R scripts and Problems Solutions, Stat 470 Exam Spring 2006. ------ (1) d_y = 2000*1.003^(y-20), 20<=y<45, l_{20}=850000 l_{y+1} = 850000-(2000/.03)*(1.003^(y-19)-1) l_{45} = 798157.9 > lvec <- 8.5e5-2000*cumsum(1.003^(0:24)) EZ = > (sum((0:24)*2000*1.003^(0:24))+25*798157.9)/8.5e5 [1] 24.21662 ### = sum(lvec)/8.5e5 ------ (2) (a) (lvec[10]-lvec[16])/8.5e5 ## = 0.01465648 (b) -(d/dy) log (l_{37} - y*1.003^17*2000) |_{y=.5} = (1.003^17*2000)/(lvec[17]- .5*1.003^17*2000) [1] 0.002584983 > (2*1.003^17/850)/(1-(2/850)*((1.003^17-1)/.003+.5*1.003^17)) [1] 0.002584983 ------ (3) (i/i^(4))*2e5*A_{22:10}^1 + 1e5*v^{23}*l_{45}/l_{22} = (.04/(4*(1.04^.25-1)))*2e5*1.003^2*(2000/lvec[2])* (1-(1.003/1.04)^10)/(1.04-1.003) + 1e5*1.04^(-23)*lvec[25]/lvec[2] [1] 42243.34 ## = 3964.855 + 38278.48 ----- (4) Formula = 1e4*(10*i^(4)/(1-v^10)-1) ----- (5) Amount of premium = 1e5*1.03*A_{25:25,.05}/a"_{25:25,.05} So lump-sum amount borrowed = X = P*a"_{25,.06} At 6%, net amount borrowed at end of endowment insurance has expected time-0 present value P*a"_{25:25,.06} The amount received from the endowment insurance, re-valued at 6% present value, is 1e5*A_{25:25,.06}. Thus the net gain, in terms of expected present value at 0 using 6% interest rate, is: 1e5*A_{25:25,.06} - P*a"_{25:25,.06} = 1e5*A_{25:25,.06}*( 1 - 1.03*(A_{25:25,.05}/a"_{25:25,.05})* (a"_{25:25,.06}/A_{25:25,.06}) ) Recall identity A_{25:25,i} = 1 - d a"_{25:25,i} , to get formula 1e5*A_{25:25,.06}*( 1 - 1.03*(1/a"_{25:25,.05} - .05/1.05)/ (1/a"_{25:25,.06} - .06/1.06) ) With PV at 5%, the problem is quite a bit harder, because the time-0 PV of the net amount borrowed at end of endowment insurance is not easily expressible in standard actuarial functions. ----- (6)(a) _{20}p*_{45} = (_{20}p_{45})^1.5 = exp(-1.5*(.025*15+.05*5)) [1] 0.3916056 (b) ((1-exp(-.0375))/(1-exp(-.025)))*((1-(1.06*exp(.0375))^(-15))/ (1-(1.06*exp(.025))^(-15)))*(1.06-exp(-.025))/(1.06-exp(-.0375)) [1] 1.393792 ---- (7) A_{55:10}^1 = > (1-exp(-.025))*(1-(1.06*exp(.025))^(-5))/(1.06-exp(-.025)) + (exp(-.025)/1.06)^5*(1-exp(-.05))*(1-(1.06*exp(.05))^(-5))/(1.06-exp(-.05)) [1] 0.2228881 A_{45:20}^1 = > (1-exp(-.025))*(1-(1.06*exp(.025))^(-15))/(1.06-exp(-.025)) + (exp(-.025)/1.06)^15*(1-exp(-.05))*(1-(1.06*exp(.05))^(-5))/(1.06-exp(-.05)) [1] 0.2616817 a"_{55:10} = > (1-(1.06*exp(.025))^(-5))/(1-(1.06*exp(.025))^(-1)) + (exp(-.025)/1.06)^5*(1-(exp(.05)*1.06)^(-5))/(1-(1.06*exp(.05))^(-1)) [1] 6.948886 a"_{45:20} = > (1-(1.06*exp(.025))^(-15))/(1-(1.06*exp(.025))^(-1)) + (exp(-.025)/1.06)^15*(1-(exp(.05)*1.06)^(-5))/(1-(1.06*exp(.05))^(-1)) [1] 10.09511 ## So answer is: 0.2228881 - (0.2616817/10.09511)*6.948886 [1] 0.04276165 ---- (8) Each monthly payment is 2e5/((1-1.05^(-30))/(1.05^(1/12)-1)) [1] 1060.110 ## Next figure the balance after end of 13th year: > 2e5*(1-1.05^(-17))/(1-1.05^(-30)) + 1060.110*(1.05^(5/12) + 1.05^(1/4)) [1] 148833.8 ## Now the balance after end of 14th year: > 2e5*(1-1.05^(-16))/(1-1.05^(-30)) + 1060.110*(1.05^(17/12) + 1.05^(5/4)) [1] 143265.2 ## So the total payments made for the year minus the paid-down ## principal is > 12*1060.110- (148833.8-143265.2) [1] 7152.72 ### Avg balance for year is around 146,050, so year's interest should ### be around 141,630*.05 = 7302.48 OK ---- (9) Insurance net single risk premium involves exact time-varying benefit > (20/25)*(.15/20)*(1/.976)*integrate(function(t) (4e5/(t+24))*1.05^(-t),0,20)$value [1] 998.5825 ### mult. by 1.02 to get 1018.554 with loading Using assumption (i) interpolation: > a"^(2)_{24:20} = .5*sum( 1.05^(-(0:39)/2)*(1-.024-0.5*(0:39)*.15/25)/.976 ) [1] 12.27994 ### Another way: a"_{24:20} = sum( 1.05^(-(0:19))*(1-.024-(0:19)*.15/25)/.976 ) [1] 12.44959 ### Now use adjustment formula with alpha(2) = (.05^2/1.05)/(4*(1-1.05^(-.5))*(sqrt(1.05)-1)) ## = 1.000149 beta(2) = (.05-2*(sqrt(1.05)-1))/(4*(1-1.05^(-.5))*(sqrt(1.05)-1)) [1] 0.2561738 ### Alternate evaluation of a"^(2)_{24:20} = > 12.44959 * 1.000149 - 0.2561738*(1-1.05^(-20)*(1-.15*24/25)/.976) [1] 12.27995 ### Agrees closely with previously found value. ### So level premium each half-year = > 1018.554/(2*12.27995) ### = 41.472 ---- (10) Actual probabilities are based on mu(t)=.03 for all t: > exp(-.03*1.7) - exp(-.03*3.7) [1] 0.05533992 Balducci is based on formula _hp_k = p_k/(1-(1-h)*(1-p_k)) > (exp(-.03*2) - exp(-.03*4))/(1-.3*(1-exp(-.03))) [1] 0.05533471 ### The two answers are incredibly close ! ---- (11) Want root of the function of x: > fbal <- function(x) sum( (10+20*(0:10))*((1+x)^(10-(0:10)) - exp(.7-.08*(0:10)+.001*(0:10)^2)) ) > uniroot(fbal, c(0,.2))$root [1] 0.06788428 ---- (12) median = 40 + (log(2)/8.557e-7)^.25 [1] 70.00033 mean = 40 + .25*gamma(.25)/(8.557e-7)^.25 [1] 69.80168 > 40+ integrate(function(t) 4*t^4*8.557e-7*exp(-8.557e-7*t^4), 0, Inf)$value [1] 69.80168