.01, we have
01 x ¢c = Opv A2$v + Q30v A4~bv + &e.
Substitute the value of dv thence obtained, and we
XXViii APPENDIX THE FIFTH.
have a method of finding the value of the required annuity, which may be described in the following
RULE.
Take out the value of an annuity of 11. at the given rate of interest, and at several successive higher rates : take the successive differences, the difference of the differences, and so on. To the first difference add half of the second difference, one-third of the third, and so on : the sum of these, multiplied by the amount of 1001. in one year at the first named rate, is the value of the annuity required.
I take examples from the Northampton tables, at 4 per cent., because Mr. Morgan has given a table of the annuities required, which will serve to find verifications. First suppose the age to be 5 years.
Annuity 4 p. c. 17'248 2'421
- 5 14.827 1 865 '556 164
- 6 12.962 392 '060
- 7 11.489 1 473 '288 104 -- 8 10.304 1185
2.421 + A of 556 + of 164 + 'q of 060 = 2.769 2.769 x 104 = 288.0 answer : in Morgan 288.4
Next suppose the age to be 80 years.Annuity 4 p. c.3.6433.515100798 5 -- 6 3 394121008 7 3.281.1130068 3.174107128 + A of 007 = '132 ; 132 x 104 = 13.7 answer13.8 in Morgan.
PROBLEM. A life annuity is ,'m at the end of the first year, and diminishes n every year, until nothing is due, after which it ceases entirely. Required its present value.
RULE. When n is so small, that the annuity cannot
ON UNIFORMLY CHANGING ANNUITIES. XXIX
be extinguished during the tabular life of the party, from the value of an annuity of '(m+n) subtract n times that of an increasing annuity of 11. found as already described. But when the annuity can be extinguished during the life of the party (say in t years exactly, so that m=nt), then to the preceding result add n times the value of an increasing annuity of I1. on a life t+ 1 years older than the party, multiplied by the chance of his living t+1 years, and by the present. value of X1 due t + 1 years hence.
PROBLEM. Required the present value of lm to be received at the end of the year in which A dies, if in a year, or , '(mn) if in the second year, and so on.
RULE. When n is so small that the sum insured cannot be extinguished during the tabular life of the party, to the value of a perpetuity of I'm, add that of an increasing annuity of Win,2n, in, &c., and subtract the value of a simple life annuity, of which the yearly payment is :m, increased by the product of a perpetuity due, and the value of a simple annuity of
: divide the difference by the value of a perpetuity due, and the quotient is the present value required. But if the insurance be extinguished in t years, or if m=nt : find the product of an annuity due of £1 on a life t+1 years older than the given life, and of a perpetuity ; subtract the value of an increasing annuity of 11. on that life, and having multiplied the difference by the chance of the first life surviving t+ 1 years, and by the present value of n due t + 1 years hence, add the result to the dividend in the first part of the rule, before dividing by the value of a perpetuity due of 1.
Various other questions will present themselves, which can be easily reduced to practice by aid of the expeditious rule for finding the values of increasing annuities. This rule may be applied to the Carlisle tables (for which Mr. Milne has deduced the values of annuities on single lives, at rates of interest from S to 8 per cent.,
XXX APPENDIX THE FIFTH.
both included), and also to joint lives, though not with so much correctness, on account of the tables not containing so many rates of interest.
The following is an instance in which a deduction from the calculus of differences will supply in a rough manthe deficiencies of tables. There are none of these for determining the mean duration of the joint existence of two lives, but the defect may be supplied with sufficient accuracy for many purposes, and particularly at the middle and older ages, by the following RULE. Let (3), (4), &c. stand for the values of an annuity on a single life, or on two joint lives, at 3, 4, &c. per cent.: from twice (3) subtract (6) and reserve the remainder : from (4) subtract (5), and having halved the remainder, to it add the tenth part of (5), and multiply the result by 9. Subtract the last product from the reserved remainder, and multiply the difference by 10. The result increased by 5 in the case of a single life, or by 25 in that of two joint lives, will be something under the mean duration required. For example, and to take a very unfavourable case, let the Carlisle table be used, the life being 10 years old.
(3) = 23.512 (4) = 19.585
2 (5) = 16.669
47.024 2) 2.916
(6) = 14.448
1.458
32.576 is (5) 1.667
28.125
3.125
4.451 9
10 28.125
44.51 5
45.01 the truth being 48'82