# 2.5 Life tables, force of mortality and standard life insurance products

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1 Soluions 5 BS4a Acuarial Science Oford MT Life ables, force of moraliy and sandard life insurance producs 1. (i) n m q represens he probabiliy of deah of a life currenly aged beween ages + n and + n + m, i.e. as required. F (n + m) F (n) n+m n f ()d n+m n p µ + d (ii) p ep{ 1 µ +d} and 1 µ +d µ +.5. Hence p ep{ µ +.5 } and µ +.5 log(p ). Clearly also µ.5 log(p 1 ). Therefore µ 1 2 (µ µ.5 ) 1 2 (log(p ) + log(p 1 )). (iii) Use F () / o ge µ f() F () d d log( F ()) d d (log( 1 )) 2(1 ), and hence µ 84 1/ (i) We have where now p (1) + µ (1) ep + µ (1) y dy, y dy Bc c z dz Bc e z log(c) dz Bc log(c) (c 1). This leads o he form g c (c 1) for B/log(c) log(g) as required. (ii) This is similar, wih he era facor ep{ Ady} s for s e A. (iii) Jus calculae / p ep{ µ y dy} ep{ A( log() )} (/e) A.

2 Soluions 5 BS4a Acuarial Science Oford MT (iv) We have hree equaions in hree unknowns s, c and g: 1p 3 s 1 g c3 (c 1 1) 4 3 : α p 4 s 1 g c4 (c 1 1) 5 4 : β p 5 s 1 g c5 (c 1 1) 6 5 : γ Firs eliminae s by aking raios, aking log hen gives and c 3 (c 1 1) 2 log(g) log(β/α) c 4 (c 1 1) 2 log(g) log(γ/β) and again aking raios yields c 1 log(γ/β)/ log(β/α) and so c Then he log(β/α)-equaion gives log(g) So B log c log g Finally, he firs equaion gives s and hen s , so ha A log s µ (2) 2µ (1). Think of µ (1) o be from a general life able, bu of a populaion wih a cerain disease or oher risks causing heir moraliy o be higher han in ha lifeable. One way o epress his increased moraliy is by so-called proporional hazards, here proporionaliy facor 2. (i) n p (2) ep{ 2 n µ(1) +sds} ( n p (1) ) 2 as required. (ii) Under he Gomperz law np (2) g 2c (c n 1) n p (1) +a g c+a (c n 1) if and only if c a 2, i.e. a log(2)/ log(c). This is saying ha, under he Gomperz law, he model wih moraliy doubled is he same as a model wih age shifed appropriaely. The laer is anoher way of epressing he fac ha a populaion wih higher moraliy may behave like an older populaion. For general moraliy laws, hese wo models are differen. 4. (i) We spli ino deah benefis during he period and survaval o end of period, hen use q +k 1 p +k : A :n kp q +k v k+1 + n p v n k n 2 kp v k+1 kp p +k v k+1 k k vä :n jp v k j1 vä :n a :.

3 Soluions 5 BS4a Acuarial Science Oford MT The firs erm pays one a benefi of 1 a he end of every year 1,...,nprovided alive a he beginning. The second erm akes away he benefi of 1 if alive a he end of he year, ecep in year n. Wha remains is he paymen a he end of he year of deah (or n if earlier), and his is he cash flow of an endowmen assurance, as required. (ii) We calculae ä d(iä) kp v k d kp v k (k +1) k k kp v k (1 (1 v)(k +1)) k kp v k+1 (k +1) k j+1p v j+1 (j +1) j ( k p k+1 p )v k+1 (k +1) k kp q +k v k+1 (k +1)(IA). k (IA) pays 1 for each beginning of a year we were alive, paymen is a he end of he year of deah. ä pays one for each beginning of a year we were alive, paymen is a he beginning of each of hese years. d(iä) akes away he ineres we earn on he early paymens so ha he accumulaion a he ime of deah is (IA). 5. (a) Clearly b a q +a 1 b a p +a equals 1 b p / a p if and only if b a p +a a p bp. This is rue by he consisency condiion, in fac, his is he consisency condiion. (b) Deahs uniformly disribued over he age year (, + 1) means ha which is he same as In paricular, we have P(a T b T 1) b a for all a b 1, P( + a T + b T +1)b a. P(T T 1) P(T ) P(T 1) q q p 1 q 1 q, and so, by (a), b aq +a 1 1 bq 1 aq (b a)q 1 aq.

4 Soluions 5 BS4a Acuarial Science Oford MT (i) The condiion is P(T y ) P(T (y )+ T > (y )) for y. (ii) (a) µ + lim ε 1 ε P(T + + ε T >+ ). (b) µ + lim ε 1 ε P(T + ε T >). Now use he consisency condiion o ge as ε. µ + 1 ε P(T + ε T >) P( <T + ε) εp(t >) P( + <T + + ε T >) εp(t >+ T >) P( + <T + + ε)/p(t >) εp(t >+ )/P(T >) 1 ε P(T + + ε T >+ ) µ +, (iii) By definiion p P(T >). We defined he force of moraliy as infiniesimal survival probabiliy and can wrie his in erms of p as µ + lim ε p +ε p ε p p p (log( p )) This differenial equaion is solved by simple inegraion log( p ) µ +s ds + c p e c ep From p P(T ) 1, we ge c, hence he resul. µ +s ds. 7. (a) (i) C 1 ((n, B)) where B I(K n). (ii) C 2 ((k, B k )) k 1 where B k I(k K +1). (iii) C 3 (k, B k ) k1,...,n, wih B k as in (ii). (iv) C 4 (k, B k ) k1,...,n, wih B k I(k K +1 n)+i(k n K). (b) (i) A 1 (δ) E(NPV 1 (δ)) e δn P (K n), V 1 V ar(npv 1 (δ)) E(NPV 2 ) E(NPV) 2 A 1 (2δ) (A 1 (δ)) 2 (ii) A 2 (δ) E(NPV 2 (δ)) k1 e δk P (K +1k) E(e δ(k+1) ), V 2 V ar(npv 2 (δ)) A 2 (2δ) (A 2 (δ)) 2 (iii) A 3 (δ) E(NPV 3 (δ)) n k1 e δk P (K +1k) E(e δ(k+1) 1 {K } ), V 3 V ar(npv 3 (δ)) A 3 (2δ) (A 3 (δ)) 2 (iv) A 4 (δ) E(NPV 4 (δ)) k1 e δk P (K +1k)+e δn P (K n 1) E(e δ(min(k+1,n)) ), V 4 V ar(npv 4 (δ)) A 4 (2δ) (A 4 (δ)) 2 (c) C 4 C 1 + C 3, A 4 A 1 + A 3, V 4 V 1 + V 3 +2Cov(NPV 1 (δ),npv 3 (δ)) V 1 + V 3 2A 1 A 3 since NPV 1 NPV 3 always.

5 Soluions 5 BS4a Acuarial Science Oford MT (d) Only he pure endowmen can sill be epressed in he form asked for in par (a): C 1 ((n, B)) where B 1(T n). For (ii)-(iv), he ime of he paymen is now a coninuous random variable. The epressions of for ne premiums are essenially he same as in he discree case, bu replacing he discree probabiliy mass funcion of K + 1 by he densiy funcion of T, and inegraing raher han summing. For eample for par (ii), Ã 2 (δ) e δ f()d. In each case he variance again has he form V A(2δ) A(δ) (a) We have (b) Recall ha T NPV [,T](δ) ρ()e δ d E NPV [,T](δ) E 1 {T>}ρ()e δ d P (T >)ep E(1 {T>})ρ()e δ d P (T >)ρ()e δ d. µ T (s)ds, since d d log P (T >)) d d log F T () f T () F T () µ T () and log P (T >) log 1. So wih he given form of µ T we have P (T >)ep µ T (s)ds e.1 5 e.5 ( 5).2 5 < 1 and he answer is given by solving he inegral in (a) o give E NPV [,T](δ) e.1 e δ d e.5 ( 5).2 e δ d (c) Here he cash flow is C 2 (T, 1 {T 1}4) and we calculae he probabiliy densiy funcion of T from he survival funcion: f T () d.1e d P (T.1 5 >).2e.5 ( 5).2 5 < 1. Then E (NPV(δ)) 4E e δt 1 {T 1} 4 1 e δ f T ()d

6 Soluions 5 BS4a Acuarial Science Oford MT a)(1.4) b) (1.4) p [35] (1.4) 5 5(1 q [35] )(1 q [35]+1 )(1 q 37 )(1 q 38 )(1 q 39 ) p [6] (1 q [6] )(1 q [6]+1 )(1 q 62 )(1 q 63 )(1 q 64 ) (1.7) 5 X 5 p [6] 1 X 1(1.7) 5 / 5 p [6] a) ä v k kp 1+vp k b) ä :n v k kp k c) vä a v p +j 1 )A n k1 k1 v k kp k v k 1 k 1p +1 1+vp ä +1 v k kp +1 v n np a :n +1 A 1 :n. v j jp j1 v j ( j 1 p j p ) j1 v j j 1p (1 j1

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