Stat 470 Numerical Solutions to HW 4 problems
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(1) (a) Expected age at death = 59.63
     (b) Median age at death = 57.23

(2)  mu = 4.094345, sigma^2 = 0.4462871
S(50) = .60754, S(82) = .32004

(3) Solve 3 eq'ns: must solve a nonlinear equation in  c, giving 
c =  1.07975 .
To do this in R, use the root-solver "uniroot". Otherwise do it 
by the method of bisections, or a graphical method with 
zooming-in.

Eq'ns are : (B/log(c))*c^50*(c^15-1)^2 = log(20/9)
             15 A + log(10/9)/(c^15-1) = log(3/2)
which leads to:
    (50/15)*(log(3/2)-log(20/9)/(c^15-1) + log(20/9)*
        (1-c^(-50))/(c^15-1)^2 = log(4/3)

> uniroot( function(y) log(4/3) - (50/15)*(log(3/2)-
        log(20/9)/(y^15-1)) - log(20/9)*
        (1-y^(-50))/(y^15-1)^2, c(1.001,1.3))$root
[1] 1.079746


(12)  Use the equation relating the two force of mortality at 
ALL ages x: note that you are given  S*(x+6)/S*(x) = S(x+9)/S(x)  
for all integers x.

We know that for Makeham-Gompertz 
models with force of mortality mu(t) = A+Bc^t,
the integrated hazard = -log S(t) is given by 
the formula  At + (B/log(c))(c^t-1)
which means that the formula for the proability 
of surviving k years from integer age x is
S(x+k)/S(x) = exp(-Ak - (B/log(c))(c^x)(c^k-1))

Now what we are given in the problem is that for 
every integer age x, the model with parameters 
A*,B*,c gives the probability of surviving 6 years 
to be the same as for surviving  9 years in the model 
with parameters A,B,c. The equation, holding for all x,
is

6A* + (B*/log(c))(c^x)(c^6-1) = 
9A + (B/log(c))(c^x)(c^9-1)

Writing the same equation with x replaced by x+1 and 
subtracting the two equations gives

(B*/log(c))(c^(x+1)-c^x)(c^6-1) = (B/log(c))(c^(x+1)-c^x)(c^9-1)

Cancelling terms in this last equation gives

B* = B (c^9-1)/(c^6-1)

and then substituting back into the original equation gives

6A* = 9A or  A* = 1.5 A


(14) If z = 1-d^(12)/12, then 
300*z^(-36)*exp(log(24)/3-log(12)/3) = 574
so d^(12) = 12*(1-((574/300)*exp(-log(24)/3+log(12)/3))^(-1/36))
= 0.1384609

(16) In this problem, the "face amount" according to the book's 
definition is the amount which will be paid at maturity. With 
this definition, the value for the bond at time t=15 
is 1e4*(1.07^(-15)).

If the 10,000 is instead taken as the purchase price (which is the way
I originally intended the problem to be done, then the price of the
bond at year 15 should be: 1e4*1.06^30*(1.07^(-15)).

Credit will be given for either of these two interpretations.