ACTUARIAL EQUIVALENCE S. David Promislow ABSTRACT

Transcription

1 ACTUARIAL EQUIVALENCE S. David Promislow ABSTRACT This paper is a continuation of ideas introduced in Accumulation Functions" ARCH Criteria for actuarial equivalence with respect to a general two variable accumulation function are developed. 1. Introduction A basic and elementary fact of interest theory, is that when comparing two transactions with respect to compound interest, we need only consider an arbitrary single point of time. If the transactions have the same value at that point they will be actuarially equivalent. Moreover, it is easy to see that this principle does not hold in the simple interest case. We can ask the question of why this is true and attempt to produce the ultimate generalization of these results. This investigation was begin in [1] and we will begin by reviewing briefly some of the definitions of that work. The reader is referred to the original paper for more details. An accumulation function is a two variable positive.valued function, a, defined for all s, t in some set. B of the real line, such that a (t,s) d(s,t)-l for all sand t In B. (1.1) This means that necessarily a(s,s,) = 1, for all s in B (1.2) The interpretation is that for s ~ t, a (s,t) is the accumulated amount at time t of an original investment of 1 at time s, while for s ~ t, a (s,t) is the amount required at time s in order to

2 produce 1 at time t. This is based on the assumption that investments of amounts other than 1 yield proportional returns. In [1] B was restricted to be the nonnegative reals. It will be more convenient in this work to allow a general subset. When necessary to distinguish the particular subset we will refer to an accumulation function on the set B. An accumlation function is said to be Markov if, for some r in B a(s,t) = a(r,t) a(r,s) for all s, t In B ( 1.3) It is easy to see that if (1.3) holds for some r in B, then it will in fact hold for all r in B. An accumulation function is said to be stationary if, given and h where s, s+h, t, and t+h all are in B then s, t a(s,s+h) = a( t.t+h) (1.4 ) Suppose B is the nonnegative reals. The only continuous accumulation function which is both Markov and stationary is the compound interest function a () s,t = (1+ for some i > -1 1 ) t-s Consider a tr~n$/lction where all Si are in B. Here, (K i, Si) denotes a payment of K j at time Sj (As in [1]. we are considering only finite discrete transactions in this paper.) For any t in B we define

3 Valt(T) = n LKi a (Si,t» i=l and two transactions Sand T are said to be actuarially equivalent with respect to the accumulation function a if Valt(T) = Valt( S),for all t in B. ( Normally we will suppress the qualifing words with respect to a" if there is only one accumulation function under discussion. ) In [1),two extreme cases are considered. THEOREM 1.1. The accumulation function is Markov if and only if the following holds: For all transactions Sand T, Valt(S) = Valt(T) for some t in B implies that Sand Tare actuarially equivalent. The proof is straightforward and can be found in [11. The significance of the theorem is that it establishes the Markov property as characterizing the behaviour which we mentioned above for the compound interest case. The exact opposite behaviour is shown in the following.. EXAMPLE 1.2. [1, Appendix J accumulation function a(s, t) = 1 + i( t-s) For the simple interest s < t, 1 x 0, two nonidentical transactions cannot be actuarially equivalent

4 ( Transactions are identical if the payments have exactly the same amount at each time s, with missing payments considered to be of zero amount. ). As we mentioned in (1),a central problem of actuarial mathematics is as follows. Given two transactions with unknown parameters, solve for the parameters in order to make the transactions actuarially equivalent. For non-markov functions we cannot simply equate values at a single point to produce actuarial equivalence and this problem may be impossible to solve. We suggested in [1] the possibility of redefining - actuarial equivalence-, but on reflection we feel that this is not feasible. The given definition seems the only natural way to define this concept. We are able however to produce some criteria for actuarial equivalence which generalize Theorem 1.1, and can assist with this basic problem. The main results are Theorems 3.1 and 3.2 below. The extent to which Example 1.2 can be generalized is discussed in section 5. These results all hinge on a certain equivalence relatation which we define on B. This is the content of section 2. 2 The relation We define a relation - on B by r- s if a(r,s,> a(s,t) = a( r,t) all t in B. PROPOSITION is an equivalence relation. Proof. From (1.2). r - r for all r in B,so - is reflexive. If r - s, then for all t in B,., (s,r) -., (r,s)-1 -., (s,t) /., (r,t) for all t,which shows that s - r. So - is symmetric

5 To show transitivity, we note that if all t in B r - s, and s- u, then for a(r,u)a(u t) [a( r,s) a (s,u)] a(u,t) ], since r - s, = a(r,s) a(s,t) since s - u = a (r,t), since r - s. which shows that r - t and completes the derivation. It is then well known B is partitioned into pairwise disjoint subsets, known as - - equivalence classes, such that rand s are in the same subset if and only if r- s. A subset of B obtained by choosing exactly one point from each equivalence class is called a - - cross section. To say that r - s means the following. Take any third point t, and consider an investment made at the earliest time of the three.. If the proceeds are taken out from the fund at the middle of the three points and then immediately reinvested, the amount accumulated by the latest of the points will not be affected. EXAMPLES 2.2 (a) It follows directly from the definitions that a (s,t) is Markov iff there is exactly one - - equivalence class. (b) Take B to consist of more than 3 points. Consider the simple interest function, a (s, t) = 1+ i(s-t),i)( O. Then r - s implies that r = s. That is, the equivalence classes consist of just the single points. This foiiows immediately from the fact that (l+a)(l+b) = 1 + a + b implies a or b = o. (2.1) (c) Suppose that it is a function of t in B and Let

6 ,,(s,t) a (1 + it )t-s for s ~ t, In this case,accumulation is at compound interest, but the interest rate depends on the maturity date t. Take any r, (other than the smallest element of B if such exists.) For r < s, r - s iff it is constant on { t B : t ~ r} If this holds a direct calculation shows that r - s. Conversely suppose r - 5, For t ~ 5, ( 1 + is)s-r d( r,s) d(r,t)/ d(s,t) (l+it)s-r showing that For r < t s s it = is. ( 1 + it)t-r = d( r,t) = d(r,s)/ d(t,s) = (l+is)t-r showing again that Finally, choosing any u < r, we similarly deduce that (l+i r )r-u = (1 +is) r-u, and hg!ncg! ir = is. We see then that the equivalence classes reduce to single points, except I provided that it stabililizes at a constant valug! as t increases, in which case we will get one nontrivial equivalence class at the right tail of B

7 3. Main results We now need some additional notation. If A is any subset of B, we let aa denote the restriction of the function a to the set A. That is, da (s,t) = a( s,t) for all s,t, in A Similarly for any transaction T J T A denotes the restriction of T to the set A. That is.all paym~nts at times not in A are set equal to O. We let -A denote the equivalence relation - defined with respect to the function aa. THEOREM 3.1 (a) If A is any - equivalence class then aa 15 Markov. (b) Let Sand T be any transactions. If SA and TA are actuarially equivalent with respect to aa for all - - equivalence classes A J then S and Tare' actuarially equivalent with respect to a. Proof. (a) Take any rand s in A. Since the given condition in the definition of - holds for all t in B. it obviously holds for all t in A and necessarily r - A s. (b). Let t be any point in B. Consider any equivalence class A and fix a point r in A. Then for any s in A, we have r-s and so It follows easily that a(s t) = a(s,rl a(r,t) (3.1)

8 Hence if to aa TA and SA are actuarially equivalent with respect and from (3.1) we get that The conclusion following by summing the last equation over all - - equivalence classes. THEOREM 3.2 The transactions Sand Tare actuarially equivalent if and only if, given any - - cross section X, Valt(S) = Valt(T), for all t in X. Proof. Obviously actuarial equivalence implies the stated condition. Conversely, assume this condition. Let t be an arbitrary point in B, and let 0 denote the equivalence class of t. For each equivalence class A, we let SA denote its representative in X. Then clearly t - sd, so for any class A, Letting v A denote VaisA ( T A), we have (3.2) Valt(T) = Ia(sA,t)VA. A Taking reciprocals in (3.2) and substituting

9 Valt(T) = a (SD,t) I a(sa,sd)va A = a (SD,t) Val SD ( T) from which the conclusion easily follows. Note that when we apply our results to the Markov case, Theorem 3.1 is trivial since there is only one class, while Theorem 3.2 IS a direct generalization of Theorem 1.1. As an illustration of how we may use these results, suppose we are providing a stream of payments and we want to arrive at a actuarially equivalent set of net premiums to charge.. If the accumulation function is non-markov then we may not be able to do this with a single premium. However Theorem 3.1 tells us that we can always accomplish our task with k premiums where k is the number of equivalence classes intersected by the times of the given payments. In each such class A we can choose any point t and charge Valt(TA) as a net premium payable at time t. We now give an example to show that the converse of part (b) of the Theorem 3.1 is not true. EXAMPLE 3.3 Let B = [0, 00) a( s,t) = 1 if s - and t- 1, if o ~ t ( 10 a(o,t) = 4/3 if 10 ~ t ( 20 _ 2 if 20 ~ t and of course a(s,o) is defined by taking reciprocals. Let T = {( 1,0), (-2,10), ( 1, 20) }

10 It is easy to check that the - - equivalence classes are ( [0,10),[ 10, 20}, [20, 00) } Now VaIO(T) = (1-2(3/4)... 1/2 ) = 0 and clearly Valt(T) = 0 for all t ~ 0, since the sum of payments is zero. Hence T is actuarially equivalent to the zero transaction but this is not true for the restriction of T to the equivalence classes. 4. The extreme non-markov case and reciprocal matrices The next question is to seek generalizations of Example 1.1. What conditions on a will ensure that that two nonidentical transactions cannot be actuarially equivalent? By Theorem 3.1 a nt?cessary condition is that the equivalence classes reduce to single points. Any Markov function will clearly allow distinct but actuarially equivalent transactions on a set of more than one point This condition will not.be sufficient however as indicated by the failure of the converse to part (b) of the theorem. On the other hand, in the simple interest case we have a much stronger result than the fact that two distinct points are not equivalent. In fact for r,5, and t all distinct (2.1) shows that a(r,5,> a(s, t) ~ a(r,t) (4.1) It is then reasonable to conjecture that condition (4.1) implies the non-actuarial equivalence of two distinct transactions. Even here the result is not clear, It involves an interesting problem in linear. algebra which we will now describe. A square matrix C = ( Cij) with positive entries reciprocal if 15 said to be Cji = (cij) -1, for all i and j

11 This concept was introduced by Saaty in [2] where it forms a major mathmatical tool in his analytic hierarchy process, a method for ranking various alternatives Saaty defines such a matrix to be consistent if = cik for all i, j, and k., since this condition arises from ranking in a consistent manner. It is easy to see that a consistent matrix is of rank 1. In fact each row is a multiple of any other row. It therefore has a unique (up to scalar multiplication) nonzero eigenvector which establishes the relative weight of the alternatives. For our purposes we are interested in matrices in which the consistency equation never holds, except when there are two (or three) of the subscripts equal,in which case equality is implied by the reciprocal property. Accordingly we will define a reciprocal matrix to be extremely inconsistent if for all i < j < k Suppose then that a is an accumulation function satisfying (4.1). As in the appendix of [1], we choose any transaction su<;:h that Valt (1) = 0 for all t in B (4.2) and we would like to show that T is the zero transaction. Define the nxn matrix C =. (Ci) by Then C is a, reciprocal, extremely inconsistent matrix. If X is the vector (K 1, K2,... Kn) we apply (4.2) with t = Si,

12 i = 1,2,... n to deduce that ex = 0 So the desired result would follow if we could show that a receiprocal exremely inconsistent matrix is nons~ngular. In other words, when we move from consistency to extreme inconsistency do we move from the minimum rank of one to the full rank of n.? We conjecture that the result is true, but so far have been able to show it only for n = :3 or 4,( it is true vacuously for n = 2). 5. Stationary function5 For stationary functions, the equivalence classes can be expressed in terms of a natural partition which arises in group theory. Assume then that a is a stationary function on R, the whole real line. (This will also cover the case where B = [0, 00), since in that case there is clearly a unique stationary extension to It) Following the notation in [1],we let act) denote a(o,t). Then the accumulation is completely defined by Note that a (s,t) = a( t-s) (5.1) t2 (t) (5.2) and conversely any positive valued function a defined on R which satisfies (5.2) defines a stationary accumulation function by (5.1).Now, given such a function let H = {t R: a (t) d (x).. d (t+x) for all x R }

13 LEMMA 5.1 (a) H is a subgroup of the additive group (R, +). Moreover, a restricted to H is a homomorphism from (R, + ) to themultiplicative group of nonnegative reals. (b) If r+s E: H, then a(r)a(s) = a(r+5) Proof (a). Clearly a (0) = 1, in H, and any x in R, so 0 is in H. Now given sand t a( s-t +x) = a ( 5) a (-t+x) since 5 E: H 4(5) 4(t-x)-1 by (5.2) = a(s) [ act) a(-x)r 1 since t E: H = a(s) a(-t) a(x) using (5.2) agam = a(s -t) a(x) since s H. So we see that s-t H, completing the proof that H is a subgroup. The second statement in (a) is evident. (b) If r+s H. ThG!n a(r) = a(r+s -s) = a(r+s)a( -5),and WG! invokg! (5.2). Is is instructive to consider the extreme cases of this lemma. If a is a homomorphism then clearly H is all of R and there is only one --equivalence class. This corresponds to the compound interest case act) = (1 +i)t. On the other hand,in the simple

14 interest case, H consists only of 0,,as shown by part (b) of Example 2.2, and the equivalence classes reduce to single points. In general we obtain the following. THEOREM 5.1 For a as above, r - s iff r-s E: H. That is,the equivalence classes are the cosets of H. Proof. This follows directly from the definition of - (5.1) and Lemma 5.2. As an illustration, we will completely classify all stationary functions with H = l, the integers. Let i = a(l) -1, ( the one year interest rate) Since a is multiplicative on l, we easily deduce that a(n) = (1 +i)n, for all n m 1. The function a on (0,1), smce is then a completely determined by its value for t = n+r, with n E: land 0 < r < 1 a(n+r) = (1 +i)n a(r) (5.3) From part (b) of Lemma 5.2, a necessary condition for a is that a (1-r) = (1 +j) a (r)-1, 0< r < 1 (5.4) (Note that (5.4 }implies that a (1/2) must

15 Condition (5.4) is also a asufficient condition for (5.2) to hold. For n Z and < r < 1, we use (5.3) and (5.4) to deduce that a(-[n+rd = a( -n-l + [1-r] ) To summarize, all stationary accumulation functions for which H contains Z are obtained by,choosing an interest rate i, choosing a function a on (0,1) satisfying (5.4), and then extending a by (5.3). It is of course possible in this situation for H to contain elements not in Z. If so, it must contain some element r in the interval (0,1) and it is easy to see that such an element is in H iff a(r+s) = a(r)a(s) for all s in (0,1-r) (5.5) and (5.6) a(r) a(u) = a(r -u) for all u in (O,d Hence H will exactly equal Z, if and only if, for each r in (0,1) either (5.5) or (5.6) fails. For a particular example, let be nonzero and define

16 _ 1, if 0 ~ r < 1/2 a(r) (1 +i)1/2 if r - 1/2 _ 1 +i if 1/2 < r ~ 1. Then d clearly satisfies (5 4). For 0 < r < 1/2, the condition in (5.5) fails for s 1/2,for r - 1/2, it fails for -s.. 1/4 and for 1/2 < r < 1 the condition in (5.6) fails for u = 1/2. Accumulation in this case is at compound interest except, for fractions of years less than 1/2,we round down to the nearest number of whole periods, and for fractions more than 1/2 we round up. 6. The number of equivalence cla55e5. In section 4 we gave an example of an accumulation function with three equivalence cla55es, and we can obviou51y modify thi:5 example to produce examples with an arbitrary finite number of classes. However, the accumulation function in that example was discontinuous. The behaviour is different in in the presence of continuity. THEOREM 6.1 Suppose that B is an interval and that the accmulation function d is continuous on B and non-markov. Then there are infinitely many equivalence classes. Prool ht: r 0 in B. For each t in B define the function by By definition, s - r 0 iff ht(s) = 0 for all t. That is

17 The equivalence class of r 0 = n ht -1(0). te:b Now the given condition on a makes ht continuous. (in fact we need only the weaker condition that a is continuous with respect to each variable separately when the other is fixed). Hence each ht(o) is closed and the equivalence class of r 0 is the intersection of closed sets and therefore closed. We know by the non-markov property that there is more than one class. If there were only finitely many such classes, each would be both closed and open. But, by the connectivity of B this is impossible. References [1) Promislow D.. Accumulation functions. ARCH [2] Saaty, T., The Analytic Hierachy Process l McGraw Hill (1980)

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