Question

Death can take on the following values

3, 3.5, 4, 4.5, . . . , 79.5, where death at age 3 during the first half-year is counted as 3, death at age 3 in the second half-year as 3.5 etc.

Preliminaries I: Compound interest & present values

Preliminaries II: Expected value

de Witt’s formula for life annuities

? Then he states based on the total number of chances he calculated

de Witt’s assumptions

de Witt’s assumptions (cont’d.)

? P(death = l) = 1/128 for l = 3,3.5,...,52.5;

? P(death = l) = (2/3)/128 for l = 53, 53.5, . . . , 62.5;

? P(death = l) = (1/2)/128 for l = 63,63.5,...,72.5;

? P(death = l) = (1/3)/128 for l = 73,73.5,…,79.5.

Based on de Witt’s death distribution, find the probability that death occurs before or at 62.5.

Calculate the present value of an annuity certain for 10 years (that pays 2e every year) if i = 3.5%

Answer 1

P(death occurs on or before 62.5)= P(D<=62.5)=P(D=3)+P(D=3.5)+...P(D=52.5)+P(D=53)+...P(D=62.5)

Now,

P(D=l)=1/128 for l=3,3.5,...,52.5

P(D=l)=2/384 for l=53,.,62.5

Thus the required probability is 1/128*101 (since there are total 50 numbers between 3 and 52 both inclusive and a further of 52.5) + 2/384*11 (as there are 10 numbers between 53 and 62 and a further of 62.5)

Thus the required probability is 0.8464.

Since this is an annuity certain we have the present value as v+v^2+v^3+...+v^10 = v(1-v^10)/(1-v)=(1-v^10)/i where i=0.035,v=1/1.035. Thus answer is 8.32.