Solutions to HW2, Stat 470, Spring 2006
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NOTE: numerical solutions to problems 5 and 6(b)-(c)
hsve been corrected, as of 2/24/06.
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1. ( (1+.05/4)^(4*4) * (1-.03/2)^(-2*5) )^(1/9) - 1 
[1] 0.03964303

2. The equation to solve is B(t+1) = 
1.06*B(t) - .015*B(t), since the interest 
earned between t and t+1 is .06*B(t) and 1/4 of 
this must be paid back. Thus B(t+1)=1.045*B(t) 
and we can see that B(9) = 1.045^9 *B(0) and the 
effective APR is i = .045. Thus, in this instance, 
the taxable investment in problem 1 is better.

3. 
> lx <- c(1000,850,800,760,700,620,520,400,250,75,0)
  px <- c(-diff(lx)/1000,0)

Unconditional expectations:
> sum(1000*(10-(0:10))*px)
[1] 5025
> sum(10000*(1.04^(-(1:11)))*px)
[1] 7965.06

Conditional expectations:
> sum(1000*(10-(3:10))*px[4:11])/(lx[4]/1000)
[1] 3625
> sum(10000*(1.04^(-(4:11)))*px[4:11])/(lx[4]/1000)
[1] 7506.415

4. In this problem we can integrate t times
f(t) = 3(t-39)^(-4) to get an approximate answer (but 
note that there is a tiny (= (90-39)^(-3) = 7.539e-06)
probability left that a population member must be killed off
instantanteously at 90), ie, f(t) integrates only to 
1-7.539e-6. A way to avoid this difficulty is to use 
the formulas  E(T) = int_0^omega S(t) dt  or
E(T-x | T>=x) = int_x^omega S(t) dt/S(x).

(a) (S(60)-S(80))/S(50) = .1244
(b) The formula for expectations we have been using gives:
> integrate(function(t) 3*t*(t-39)^(-4),60,90)$val*(60-39)^3   
[1] 62.43639
## To be 100% precise, we must add to this 90 times the conditional 
## probability of being killed off just at the 90th birthday:
> 7.538579e-06*90*(60-39)^3  
[1] 6.2833
### So exact answer is 68.71972, which can be obtained 
###   directly, using the formula given above, as:

 60 + (60-39)^3 * int_60^90 (t-39)^(-3) dt =
 60 + (60-39)^3 * 0.5*((60-39)^(-2)-(90-39)^(-2)) = 68.72

(c) The conditional probabilities for whole-number ages 
at death z=60,61,...,89  are (S(z)-S(z+1))/S(60), with 
the probability for 90 given by S(90)/S(60),
so we get the answer :

> (sum((60:89)*(-diff(((60:90)-39)^(-3)))) + 
     90*((90-39)^(-3)))*(60-39)^3
[1] 68.26618

### The exact formula, analogous to the one above but now 
###  gotten via summation (rather than integration) by parts, is
### 60 + summation from 61 to 90  of S(z)/S(60), =
> 60 + sum(((61:90)-39)^(-3))*(60-39)^3
[1] 68.26618


5. (a) Integral of r(t) from 0 to 10 is sum of integrals from 
 0 to 5 and from 5 to 10, 
= .01*(5*5-.1*25) + .01*(2*5 + .2*(10^2-5^2)) = 0.475
## So the balance after the 10 years is 10000 multiplied by
> exp(0.475)
[1] 1.608014
### and to find internal rate of return, equate this to (1+i)^10
### so i = 
> 1.608014^.1-1
[1] 0.048646

(b) Accumulation factors (= balance to which $1 accumulates) are:
 from 0 to 3, exp(.01*(5*3-.1*9)) = 1.151425
 from 0 to 6, exp(.01*(5*5-.1*25)+.01*(2*1+.2*(6^2-5^2))) = 1.306040
 from 0 to 10,  1.608014  as found above.
### So the answer to this problem is:
> 2000*1.608014 + 2000*1.608014/1.151425 + 2000*1.608014/1.306040
[1] 8471.54

(c) Take answer to (b) and subtract the amount accumulated by $400
over the last 5 years, ie subtract
> 400*exp(.01*(2*5+.2*(10^2-5^2))) ### = 513.6102

## So answer in (c) is 8471.54-513.61 = 7957.93


6. (a) Integrate D(t)*(1.08^(10-t)) from 0 to 10 to get
> (1/log(1.08))*(3000*(1.08-sqrt(1.08))+2000*(sqrt(1.08)-1))*
     (1.08^10-1)/.08
[1] 37791.33

### Compare this with 2500 deposited per year, which is the  
###  average, compounded from times .5, 1.5, ..., 9.5, or
###  2500*sqrt(1.08)*(1.08^10-1)/.08 = 37637.19

(b) Accumulation factor as function of time is
A(t) = exp(.01*(6*t+.25*t^2)). So answer involves integrals 
of this, which must be numerical, making the whole problem messier

> Dfcn <- function(t) ifelse (t-floor(t) < .5, 3, 2)
> integrate(function(t) Dfcn(t)/exp(.01*(6*t+.25*t^2)),0,10,
     subdivision=500)$val* exp(.01*(60+25))*1000
[1] 41232.23

(c) Subtract from (a), 500*sqrt(1.08)*(1.08^10-1)/.08 = 7527.439
So new answer for case (a) is 37791.33-7527.44 = 30263.89

Subtract from (b) 
> 500*exp(.06*10+.0025*10^2)*sum(exp(-.06*seq(.5,9.5,1)-
      .0025*seq(.5,9.5,1)^2))                                                
[1] 8212.315