= K60+1 10;000v ▯ Pa5 L60 ▯ 5 K70+1 L 20= 10;000v since no more premiums are due For gross, ( K60+1 g 10;000v ▯ 0:99PaK 601 K60 ▯ 4 L10= K60+1 10;000v ▯ 0:99Pa5 K60 ▯ 5 If policy value basis is the same g g 10V = E[L 10 50> 10] = 10;000A 60:99Pa 60:5= 4568:085▯0:99▯377:41▯4:22367 = 2989:97 The gross for Lg is the same as the net. 20 20Vg = E[L jT 50 > 20] = 10;000A 70= 5861:87 20 Now instead, assume the policy value basis is: same mortality, same expense, 5% g interest. Then10V = 10;000A 60:5%▯ 0:99 ▯ 377:41 ▯a60:5 5107:311 ▯ 0:99 ▯ 377:41 ▯ 7 g 4:29763 = 3501:56 > 10V5%. More cautious assumption is going to result in higher policy g values. Similarly20V = 10;000A 70:5%= 7687:99 Back to Net, if we use policy value basis of 5%, we need to calculate P (arti▯cial premium).P’ is the theoretical premium that would have been charged at time 0 if we had used the policy value basis, and no expenses. We need 10;000A 50 = P 50:155%. Then 0 3908:23 P = 9:764268 = 400:26 n 0 n Then 10 = E[L j10 50 > 10] = 10;000A 60 ▯ P a60: = 3387:15; 20V = same = 7687:99 Again if policy value basis is premium basis, 20V n = 5861:87 n 00 10V = 10;000A 60▯ P a60:5 where 00 10;000A 50 P = 50:15 Why arti▯cial premiums? Consider E[L jT10 50] where there were never any expenses is equal to SA ▯ Pa = SA ▯ SA 60:6% 60 60:5 60:5%  60:55% 50:56% There is no way to simplify this because the interest rates don’t match. Theoretically, it’s nicer if that the basis used to calculate P in the policy value is the same. i.e SA E[L 10 60 > 10] = SA 60▯ 50a 60:5 50:15 Take a general case of $1 whole life issued at (Xt. V = E[t jX > t] = A X+t ▯ Axx+t x If everything is on the same basis 1 ▯ dx x+t = (1 ▯ d x+t) ▯ ( )x+t= 1 ▯ x x Similarly for endowment insurance. A x:n x+t:n▯t tV = A x+t:n▯t ▯ x+t:n▯t = 1 ▯ ax:n x:n We can only use these simpli▯cations if with the same basis and premium assumption with no expenses. 8 2.1 Retrospective Policy Value If the policy value basis is equal to the premium basis and the experience matches the assumptions (actual interest, mortality, and expenses are as predicted) then we can also express t in terms of cash ows from time 0 to time t. 0V + EPV at 0 of premiums in (0;t) ▯ EPV at 0 of bene▯ts in (0;t) tV = tE x Compare to prospective policy value tV = EPV at t of future bene▯ts ▯ EPV at t of future premiums We don’t actually use retrospective policy values much. We use asset shares instead which are based on the actual experience in time 0 to t. Proof. 0V = EPV B ▯ E0V P 0 = EPV at 0 of (0, t) bene▯ts + EPV at 0 of (t;1) bene▯ts ▯ EPV at 0 of (0, t) premiums ▯ EPV at 0 of (t;1) premiums = EPV at 0 of (0, t) bene▯ts ▯ (0;t) premiums + (policy value at t brought back to time 0 for interest and survival = E Vt)xt Solving for t gives the result. Recursion: we know A = xq + vpxA x x+1 and  x 1 + vp ax x+1 tV = EPV at t of bene▯ts ▯ Premiums = EPV at t of (t;t + s) bene▯ts ▯ premiums + EPV at t of (t + s;1) bene▯ts ▯ premiums = EPV at t of (t;t + s) bene▯t ▯ Premiums + E x x+tt+sV In particular , let s = 1, tV = EPV at t of next year of bene▯ts ▯ premiums + vp x+tt+1 V In general terms, let P =tPremium at t, e = pretium-related expenses at t, S t+1 = sum insured payable at if K =xt at t + 1, E t+1 = bene▯t-related expenses at t + 1, i =t interest rate earned in (t;t + 1). then we can obtain the recursion t V + P t e t(1 + i t = q x+t(S t+1+ E t+1 ) + Px+tt+1V Policy value at t plus net income at t accumulated for 1 year is exactly enough to provide the death bene▯t for those who die and the t + 1 policy value for survivors. 9 Proof. Assume whole life, no expense, constant interest rate i. tV = SA x+t▯ Pax+t t+1V = SA x+t+1 ▯ Pax+t+1 but we also know A = vq + vp A x+t x+t x+t x+t+1 x+t = 1 + v x+t x+t+1 so t = S(vq x+t+ vp x+tA x+t+1) ▯ P(1 + vpx+t x+t+1 ) t + P = Sq x+tV + vp x+t(SA x+t+1 ▯ Pax+t+1 ) divide both sides by v. Hence we get the recursion formula. In general for any contract, we can always derive the recursive relationship by splitting out the cash ow in the next year (or 1 year) from the cash- ow from the onwards. m What we have at t + what we get at t, accumulated for one period must equal what we need to provide then. At time t + 1 we must provide ▯ Policy value for t + 1 t+1V ) ▯ enough extra to increase that to the bene▯t payable if the life dies The extra amount is S t+1 + E t+1▯ t+1V which is called NAAR (net amount at risk) or DSAR (death strain at risk). The NAAP can be thought at as the sensitivity to t+1 mortality in the year t ! t + 1. Example: 5-year discrete endowment insurance to (50) usual mortality and 6% interest, no expenses. P was 1735.55 and V1was 1727.95. (Calculated between 10000A 51:4▯Pa 51: but we could also get 1 using recursion ( V + 1735:55)(1:06) = q (10000) + p V 0 50 501 so 1 = 1727:95. Also then NAAR = 10100 ▯ V = 8212:05. Then for time 2, (1V + 1735:55)(1:06) = q5110000) + p 512 Hence V2= 3578:16 and NAAP = 6422:81. Policy value is higher than NAAR is lower in the second year. Similarly NAAR = 4436:57; V = 5563:43; V = 7698:41;NAAR = 3 3 4 4 2301:59;5V = 0 but 5▯ V = 10000, so over the entre contract, reserve goes up and NAAR goes down. For a 5-year term insurance instead, P = 146:16;1V = 20:14; 2 = 31:69; V3= 33:27;4V = 23:31; 5 = 0 = 5▯V ; NAAR’s are huge compared endowment insurance. 10 Term insurance is much more sensitive to mortality risk then endowment insurance. Cash ow more frequent than usual. We could calculate (or approximate using UDD), the premium and the policy value at any payment date. At time t + s (t 2 Z;0 ▯ s < 1), V = EPV at t + sof future bene▯ts ▯ premiums (e.g.) = SA (m) ▯ Pa(m) t+s x+t+s x+t+s but we don’t usually have A’s and a  for non-integer ages. This is where recursions are useful. (tV + P t e t(1 + i t1=m = 1 qx+t(S 1 + E 1 ) + 1p x+t( 1 V ) m t+ m t+m m t+m so that gives ust+ 1V then m 1 (t+ 1V + P t+ 1 ▯ et+ 1)(1 + i)m = 1 qx+t+ 1)(St+ 2 + E t+ 2) + 1 px+t+ 1(t+ 2V ) m m m m m m m m m m For endowment insurance, NAAR decreased over time (same for whole life). For term, NAAR is always large. Endowment insurance is less sensitive to mortality risk than term insurance is. On the other hand, endowment insurance is more sensitive to interest rate risk than term insurance is, because the reserves hold are much larger. In any year of a contract at time t, suppose there are N policies in force. The insurer has NtV in reserves (they get N(P ▯e )) The expected extra amount needed at time t+1 to pay death bene▯ts t t is Nq NAAR x+t t+1 The actual amount needed is the actual number of dollars ▯ NAAR t+1.The di▯erence Nq x+tNAARt + 1(actual number of dollars ▯ Nq x+t) is the mortality loss (gain) in the 1 year t ! t+1. We looked at m ly payment contracts and everything is the same as annual. Recursion was 1 t+ 1 V + P x+ 1 ▯ ex+ k)(1 + i)m = 1 qx+t+ k(S t+k+1+ E t+ k+1+ 1 Px+t+ k (t+k+1V ) m m m m m m m m m m But if bene▯t and premium payment frequencies are di▯erent, we need to leave out the corresponding term from the recursion on dates where no payment is made. e.g. premiums semiannual, bene▯ts monthly 1 (tV + P t(1 + i)12 = 1qx+t S 1 + 1px+t( 1V ) 12 t+12 12 t+ 12 1 ( 1 V + 0)(1 + i)12 = 1q 1 S 1 + 1p 1( 2V ) t+ 12 12 x+t+ 12 t+12 12 x+t+ 12 t+12 ▯▯▯ What if m ! 1 and we have continuous payment. No new principles! At any time t, t = EPV at time t of future bene▯ts ▯ premiums given in force at time t We don’t need to worry about when bene▯ts or premiums are payable since it’s all contin- uous T xs continuous =) P(T = t)x= 0. 11 Example Whole life, no expenses, continuous ▯ tV = SA x+t▯ Pa▯x+t Tx L jT ▯ t = Sv Tx+t▯ Pa▯ = Sv Tx+t▯ P( 1 ▯ v ) = (S + P )vTx+t▯ P t x Tx+t ▯ ▯ ▯ Hence we can ▯nd P(L >tljT ▯ x) P P T P T l +▯ P((S + )v x+t ▯ > l) = P(v x+t > P) ▯ ▯ S + ▯ P l + ▯ = P(▯▯T x+t > ln( P )) S + ▯ P P ln(l + ▯) ▯ ln(S + ▯) = P(T x+t < ) =k▯qx+t ▯ We could do a similar procedure for a discrete contract but it’s not as nice. Also V ar(L jT ▯ t) = (S + P ) ( A ▯ A▯2 ) t x ▯ x+t x+t This approach also works for endowment insurance. But not for term, deferred, or other cases where the RV governing the premiums is di▯erent from the RV for bene▯ts. Similarly, for n-year endowment insurance with premiums for n. Let H x+t = minfT x+t;n ▯ tg H x+t P 2 2 ▯ ▯2 and then LtjTx▯ t = Sv ▯Pa▯H x+tso V ar(LtjTx▯ t) = (S+ )▯( A x+t:n▯t▯A x+t:n▯t) and ▯ ▯ ▯ P(L t ljT x t) = P(H x+t ▯ k ) = P(T x+t ▯ k and n ▯ t ▯ k ) ( k q if k ▯ n ▯ t Hence it is ended up with x+t 0 if k > n ▯ t But if bene▯ts and premiums have di▯erent RVs, it’s more complicated. For example, term insurance ( T Sv x+t ▯ Pa▯Tx+t Tx+t ▯ n L tTx▯ t = 0 ▯ P▯n Tx+t > n deferred ( 0 ▯ Pa▯Tx+t Tx+t < n ▯ t L tT x t = Tx+t v ▯ Pa▯ n▯t Tx+t ▯ n ▯ t 12 Then for V ar(L jt x t) we could need V ar(PV bene▯ts);V ar( PV Premiums) and their covariance Lt= SZ ▯tPY jT t tx Then 2 2 V ar(L tT x t) = S V ar(Z jTt▯ x) + P V ar(Y jT ▯ t) x 2SPCov(Z ;Y jT ▯ tt t x Cov(Z ;t )t= E[Z Y t t▯ x] ▯ E[Z jT ▯tt]x[Y ▯ t] t We can now ▯nd the policy value (mean of L )tor any payment date (integer t for annual, 1 1 multiple of m for m ly, and any real t for continuous contract. But what about in between payment date? No new principles! Still PV at time t of future benefits ▯ premiums. But (m) A x:tor ax:t don’t exist if = Z nor do A (m)or a for t not a multiple of 1 . x:t x:t m We can use the policy value at the nearest payment date. Two approaches 1. start with V , discount back (1 ▯ s) of a year for interest and survival; adjust for t+1 any income or outgo due to events in (t + s;t + 1]. 1▯s 1▯s t+sV = t+1V v 1▯s x+t+s + Sv 1▯s x+t+s 2. start with t and accumulate for S of a year for interest and survival and adjust due to events in (t;t + s] (tV + P t Sv 1▯ss x+t v1▯s t+sV = ▯ sE x+t s x+t equivalently s 1▯s (tV + P t(1 + i) S s x+tV t+sV = ▯ spx+t] s x+t In both cases we use an adjacent policy value, bring it to the correct time, and adjust for what did/did not happen in the time between. Example Use illustrative life table 6%. Whole life insurance for (40). Fully discrete S = 1000. 1000A40 P = 40 , Find 20:25V . 21V = 1000A 61 ▯ Pa61 = 264:061, 20V = 247:78. V = 264:061v 0:75 p + p 1000v 0:75= 260:065 20:25 0:75 60:25 0:75 60:25 p60 1▯0:01376 Under UDD 0:75 60:25= 0:25 60 1▯0:25▯0:0376= 0:989644 and 0:75 60:25 0:010356 or V = (20 + P) ▯ 0:25 60000v = 260:065 20:25 v0:250:25 60 13 Under UDD 0:25 60= 1 ▯ 0:25 ▯ 0:01376 = 0:99656 and 0:25 60= 0:00344 V assumes that the P at time 20 is in the future V assumes it is in the past. Over 20 20+▯ an entire whole life contract, the overall trend is increasing but there are discontinuities at every payment date. We can approximate t+sV by using interpolation between V tnd t+1V . But we need to take into account the discontinuities caused by the premiums (tV + P)(1 ▯ s) + ( t+1V )s = t+sV In our example, 20:25V ▯ 0:75( 20+ P) + 0:25( V 21= 260:016 For more accuracy, we can also incorporate interest in our interpolation. V = ( V + P)(1 + i) (1 ▯ s) + ( V v1▯s)s t+s t t+1 In our example 1 V = 0:75( V + P)(1:06) 0:25+ 0:25( V )( ):75= 260:04 20:25 20 21 1:06 2.2 Thiele Di▯erential Equation k 1 We can calculate Vtfor any t in continuous integer t in annual t + m in m ly. Any time t in annual or 1 ly, we also have 2 approximations. Discrete contracts have discontinuities m due to payments, but not for the continuous case. We can derive a di▯erential equation to ▯nd the rate of change of Vt. We can then use the DE to approximate the change in V t for a small interval (t;t + dt). Use the principle that t = EPV of bene▯ts ▯ premiums Z Z 1 Rt+u 1 Rt+u tV = (St+n + E t+n)e▯ t ▯sdsu x+t ▯x+t+u du ▯ (P t+u▯ e t+u)e▯ t ▯sdsu x+tdu 0 0 ▯▯u NotRt+uis allows ▯ to be a function of time if ▯ s = ▯;8s then it’s e . But if not, e▯ t ▯sds = v(t+u). Substitute r = t + u and du = dr, t = r ▯ u. u = 0 ! r = v(t) t;u = 1 ! r = 1. r x u x+t : t xu x+t = t+u x = pr x p u x+t = t x Z Z 1 v(r)r x 1 1 tV = ((Sr+E )r x+r▯(P ▯r ))r dr = ( ((Sr+E )r x+r▯(P ▯r )rv(r) pr x t v(t)t x v(tt x t Di▯erentiate t v(t) t xith respect to t and it will be equal to v(t)t x(pt▯ et▯ (S t E )▯t x+t) 14 ▯▯t t If we assume compound interest, v(t) = e = v d ▯▯t ▯▯t ▯▯t d e t xt = (▯e t x(▯ + ▯ x+t ))tV + e t x( t ) dt dt equate, cancel out e ▯▯t tp ▯ x d V = ▯ V + (P ▯ e ) ▯ (S + E ▯ V )u dt t t t t t t t x+t Above represent the rat eof change of policy value at t. ▯ V is the amount held x instant t interest rate and (P ▯ t ) it the premium income instant rate minus expenses. For last part, if death occurs, provide NAAR to pay bene▯ts. If we also have boundary conditions for the DE ( V0=?) then we can identify all the contract details. There is a one to one relationship between contracts and Thiele’s DE’s. Example: 1. dttV = ▯(1000 ▯ V )▯t x+t + ▯tV with V 0 P. This is a whole life insurance with single premium of P at time 0. 2. d tV = ▯(10000▯ V )▯t x+t +▯ Vt+P if 0 ▯ t < n. V = t for t ▯ n and lim ▯ tV = dt t!n n ▯V = 20000. continuously paid P throughout n years and bene▯t of 10,000 if dies before n , bene▯t of 20,000 if they survive. Identify DE from contact details and identify contract from DE and boundary condi- tions. If the De was identical except n▯V = 0, it would be a term insurance contract with 10,000 on death within n. 3. d V = (S ▯ V )▯ = ▯ V ▯ X with V = P. There is a single premium paid at dtt t x+t t 0 time 0 and death bene▯t of S paid on death. X is an annuity payment. Using Thiele to approximate V , stnce a derivative is d t+h V ▯ Vt tV = lim dt h!0 h We can choose a small h and then use the right side of the DE to approximate the change in policy value over any interval of length h for a continuous contract. This gives us: (1 + ▯h) Vt+ (P ▯ t )h t t+h V + NAAR h▯ t x+t (more accurate if h is smaller) It is essentially the continuous recursion for a small step size h. We can isolate V or t t+h V (whichever we don’t know) and solve. We need a starting point (usually either 0 or time n in a term/endowment insurance contract) and then we can work iteratively to ▯nd the t at any time that is a multiple of h. 15 Example: $10000 10endowment insurance contract to (40), 6% interest Usual Makeham mortality ▯x+t = A + Bc x+t where A = 0:0001;B = 0:00035;c = 1:075 We know the DE is d tV = ▯ t + P ▯ (10;000 ▯ V )t 40+t dt ▯ V = 10;000 10 where ▯ = log(1:06). However for P, we will set P so that V 0 0. Each step, t+h V ▯ t ▯ ▯ t + P ▯ (10;000 ▯ V t ▯ ▯ 40+t h Solving for t , V ▯ Ph + 10;000h ▯ ▯ t = t+h 40+t h▯ + 1 + h▯ 40+t 2.3 Advanced topic: Asset shares and analysis of surplus Retrospective policy value(based on an assumptions) is R 0V + EPV at 0 of premiums in (0;t) ▯ EPV at 0 of bene▯ts in (0;t) t = tE x P If policy value basis = premium basis and equivalence P is used, then it equals V t = EPV of future bene▯ts ▯ EPV of future premium. It represents the amount the insure needs ti have per policy to cover future obligations. Asset share is similar but based on actual experience in time (0;t) and represents the amount the insurer actually has per policy. If AS V t wet have a pro▯t. AS =tV it possible but it would only happen if experience exactly matched assumptions (unlikely). Example: Usual Makeham model, 6% interest 15-year deferred discrete whole life insur- ance of $100, 000 is issued to (50). Death bene▯t in deferred period is return of premiums without interest. Expense, 15% of ▯rst P and 2% of other premiums and $100 on payment of bene▯t. First, calculate P using equivalence principle EPV premiums = EPV bene▯ts + Expense Pa = 100;000 ▯ 100 A 50+ 13%P + 2%Pa  + P(IA) 1 + 100A 50 50:15 15j 50:5 50:15 Hence P = 2038:16. Next ▯nd V (s5me basis) 16 ▯ retrospective ▯ prospective ▯ recursion 5V = 11;612:70 Retrospective: 0:98Pa ▯ 0:13P ▯ P(IA) 1 ▯ 100A 1 50: 50:50 50:5 = 11;612:70 5E 50 Prospective: P(IA) 55:10+ SPA 55:10 + 100;000 10A 55 65 + 0:02P a :5510 +100A 55▯ Pa 5510 Recursively: (0V + P ▯ 0:13P)(1 + 0:06) = q (P 50100) + P 501V ( V + P ▯ 0:02P)(1:06) = q (2P + 100) + p V 1 51 512 Keep doing 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 expense were 10% of ▯rst P, 1% of rest, and $50 on death. Actual mortality was q = 0:014 xor x = 50;51;▯▯▯ ;54. Then we can calculate AS = 11;579:98. In total, pro▯t of 365.28 per policy. t amount policy expenses acc value amount paid at remaining survivors AS t at t▯1 value at at at t t at t▯1 t ▯ 1 1 0 P ▯ 1 10%P ▯1 (a1+ P ▯1 (P +50)▯0:014 accv1 ▯ db1 1▯0:014 rem 1 surv1 1 e1)(1 + 6%) rem 2 2 rem 1 P ▯ 1%P 2 (a2+ P ▯2 (2P + accv 2 db 2 surv 11▯ surv2 surv 1 e2)(15:5%) 50)(surv 1 ▯ 0:014) (0:014) . . . . . . . . . . . . . . . . . . We have a surplus of AS ▯ V5= 365:28. Interest was higher, expense were lower, mortality was close. Analysis of surplus can tell us how much of the surplus (or loss) is caused by each of: expenses, mortality, interest. The idea is to change one factor at a time from assumed to actual. Order matters!. 17 V (all assumed ) =) AS expense(actuall expense assumed, mort, int) t 5 mortality expense =) AS 5 (actuall expense, mort assumed int) =) AS 5 expense mortality expense expense AS 5 =) 5V = e▯ect of expense and AS t ▯AS 5 = e▯ect of mortality. mortality expense Lastly AS 5 AS 5 = e▯ect of interest. By calculation, 365 surplus = 250 expenses ▯ 10 mortality + 125 interest. 2.4 Advanced topics: Contracts where bene▯t is a% of V t Sometimes the bene▯t in an insurance contract be a percentage of the policy value at the time of death. We can’t calculate P directly so we have to use recursion (or Thiele). Example: fully discrete, death bene▯t is c V for t = 0;1;▯▯▯ ;n▯1, premium P payable t+1 t at times t = 0;1;▯▯▯ ;n ▯ 1. Bene▯t in the very last year isnV = S. Recursion: ▯ (tV + P)(1 + i) = q x+t(ct+1V ) + px+t t+1V ) = px+tt+1V ▯ where p x+t = px+t + cqx+t ▯ px+tv t+1V ▯ t = P multiply by E = v p p ▯ ▯ ▯▯▯p ▯ t x x x+1 x+t▯1 t+1 ▯ ▯ ▯ v t+1pxt+1V ▯ EtVxt P E t x Sum from t = 0 to n ▯ 1, middle terms cancel. X▯1 nE ▯nV ▯ E0 ▯0V = P v t ▯ x x x t=0 then V + Pa▯ t = 0 x:n t x If V = known number, then n ▯ n VnE x P = ▯ x:n If c = 1, Pat ▯ tV = t (px= 1) v 18 2.5 Advanced topics: policy alternations Sometime ofter inception, the policyholder may request to ▯ cancel (lapse) the policy. ▯ change premium terms ▯ Change the sum insured ▯ change the bene▯t type (from whole life/endowment insurance to something else) + By law, if the policy has been in force for 2 years, the insurer must provide some surrender value if the policy is lapsed. The surrender value (or cash value) can be ▯ agreed upon ahead of time (at t = 0) ▯ some calculation based on the policy value at time t For example some% of V (
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