155. Ifa always denote the probability of the happening of an event, and b the probability of its failing, then the fraction (a+b) will express the probability of its happening n times in succession, and probability of its failing n times successively. EXAMPLES. b (a + b) the 1. An urn contains 15 white and 11 black balls, what is the probability of drawing first a white and then two black balls. 11 Ans. 104 2. In how many trials may a person undertake, for an even bet, to throw an ace with a single die? Ans. 3.776. 3. Among 32 counters, 14 are red and 18 white, what is the probability of drawing 4 red ones in succession, and also the probability of drawing 6 white ones successively? 156. To find the probability of a person of a given age (m) living any number of years (n). Let Pm, denote the probability required, the number of persons living at the age m according to the Tables of Mortality, and + the number living at the age m +n; then lm + is the number of chances of living n years, and I is the whole number of chances; hence (152) the probability of a person A, aged m years, living n years is + " m + n λα. The probability that A will be dead at the end of n years is 1 Thus, by the Northampton Tables, of 11650 persons born, 2936 survive 49 years, and 2612 survive 53 years; hence the probability that a person aged 49 years will complete the age of 53 is 157. To find the probability that two persons, A and B, will live n years. Let m, denote the age of B, and, as in the last proposition, find Then the probability that both A and B will be alive at the end of n years is the product of the separate probabilities of A and B being alive at the end of n years; hence it is The probability that both A and B are dead, is (1 — λ„) (1 — X',). If there be more lives than two, as A, B, C, it may be shown, in a similar manner, that the probability of all three being alive at the end of n years is A, A', A′′, and the probability of their joint existence failing in n years is 1 An An, whilst the probability of their being all dead, is (1 − λ„) (1 — X′n) (1 − X′′„). The probability of two, at least, out of three persons, A, B, C, being alive at the end of n years will be found in the following manner : The probability that all are alive is n. A and B are alive and C dead, is λ„ X′„ (1 — X′n)• A and C are alive and B dead, is λn λ′′, ( 1 — λ„'). B and C are alive and A dead, is X', X′′n (1 — λ„). Hence the probability of two, at least, of these three persons surviving n years is the sum of these four probabilities, and it is there The probability that one, at least, of these three persons will be alive at the end of n years is evidently 1 − (1 − x) (1 — X′„) (1 — X''n)• 158. To find the probability of a life failing in any particular year. 78 Let m denote the age of the person, and qm, the probability that the person will die in the 7th year from the present time; then, by the Tables, the number now aged m who survive n1 years, or enter upon their (m + n)th year, is 7m+n-1, and the number who complete their (mn)th year is m+ the difference between these is the number who die in the nth year, and this difference, divided by the number living at the age of m years, gives the probability of a person aged m dying in the th year from this time. Hence we have n; average 159. The number of years' expectation of life of a person whose prospect of longevity is the same with that of persons of the same age is the number of years enjoyed by each person. Thus, in the Northampton Tables, opposite to the age 48, we find 19, and this signifies that persons aged 48 may expect to live 19 years longer. Suppose that I persons are alive at the age m, m+1 are alive at the age m + 1, then l„ die in their (m+1)th year, and + survive their (m + 1)th year. Now, if we suppose those who complete their mh year, but die beforecompleting their (m + 1)th year, to die at equal intervals therein, so that for every one who dies before the expiration of a half of the year some other will survive so much more than the half-year, then each person who dies in the year survives, upon an average, one-half of that Im − lm + 1 year. Hence is the number of years enjoyed by all those who die in the (m + 1)th year; add to this the number 7m+1, who complete their (m + 1)t year, and we have 1+1+2 (Im − lm + 1) 1 lm 2 = }} (m + 4 + 1) = the number of years enjoyed in the first year by appears that m these 7 persons or the survivors. In a similar manner, it 1 1 1⁄2 (m + 1 + lm + D) ; } (m + 2 + 1m+3); } (lm +3 + lm + 4), etc., 12 2 is the number of years that will be enjoyed in the 2nd, 3rd, 4th, etc., years by these survivors. Hence the total number of years enjoyed by these persons until they all cease to exist is This expression, divided by lm, the number of persons amongst whom this quantity of existence is divided, gives em, the expectation of life of a person aged m; hence 1 Im 2 + λ + λ + + 1 + Im + 2 +2 Im etc. Hence, to find the expectation of life, divide the sum of the number of those who complete each age above the given one by the number living at the given age, and to the quotient add half unity. The following is the method of calculating a table of expectations, according to the Northampton rate of mortality: 196 = 1 1 The duration of life that a person has the present expectation of enjoying after a given period (t years) is found by multiplying the expectation at the advanced age by the probability the person has of attaining that age. The expression is, therefore, 160. To find the present value of an annuity of 11. payable at the end of every year during the existence of a single life. Let m denote the age of the person during whose life the annuity is to be continued, and am the present value of the annuity; let also v = 1 1 be the present value of 17. due at the end of one year; 1+r R then the present value of the first year's payment of the annuity is found by multiplying by the probability of the person living one year; the present value of the second year's payment by multiplying the present value of 17. due at the end of two years by the propability of the person living two years, and so on. Hence the whole value required is am=vpm, 1+ v2 Pm, 2 + v3 Pm, 3 + v* Pm, 4 + ̧ v \ m + 1 + v2 lm + 8 + vi3 la + 8 + v1 lm + + + +4 = .... to the end of the tables .... . (1). Hence a lv lm + 1 + v2 lm + 2 + v3 lm + 3 + v1 lm + s + . · · Also, the present value of the annuity on a life one year older is vlm + 2 + v2 lm + 3 + 23 lm + + + v2 lm +5 + v Am + 1 = lm + 1 consequently we have = am lm v lm +1 by (1'); amlm = vlm +1 (1 + ɑm + 1), or ɑm = v \1 (1 + Am + 1) (2). Hence it appears that the present value of an annuity at any age may be deduced from the value at the age one year older; and if we commence at the oldest age in the table at which the value of the annuity is 0, and proceed through all the other ages to the time of birth, a Table of the present values of Annuities" will be formed. Thus by the Northampton rate of mortality, of 11650 persons born, 1 survives 96 years, 4 survive 95 years, 9 survive 94 years, 16 survive 93 years, and so on; hence, calculating at 3 per cent., we get (1 + a‰o) = 1 × ·970874 × (1 + 0) = ·2427. 1.03 In a similar manner the present value of an annuity which depends on the joint existence of two persons aged m and m, respectively is v lm + 1 lm, + 1 + v2 lm + 2 lm + 2 + v3 lm + 3 l m or, as before, ɑm, my, my ASSURANCES ON LIVES. (3); (4). (5). 161. When an engagement is entered into to secure the payment of a sum on the death of a person, in consideration of a single or annual payment, the transaction is called an assurance on the life of that person, and the stipulated payment is called the premium. 162. To find the present value (P) of 11. to be paid at the end of the year in which a person shall die. The probability of his dying in the first year is 1 - A1, by (156), and the value of the first year's expectation is v (1-λ1). Again, the probability of his dying in the second year is A,-λ, by (158), and the value of the second year's expectation is v2 (A, — λ), and so on. But the whole value of the expectation is the sum of all these contingent values; therefore the present value is - P = v (1 − x,) + v2 (λ, − λ2) + v3 (λq − λg) + = v + v (v λ, +22 λ3 + 223 λ3 +...)-(v), + v2 λg + v3 λz +...) =V- (1 v) am by equation (1) Art. (160) (1), or P = 1 - (1 − v) (1 + am), which is well adapted for computation. -- Similarly, the single premium, or present value of an assurance on the failure of the joint existence of two lives is r− (1 − r) ɑm, on for any number of lives. and so 163. To find the annual premium (p) that must be paid to secure \l. at death. Here we must find the present value of 17. by the preceding proposition, and then find how much must be paid annually to form an equivalent to the present value. Now since the annual payment is made at the beginning of each year, there will be one payment more than for an annuity; hence (160) we have and consequently p (1+a) is the present value of 17. at the end of the year in which the assured dies. But by the last proposition that value is P = v − (1 − v) a„ ; whence p(1a) = P, or p = P 1+ am = v − (1 − r) am 1 − (1 − v) (1 + am) = 1 1+ am 1+am (1 - v). Similarly the annual premium on two lives will be found to be The annual premium is equal to the single premium divided by the annuity on the given life increased by unity. 164. To find the single and annual premiums of a temporary t assurance of a single life. Let am, denote the value of an annuity of 17. payable at the end of every year during t years only, instead of the whole life; then (160) we have = v (1 − λ, ) + v2 (λ, − λ2) + v3 (λ2 − λ3) + + v2 (^_^) = v + v (v λ, + v2 λ2 +. − (v λ1 + v2 λy + 23 λ3 + . . . . + v' \;). v1 − Hence P+ v (Am, t − v' λ,) — am, t = v(1 − v' λ) − (1 − v) am, t• Now since the number of annual payments will be t, consisting of an immediate payment, and of a temporary annuity for t1 years, the single premium must therefore be divided by In exactly the same manner we can find the single and annual premiums to secure a sum payable at the end of the year in which any number of joint lives shall fail, provided the event happen within t years. For further information on this important subject we must refer to the valuable Treatise on Annuities in the Library of Useful Knowledge. EXAMPLE. Find the single and annual premiums that would be required to |
