s2005 - TEST — ACTSC 331, SPRING 2005 Family Name: Given...

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Unformatted text preview: TEST — ACTSC 331, SPRING 2005 Family Name: Given Name: ’ ID No. ‘I Date: Thursday, July 7, 2005 Time: 11:30 — 12:50 ' Total Questions} 6 Total Marks: 70 Total Pages: 9 (including this cover page) Aids: N on—Programmable Calculators INSTRUCTIONS 1. Print your name and ID. number in the space provided on this page. 2. Full solutions are required. Part marks will be awarded for partial solutions. 3. All your work should be done in this exam book. You may use backs of these sheets as scratch paper. ' l. iofg/‘g’ t5” 9/ L. Test page 2 1. For a fully‘whole life insurance on (50), you are given that 6 = 0.08, p54 2 0.975: Z215 24800, and 15V = m + 14V. / 7 r ,7 g, , Calculate 15V. ‘ L, ,‘ a n [10 marks] 2/ w "" x \l ‘I 7E “~» {5w " W'N imh *3 ‘x/Mx/“M/ {a — m? 7m fig»; « {Wm HiwstV : 0323‘??? (mv 4; W}: » Wank/U - om “a: v (MW) “f‘ 0‘“: 134% (7m ~W} my \5\/ ’3 (430‘3flsvfim OMYBO’IW 7L“ {3"‘2/ “(41/ MW-_M «M A. Test 2. For a population Comprised of smokers and nonsmokers, your are given -i_' ’1'! a] o For smokers. mortality follows = 8-5 —— .1'. U s (1' S 8-3. r-"',';‘~' d3 . . lg: _ " “ o For non—smokers, mortality iollovls if = 90 ~ 13,, 0 S a; g 95. 30 ’15 ,L :o *33 o (60) is a smoker and (65) is a non—smoker. o (60) and (65) are independent lives. (a) Calculate the probability that exactly one of (60) and (65) will survive 15 years. W I“, y f [10 marks] anmmjciéwlw) l - Waxing“ WVGMQ' 7‘" “gm? " Wlifi‘ —« rv - wn—w rm w) w is gagging lazé‘filwég’l \8 5m) WWW : a: W [Li a. 2" a V y M ‘ W , s w I A 3° ‘13 an ill isllmslislwl l _ , 1 r v; KS“ ’ Li '2 QJLIOMM) 4‘” ;L \« {Vial/K l"zs {no 0 9‘56 2 ",9 :: me}? Jr 0% «0922314: :7 0613 72‘ L/ \ _ Ow)" {’Mflm) >23,m§) {if} ffrfTW/‘S/S, many} ‘ is” %~,§iés‘l‘is’€l.w afig : 13% ,i 4» 53'1‘ ~l+ 23 a; ' ’S - g s” Test page 4 (b) Calculate E[T(60 : 65)]. [10 marks] éFH6v$mj , {28 V‘ 0 A; )0 1 £9 Ms M. r13 25% 31—4 ‘2, / 3 )0 ZS :9 M “Em \ ‘\ ‘36 %3”W§ _ ,3, -1; ,g _ Kb 78% Z + 3 J“ Test page ,5 3. For two independent lives (40) and (-50). you are given: 0 For (40). mortality follows “(—10 + t) = 0.06, t 2 O. o For (‘50). mortality follows £1. = 90 — 0 g a: g 90. 14.9. F/ “M,” If (-10) and (50) are subject to a n fatal sh to have an exponential distribution with the mean of 20. Calculate the expectation of the time of the second death. [10 marks] )3; / . , Q t n t 5,“. a. l; J . ‘ n Ema: v » t, ElTMoIS‘M) c ’1‘ 493i re ' ' - M " «P ,4” E . ‘ [Sj‘ll’wy’lgfl‘www — ST’lwoll/l ET‘WS’WW” 1°15 633:”, w 90 (l W KO Wm + a \ Lla 1’ \k"%‘ [4, " all,“ v Q‘Ws .‘ J- gent. ’ ~~ 0H7 L-. QKQKD't , ‘44,“ {pent /,cx05l out qswt I is. E; Jr 40 6, ’ 6, f” M 40 w' 4, .4 i» >0 ” )9, / (f0 § w x. «Oxll’t '" 1—{3 Pi} .t \ l (to ' hi, v- , ’QQC 0‘ ’_V\‘ ’ lg J“ it Hi Wold -~ me ‘~ we ._. QUv—ootf,-gtgfi ‘ 'u 4‘4; W est; [0C '«Hl‘fl‘t’le "at .. [w star—t} g-Mf »t <*’ k Q dc Test page 6 4. A life insurance is issued to (9:) and (y) and pays 1,500 at- the moment of th rst death and 3500 at £1.16. rentingfjghg Riga}; .Qvanihesemndmleatn. Assume that Eta: = 10, c‘zy = 12, flay = 15, am = 14, and 6 = 0.02. Calculate the actuarial present value at time 0 of the icleath ben® [10 marks] % ‘\ art/v mum ...‘., »' ~" {ax/l m, «2M 24min \/ ll EEWW_______________________P§§£_7 5. An annuity is payable ébntiinm’ at the rate of 150 per year While both and (y) are alive and 100 per year while one of (w) or (y) is alive and other is dead. Let Z be the present value at time 0 of the Assume that (x) and (y) are independent lives and have the same constant mortality force of 0.02. In addition. the force of interest is w/‘lfi y I; “NA, Q ._ 1 £33 (2:) Calculate E(Z). 7’5 :7 (Ml [4 marks] (:59 ‘i' hf) {SQ w 3 a} “30 lav—IL ML.WWWMMWL.~ ...... ml ........... mwfl Tm} "m 7 x E ill f \W at? ‘t SQ Km 26% +563, Own} l l figgumelo + thwifimm Tm! 1"”; .f“ I? I (1 ‘ ,\ t r A?” "‘3 ‘5 +«500fm: - m ; k " J j: q ( 8 "9 w ’5 ;-O§2i ,Gufl‘f 1‘) " ') } 1 1 v Z 4*M";€ 9‘ ‘6 3 V M . V / M V a \J' -mim (J) Ca cu ate cu( )- (5 4+ Ea [6 marks] I" «a» V CQ’V/ mo ,1 ,. \MU "* {000,9 0 <2 .- 1 '" ’ 3.; Iv? 3 S j + 2300£El§ I01] HQ,\N\3Y(,\2§ H, nguvsflfl (0X Wmmw CWGA WWW L36 flgA‘th/e "T‘ V xix/(z): we“ Maw) HWWWWH ZWWW (“mic fl???) e-V 2 . 7" i fl‘ 11. h f r~ r “L 2/3973“ A“) +qu X m MW) ‘f’ZNGU ms 153% ’00“; f 9‘3“ Wm) 1 NO 7‘ 4”???“ ‘ ’ ‘ a 4% T FNMA " + {02 3 a L_____________________P_ag§_9 6. For a fully discrete n-year endowment insurance on (as), let kV denote the benefit reserve at the end of year It. The death benefit in the k—th policy year is equal to k + kV for k = 1,2, ...,n. The benefit premium in the k-th policy year i®for k_=_1 2 n. A maturity value of n will be paid to at the end of n years if survives to that time. Assume that qz is a constant for ., mgmand‘aeflgualsmtmhkemgiuterestmga‘temojmék where q, is the probability that (11:) dies in one year. Show that kV = k f9: E»:_:flpljflr.d_._lfl, I,“ 410 marks] 13k 7— i’ (< \/ hm = K+l +~g+l\/ {Rt emit :€'fi ‘ ' I MV— Tlhc lbkh’heV) 071% 4* mu, Fm ’ I, evifih‘ . X « wit} VQMU “V 1 fl 7“ :2 7L“: lh+l)V‘OLx+t+(Vh+i\/'21Vl ~utsvi: \wn<lw®v“lww+\fllvwrmvlwww w AA K ' e réwk=ewwwtt+vw TY M 1: gumovl‘ilm-e‘”) + x/Km/ Elm/villi) xlx/m 0 ~~~ T interim-n {E‘W‘m V - "04 7K (1m :: ng=§vah Wad/0W) lhi"l\/W‘<l“€7lk) 3 Ram“ Va?" amalgam/2v } v H l A a \/' dmfl'v‘ “WV‘I'HW mm ‘1' mm ’Vl‘“ J” l‘ u 1’ A w I (mama? 7H M “‘9 \ka .1 k\ 0‘7" fit flit-Ni WW deflate -i<\/“+v iv Elm” bhqfihV 73H 2 [14mg Alle : x \<\/ :. \Q‘ We . W remw h =o~ rm Ttlhhflrg «Hi My _hv \ 73%;“ = “l gilafiluhfi+uflnv "- iébuk'i‘fn : film’mn” fl arr a: 4x“ v givwi‘lm n lash— « Limmuwa-v‘il/ , 11k ...
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s2005 - TEST — ACTSC 331, SPRING 2005 Family Name: Given...

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