CHAPTER VI VALUATION OF POLICIES 43. General Methods for Finding Policy Values.—We have seen how terminal reserves arise, either from the payment of single premiums or level annual premiums. The value of a policy at any time depends .upon the terminal reserve at that time. Thus in Sec. 35, we found the terminal reserve at the end of the fifth year to be $3,243,024. This reserve belongs to the 89,032 living persons at that time. The value of each of the $1000 policies outstanding is, therefore, 3,243,024/89,032 = 36.4, or approxi- mately $36. Again, in Sec. 39, with an annual premium, we found the terminal reserve at the end of the fifth year to be $64,000, or approximately $0.70 for each $1000 policy outstanding. It must be remembered that these policies are 10-year term policies and hence the small value of each. The symbol nVx is used to denote the value of an insurance of $1 at the end of the nth year, x being the age of the person to whom the policy was issued at the time of issue. There are four principal methods for finding the numerical value of nV,, namely: 1. Prospective method. 2. Retrospective method. 3. Accounting, or accumulation method. 4. Valuation in terms of cost. 44. Prospective Method.—The prospective method is based upon the following principle: The value of a policy at any time must equal the net single premium for future insurance less the value at that time of future premiums. 61 62 MATHEMATICS OF LIFE INSURANCE ;§44 Thus, if (a;) pays the single premium A, for his insurance of $1, the value of his insurance at the end of the nth year is nVx = A,+», (1) since there are no future premiums. If (x) pays the annual premium P,, the value of his policy at the end of the nth year will be nV^ = Ax+n — P, a,+,, (2) since P, a, +„ is the value of future premiums at age x + n. If the policy is a (-payment policy (0 re), then n7, = A,+, - (P, ,_,a,+,, (3) since there are but t — n future premiums to be paid. In case t<n, Eq. (1) holds, since there are no future premiums. Equations (2) and (3) may be somewhat simplified by replacing A, +n by its value in terms of an annual premium. Thus, Aa,+n = P»4.n a,+» == t-nPx+n [l-na.i+n- Equations (2) and (3) then become »7, =(?.+„- P.)&^n, (4) nV", = (.-A-), - (P,) |,_,a,-i,. (5) In words, the value of the policy must be the difference between what (x) would pay at age x + n and what he does pay. As a numerical example, suppose (a;) has an ordinary whole-life policy of $1000, issued at age twenty, and wishes to know its value at the end of the tenth year. We use Eq. (4) and have wVw = (Pso - ?2o) . aso = (0.01719 - 0.013477) X 19.605 = 0.07279. (6) The value of the policy at age thirty is, therefore, ,1ji72.79. It is hardly necessary to remark that the computation of nVx is effected by means of commutation symbols. Exercises 1. Calculate the value of a $1000, twenty-payment life policy to a person of age twenty-eight at the end of the tenth policy year. Also at the end of the twentieth policy year. 2. What is the value of a $5000, twenty-year endowment policy to a person of age thirty-five at the end of the fifth policy year? At the end of the twentieth policy year? §451 VALUATION OF POLICIES 63 46. Retrospective Method.—The retrospective method for computing the numerical value of nVx is based upon the following principle: The value of a policy at any time is equal to the accumulated value of the premiums paid since the issue of the policy less the net single premium for a term insurance covering the time elapsed since the issue of the policy. The value of nV. at the time of issue of the policy is nE, nV,, the present value of the premiums to be paid in n years is P, |»a*, and the net single premium for a term insurance for n years is |,A,. Hence, we have, n TI- n I 11 T7- vx ' "a:c — nAa;) ,., > nEx nVx = Px \nS.x - \nAx, Or nVx = —————p,—————— (1) n"r For a (-payment life policy (On), v v-P\»-\AnrV- ^lpx'^ ~ ^nAX)- (2) n-"» 'n' x ss tl x ' \n°.x nAi, Orn V x — Ei '.-/ n^x If, in these equations, we replace |nA, by its value Px:n\ |»a«» we get the somewhat more convenient equations, V - (P — p -\ . 1"^. f3) nV x — vx '-x:n\i p --' n-^x and
For example,
nVx = (.Px - Px-^ ^(4) For example, .^^(P^-Pzo:^).11^20- (5) 10" 20 But we have found Pw = 0.013477, Ps^ = 0.00776 (Sec. 38) ioa,o = 8.325 (Sec. 32), and 10^20 = 0.65384 (Sec. 29). Substituting in Eq. (5), 10^20 = 0.07279, which agrees with the result found in the preceding section. Exercise Solve Exercises 1 and 2 of the preceding section by means of the retrospective method of valuation and compare with the results found by the prospective method. 64MATHEMATICS OFLIFE INSURANCE (§46 46. Equivalence of Prospective and Retrospective Methods.— It is desirable to prove that the two methods for computing nVx, considered in the last two sections, will always give the same result irrespective of age or time or rate of interest. That is to prove that A —P -a — (P»la»~ 1»A.») /,,, Ax+n — fx a»+n — ———————„—————— (1) n^x is an identity for all values of x and re. To do this we should be able to reduce the right member of Eq. (1) to the left member without altering the latter (or vice versa). From Sec. 32we have|«a, = a, - «^ a.+,,, and, from Sec. 34, |nA. = A, — nEx- A^+n. Hence, (P,|.a, - |,.A,) _ P,(a» - nE, a^.) - (A, - ,E, Ax+n) "Ex nEl————————— = PX&XJ~A" + "E' A^" - P* nEx a,+» nEx——— But P. a,= A, (Sec. 36). Therefore the right member of Eq. (1) reduces to the left member, as it should. We have here proved the equivalence of the two methods for finding ,7. only in case of ordinary whole-life policies, but the same reasoning applies whatever the form of the policy. 47. Accounting or Accumulation Method.—The two methods considered for finding the value of »F, are useful when it is desired to find the value of a policy at some particular time. But when it is necessary to tabulate the value of the policy at the end of each year for a period of years, neither the prospective nor the retro- spective method is especially convenient. We then proceed as follows. At age x there are I, people each paying the premium P,. The fund created is, therefore l.Px. This fund amounts to lxPx(l + i) at the end of the first year. Since there are dx persons dying during the year, the death claims amount to rf, dollars. The terminal reserve is, therefore, ^P,(l + i) - d.. This reserve must be credited to the ;,+i policies still outstanding. Hence each policy has a value credited to it of ^ , mi + i) - d. lx+1 per dollar of insurance. §48] VALUATION OF POLICIES 65 At the beginning of the second year, each policyholder contributes to the fund the premium Px and also the value of his policy iVx. Hence, the fund is now li+i(Px + iV»). This fund goes on interest for 1 year, and the death claims subtracted from the amount leave a terminal reserve represented by h+i(.P. + i7«)(l + i) - <Ui. Distributing this reserve among the L+2 policies still outstanding, we have ^ = ^p' + 'y^1 + ^ - dx^ (2) lx+t Proceeding in a like manner, we find v ^(P»+2^)(l+z) -d^, /„. S'x = —————————————7——————————————— \d) H+3 V _ ^+3(P« + 3^)(1 + i) - d^ {A\ 4'x ~ 7 \ft) (1+4
-. lx+n-l(Px + n-l7»)(l + i) — d,+n-l ,.„,. nV i ———————————————,————————————————— \0) tt+n These equations enable one to compute terminal reserves and policy values year by year and to tabulate the results. 48. Faclder Valuation Symbols.—Equation (1) of the preceding section may be written y _ L(l + i) p d. ,.. iVz=—,——'*—;— (1) tl+l rx+l If we agree to write t L(l +0 _i 7 r dx ,„. Ui for ———- and fc, for ,—> (2) (*+1 lx+1 the series of equations in the preceding section become i7, = M.P, - &,, (1) 27, = M,+i(P, + iV^) - A;,+i, (2) sV, = M,+2(P, + 27.) - fc,+2, (3) and, in general, nVx = M.+,-l(P, + n-l7i) — fc+n-l. (4) The symbols Ux and fc, are called Tackier valuation symbols because first introduced by the actuary David Parks Fackler. 66
MATHEMATICS OP LIFE INSURANCE
They can be computed for all ages and for various rates of interest and tabulated once for all (Table III). As a matter of fact, k, does not depend upon the rate of interest, and is, therefore, the same for all rates of interest. The Fackler valuation symbols can be expressed in terms of commutation symbols. Thus, ^ Ul + i) ^ _l^_ _ v'l, _ D. l^i vl^ v^l^~ D^ w and kx = A = S'- = ^ ^- sw- 33)- (6) <-x+l V ti+i Ux+1 Or the valuation symbols can be expressed in terms of probability functions as follows:
(5)
Ux =
W + i) _ (1 + i)1 + i
and
k,=
tt+1
l^/l. dt _ l,+i I, ' I, "
p*
5*. P.
(7) (8)
The following table illustrates the use of the valuation symbols in computing policy values for an ordinary whole-life policy for $1000 issued to a person of age twenty.
§49] VALUATION OF POLICIES 67 This table exhibits the policy values year by year and can be continued as far as may be desired. In particular, the value at the end of the tenth year agrees practically with the result found in Sees. 44 and 46, as it should. Exercises 1. Check the tabular values of Mso and kso. 2. Compute P30:io| and prepare a table like that in this section, assuming $1000 of insurance and interest at rate 0.035. 49. Cost of Insurance.—The policy value, or reserve, ,7, is held by the insurance company but has been contributed by the insured and is, therefore, credited to his policy. In case of death, the company must pay $1 to the estate of the insured. Hence the amount at risk at the end of the roth policy year is 1 - ,7,, per $1 of insurance. Thus, for example, we have found in the preceding sections io72o = 0.07279. The amount at risk at the end of the tenth policy year is, therefore, 0.92721. The cos( of insurance is the mathematical expectation of having to pay the amount at risk. The age of the insured at the begin- ning of the nth policy year is x + n — 1 and the probability of his dying during the year is d,+n_i/;.r+n-i. Hence, the cost of insurance for the nth year is nK^^-^d-nV.). (1) d+n-l For example, if the insured is twenty years of age at the time of issue of his policy, he will be twenty-nine at the beginning of the tenth policy year and the probability of his death during the year is 0.008345 (Table II). Hence the cost of insurance for this yearis 0.008345 X 0.92721 = 0.007737, or $7.74 per $1000 of insurance (American Experience Table, Sy-t per cent). 50. Cost in Terms of Premium.—In Sec. 47 (Eq. (5)), we found V ' - ^-l(P» + n-lV»)(l + i) - d.+n-l. /^ n y x — 7v / fx+n 68 MATHEMATICS OF LIFE INSURANCE f§51 Substituting this value of nVx in Eq. (1) of the preceding section, we have, ,., rf,+»-iri lx+n-i(Px + n-iFJCI + i) - <^+.-i T »"* ~~i———I - ~ 71 l.t+n-1 L rz+n -' dx+n-1 Vix+n - L+n-l(P» + n-lVx)(l + Z') + ^+n-l'|. /^ — -———— ———,\"1 lx+n-1 L '"s+t ' But;i+n + dx+n-i = L+n-i and, hence, ^ = ^^ [l - (P. + »-i7.)(l + i)]. (3) fras+n Equation (3) admits of the following interpretation: At the beginning of the rath policy year each insured person furnishes Px + n-\Vx per dollar of insurance, which amounts to (.Px + n-iVx) (1 + i) at the end of the year. Hence, for each death during the year there must be a fund of 1 - (P, + n-i^) (1 + i) over and above the amount contributed by the deceased to pay the death claim. Since there are d»+n-i persons dying during the year, there must be a total fund of d,+,-i[l - (P. + n-iV.)(l + z)l to pay all death claims. This fund, distributed among the l^.n living policyholders at the end of the year, gives the cost of insurance per individual per dollar of insurance. Hence, Eq. (3). Since d^+n-i/lx+n = fc+»-i, Eq. (3) may be written . A = [1 - (P. + n-iT^) (1 + i)]fc.+n-i. (4) 61. Valuation in Terms of Cost.—Resuming the Equation „ Z,+.-l(P»+ n-lV,)(l + i) - rfx+n-1 nVx = ———————————————7————————————————' t-c+n and replacing ;,+B-I by its value ;,+„ + dx+n-i, we get „ (;.+„ + d.+n-i)(P. + n-iVxW + t) - d.^ nV x —'——————————————————'» ix+n _ lx+n(Px + n-lV.W +i)+ ^+n-l(P» + n-l^)(l + i) - ^+n-t _ lx^(.Px + n-lV.W + l) - [1 ^\Px + n-l^)(l + i)] rf^n-1 = (P. + n-lV^l + i) - nKx. (1) §511
VALUATION OF POLICIES
69
The expression 1 — (P, + n-iV,) (1 + i), may be called the residue of insurance over amount furnished by the insured. Equation (1) gives a convenient method for computing policy values year by year, as illustrated by the following table;
A study of this table shows how Eq. (1) is applied to determine policy values. The slight variation in cost is to be noted at these early ages. The importance of fully understanding methods for determining policy values or reserves cannot be overemphasized. The sur- render value of the policy depends upon the reserve. The amount of a loan depends upon the policy value. Insurance companies are required by law to hold reserves acquired by the overcharge during the younger ages in order to meet demands when the level premium becomes insufficient. Companies which do not have requisite reserves are insolvent. Exercises 1. A was thirty-three years of age when granted a $1000, 10-year endowment policy. Determine the amount at risk at the end of the fifth policy year and the cost of insurance for this year. 2. Make a table illustrating policy values for the policy in Exercise 1, using the method in this section.
70 MATHEMATICS OF LIFE INSURANCE [§52 52. Preliminary Term Valuation.—Reserves computed by the methods explained in the foregoing sections may be called net reserves. Net reserves are usually higher than those prescribed by law and hence a practice has arisen of lowering the net reserves, which is called preliminary term valuation. By full preliminary term valuation is meant the practice of dividing a policy, issued to a person of age x, into a 1-year term policy followed by a policy of like kind issued to a person of age .c + 1, and of computing net premiums and net reserves on these two policies. Thus, if (a) has an ordinary whole-life policy of 1, valued by the full preliminary term method, his first net premium is considered to be vd,/l, (Sec. 36) and the reserve at the end of the first year to be zero, since he is regarded to have paid a net premium just sufficient to meet the death claims at the end of this first year. The net premium for any future year is considered to be Pc+i, and the reserve at the end of the nth policy year to be n-iVi+i; that is, the (n — l)st net reserve on an ordinary whole- life policy of 1 issued to a person of age x + 1. Since nVx = A,+n — Px dr+n and n-iVx+i = A,+n — Px+i a*+n (c/. Sec. 44), the net reserve at the end of the nth policy year is lowered by the difference nVx - n-lVx+l = (P,+l - P,)a,-hl. Since a,+n approaches unity as n increases, the full preliminary term reserve on an ordinary whole-life policy is always less than the net reserve. This plan does not, of course, effect, the actual net premium charged to (a;), which is Px. At the beginning of the first year, therefore, there is a net fund denoted by ?„ — — freed for defray- ic ing the extra expenses incident to placing the policy. This explains the purpose of the plan. The following table exhibits the method of full preliminary term valuation on an ordinary whole-life policy of $1000 to a person of age twenty. The first net premium is taken to be $1000 X vdm/lw = $7.54 (Sec. 36) and the net premium for any other yeartobeP,+i == $13.77. §53]
VALUATION OF POLICIES
71
This table should be compared with the tables exhibiting net reserves in preceding sections. The same plan is employed for other policies. Thus, for a t-payment life policy to (x) of 1, the first net premium is considered to be vdx/lx, the same as for an ordinary whole-life policy, and the reserve at the end of the first year to be zero. Each of the remain- ing ( — 1 net premiums is regarded to be i-iPi+i, and the reserve at the end of the nth policy year (re < () is taken to be n-lVx+l = Ax+n — t-lPx+l ll-ndt-l-n. Since the net reserve is »7, = Ax+n — tPx {t-ndx+n, the full preliminary term reserve is less than the net reserve by (.t-lPx+l ~ <Px) ' l-nd.t+ir But here |<-n<t»+n approaches zero as n approaches (. Hence the full preliminary term reserve reaches the net reserve at the end of the premium paying period, as it should. Exercise 1. Use full preliminary term method of valuation on the policy in Exercise 1, Sec. 61, and make a table like the one in this section. What should be the value at the end of the term? 63.Modified Preliminary Term Valuation.—Full preliminary term valuation may release a greater fund than necessary for placing expenses, especially if the policy is an expensive one. For
72 MATHEMATICS OF LIFE INSURANCE [§53 example, a ten-payment life policy of $1000 to a person of age twenty requires a net annual premium of $1000 X itPzo = $34.23. If the policy is valued by the full preliminary term method, a fund of $34.23 — $7.54 = $26.69 is immediately released for use of the insurance company. If this fund is greater than the policy's share of placing expenses, the difference might conceivably be spread over the premium-paying period, say, in the form of an annuity increasing the surplus dividends return- able to the policyholder. But, for various reasons, the practice isquite different. A standard policy is selected as a basis and reserves computed on it by full preliminary term valuation. Sup- pose the given policy is a (-payment policy of 1 to (x). The net reserve on this policy at the end of the premium-paying period is Ax+f Let i-i^^i represent the full preliminary term reserve on the standard policy at the end of the premium-paying period. The difference A,+. - .-iV^i is then considered as an endowment for ( years on the life of (a;) whose present value is ,E. (A.+. - ,-i7^\). The net annual premium for this endowment is , _ ,E,(A^. - ,_i7^)
|ia. The reserve on the (-payment life policy at the end of the nth policy year (n < t) is then taken to be the reserve on the standard policy plus the accumulated premiums on the endowment for n years and is represented by
, ,y(.) + ^. (-It^+l-t- ^
Or, if P"i stands for the annual premium after the first year on the standard policy, one may consider the first net premium on the (-payment policy to be — + z and each of the remaining ( — 1 premiums to be -P' + z. For example, if the standard policy is an ordinary whole-life policy of $1000 to a person of age twenty, the full preliminary term reserve at the end of the tenth year was found to be practi- cally $67. The net reserve on a ten-payment life policy of $1000 §53] VALUATION OF POLICIES 73 to the same person at the end of the tenth year is Ago = $337. The difference, $270, considered as an endowment to (x = 20) for 10 years has a present value denoted by io2?2o $270 and a net annual premium equal to UEM1$270 = $21.21. 110&20 Accumulating this annuity-due for 1, 2, 3, . . . to 10 years and adding the amounts to the reserves shown in the table (Sec. 52), one gets the modified preliminary term reserves $22.12, $51.66, $82.42, . . . ,$336.96, the last being practically A so = $337. Or, if one considers the first net premium to be $7.54 + $21.21 = $28.75 and each of the remaining nine premiums to be $13.77 + $21.21 = $34.98, and computes reserves as was done in the pre- ceding section, he will find the same results as noted above. Clearly, any policy may be adopted as the standard policy. If the standard policy is an ordinary whole-life policy, other policies are said to be modified on an ordinary whole-life policy, or valued by the straight modified preliminary term method. If the standard policy is a twenty-payment life policy, other policies with larger premiums are said to be modified on a twenty- payment life policy. Insurance companies using a twenty-payment life policy as a standard and valuing policies with lesser premiums by the full preliminary term method are said to use the Illinois standard method of valuation. The two methods here mentioned are of most frequent use. Other methods have been devised, but it must be said that any method which lowers the net reserves on a policy also lowers its surrender values and thus effects policy options dependent upon surrender values. Exercises 1. Make a complete table for the first 10 years for the policy dis- cussed in this section, taking $28.75 for the first net premium and $34.98 for each of the nine remaining premiums. 2. Use the modified preliminary term method of valuation, Illinois standard, on the policy in Exercise 1, Sec. 61, and make a table of values.