# 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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## This note was uploaded on 10/30/2009 for the course ACTSC 331 taught by Professor David during the Spring '09 term at Waterloo.

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

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