Solutions to HW6, Stat 470
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Ch4 #4: 
(a) > .025*sum(1.06^(-(1:40)))
[1] 0.3761574
### Should agree with .025*(1-1.06^(-40))/.06 : OK

(b) 20-yr term ins:
> .025*(1-1.06^(-20))/.06 
[1] 0.2867480
    30-yr endowmt ins:
> .025*(1-1.06^(-30))/.06 + .25*1.06^(-30)
[1] 0.3876483

#5. This one is simple algebraic proof of basic 
formula, followed by manipulation:

c_x = sum_{k=1}^{Inf} p_x * (1+(k-1)){k-1}p_{x+1} q_{x+1+k-1}
    = p_x * (1 + c_{x+1}) 

#12 NB for ages x,t >=30, mu(t) = 3/t  and p_{x+s} 
       = exp(-3 log((x+s)/x)) = (x/(x+s))^3
    and (S(70+y)-S(71+y))/S(70) = (70/(70+y))^3-(70/(71+y))^3

e_{70}^o 
> sum(((70/(70+(0:5000)))^3-(70/(71+(0:5000)))^3)*(0:5000))
[1] 34.5

Abar_{60}
 = int_0^Inf (3*60^3/(60+s)^3)*1.09^(-s) ds 
> integrate( function(s) 3*(60^3/(60+s)^4)*1.09^(-s), 0,Inf)$val
[1] 0.3415603

#13  3p95 = (1-.3)*(1-.4)*(1-.5) = .21
So for impaired lives, take .21^3 = 0.009261.

#16  Algebraic formula , superposition of 1e5* 18E0 and 5e5*60E0
and 2e5* (18 yr term life insurance on newborn).


Ch. 5 #12. 
f = .5p60 - p60* .5p61 = (1-.5*.3) - .7*(1-.5*.4)
  = .29
while  
g = (1-.3)/(1-.5*.3) - .7*(1-.4)/(1-.5*.4)
  = 0.2985294
### So answer is round(1e4*(0.2985294-.29)) = 85