AS/MATH2580 A&B Chapter 9 Notes prepared by R.L.W. Brown
Introduction

Chapter 9 of Mathematics of Finance by Zima-Brown is entitled Life Annuities and Life Insurance. Life annuities are annuities in which the payments are made only while the person making (or receiving) the payments is alive. In other words, the payments cease when the person dies. Life insurance is, typically, a single payment made, on the death of a person, to that person's named beneficiary. In both life annuity and life insurance calculations it is necessary to have information about human mortality.

Zima-Brown has the Canadian Institute of Actuaries 1982-88 Mortality tables on pages 325 to 327. These are tables of qx = the probability that a person aged x dies while still aged x. For example, the probability that a male smoker aged 20 dies while aged 20 is q20 = 1.31/1000 = 0.00131 according to the table on page 325.

Here is a more complicated probability calculation using the CIA tables of qx in the text:

Example 1: What is the probability that a male smoker aged 20 lives to reach age 30?

Solution: This is the probability that he does not die at age 20, and does not die at age 21, and ..., and does not die at age 29. Thus the answer is (1-q20)(1-q21)...(1-q29) = 0.987857. To get this result with a calculator requires 10 table lookups, 10 subtractions, and 9 multiplications.

In these notes, you will find an alternative and more traditional approach to mortality calculations. One result will be much less calculator work to get answers. Another result is conceptual: the Zima-Brown text does not indicate how the probabilities qx are calculated so it is hard to judge what they really mean. In the traditional approach given here, the methodology is clarified.

Building a Mortality Table

When someone dies, records are kept about the death including the person's age at the time of death, whether the person was male or female and (possibly) whether the person was a smoker or nonsmoker. Over a period of years, data builds up about the number of people who die at various ages. For a given population of interest (say all people, or all males, or all female nonsmokers) let dx be the number who die while aged x during the measurement period, say 1982-88. Then a table of values of dx for x = 0,1,¼,105 is the basic mortality table information. The final age of 105 is chosen somewhat arbitrarily based on several practical considerations: that there is no exact upper bound on life, the table should end somewhere, the data is too thin above this age, there are too few people above this age to make much of an impact on the business of life annuities and life insurance.

The Life Table

But the dx table by itself is still difficult to use. So the sum of all the dx entries is taken from x = 0 to 105 and this number l0 is considered to be the number of people in a hypothetical population of people of age 0. While they are age 0, d0 of them die leaving l1 = l0-d0 people of age 1 after 1 year. This happens each year. At the start of year x there are lx people, dx of them die during that year, and there are lx+1 = lx-dx people at age x+1.

The table of lx values is much easier to use in computations than either the dx values or the qx probabilities. For example, given a table of lx values, we can compute dx = lx-lx+1 and qx = dx/lx. So the life table can be used to get the dx and qx values easily. As the following example shows, the use of the life table simplifies other calculations greatly.

Here is the probability calculation of the Introduction above using the Life Table:

Example 1: What is the probability that a male smoker aged 20 lives to reach age 30?

Solution: There are l20 = 9914043 people living at age 20 and l30 = 9793655 people still living at age 30 so the probability of surviving from age 20 to age 30 is 9793655/9914043 = 0.987857. This calculation requires two table lookups and one division.

Probabilities

The notation introduced in Section 9.2 for certain mortality probabilities can be related to the life table lx notation in a simple way. For example the probability of a person aged x surviving to age x+n is npx = lx+n/lx. And the probability that a person aged x dying before age x+n is nqx = 1-npx = 1-lx+n/lx.

Example 1: On New Year's Eve, 2000 four friends agree to meet again on New Year's Eve, 2050. The four friends are a female nonsmoker aged 21, a male nonsmoker aged 24, a female smoker aged 25, and a male smoker aged 29. What is the probability that all four will survive to New Year's Eve, 2050?

Solution: We assume the mortalities of these people are independent so that the probability that all four survive is the product of the probabilities that each survives. That product is (using the life tables for the four different populations) (l71/l21)(l74/l24)(l75/l25)(l79/l29) = (8389318/9947529)(6885521/9877213)(6243596/9930430)(3522943/9805519) = 0.133.

Remark: We are using mortality data from 1982-88 to predict survival probabilities out to the year 2050. Like most predictions of the somewhat distant future, this one is subject to some questionable assumptions! Can we assume the the mortality data measured from people who died between 1982 and 1988 will continue to apply for the next half century? Probably not. There may be new developments in medicine that prolong lives. Or there may be new diseases and environmental hazards that shorten lives. We just don't know. But life insurance companies must estimate the future mortality every time they sell an insurance policy or a life annuity. So they do the best they can using mortality information now available.

Pure Endowments

If a person now aged x wants to buy a pure endowment that will pay \$R when (and if) the person reaches age x+n, what should the cost be now? (This cost is called the net single premium.) Let the cost now be X. If lx people each pay X now then the accumulated amount after n years will be lxX(1+i)n. On the other hand, each of the lx+n people who survive to age x+n must be paid R for a total of lx+nR. Setting these amounts equal gives X = R(1+i)-n(lx+n/lx). In the notation of the text, X = (R)(nEx).

Although life insurance companies rarely sell endowments anymore, the calculation of endowment values might arise in another manner.

Example 1: A male smoker aged 67 wants to borrow \$10000 now and pay it back with interest with a single payment at the end of 3 years. You feel that he will repay the loan if he is alive at 70 but you doubt that the loan will be repaid if he dies before age 70. Normally you would charge j1 = 12% for such a loan. What rate would be economically equivalent in this case, using CIA 1982-88 mortality?

Solution: The amount to be paid back in 3 years becomes an endowment R with net single premium \$10000. So 10000 = R(1.12)-3(l70/l67) and R = 10000(1.123((l67/l70) = (14049.28)(6961828/6201631) = \$15771.44. Let 10000(1+i)3 = 15771.44 and solve to get i = 16.4%.

Commutation Symbols

It is traditional to combine the interest rate and the life table into a table of commutation symbols Dx defined as follows:

 Dx = lx(1+i)-x
With this notation, the formula for the net single premium to a person aged x for an endowment of \$1 payable in n years is nEx = Dx+n/Dx. Thus given a Dx-table we can compute net single premiums for endowments by division without having to compute powers.

Example 1: Solve the example above about a 3-year loan of \$10000 to a male smoker aged 67 using commutation symbols.

Solution: The repayment R becomes an endowment at age 70 with present value \$10000. So 10000 = R(D70/D67) and R = 10000(D67/D70). Using tables of Dx at interest rate 12% we calculate R = \$15771.44. Then we compute i = 16.4% as above.

Life Annuities

A life annuity is just a sequence of pure endowments. For example, a whole life annuity immediate to a person aged x with payments of \$1 is a sequence of \$1 endowments payable at ages x+1,x+2,¼. So the net single premium is ax = (Dx+1/Dx)+(Dx+2/Dx)+... = (Dx+1+Dx+2+¼)/Dx.

It is traditional to define another commutation symbol Nx by

 Nx = Dx+Dx+1+Dx+2+...
and with this notation, ax = Nx+1/Dx.

Example 1: Find the net single premium now to a female smoker aged 32 for a whole life annuity with annual payments of \$40000 starting at age 33 if j1 = 5%.

Solution: We will need a table of commutation symbols Dx and Nx at interest rate i = 5%. This can be obtained from a spreadsheet model. Then the first net single premium = 40000N33/D32 = 40000(35740281.35/2075656.14) = \$688751.49.

Example 2: What would the net single premium now be in Example 1 if the payments were to start at age 65?

Solution: For the deferred `retirement' life annuity, the net single premium at age 32 is 40000N65/D32 = 40000(3934925.33/2075656.14) = \$75830.00.

Example 3: For the retirment annuity of \$40000 starting at age 65 of Example 2, what would be the net annual premium payable now at age 32 and every year until and including age 64?

Solution: The annual premiums form a temporary life annuity with payments at ages 32 to 64. We can represent this as a whole life annuity starting at age 32 minus a whole life annuity starting at age 65. Let R be the annual premium. Then the present value of the annual premiums should equal the present value of the benefit. Hence R(N32-N65)/D32 = 40000(N65/D32) and this gives R = 40000N65/(N32-N65) = 40000(3934925.33)/(37815937.48-3934925.33) = \$4645.59 per year ages 32 through 64.

Remark: If she dies before age 65, all the premiums she has paid will be kept in the pool of money needed to pay the retirement annuities of the others in the plan who do survive to age 65. If she survives to age 65 and starts to draw her \$40000 annuity then she is benefiting from those who entered the same plan but died before they could claim any benefits.

Example 4: For the retirement annuity of Example 3, suppose that the annual premiums paid at ages 32 through 64 would be returned with accumulated interest at j1 = 5% to the woman's estate in case of her death before age 65. What would her annual premiums be for this \$40000 per year retirement life annuity?

Solution: This question sounds complicated because it has a death benefit added to the previous question. But in fact it is simple! When you make regular deposits into a bank account (as opposed to regular premiums paid to an insurance company) your accumulated account is paid to your estate in case of your death. That is what is described in this problem. Hence, we can assume that she makes annual deposits into a bank account at ages 32 through 64. If she survives to age 65, her accumulated payments are used at age 65 to purchase a whole life annuity of \$40000 per year. Otherwise her bank account goes to her estate rather than remaining in a pool for the surviving plan members. Therefore, the annuity of premiums is actually a certain annuity of the kind studied in Chapter 3. Thus Rs33|0.05(1.05) = 40000(N65/D65) and R = \$5430.60 per year ages 32 through 64.

Life Insurance

Consider a life insurance policy taken out by a person now aged x that pays \$R to the person's estate in case the person dies during the one-year period when they are of age x+n. (Yes, it is true that such a 1-year policy is rather rare; but the reason that we study 1-year policies is that a more general policy with a term of many years is just a sequence of 1-year policies.) The standard assumption is that, in case of death in a given year, the benefit is paid at the end of the year of death. Accordingly, the payment of R, if made, will be made n+1 years from now. Let the net single premium payable now be X. Then you collect Xlx now and at the end of n+1 years you must pay Rdx+n. Hence Xlx = Rdx+n(1+i)-(n+1) and X = R(1+i)-(n+1)(dx+n/lx).

Example 1: A male nonsmoker now aged 46 wants a 1-year life insurance policy that will pay \$100000 in case he dies while at age 48. If annual interest is at 5%, what net single premium should he pay now?

Solution: The net single premium is 100000(1.05)-3(d48/l46) = 86383.76(23660/9658595) = \$211.61.

It is traditional to define commutation symbols Cx as follows:

 Cx = dx(1+i)-(x+1)
With this definition and using a table of Cx values, the calculation of the net single premium in Example 1 above becomes 100000C48/D46 = 100000(2166.4021/1023778.96) = \$211.61.

In a manner similar to life annuities, life insurance with a term of more than one year can be computed as a sum of one-year policies. To make this easier, the traditional approach is to introduce another commutation symbol Mx defined as follows:

 Mx = Cx+Cx+1+Cx+2+...
With a table of values of Mx we can now compute a wide variety of life insurance policy solutions.

Example 2: A male nonsmoker now aged 31 wants to purchase a whole life insurance policy that will pay \$100000 to his heirs whenever he dies. What is the net single premium payable now if interest is 5% per annum?

Solution: The net single premium is 100000M31/D31 = 100000(252630.577/2163750.1) = \$11675.59.

Example 3: The male nonsmoker now aged 31 decides that the whole life policy is too expensive and is considering a \$100000 term life policy that covers him from his current age for the next 25 years until his children are grown. What is the net single premium if interest is 5% per annum?

Solution: We take a whole life policy starting at age 31 and subtract a whole life policy starting at age 31+25 = 56. Thus the net single premium is 100000(M31-M56)/D31 = 100000(252630.577-204079.327)/2163750.1 = \$2243.85.

Example 4: Of course it is possible (usual!) to pay annual premiums for life insurance instead of single premiums. What would the annual premiums be in Example 2 if these premiums are required as long as the man is living?

Solution: Let R be the size of each annual premium. Then these premiums form a whole life annuity. The present value of that whole life annuity should be the same as the present value of the whole life insurance benefit. Hence RN31/D31 = 100000M31/D31 and R = 100000M31/N31 = 100000(252630.577/40133508.1) = \$629.48 per year for life.

Example 5: What are the annual premiums for the 25-year term insurance of Example 3? (These premiums will be paid annually starting at age 31 and continuing until 25 premiums have been made or until the man dies, whichever happens first.)

Solution: R(N31-N56)/D31 = 100000(M31-M56)/D31 so R = 10000(M31-M56)/(N31-N56) = \$153.35 per year, if alive, for 25 years.

File translated from TEX by TTH, version 1.98.
On 29 Feb 2000, 16:07.