Annuities under random rates of interest
Abraham Zaks
∗Department of Mathematics, Technion-Israel Institute of Technology, 32000 Haifa, Israel
Received 1 December 1998; received in revised form 1 June 2000
Abstract
We investigate the accumulated value of some annuities-certain over a period of years in which the rate of interest is a random variable under some restrictions. We aim at the expected value and the variance of the accumulated value, and we suggest two methods to derive these moments. In some cases both methods have similar difficulties, while in other cases one method is significantly preferable in terms of the simplicity of calculations. The novelty of our second approach lies in the fact that we find recursive relationships for the variance of the accumulated values, and solve these relationships, whilst previous investigators first obtained the second moment. © 2001 Elsevier Science B.V. All rights reserved.
Subj. Class: IE50; IE51
Keywords: Mathematics of finance; Random rates of interest; Random variable; Independent variables; Expected value; Variance; Annuities; Future value
1. Introduction
Many attempts have been made to investigate stochastic interest rate models and to evaluate their impact under various hypotheses: see Pollard (1971), Boyle (1976), Wilkie (1976), Westcott (1981) and McCutcheon and Scott (1986) to mention just some of the contributions. We investigate the accumulated value of some annuities-certain over a period of years in which the rate of interest is a random variable under some restrictions. We aim at the expected value and the variance of the accumulated value, and we suggest two methods to derive these moments. In some cases both methods have similar difficulties, while in other cases one method is significantly preferable in terms of the simplicity of calculations.
The novelty of our second approach lies in the fact that we find recursive relationships for the variance of the accumulated values, and solve these relationships, whilst previous investigators first obtained the second moment.
2. The case of a fixed interest rate
Suppose that the yearly rate of interestj is fixed through the period ofnyears. We recall that the yearly rate of discountd is given by
(1+j )d =j (2.1)
∗Tel.:+972-4-829-4028/972-4-822-3108; fax:+972-4-832-4654.
E-mail address: [email protected] (A. Zaks).
and the yearly present valuevis given by
(1+j )v=1, (2.2)
so that we have the equalities
d =jv, (2.3)
v+d =1. (2.4)
We assumek≤nthroughout, unless otherwise specified. The accumulated value afterkyears (denoteds¨k¯|j) of
an annuity-due ofkyearly payments of 1 is known to be given by the formula
¨
sk¯|j =
(1+j )k−1
d (2.5)
and we note that
¨
sk¯|j =(1+j )
k+ · · · +(1+j )=(1+j )(1+ ¨s
k−1|j). (2.6)
The accumulated value afterkyears of an increasing annuity-due ofkyearly payments of 1,2, . . . , k, respectively, is denoted(Is)¨ k¯|j, and is known to be given by the formula
(Is)¨ k¯|j = ¨
sk¯|j−j
d . (2.7)
We also note that
(Is)¨ k¯|j =(1+j )k+2(1+j )k−1+ · · · +k(1+j )=
1+j
(Is)¨ k−1|j+k. (2.8)
The accumulated value afterkyears of an increasing annuity-due ofkyearly payments of 12,22, . . . , k2, respectively, is denoted by(I2s)¨ k¯|j.
Theorem 2.1.
(I2s)¨ k¯|j =
2(Is)¨ k¯|j− ¨sk¯|j−k2
d (2.9)
Proof. By definition,(I2s)¨ k¯|j =(1+j )k+22(1+j )k−1+ · · · +k2(1+j ). A straightforward calculation of (I2s)¨ k¯|j −v(I2s)¨ k¯|jleads to the following equation:
d(I2s)¨ k¯|j =(1+j )k+ · · · +(2k−1)(1+j )−k2, from which we obtain
d(I2s)¨ k¯|j =2[(1+j )k+ · · · +k(1+j )]−[(1+j )k+ · · · +(1+j )]−k2
and the result follows by using (2.6) and (2.8).
We observe that
(I2s)¨ k¯|j =(1+j )k+22(1+j )k−1+ · · · +k2(1+j )=(1+j )((I2s)¨ k−1|j+k2). (2.10)
Corollary 2.1.
(I2s)¨ k¯|j =
(1+v)(s¨k¯|j+k2)−2k−2k2]
d . (2.11)
(Whenever the value of the rate of interestj is clear from the context, it is customary to omit it from the symbols and denote the corresponding values ass¨k¯|,(Is)¨ k¯|and so on.)
Similar arguments may be used to show that, for givenm=3,4, . . ., a recursive formula of type (2.9) can be derived for(Ims)¨ k¯|j, which is defined as the accumulation of an annuity-due ofkpayments of 1m,2m, . . . , km.
3. Single payments and series of level payments
Let us suppose that the yearly rate of interest in thekth year isik. We assume that, for eachk, we haveE(ik)=j and Var(ik)=s2, and thati1, . . . , inare independent random variables. We write
E(1+ik)=1+j =µ, (3.1)
E[(1+ik)2]=(1+j )2+s2=1+f =m, (3.2)
from which it follows that
f =2j+j2+s2, (3.3)
Var(1+ik)=m−µ2. (3.4)
We definerto be the solution of
1+r=1+f
1+j . (3.5)
That is,
r= f −j
1+j (3.6)
or, using (3.3), we have
r=j+ s
2
1+j. (3.7)
For ak-year variable annuity-due with payments ofc1, . . . , ck, respectively, we denote the accumulated value after kyears byCk. We assume thatc1=1 and we write
E(Ck)=µk, (3.8)
E(Ck2)=mk. (3.9)
It follows thatµ1=µ,m1=m, and
Var(Ck)=mk−µ2k. (3.10)
Example 3.1. Consider the case whenc1 = 1, c2 = c3 = · · · = ck = 0. This is, in fact, the case of a single investment at the beginning of the first year. In this case we have
From this we easily find that (see, e.g., Kellsion (1975, Ch. 11))µk =µk−1µand hence
µk =µk. (3.12)
We also havemk =mk−1m, and hence
mk=mk. (3.13)
Therefore
Var(Ck)=mk−µ2k =((1+j )2+s2)k−(1+j )2k. (3.14)
In particular, afternyears we have
E(Cn)=(1+j )n and Var(Cn)=((1+j )2+s2)n−(1+j )2n. (3.15)
Example 3.2. In this example we assumec1 = · · · = cn = 1; i.e.,Ck is the accumulation afterkyears of an annuity-due ofkyearly payments of 1. In this case we have
Ck =(1+ik)(1+Ck−1) for k=2, . . . , n. (3.16)
From this we find by straightforward reasoning, that
µk =µ(1+µk−1), (3.17)
mk=m(1+2µk−1+mk−1). (3.18)
Applying (2.6) to (3.17) we obtain the following result.
Theorem 3.1. IfCk denotes the future value after k years of an annuity-due of k yearly payments of 1 and if the
yearly rate of interest during the kth year is a random variableiksuch thatE(1+ik)=1+jand Var(1+ik)=s2,
andi1, . . . , inare independent variables, thenµk =E(Ck)= ¨sk¯|j; in particular,µn=E(Cn)= ¨sn¯|j.
The expected value ofCkis given thus in Theorem 3.1, so that using (2.5), we have a formula to computeE(Ck). To derive the value of Var(Ck)we can use (3.10) for which we need the value ofE(Ck2)=mk. The value ofmkis given recursively in (3.18), but we wish to find a closed form. We shall use induction, here and in the other examples. We note that one could use difference equations, as in Kellsion (1975).
Lemma 3.1. Under the assumptions of Theorem 3.1 we have
mk=(m+ · · · +mk)+2(ms¨k−1|j +m 2s¨
k−2|j+ · · · +m k−1s¨
¯ 1|j). Proof. We proceed by induction. Let
M1k =m+ · · · +mk, (3.19)
M2k =ms¨k−1|j+ · · · +mk−1s¨1|¯j, (3.20) so we must show thatmk = M1k+2M2k. Whenk = 2, this follows by (3.18), sinceµ1 = ¨s1|¯j andm1 = m.
Assuming our result is true for a givenk(2≤k≤n−1), it follows by formula (3.18) that it is also true fork+1.
This completes the proof by induction.
Since, by (3.2),m=1+f, we apply (3.11) to find that
We recall the value ofras given in (3.5), and that ofmas given in (3.2), and apply (3.11) to derive
By applying the value ofdfrom (2.1), we easily derive
M2k = Lemma 3.2. Under the assumptions of Theorem 3.1, we have
mk=
the equality follows as stated.
We have thus reached a formula forE(Ck2).
We have thus reached the following theorem.
Theorem 3.2. Under the assumption of Theorem 3.1,E(Ck)= ¨sk¯|j and
Var(Ck)=
2(1+j )k+1s¨k¯|r −(2+j )s¨k¯|f −(1+j )s¨2k|j+2(1+j )s¨k¯|j]
j . (3.25)
4. Varying series of payments
A natural extension is the case in whichci =ifori=1, . . . , nwhich corresponds to the increasing annuity-due ofnpayments of 1,2, . . . , n. In this example, we have
Ck =(1+ik)(Ck−1+k). (4.1)
From this we derive, by straightforward reasoning,
µk =µ(µk−1+k) for k=2, . . . , n, (4.2)
mk=m(mk−1+2kµk−1+k2) for k=2, . . . , n. (4.3)
Sinceµ=1+j =(Is)¨ 1|¯japplying (2.8) to (4.2) we obtain the following theorem.
Theorem 4.1. IfCk denotes the future value after k years of an increasing annuity-due of k yearly payments of 1, . . . , k, and if the yearly rate of interest during the kth year is a random variable,ik, so thatE(1+ik)=1+j
and Var(1+ik)=s2, and so thati1, . . . , inare independent variables, then µk =E(Ck)=(Is)¨ k¯|j for k=1, . . . , n
and, in particular,µn=E(Cn)=(Is)¨ n¯|j.
To overcome the inconvenience of the recursive equation for mk (k = 2, . . . , n), we proceed as in Example 3.2.
Lemma 4.1. Under the assumption of Theorem 4.1, we have
mk=(mk+22mk−1+ · · · +k2m)+2(2mk−1(Is)¨ ¯1|j + · · · +km(Is)¨ k−1|j).
Proof. Using (4.3), the proof follows by induction, using straightforward computations (like those in Lemma 3.1).
We write
M1k =mk+22mk−1+ · · · +k2m, (4.4)
M2k =2mk−1(Is)¨ 1|¯j+ · · · +km(Is)¨ k−1|j. (4.5)
So we havemk=M1k+2M2k.
Sincem=1+f, we apply (2.10) to obtain
M1k =(Is)¨ k¯|f (4.6)
M2k=
Hence we have the following lemma.
Lemma 4.2. Under the assumptions of Theorem 4.1, we have
M2k =
(1+j )k+2(Is)¨ k¯|r −(1+j )(Is)¨ k¯|f −j (1+j )(I2s)¨ k¯|f
j2 . (4.7)
Hence we have the following theorem.
Theorem 4.2. Under the assumptions of Theorem 4.1, we have
mk=
Proof. By (2.7) and Lemma 3.3, we have
so that
(Is)¨ 2¯ k|j =
(Is)¨ 2k|j−2(1+kd)(Is)¨ k¯|j −k2
d2 .
Summarizing, we obtain the following theorem.
Theorem 4.3. Under the assumption of Theorem 4.1,E(Ck)=(Is)¨ k¯|j and
Var(Ck)=
2(1+j )k+2(Is)¨ k¯|r−(2+2j −j2)(Is)¨ k¯|f −(j+j2)(I2s)¨ k¯|f j2
−(Is)¨ 2k|j−2(1+kd)(Is)¨ k¯|j−k 2
d2 . (4.9)
We have thus established formulae to evaluateE(Ck)and Var(Ck)in terms of future values of increasing annuities due for periods ofkand 2k, and in terms of rates of interestj, f andr.
The next example is the decreasing case, i.e., the case in whichci =n−i+1 fori=1, . . . , n. In this example, we have
Ck =(1+ik)(Ck−1+n−k+1). (4.10)
From this we derive, by straightforward reasoning,
µk =µ(µk−1+n−k+1), µ=1+j, (4.11)
mk=m(mk−1+2(n−k+1)µk−1+(n−k+1)2), m=1+f. (4.12)
Observe thatµ1=n(1+j ),m1=n2(1+f )and Var(C1)=n2s2, and recall that(Is)¨ k¯|j+µk =(n+1)s¨k¯|j.
Hence Theorem 4.4 follows.
Theorem 4.4. IfCk denotes the future value after k years of the decreasing annuity-due of k yearly payments of n, n−1, . . . , n−k+1, and if the yearly rate of interest during the kth year is a random variable ik, so that E(1+ik)=j and Var(1+ik)=s2and so thati1, . . . , inare independent variables, then
µk =E(Ck)=(n+1)s¨k¯|j−(Is)¨ k¯|j for k=1, . . . , n (4.13)
or, equivalently,
µk =
n−1
j
¨
sk¯|j+ k
d. (4.14)
Instead of findingmk, we shall find a recursive formula for Var(Ck)and hence an explicit expression for Var(Ck).
Var(Ck)=mk−µ2k=m(mk−1+2(n−k+1)µk−1+(n−k+1)2)−µ2k and we derive
Var(Ck)=m(mk−1−µ2k−1)+m(µk−1+n−k+1)2−µ2k =mVar(Ck−1)+
1+f (1+j )2µ
2 k−µ2k =mVar(Ck−1)+
s
1+j
2
µ2k, setting C0≡0.
Lemma 4.4. Under the hypothesis of Theorem 4.4, we have summarize the results in the below theorem.
Theorem 4.5. Under the assumption of Theorem 4.4, we have
As in the previous example, we can proceed directly to evaluate the variance of Ck, noting that µk =
One can easily verify that this equation holds in the case of the fixed annuity as well as in the case of the increasing one
It follows that (compare with the previous example)
Var(Ck)=
s
1+j
2
[mk−1µ21+ · · · +µ2k] and, using Lemma 3.3, we obtain
Var(Ck)= So, using the above notations, we can summarize as follows.
Acknowledgements
This research was supported by the Fund for the Promotion of Research at the Technion, and by the Research Fund of the Israeli Insurance Companies. The author would like to thank the referee for his many remarks and helpful suggestions.
References
Boyle, P.P., 1976. Rates of return as random variables. Journal of Risk and Insurance 53 (4), 693–713. Kellsion, S.G., 1975. Fundamentals of Numerical Analysis. Irwin, Homewood, IL.
McCutcheon, J.J., Scott, W.F., 1986. An Introduction to the Mathematics of Finance. Butterworth/Heinemann, London. Pollard, J.H., 1971. On fluctuating interest rates. Bulletin de l’Association Royale des Actuaires Belges 66, 68–97.
Westcott, D.A., 1981. Moments of compound interest functions under fluctuating interest rates. Scandinavian Actuarial Journal 4, 237–244. Wilkie, A.D., 1976. The rate of interest as a stochastic process — theory and applications. In: Proceedings of the 20th ICA, Vol. 1, Tokyo,
