(1). Show from first principles that for all integers x, n, and all fixed interest-rates and life-distributions
ax:ne = ¨ax:ne − 1 + vnnpx
(2). Show from first principles that for all integers x, and all fixed interest-rates and life-distributions
Ax = v ¨ax − ax
Show further that this relation is obtained by taking the expectation on both sides of an identity in terms of present values of payment-streams, an identity whoch holds for each value of (the greatest integer [T ] less than or equal to) the exact-age-at-death random variable T .
(3). Using the same idea as in problem (2), show that (for all x, n, interest rates, and life-distributions)
A1x:ne = v ¨ax:ne − ax:ne
(4). Suppose that a life aged x (precisely, where x is an integer) has the survival probabilities px+k = 0.98 for k = 0, 1, , . . . , 9. Suppose that he wants to purchase a term insurance which will pay $30,000 at the end of the quarter-year of death if he dies within the first five years, and will pay
$10,000 (also at the end of the quarter-year of death) if he dies between exact ages 5, 10. In both parts (a), (b) of the problem, assume that the interest rate is fixed at 5%, and assume wherever necessary that the individual’s distribution of death-time is uniform within each whole year of age.
(a) Find the net single premium of the insurance contract described.
(b) Suppose that the individual purchasing the insurance described wants to pay level premiums semi-annually, beginning immediately. Find the amount of each semi-annual payment.
(5). Re-do problem (4) assuming in place of the uniform distribution of age at death that the insured individual has constant force of mortality within each whole year of age. Give your numerical answers to at least 6 significant figures so that you can compare the exact numerical answers in these two problems.
(6). Using the exact expression for the interest-rate functions i(m), d(m) respectively as functions of i and d, expand these functions in Taylor series about 0 up to quadratic terms. Use the resulting expressions to approximate the coefficients α(m), β(m) which were derived in the Chapter.
Hence justify the so-called traditional approximation
¨a(m)x ≈ ¨ax − m − 1 2m
(7). Justify the ‘traditional approximation’ (the displayed formula in Exer-cise 6) as an exact formula in the case (i) in the limit i → 0, by filling in the details of the following argument.
No matter which policy-year is the year of death of the annuitant, the policy with m = 1 (and expected present value ¨ax) pays 1 at the beginning
of that year while the policy with m > 1 pays amounts 1/m at the beginning of each 1/m’th year in which the annuitant is alive. Thus, the annuity with one payment per years pays more than the annuity with m > 1 by an absolute amount 1 −(Tm−[T ]+1/m). Under assumption (i), Tm−[T ] is a discrete random variable taking on the possible values 0, 1, . . . , (m −1)/m each with probability 1/m. Disregard the interest and present-value discounting on the excess amount 1 − (Tm− [T ])/m paid by the m-payment-per year annuity, and show that it is exactly (m − 1)/2m.
(8). Give an exact formula for the error of the ‘traditional approximation’
given in the previous problem, in terms of m, the constant interest rate i (or v = (1 + i)−1), and the constant force µ of mortality, when the lifetime T is assumed to be distributed precisely as an Exponential(µ) random variable.
(9). Show that the ratio of formulas (5.14)/(5.13) is 1 whenever either qx+k or i is set equal to 0.
(10). Show that the ratio of formulas (5.15)/(5.13) is 1 whenever either qx+k or i is set equal to 0.
(11). For a temporary life annuity on a life aged 57, with benefits deferred for three years, you are given that µx = 0.04 is constant, δ = .06, that premiums are paid continuously (with m = ∞) only for the first two years, at rate P per year, and that the annuity benefits are payable at beginnings of years according to the following schedule:
Year 0 1 2 3 4 5 6 7 8+
Benefit 0 0 0 10 8 6 4 2 0
(a) In terms of P , calculate the expected present value of the premiums paid.
(b) Using the equivalence principle, calculate P numerically.
(12). You are given that (i) q60 = 0.3, q61 = 0.4, (ii) f denotes the probability that a life aged 60 will die between ages 60.5 and 61.5 under the assumption of uniform distribution of deaths within whole years of age, and (iii) g denotes the probability that a life aged 60 will die between ages 60.5 and 61.5 under the Balducci assumption. Calculate 10, 000 · (g − f) numerically, with accuracy to the nearest whole number.
(13). You are given that S(40) = 0.500, S(41) = 0.475, i = 0.06, A41 = 0.54, and that deaths are uniformly distributed over each year of age. Find A40 exactly.
(14). If a mortality table follows Gompertz’ law (with exponent c), prove that
µx = Ax
.a0x
where Ax is calculated at interest rate i while a0x is calculated at a rate of interest i0 = 1+ic − 1.
(15). You are given that i = 0.10, qx = 0.05, and qx+1 = 0.08, and that deaths are uniformly distributed over each year of age. Calculate A1x:2e . (16). A special life insurance policy to a life aged x provides that if death occurs at any time within 15 years, then the only benefit is the return of premiums with interest compounded to the end of the year of death. If death occurs after 15 years, the benefit is $10, 000. In either case, the benefit is paid at the end of the year of death. If the premiums for this policy are to be paid yearly for only the first 5 years (starting at the time of issuance of the policy), then find a simplified expression for the level annual pure-risk premium for the policy, in terms of standard actuarial and interest functions.
(17). Prove that for every m, n, x, k, the net single premium for an n-year term insurance for a life aged x, with benefit deferred for k years, and payable at the end of the 1/m year of death is given by either side of the identity
An+k mxe − Ak mxe = kEx An mx+ke
First prove the identity algebraically; then give an alternative, intuitive ex-planation of why the right-hand side represents the expected present value of the same contingent payment stream as the left-hand side.
5.6 Worked Examples
Overview of Premium Calculation for Single-Life Insurance & Annuities Here is a schematic overview of the calculation of net single and level pre-miums for life insurances and life annuities, based on life-table or theoretical survival probabilities and constant interest or discount rate. We describe the general situation and follow a specific case study/example throughout.
(I) First you will be given information about the constant assumed in-terest rate in any of the equivalent forms i(m), d(m), or δ, and you should immediately convert to find the effective annual interest rate (APR) i and one-year discount factor v = 1/(1 + i). In our case-study, assume that
the force of interest δ is constant = − ln(0.94)
so that i = exp(δ) − 1 = (1/0.94) − 1 = 6/94, and v = 0.94. In terms of this quantity, one immediately answers a question such as “what is the present value of $1 at the end of 712 years ?” by: v7.5.
(II) Next you must be given either a theoretical survival function for the random age at death of a life aged x, in any of the equivalent forms S(x + t), tpx, f (x + t), or µ(x + t), or a cohort-form life-table, e.g.,
l25 = 10,000 0p25 = 1.0 l26 = 9,726 1p25 = 0.9726 l27 = 9,443 2p25 = 0.9443 l28 = 9,137 3p25 = 0.9137 l29 = 8,818 4p25 = 0.8818 l30 = 8,504 5p25 = 0.8504
From such data, one calculates immediately that (for example) the probabil-ity of dying at an odd attained-age between 25 and 30 inclusive is
(1 − 0.9726) + (0.9443 − 0.9137) + (0.8818 − 0.8504) = 0.0894 The generally useful additional column to compute is:
q25 = 1 −1p25 = 0.0274, 1p25 − 2p25 = 0.0283, 2p25 − 3p25 = 0.0306
3p25 − 4p25 = 0.0319, 4p25 − 5p25 = 0.0314
(III) In any problem, the terms of the life insurance or annuity to be purchased will be specified, and you should re-express its present value in terms of standard functions such as ax:ne or A1x:ne. For example, suppose a life aged x purchases an endowment/annuity according to which he receives
$10,000 once a year starting at age x+1 until either death occurs or n years have elapsed, and if he is alive at the end of n years he receives $15,000.
This contract is evidently a superposition of a n-year pure endowment with face value $15,000 and a n-year temporary life annuity-immediate with yearly payments $10,000. Thus, the expected present value (= net single premium) is
10, 000 ax:ne + 15, 000npxvn In our case-study example, this expected present value is
= 10000 ¡0.94(0.9726) + 0.942(0.9443) + 0.943(0.9137) + + 0.944(0.8818) + 0.945(0.8504)¢ + 15000(0.945 · 0.8504)
The annuity part of this net single premium is $38,201.09 , and the pure-endowment part is $9,361.68 , for a total net single premium of $47,562.77 (IV) The final part of the premium computation problem is to specify the type of payment stream with which the insured life intends to pay for the contract whose expected present value has been figured in step (III). If the payment is to be made at time 0 in one lump sum, then the net single premium has already been figured and we are done. If the payments are to be constant in amount (level premiums), once a year, to start immediately, and to terminate at death or a maximum of n payments, then we divide the net single premium by the expected present value of a unit life annuity
¨ax:ne. In general, to find the premium we divide the net single premium of (III) by the expected present value of a unit amount paid according to the desired premium-payment stream.
In the case-study example, consider two cases. The first is that the pur-chaser aged x wishes to pay in two equal installments, one at time 0 and one after 3 years (with the second payment to be made only if he is alive at that time). The expected present value of a unit amount paid in this fashion is
1 + v33px = 1 + (0.94)30.9137 = 1.7589
Thus the premium amount to be paid at each payment time is
$ 47, 563 / 1.7589 = $27, 041
Alternatively, as a second example, suppose that the purchaser is in effect taking out his annuity/endowment in the form of a loan, and agrees to (have his estate) repay the loan unconditionally (i.e. without regard to the event of his death) over a period of 29 years, with 25 equal payments to be made every year beginning at the end of 5 years. In this case, no probabilities are involved in valuing the payment stream, and the present value of such a payment stream of unit amounts is
v4a25e = (0.94)4(.94/.06) (1 − (0.94)25) = 9.627 In this setting, the amount of each of the equal payments must be
$ 47, 563 / 9.627 = $ 4941
(V) To complete the circle of ideas given here, let us re-do the case-study calculation of paragraphs (III) to cover the case where the insurance has quarterly instead of annual payments. Throughout, assume that deaths within years of attained age are uniformly distributed (case(i)).
First, the expected present value to find becomes 10, 000 a(4)x:ne + 15, 000 Ax:ne1 = 10000³
¨a(4)x:ne− 1
4(1 − vnnpx)´
+ 15000 vnnpx
which by virtue of (5.6) is equal to
= 10000 α(4) ¨ax:ne − (1 − vn npx) (10000β(4) + 2500) + 15000 vnnpx
In the particular case with v = 0.94, x = 25, n = 5, and cohort life-table given in (II), the net single premium for the endowment part of the contract has exactly the same value $9361.68 as before, while the annuity part now has the value
10000 (1.0002991) (1 + 0.94(0.9726) + 0.942(0.9443) + 0.943(0.9137) + + 0.944(0.8818)) − (6348.19) (1 − 0.945(0.8504)) = 39586.31 Thus the combined present value is 48947.99: the increase of 1385 in value arises mostly from the earlier annuity payments: consider that the interest on the annuity value for one-half year is 38201(0.94−0.5− 1) = 1200 .