Stat 470, Spring 2006

Homework Set 7, 8 Problems Due Monday 5/8/06
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(1) Estimate = 5.5/7 = 0.7857143
    CI = 5.5/7 + c(-1,1)*1.96*sqrt((5.5/7)*(1-5.5/7)/7e4)
[1] 0.7826745 0.7887540
CI for the endowment: .Last.value*(1.05^(-15)*5e4)
[1] 18823.99 18970.21

Ch.4 #10: (a) 30p10 = exp(-.001*10-.004*20) = 0.91393
    (b) e_{50}^o = sum(exp(-.5e-4*((51:500)^2-50^2))) = 87.137
> sum((0:949)*(-diff(exp(-.5e-4*((50:1000)^2-50^2)))))
[1] 87.13686

### NOTE that this survival law does allow very long lifetimes
(eg 250p50 > 1/80)

Ch4 #15: This is a theoretical problem. Benefit (present value 
as of end of year of death) if death occurs in period (k,k+1)) 
is  a"_{N-k}. Prob for life aged x of death in policy interval 
[k,k+1) is 1/(w-x). So 
B(x,n,N) = (10/(w-x))*sum_{k=0}^{n-1} v^{k+1} a"_{N-k} = 
(10/(w-x))*sum_{k=0}^{n-1} v^{k+1} (1-v^{N-k})/d  and using d=iv, 
get 
B(x,n,N) = (10/(w-x))*sum_{k=0}^{n-1} (v^k-v^N)/i 
    = 10*((1-v^n)/(1-v) - n v^N)/(i*(w-x))

Ch5 #4: (a) (i/i^(4))*.02*(sum_{k=0}^4 (.98/1.05)^k + 
      2*sum_{k=0}^9 (.98/1.05)^k)/1.05 = 
  (.05/(4*(1.05^.25-1)))*.02*1e4*( (1-(.98/1.05)^10) + 
       2*(1-(.98/1.05)^5))/(1.05-.98)  =  3148.504
## Reasonable because we get 30,000 wp 1-.98^5 and 10,000 wp
.98^5-.98^10, roughly 1e4*(3*0.096 + 1*0.0868)*1.05^(-.5*(3*5+1*10)/4)
      = 3217.978
(b) alpha(2) = (.05^2/1.05)/(4*(sqrt(1.05)-1)*(1-sqrt(1/1.05))) 
= 1.000149, beta(2) = (.05-2*(sqrt(1.05)-1))/(4*(sqrt(1.05)-1)*
    (1-sqrt(1/1.05))) = 0.25617 .
For premium, divide answer in (a) by 
2*(1.000149*sum_{k=0}^9 (.98/1.05)^k - 0.2561738*(1-(.98/1.05)^10))
 = 2*(1-(.98/1.05)^10)*(1.000149/(1-.98/1.05) - 0.2561738) =14.69853
SO ANSWER IS :  216.794

Ch5 #5: (a) 1e4*sum_{j=0}^39 (.98/1.05)^(j/4)*(1-.98^.25)
  + 2e4*sum_{j=0}^19 (.98/1.05)^(j/4)*(1-.98^.25) = 
  1e4*((1-.98^.25)/(1-(.98/1.05)^.25))*
     (1-(.98/1.05)^10 + 2*(1-(.98/1.05)^5)) = 3187.388
(b) Again get result by dividing (a), this time by
  sum_{k=0}^{19} (.98/1.05)^(k/2) = (1-(.98/1.05)^10)/
     (1-(.98/1.05)^.5) = 14.69815  leading to 
  Premium = 3187.388/14.69815 = 216.856.

Ch5 #8: When T is Expon(mu), let r = exp(-mu) and find exact formulas:
a"^(m)_x = (1/m) sum_{k=0}^Infty (r*v)^(k/m) = 1/(m*(1-(r*v)^(1/m)))
and in particular case m=1:  a"_x = 1/(1-r*v)

Exact formula for error = 1/(m*(1-(r*v)^(1/m))) - 1/(1-r*v) - (m-1)/(2*m)

Ch5 #11: Here we are within interpolation assumption (ii).
(a)  Expected present value of premiums paid is Pbar * 
int_0^2 exp(-(.04+.06)*t) dt = 10*(1-exp(-.2)) = 1.812692

(b) Expected present value of benefit = 
sum(c(10,8,6,4,2)*exp(-.1*2:6)) = 21.65952
So risk premium using (a) is Pbar = 21.65952/1.812692 = 11.94881

Ch5 #14: Abar_x = int_0^Infty exp(-(B/log(c))*c^x*(c^t-1))*(v^t)*
    B*c^{x+t} dt, while

abar'_x = int_0^Infty exp(-(B/log(c))*c^x*(c^t-1))*(v'^t) dt
        
So, factoring out mu_x = B*c^x from the first expression and
expressing  v'^t = ((1+i)/c)^(-t) = (vc)^t , we find

Abar_x = mu_x* int_0^Infty exp(-(B/log(c))*c^x*(c^t-1))*((vc)^t) dt
       = mu_x* abar'_x .