HWSet 5 solutions.

(1) We are saying  
      lambda*(40^gam-20^gam) = -log(92315/95840)
      lambda*(60^gam-20^gam) = -log(75221/95840)
So  (taking ratio of 2nd eqn over 1st)
(60^gam-20^gam)/(40^gam-20^gam) = log(75221/95840)/log(92315/95840)

and we solve that numerically :

> uniroot(function(g) (60^g-20^g)/(40^g-20^g) - log(75221/95840)/
     log(92315/95840), c(0.01, 6))$root
[1] 4.509704

## So this is the root gamma, and lambda = 
> -log(92315/95840)/(40^4.509704-20^4.509704)
[1] 2.335627e-09
## As a check, we find:

> exp(-lam*(40^gam-20^gam))
[1] 0.96322
> 92315/95840
[1] 0.96322
> 75221/95840
[1] 0.7848602
> exp(-lam*(60^gam-20^gam))
[1] 0.7848588

## Then the predicted value of 10p20 = 
> exp(-lam*(30^gam-20^gam))
[1] 0.991051
> l30 <- 95840*exp(-lam*(30^gam-20^gam)) ## = 94982.32

## Now the prediction interval:

> l30 + c(-1,1)*1.96*sqrt(l30*(95840-l30)/95840)
[1] 94925.18 95039.47

BUT the observed number is 94295 which does not fall
within the interval, suggesting that the Weibull 
assumption is WRONG for the Illustrative Table.


Next: l_70 = 53484, leading to CI for l_70/l_60
(treating l_60 as though known in advance) of

> 53484/75221 + c(-1,1)*1.96*sqrt(53484*(75221-53484)/75221^3)
[1] 0.7077855 0.7142642

(The following comment is additional, NOT part of the 
assigned problem.)

The quantity 10p60 computed using the (lambda,gamma) found in (a),
would have been
> exp(-lam*(70^gam-60^gam))
[1] 0.7827343
### So this problem part also gives evidence that the Weibull 
fitted parameters from part (a) do not fit the data in the Table.

Ch.3 #8: Let X = 1200 a"_{1}^{12} 
               = 100 (1-v)/(1-v^(1/12)) = 1163.579
Then the problem is to find X times the PV of 
an increasing 10-year annuity-due of amount X 
  = 1200 a"_{1}^{12} * Ia"_{10}
  = 1163.579 * a"_{Inf} * (a"_10 - 10 1.07^(-10))
  = 1163.579 * (1/(.07/1.07))*(1.07*(1-1.07^(-10))/.07- 
                 10 *1.07^(-10))
  = 43251.25

If you had interpreted the annuity in the problem as an 
annuity immediate payable monthly, increasing in 
successive years, then the answer would be: 43251.25/1.07^(1/12)
= 43008.08,  and if you explained what you were doing then you
will get full credit.

NB this makes sense because the total cash amount 
received over the 10 years is 1200*10*11/2 = 66000
The increasing 10-year annuity-due of 1 per year has 
value 37.17 as compared with the total cash payment = 55.

Ch.3 #10:
First, the cumulative hazard function obtained by integrating
mu(t) is H(t) = 2*log(1+t) - 2*log(1-t/100), corresponding to 
S(t) = exp(-H(t)) = ((1-t/100)/(1+t))^2
ANS. The expected number of deaths is large, ie = 
> 1e4*(S(1)-S(4)) = 1e4*((.99/2)^2 - (.96/5)^2) = 2081.6

Ch.4 #1:  NB 7/12 = 0.5833333.

ALSO NOTE: the solutions I posted initially to parts (a) of 
(i) and (ii) in this problem were wrong because I forgot to 
divide by S(t) in my conditional expectations. Here are the 
correct ones:

(i) 
> integrate(function(t) exp(-.00634*(20+t)^1.2),0,Inf)$value/
       exp(-.00634*(20^1.2))
[1] 57.66773  ## ans to (a)
> (1-exp(-.00634*((40+7/12)^1.2-40^1.2)))/(1-exp(-.00634*(41^1.2-40^1.2)))
[1] 0.5846718 ## ans to (b)

(ii) integrate(function(t) 1-pnorm((log(20+t)-log(50))/.325),
       0,Inf)$value/(1-pnorm((log(20)-log(50))/.325))
[1] 32.79475  ## ans to (a)
> (pnorm((log(40+7/12)-log(50))/.325) -
pnorm((log(40)-log(50))/.325))/(pnorm((log(41)-log(50))/.325) -
pnorm((log(40)-log(50))/.325))
[1] 0.5803664 ## ans to (b)

(iii) (a) int_0^30 exp(-.015*t) dt + exp(-.45) int_0^Inf exp(-.03*t)dt
 = (1/.015)*(1-exp(-.45)) +  (1/.03)*exp(-.45) = 45.41239
(b) (1-exp(-.015*7/12))/(1-exp(-.015)) =  0.5851555


Ch.4 #2:  
(a) 1000* int_0^infty .05*exp(-(.05+log(1.07))*s) ds  
    = 1000 *.05/(.05+log(1.07)) = 424.96
(b) 1000*sum_{k=0}^{infty} 1.07^(-k-1) * (1-exp(-.05))*exp(-.05*k))
  =  (1000/1.07)*(1-exp(-.05))/(1-exp(-.05)/1.07) = 410.6284
## (b) has to be less, because payment is made later than in (a).

Ch.4 #3.  P*sum_{k=0}^{infty} 1.07^(-k)*exp(-.05*k) 
   = P/(1-exp(-.05/1.07)) = 21.90389*P