%% Solutions to HW1, Stat 470, Spring 2006
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1. (a) C =  0.02396067 = .02/(1-exp(-1.8)) = 1 divided by integral
from 0 to 90 of C*exp(-.02*t).
   (b) 0.09757927 = integral from 40 to 50 of C*exp(-.02*t)
                  =  (C/.02)*(exp(-40*.02)-exp(-50*.02))
   (c)  32.17697  = C* integral from 0 to 90 of t*exp(-.02*t).
   (d)  31.67864  (calculated two different ways to check !)
## The answer here is C* summation over t=0 to 89 of the quantity
t* integral_0^t exp(-.02*x) dx. After integrating, this answer 
becomes,  sum_{t=0}^89 C*t*exp(-.02*t)*(1-exp(-.02))/.02 = 31.67864.

2. (a) .58947,   (b) .56 + .56 - (.56)^2 = .8064,  
                     2*(.2)*(.95-.56) - (.95-.8)^2 = .1335
### For this second part of 2b, the idea is: for each person,
the prob of dying before 40 is 1-.8 = .2; the prob of dying 
between 30 and 60 is .95 - (.8)*(.7) = .39; the probability that 
person A dies before 40 and (independently) person B dies 
between 30 and 60 is .2*.39. But we need to add the same amount 
again to accopunt for A dying between 30 and 60 and B dying
before  40, and also subtract the probability (.95-.8)^2 that 
both A and B die between 30 and 40 (in which case both people 
satisfy both events and we have double-counted this intersection).


3.(a) 87119/95385  =  0.91334
(b) (89870-85393)/95051  =   0.047101
(c) (94295-92315+53484)/94295 =  0.5881966

4. (a) 200*(1+i)^15 = 350, so   i = .03801, i^(2) = .03766 
   (b) 2^(1/7)-1 = .1041
   (c) 3^.1-1 = .1161

5. > uniroot(function(v) (1-v^10)/(v^(-.25)-1)-20, c(0.01,.99))$root
[1] 0.857097      ### This solves for v, using Splus or R code.
    ### The function we are trying to set to 0 is :
    ### 250*v^(.25)*(1-v^10)/(1-v^(.25)) - 5000 , or equivalently
    ### (1-v^10)/(v^(-.25)-1) - 20.
> 1/.Last.value-1
[1] 0.1667290  ### This solves for i = APR interest rate = (1/v)-1

### Apparently many of you did not interpolate of use a root-
finder to get sufficient accurancy in the interest rate. See page 15,
end of Sec.1.2.1 in Chapter 1, for an indication of how to do this by
linear interpolation, after finding the value of the function for 
many values of the interest rate. 

### In particular: with f(r) defined as the difference of the two
sides of the balance equation with interest rate r:
> f(r) =  250*(1+r)^(-.25)*(1-(1+r)^(-10))/(1-(1+r)^(-.25)) - 5000
> f(.16)
[1] 114.2578
> f(.165)
[1] 28.83892
> f(.17)
[1] -54.1203
### So the interprolated interest-rate solution is
> .165 + (28.83892/(28.83892+54.1203))*.005
[1] 0.1667381   
### which is remarkably close to the true value above.


6. (a) 12000/(21*(1.05^5-1)) =  2068.283
   (b) 12000/(1.05^5+1.05^4+2*(1.05^3+1.05^2+1.05)) = 1316.939