Solutions to HW3, Stat 470, Spring 2006
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1. If X denotes the desired present value, then the 
superposition of X with the same payment steam 
advanced by 1 year and the same payment stream 
delayed by 1 year gives the uninterrupted payment 
stream from year 0 to 32 inclusive: in symbols,

X*(1+v+1/v)) = 100 * a"_{33}, or X = v*a"_{33}/a"_{3}

2.(a) > 2e4*(4*(1.05^.25-1))/(4*(1-1.05^(-10))) )
[1] 635.7242

(b) Now use payment stream 4*P* a"^{(4)}_{8.25} deferred 2 yrs.
So 2e4 = 1.05^(-2)*(4P)*(1-1.05^(-8.25))/d^(4)
> 2e4*1.05^2*(4*(1-1.05^(-.25)))/(4*(1-1.05^(-8.25)))
[1] 806.727

(c) Now 2e4 = 1.05^(-2)*(4P)*(1-1.05^(-8.25))/d^(4) 
	      - (1.05^(-5)+1.05^(-6))*P
> 2e4/(1.05^(-2)*(1-1.05^(-8.25))/(1-1.05^(-.25)) - 
          (1.05^(-5)+1.05^(-6)))
[1] 859.779

3. (a) Payment = 
> 2e5/((1-1.06^(-15))/(1.06^(1/12)-1))
[1] 1670.588
## Interest paid at 1+j/12 (j=1,...,12)
##   is (i^(12)/12)*B_{1+(j-1)/12} = 
## 2e5*(i^(12)/12)*(1-v^{15-1-(j-1)/12})/(1-v^15)
## Sum these over j=1,..,12 to get
## 2e5*(i^(12)/12)*(12-v^14*((1+i)-1)/((1+i)^(1/12)-1))/(1-v^15)
Interest for whole year = 
> 2e5*(12*(1.06^(1/12)-1)-1.06^(-14)*.06)/(1-1.06^(-15))
[1] 10938.95

## Here is a conceptually simpler method: look at the total
12* 1670.588 = 20047.06 of the payments made during the course 
of the second year (1,2]: these are devoted either to interest or
reduction of principal, and since the overall reduction in principal
is equal to B_1-B_2 = 
2e5*((1-1.06^(-14))-(1-1.06^(-13)))/(1-1.06^(-15)) = 9108.106,
we conclude that the rest, or 20047.06-9108.11=10938.95 must 
have been interest.
## Note that, while the principal-reduction payments made monthly 
can themselves earn interest for the recipient from the time when 
they are made to the rest of the year, the breakdown of each 
payment into principal and interst carries the principal forward 
correctly: the interest absorbs all of the accumulation ! 

(b) B_7 = 
> 2e5*(1-1.06^(-8))/(1-1.06^(-15))
[1] 127875.5

(c) New payment:
> 2.05e5/((1-1.0525^(-15))/(1.0525^(1/12)-1))
[1] 1634.80
### So yes, it would have been advantageous to take 
### the lowered interest rate.

4. (a) C = .05^2 = 1/400 = .0025,
h(t) = f(t)/S(t) = (t-20) e^(-.05*(t-20))/(20*t*e^(-.05*(t-20)))
     =  (t-20)/(20*t)
S(t) = .05*t*exp(-.05*(t-20)), t>20

(b) .05^2*int_{40}^{infty} t(t-20) e^(-.05*(t-20)) dt/S(40)
 = .05^2*exp(-2)*int_{40} ((t-40)^2+60*(t-40)+800)*
        exp(-.05*(t-40))dt/(2*exp(-2))
 = (1/800)*(2*20^3+60*20^2+800*20) = 70


#2 in Chapter.
(a) integral of f(t) times 1e6*(1.4-t/50) from 20 to 70 
(not to 80) gives answer:  324074.1

(b) Now integrate f(t) times 1e6*(1.4-t/50)*exp(-.08*(t-20)) 
from 20 to 70 to obtain: $ 101,495.80

#3 in Chapter.  
(a) Start with S(t) = exp(-1e-3*(7*t-0.25*t^2+40*(exp(t/20)-1)))
> Sfcn <- function(t) exp(-1e-3*(7*t-0.25*t^2+40*(exp(t/20)-1)))
> round(1e5*Sfcn(0:70)) round(1e5*Sfcn(0:70))
 [1] 100000  99124  98294  97509  96768  96068  95409  94789  94207  93662
[11]  93151  92674  92230  91817  91434  91080  90754  90453  90178  89926
[21]  89697  89489  89300  89130  88976  88837  88712  88599  88496  88402
[31]  88314  88231  88151  88071  87989  87903  87811  87709  87596  87468
[41]  87322  87156  86967  86750  86502  86221  85902  85541  85135  84681
[51]  84173  83608  82982  82291  81531  80698  79789  78800  77727  76568
[61]  75320  73981  72548  71021  69399  67681  65869  63963  61967  59884
[71]  57717
(b) Sfcn(30)/Sfcn(3)
[1] 0.9057027