R scripts and Problems Solutions, Stat 470 Exam
Spring 2006.

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(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

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(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

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(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

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(4) Formula = 1e4*(10*i^(4)/(1-v^10)-1)

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(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.


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(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

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(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

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(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

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(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



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(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 !

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(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

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(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