OOne-Switch Discount Functions
Nina Anchugina Department of Mathematics, University of Auckland [email protected]
February 2017
Abstract.
Bell [9] introduced the one-switch property for preferences over sequences ofdated outcomes. This property concerns the effect of adding a common delay to twosuch sequences: it says that the preference ranking of the delayed sequences is eitherindependent of the delay, or else there is a unique delay such that one strict rankingprevails for shorter delays and the opposite strict ranking for longer delays. For preferencesthat have a discounted utility (DU) representation, Bell [9] argues that the only discountfunctions consistent with the one-switch property are sums of exponentials. This paperproves that discount functions of the linear times exponential form also satisfy the one-switch property. We further demonstrate that preferences which have a DU representationwith a linear times exponential discount function exhibit increasing impatience ([26]). Wealso clarify an ambiguity in the original Bell [9] definition of the one-switch property bydistinguishing a weak one-switch property from the (strong) one-switch property. We showthat the one-switch property and the weak one-switch property definitions are equivalentin a continuous-time version of the Anscombe and Aumann [7] setting.
Keywords:
Discount function, One-switch property, Time preferences, Linear times ex-ponential, Increasing impatience.
JEL Classification:
D90. a r X i v : . [ q -f i n . E C ] F e b n this paper we analyse the one-switch property for intertemporal preferences intro-duced by Bell [9]. The one-switch property was initially formulated for preferences overlotteries [9]. It says that the preference ranking of any two lotteries is either independentof wealth, or else there is a unique level of wealth such that one strict ranking prevails forlower wealth levels and the opposite strict ranking for higher wealth levels. For preferenceswith an expected utility representation, the one-switch property restricts the form of theBernoulli utility function. As demonstrated by Bell [9], the utility functions that satisfythis property are the quadratic, the sum of exponentials, the linear plus exponential andthe linear times exponential. The properties of these functions, and their possible appli-cations, have been extensively investigated in risk theory (see, for example, [1], [2], [11]and [12]). However, it is less well known that Bell [9] also defined an analogous one-switchproperty for preferences over sequences of dated outcomes. In this case, the one-switchproperty concerns the effect of adding a common delay to two sequences of dated out-comes: it says that the preference ranking of the delayed sequences is either independentof the delay, or else there is a unique delay such that one strict ranking prevails for shorterdelays and the opposite strict ranking for longer delays. Bell [9] claims that if preferenceshave a discounted utility representation, then the only discount functions consistent withthe one-switch property are sums of exponentials.However, in this paper we show that the one-switch property is also compatible withanother form of discount function: the linear times exponential. To the best of ourknowledge, this type of discount function has not been previously used in intertemporalcontext. As we demonstrate in the paper, linear times exponential discount functionsexhibit strictly increasing impatience (II). While strictly II has not been a very frequentexperimental observation [18], some recent empirical findings support this type of impa-tience [13, 25, 26]. We also analyse the distinction between the weak one-switch propertyand the (strong) one-switch property in the time preference context. This distinction isbased on whether weak or strict preferences are reversed at the switch point. While thisdistinction is inconsequential in the risk set-up with an expected utility representation,matters are not so clear for the intertemporal context, even assuming a discounted utilityrepresentation. We show, however, that these two properties are equivalent in a set-upanalogous to that of Anscombe and Aumann [7].The paper is organized as follows. We start by giving some preliminaries in Section 1.Section 2 is devoted to revising Bell’s characterisation [9, Proposition 2] of the discountfunctions that exhibit the one-switch property. We first discuss an ambiguity in Bell’s [9]definition of this property, and distinguish a standard (strong) version from an alternative“weak” one-switch property. We show that the discount functions consistent with the(standard) one-switch property are those which have the sum of exponentials or the lineartime exponential form. We also explore the relationship between the one-switch propertyrestricted to preferences over single dated outcomes and the impatience properties ofsuch preferences. In Section 3 we study the weak one-switch property. In the contextof expected utility preferences over lotteries, where the one-switch property refers to theeffect of wealth level on the ranking of two lotteries, the results in [9] imply that the weakone-switch property is equivalent to the standard one. In the intertemporal context, weestablish that the equivalence also holds if we endow X with a mixture set structure and1ork in an environment similar to of Anscombe and Aumann in [7]. Finally, Section 4summarizes the results. Consider preferences over sequences of dated outcomes. We work in a continuous timeenvironment throughout this paper. Points in time are elements of the set T = [0 , ∞ ),where the present time corresponds to t = 0. The set of outcomes is initially assumed tobe the interval X = [0 , ∞ ), though we will re-define X to be an arbitrary mixture set inSection 3.Let A n = { ( x , t ) ∈ X n × T n | t < t < . . . < t n } be the set of sequences with n dated outcomes. Define the set of alternatives A as follows: A = ∪ ∞ n =1 A n . The set A consists of all sequences of finitely many dated outcomes. Elements of A ⊆ A are calleddated outcomes.Consider a preference order (cid:60) on the set of alternatives A .We say that U is a discounted utility (DU) representation for (cid:60) , if U represents (cid:60) andthere exist ( u, D ), such that u : X → R is a utility function (continuous, strictly increasingwith u (0) = 0), D : T → (0 ,
1] is a discount function (strictly decreasing, D (0) = 1 andlim t →∞ D ( t ) = 0) and U ( x , t ) = n (cid:88) i =1 D ( t i ) u ( x i )for all n and every ( x , t ) ∈ A n . Necessary and sufficient conditions for a DU representationwere provided by Harvey [19, Theorem 2.1].We denote the set of positive integers { , , , . . . } by N , and the set of non-negativeintegers { , , , , . . . } by N , so N ⊂ N . For any sequence of dated outcomes ( x , t ) ∈ A n and any delay σ >
0, let( x , t + σ ) = ( x , ( t + σ, . . . , t n + σ ))denote the delayed sequence. Definition 1 ([9]) . We say that the preferences (cid:60) on A exhibit the one-switch property iffor every pair ( x , t ) , ( y , s ) ∈ A the ranking of ( x , t + σ ) and ( y , s + σ ) is either independentof σ , or there exists σ ∗ ≥ , such that ( x , t + σ ) (cid:31) ( y , s + σ ) , for any σ < σ ∗ , ( x , t + σ ) ≺ ( y , s + σ ) , for any σ > σ ∗ r ( x , t + σ ) ≺ ( y , s + σ ) , for any σ < σ ∗ , ( x , t + σ ) (cid:31) ( y , s + σ ) , for any σ > σ ∗ . It is worth mentioning that Bell’s [9] original verbal definition of the one-switch prop-erty uses the ambiguous word “preferred”, which does not specify whether the preferenceorder is used in a strong or weak sense. Bell’s Lemma 3 [9] implicitly suggests thatweak preference is intended, but this Lemma also shows that either interpretation leadsto the same restriction on expected utility preferences. Abbas and Bell [3] introduce aformal definition which is explicit about preference being strict. Therefore, we define theone-switch property using strict preference order.The one-switch property can be stated in a weaker variant, as follows:
Definition 2 ([9]) . We say that the preferences (cid:60) on A exhibit the weak one-switchproperty if for every pair ( x , t ) , ( y , s ) ∈ A the ranking of ( x , t + σ ) and ( y , s + σ ) is eitherindependent of σ , or there exists σ ∗ ≥ , such that ( x , t + σ ) (cid:60) ( y , s + σ ) , for any σ < σ ∗ , ( x , t + σ ) (cid:52) ( y , s + σ ) , for any σ > σ ∗ or ( x , t + σ ) (cid:52) ( y , s + σ ) , for any σ < σ ∗ , ( x , t + σ ) (cid:60) ( y , s + σ ) , for any σ > σ ∗ . In other words, there do not exist ( x , t ) , ( y , s ) ∈ A and σ, ε with 0 < σ < ε such that( x , t ) (cid:31) ( y , s ) , ( x , t + σ ) ≺ ( y , s + σ ) , ( x , t + ε ) (cid:31) ( y , s + ε ) , or with all strict preferences reversed.In the intertemporal context it is not known whether this alternative “weak” version isequivalent (given a DU representation) to Definition 1. This question will be investigatedis Section 3, where we adapt Bell’s Lemma 3 [9] to the temporal setting. We demonstratethat the one-switch property and the weak one-switch property are equivalent in an in-tertemporal version of the Anscombe and Aumann (AA) [7] environment similar to thatinvestigated in [5].If preferences (cid:60) on A have a DU representation ( u, D ), then the one-switch propertymeans that for any ( x , t ) , ( y , s ) ∈ A the function∆( σ ) = n (cid:88) i =1 D ( t i + σ ) u ( x i ) − m (cid:88) j =1 D ( s j + σ ) u ( y j )3ither has constant sign or else there is some σ ∗ ≥ σ ) = 0 if and only if σ = σ ∗ and ∆( σ (cid:48) )∆( σ (cid:48)(cid:48) ) > σ (cid:48) (cid:54) = σ ∗ and σ (cid:48)(cid:48) (cid:54) = σ ∗ are on the same side of σ ∗ . That is, sign (∆( σ )) = const for all σ ≥ , or elsethere exists σ ∗ such that ∆( σ ∗ ) = 0 and∆( σ (cid:48) )∆( σ (cid:48)(cid:48) ) > σ (cid:48) , σ (cid:48)(cid:48) > σ ∗ or σ (cid:48) , σ (cid:48)(cid:48) < σ ∗ and∆( σ (cid:48) )∆( σ (cid:48)(cid:48) ) < σ (cid:48) < σ ∗ < σ (cid:48)(cid:48) or σ (cid:48)(cid:48) < σ ∗ < σ (cid:48) . (1)Figure 1 provides an illustration of ∆( σ ) for preferences which exhibit the one-switchproperty and have a DU representation. Note that only the sign of ∆( σ ) is relevant. σ ∆( σ ) σ ∆( σ ) σ ∗ σ ∆( σ ) σ ∗ Figure 1: Possible behaviour of ∆( σ ) for the preferences which exhibit the one-switchproperty and have a DU representationNote that the assumed properties of u imply u ( X ) = [0 , ¯ u ] for some ¯ u > u ( X ) = R + . We say that a discount function D satisfies the extended one-switch property if thefunction ∆ : R + → R defined by∆( σ ) = n (cid:88) i =1 D ( t i + σ ) v i − m (cid:88) j =1 D ( s j + σ ) w j (2)satisfies (1) for any n, m , any t ∈ T n , s ∈ T m and any v ∈ R n + , w ∈ R m + . Lemma 3. D satisfies the extended one-switch property if and only if there exists ¯ u > such that (2) satisfies (1) for any n, m , any t ∈ T n , s ∈ T m and any v ∈ [0 , ¯ u ] n , w ∈ [0 , ¯ u ] m .Proof. “Only If”. This part is straightforward. “If”.
Suppose that there exists ¯ u > R + → R defined by (2) satisfies(1) for any n, m , any t ∈ T n , s ∈ T m and any v ∈ [0 , ¯ u ] n , w ∈ [0 , ¯ u ] m . The proofis by contradiction. Assume that there is some n (cid:48) , m (cid:48) , t (cid:48) ∈ T n (cid:48) , s (cid:48) ∈ T m (cid:48) and some v (cid:48) ∈ R n (cid:48) + , w (cid:48) ∈ R m (cid:48) + such that the function ∆ ∗ : R + → R defined by∆ ∗ ( σ ) = n (cid:48) (cid:88) i =1 D ( t (cid:48) i + σ ) v (cid:48) i − m (cid:48) (cid:88) j =1 D ( s (cid:48) j + σ ) w (cid:48) j . λ ∆ ∗ will also violate (1) for any λ >
0. Let λ ∈ (0 ,
1) be such that λ v (cid:48) ∈ [0 , ¯ u ] n (cid:48) and λ w (cid:48) ∈ [0 , ¯ u ] m (cid:48) . This is a contradiction tothe initial assumption that (2) satisfies (1) for any n, m , any t ∈ T n , s ∈ T m and any v ∈ [0 , ¯ u ] n , w ∈ [0 , ¯ u ] m .In other words, Lemma 3 states that given preferences with a DU representation, therange of u is irrelevant to whether or not the preferences satisfy the one-switch property.It follows that the one-switch property does not impose any additional restrictions onthe shape of u . In other words, given a DU representation ( u, D ) for (cid:60) , the one-switchproperty restricts only D . We therefore say that a discount function, D , exhibits theone-switch property if there is some utility function, u , such that the preferences withDU representation ( u, D ) exhibit the one-switch property. Bell [9, Proposition 8] arguesthat sums of exponentials are the only discount functions compatible with the one-switchproperty. However, we will demonstrate in this section that linear times exponentialdiscount functions also have the one-switch property.The following proposition gives the restrictions on the parameters of linear timesexponential and sum of exponential functions under which they satisfy the properties ofa discount function. Proposition 4. (a) The linear times exponential function D ( t ) = ( c + c t ) e r t is a dis-count function if and only if c = 1 , − r ≥ c ≥ and r < ; i.e., D ( t ) = (1+ ct ) e − rt ,where r ≥ c ≥ and r > .(b) The sum of exponentials function D ( t ) = c e r t + c e r t , where r (cid:54) = r , is a discountfunction if and only if • c = 1 − c , r < r < and r r − r ≤ c < , or • c = 1 − c , r < r < and < c ≤ r r − r .Equivalently, D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c > , a ≤ b/c + 1 .Proof. We need to find the parameters of linear times exponential functions and sums ofexponentials such that the following four properties are satisfied:1. D (0) = 1,2. D ( t ) > t ,3. D ( t ) is strictly decreasing, and4. lim t →∞ D ( t ) = 0. (a) Linear times exponential. Assume that D ( t ) = ( c + c t ) e r t . Obviously, D (0) = 1 ifand only if c = 1.Next, to satisfy the first and the second conditions simultaneously, that is, to have D (0) = 1 and D ( t ) > t , it is necessary and sufficient that c = 1 and c ≥ e r t > t ). 5o check whether D ( t ) is strictly decreasing, consider its first order derivative: D (cid:48) ( t ) = c e r t + (1 + c t ) r e r t = e r t ( c + r + c r t ) . Since e rt > t , the sign of the derivative depends on the sign of the linear expression c + r + c r t . Therefore, D ( t ) is strictly decreasing if and only if c + r ≤ , and c + r + c r t < t > . This condition is equivalent to requiring that one of the following two conditions is satis-fied: • c + r = 0 and c r < • c + r < c r ≤ c ≥
0, in the first case we have c = − r >
0. In the second case 0 ≤ c < − r .We can summarize both cases as follows: − r ≥ c ≥ r < . Therefore, the first three properties of discount functions are satisfied if and only if c =1 , − r ≥ c ≥ r < t →∞ D ( t ) = lim t →∞ (1 + c t ) e r t = lim t →∞ c te − r t , If r = c = 0 this limit is 1. This case is ruled out since from the previous step r < t →∞ D ( t ) = lim t →∞ c te − r t = lim t →∞ c − r e − r t = 0 . Therefore, D ( t ) = ( c + c t ) e r t satisfies all four properties of discount functions if andonly if c = 1 , − r ≥ c ≥ r <
0. Denote c = c and r = − r . Then we have D ( t ) = (1 + ct ) e − rt , where r ≥ c ≥ r > (b) Sums of exponentials. The proof is analogous to [9, Proposition 8] and is givenhere for completeness. Assume that D ( t ) = c e r t + c e r t with r (cid:54) = r .The condition D (0) = 1 is satisfied if and only if c + c = 1.We must also have D ( t ) > t >
0. Note that D ( t ) = c e r t + c e r t = e r t (cid:0) c + c e ( r − r ) t (cid:1) > t > c + c e ( r − r ) t > t > e r t > t > c = 1 − c . Substituting this expression to the inequalitywe must have c + (1 − c ) e ( r − r ) t > t >
0. Therefore, the first two propertiesof discount functions are satisfied if and only if c = 1 − c and one of the following twoconditions holds: 6i) r < r and c ≤
1; or(ii) r > r and c ≥ D ( t ) strictly decreasing. Consider its first order derivative D (cid:48) ( t ) = c r e r t + c r e r t = e r t (cid:0) c r + c r e ( r − r ) t (cid:1) . Since e r t > t >
0, the function D ( t ) is strictly decreasing if and only if c r + c r ≤ , and c r + c r e ( r − r ) t < t > . Recall that from the first two conditions we have c = 1 − c and [(i) or (ii)] holds.Therefore, D (0) = 1 , D ( t ) > t > D ( t ) is strictly decreasing if and only ifall of the following conditions hold: c r + (1 − c ) r ≤ ,c r + (1 − c ) r e ( r − r ) t < t > ,c = 1 − c , and [(i) or (ii)] holds . Note that c r + (1 − c ) r e ( r − r ) t = c r + (1 − c ) r e ( r − r ) t + c r e ( r − r ) t − c r e ( r − r ) t = ( c r + (1 − c ) r ) e ( r − r ) t + c r (cid:0) − e ( r − r ) t (cid:1) . Therefore, we must have c r + (1 − c ) r ≤ , (3)( c r + (1 − c ) r ) e ( r − r ) t + c r (cid:0) − e ( r − r ) t (cid:1) < t > , (4) c = 1 − c , and [(i) or (ii)] holds . (5)Consider case (i) of (5). Then condition (3) holds if and only if c ≥ r r − r . Given(3), condition (4) holds if and only if (1 − c ) r <
0, which is equivalent – in case (i) – to c < r <
0. Thus, in case (i), the first three properties of a discount function aresatisfied if and only if c = 1 − c , r < r < r r − r ≤ c <
1. Since r < r <
0, thefourth property is also satisfied.Next, consider case (ii) of (5). The argument is similar to case (i). The condition (3)holds if and only if c ≤ r r − r . Given (3), condition (4) holds if and only if c r < c > r <
0. Therefore, in case (ii), the firstthree properties of a discount function are satisfied if and only if c = 1 − c , r < r < < c ≤ r r − r . Since r < r <
0, the fourth property is also satisfied.In case (i) of (5) let a = c , b = − r , and c = r − r . It follows that a, b, c > r /r − r ≤ c < < a ≤ b/c + 1. We then have D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c > a ≤ b/c + 1.Analogously, in case (ii) of (5) let a = c , b = − r , and c = r − r . We thenobtain the same functional from and parameter restrictions as in case (i), that is, D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c > a ≤ b/c + 1.7he following proposition demonstrates that these two types of discount function arecompatible with the one-switch property. Proposition 5.
Suppose that the preference order (cid:60) on A has a DU representation ( u, D ) ,where • D ( t ) = (1 + ct ) e − rt , where r ≥ c ≥ and r > , or • D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c > , and a ≤ b/c + 1 .Then (cid:60) exhibits the one-switch property.Proof. The proof adapts Bell’s argument [9, Proposition 2] to the time preference frame-work. We need to prove that for any ( x , t ), ( y , s ) ∈ A the following function changes signat most once: ∆( σ ) = n (cid:88) i =1 D ( t i + σ ) u ( x i ) − m (cid:88) j =1 D ( s j + σ ) u ( y j ) . (a) Linear times exponential. Consider the discount function D ( t ) = (1 + ct ) e − rt ,where r ≥ c ≥ r >
0. Then∆( σ ) = n (cid:88) i =1 (1 + ct i + cσ ) e − rt i e − rσ u ( x i ) − m (cid:88) j =1 (1 + cs j + cσ ) e − rs j e − rσ u ( y j ) . Rearranging, ∆( σ ) = e − rσ (cid:32) n (cid:88) i =1 e − rt i u ( x i ) − m (cid:88) j =1 e − rs j u ( y j ) (cid:33) + e − rσ c (cid:32) n (cid:88) i =1 t i e − rt i u ( x i ) − m (cid:88) j =1 s j e − rs j u ( y j ) (cid:33) + e − rσ cσ (cid:32) n (cid:88) i =1 e − rt i u ( x i ) − m (cid:88) j =1 e − rs j u ( y j ) (cid:33) . Let A = n (cid:88) i =1 e − rt i u ( x i ) − m (cid:88) j =1 e − rs j u ( y j ) , and B = n (cid:88) i =1 t i e − rt i u ( x i ) − m (cid:88) j =1 s j e − rs j u ( y j ) . Then ∆( σ ) = Ae − rσ + cBe − rσ + cAe − rσ σ. This expression can be rewritten as follows∆( σ ) = e − rσ ( A + cB + cAσ ) . e − rσ >
0, the sign of ∆( σ ) equals the sign of A + cB + cAσ . Since the latter is linearits sign is constant or else changes once at a unique σ value. (b) Sums of exponentials. Consider the function D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c >
0, and a ≤ b/c + 1. Then∆( σ ) = n (cid:88) i =1 (cid:0) ae − bt i e − bσ + (1 − a ) e − ( b + c ) t i e − ( b + c ) σ (cid:1) u ( x i ) − m (cid:88) j =1 (cid:0) ae − bs j e − bσ + (1 − a ) e − ( b + c ) s j e − ( b + c ) σ (cid:1) u ( y j ) . It can be rearranged so that∆( σ ) = ae − bσ (cid:32) n (cid:88) i =1 e − bt i u ( x i ) − m (cid:88) j =1 e − bs j u ( y j ) (cid:33) + (1 − a ) e − ( b + c ) σ (cid:32) n (cid:88) i =1 e − ( b + c ) t i u ( x i ) − m (cid:88) j =1 e − ( b + c ) s j u ( y j ) (cid:33) . Denote ˜ A = n (cid:88) i =1 e − bt i u ( x i ) − m (cid:88) j =1 e − bs j u ( y j ) , and ˜ B = n (cid:88) i =1 e − ( b + c ) t i u ( x i ) − m (cid:88) j =1 e − ( b + c ) s j u ( y j ) . Then ∆( σ ) = a ˜ Ae − bσ + (1 − a ) ˜ Be − ( b + c ) σ . This expression can be factorized as follows∆( σ ) = e − bσ (cid:16) a ˜ A + (1 − a ) ˜ Be − cσ (cid:17) . Since e − bσ >
0, the sign of ∆( σ ) equals the sign of a ˜ A + (1 − a ) ˜ Be − cσ . Therefore, ∆( σ )is either constant or else changes once at unique σ value.Since Proposition 5 establishes that linear times exponential discount functions andsum of exponentials discount functions satisfy the one-switch property, they must alsosatisfy the weak one-switch property. In this section we consider preferences (cid:60) on the set A of dated outcomes. When thepreferences (cid:60) are restricted to A , then a DU representation becomes U ( x, t ) = D ( t ) u ( x )9or any ( x, t ) ∈ A . Necessary and sufficient conditions for this representation are givenin [17]. We assume that (cid:60) satisfy Fishburn and Rubinstein’s axioms [17] throughout thissection. Definition 6.
We say that (cid:60) exhibits the one-switch property for dated outcomes if (cid:60) exhibits the one-switch property on A . Obviously, if (cid:60) exhibits the one-switch property on A , it implies that (cid:60) also exhibitsthe one-switch property for dated outcomes.Consider the following notions of decreasing and increasing impatience. Definition 7 ([22]) . We say that (cid:60) exhibits [strictly] decreasing impatience , if for all ( x, t ) , ( y, s ) ∈ A such that < x < y and for all t < s : if ( x, t ) ∼ ( y, s ) then for any σ > we have ( x, t + σ ) [ ≺ ] (cid:52) ( y, s + σ ) . (6) We say that (cid:60) exhibits • [strictly] increasing impatience , if the preference in (6) is reversed; • stationarity, or constant impatience , if the preference in (6) is replaced by indiffer-ence. When preferences have a DU representation, these properties only restrict the discountfunction. The next proposition follows directly from the definition.
Proposition 8.
Suppose that (cid:60) restricted to A has a DU representation. Then (cid:60) exhibits[strictly] DI if and only if D ( t ) D ( t + σ ) [ > ] ≥ D ( s ) D ( s + σ ) , for all t, s such that t < s, and every σ > . (7) Furthermore, (cid:60) exhibits • [strictly] II if and only if the inequality in (7) is reversed; • constant impatience if and only if the inequality in (7) is replaced by the equality. The following proposition provides a complete characterisation.
Proposition 9 ([22], [6]) . Suppose that (cid:60) restricted to A has a DU representation.Then (cid:60) exhibits • [strictly] DI if and only if D ( t ) is [strictly] log-convex, • [strictly] II if and only if D ( t ) is [strictly] log-concave, The related concept of an ordinal one-switch utility function is introduced in [3]. A function f : I → R is called log-convex if f ( x ) > x ∈ I and ln( f ) is convex; and strictlylog-convex if f ( x ) > x ∈ I and ln( f ) is strictly convex. We say that a function f : I → R is [strictly] log-concave , if 1 /f is [strictly] log-convex. constant impatience if and only if D ( t ) = e − rt with r > . Proposition 9 extends [22, Corollary 1] to increasing impatience and strictly decreasing(and strictly increasing) impatience. The proof is omitted here, since it only requires aminor adjustment of Prelec’s original proof. The following lemma re-expresses the definition of the one-switch property for datedoutcomes in terms of a common advancement ( σ <
0) and a common delay ( σ >
Lemma 10.
Let (ˆ x, ˆ t ) , (ˆ y, ˆ s ) ∈ A such that < ˆ x < ˆ y , < ˆ t < ˆ s and (ˆ x, ˆ t ) ∼ (ˆ y, ˆ s ) .Then (cid:60) exhibits the one-switch property for dated outcomes only if either(i) (ˆ x, ˆ t + σ ) ∼ (ˆ y, ˆ s + σ ) for all σ ≥ − ˆ t , or(ii) (ˆ x, ˆ t + σ ) (cid:31) (ˆ y, ˆ s + σ ) for − ˆ t ≤ σ < , and (ˆ x, ˆ t + σ ) ≺ (ˆ y, ˆ s + σ ) for σ > , or(iii) (ˆ x, ˆ t + σ ) ≺ (ˆ y, ˆ s + σ ) for − ˆ t ≤ σ < , and (ˆ x, ˆ t + σ ) (cid:31) (ˆ y, ˆ s + σ ) for σ > .Proof. Let (ˆ x, ˆ t ) , (ˆ y, ˆ s ) ∈ A such that 0 < ˆ x < ˆ y , 0 < ˆ t < ˆ s and (ˆ x, ˆ t ) ∼ (ˆ y, ˆ s ). Assumealso that (cid:60) exhibits the one-switch property for dated outcomes. Therefore, by definitionof the one-switch property we have(i ∗ ) (ˆ x, ˆ t + σ ) ∼ (ˆ y, ˆ s + σ ) for all σ >
0, or(ii ∗ ) (ˆ x, ˆ t + σ ) ≺ (ˆ y, ˆ s + σ ) for all σ >
0, or(iii ∗ ) (ˆ x, ˆ t + σ ) (cid:31) (ˆ y, ˆ s + σ ) for all σ > − ˆ t ≤ σ <
0. The proof is by contradiction.In case (i ∗ ), assume that there exists µ ∗ such that 0 < µ ∗ < ˆ t and (ˆ x, ˆ t − µ ∗ ) ≺ (ˆ y, ˆ s − µ ∗ ). Let ˆ τ = ˆ t − µ ∗ > ρ = ˆ s − µ ∗ >
0. Using this notation, we obtain(ˆ x, ˆ τ ) ≺ (ˆ y, ˆ ρ ) , (ˆ x, ˆ τ + µ ∗ + σ ) ∼ (ˆ y, ˆ ρ + µ ∗ + σ ) , for all σ ≥ . This contradicts the one-switch property for dated outcomes, so (i) follows. If (ˆ x, ˆ t − µ ∗ ) (cid:31) (ˆ y, ˆ s − µ ∗ ) the proof is analogous.In case (ii ∗ ), assume that there exists µ ∗ such that 0 < µ ∗ < ˆ t and (ˆ x, ˆ t − µ ∗ ) (cid:52) (ˆ y, ˆ s − µ ∗ ). With the same notation as in the previous case ˆ τ = ˆ t − µ ∗ > ρ = ˆ s − µ ∗ > x, ˆ τ ) (cid:52) (ˆ y, ˆ ρ ) , (ˆ x, ˆ τ + µ ∗ ) ∼ (ˆ y, ˆ ρ + µ ∗ ) , (ˆ x, ˆ τ + µ ∗ + σ ) ≺ (ˆ y, ˆ ρ + µ ∗ + σ ) , for all σ > , which is a contradiction.In case (iii ∗ ) the proof is symmetric to case (ii ∗ ). The proof of the proposition can also be found in the working paper [6].
Lemma 11.
Suppose that (cid:60) has a DU representation ( u, D ) . Then (cid:60) exhibits the one-switch property for dated outcomes if and only if (cid:60) also exhibits either stationarity orstrictly DI or strictly II.Proof. “Only If”. Assume that (cid:60) exhibits the one-switch property for dated outcomes.Consider some (ˆ x, ˆ t ) , (ˆ y, ˆ s ) ∈ A such that 0 < ˆ x < ˆ y , 0 < ˆ t < ˆ s and (ˆ x, ˆ t ) ∼ (ˆ y, ˆ s ). Tosee that we can always find such a pair, suppose that 0 < ˆ x < ˆ y , ˆ t < ˆ s and (ˆ x, ˆ t ) (cid:31) (ˆ y, ˆ s ).Then it follows by the continuity of u and the fact that D is strictly decreasing that thereexists t (cid:48) ∈ (ˆ t, ˆ s ) such that (ˆ x, t (cid:48) ) ∼ (ˆ y, ˆ s ). Alternatively, suppose that 0 < ˆ x < ˆ y , ˆ t < ˆ s and (ˆ x, ˆ t ) ≺ (ˆ y, ˆ s ). Then it follows by the continuity of u and the fact that D is strictlydecreasing that there exists x (cid:48) ∈ (ˆ x, ˆ y ) such that ( x (cid:48) , ˆ t ) ∼ (ˆ y, ˆ s ).It follows by the one-switch property for dated outcomes and Lemma 10 that either Case 1. (ˆ x, ˆ t + σ ) ∼ (ˆ y, ˆ s + σ ) for all σ ≥ − ˆ t , or Case 2. (ˆ x, ˆ t + σ ) (cid:31) (ˆ y, ˆ s + σ ) for − ˆ t ≤ σ <
0, and (ˆ x, ˆ t + σ ) ≺ (ˆ y, ˆ s + σ ) for σ >
0, or
Case 3. (ˆ x, ˆ t + σ ) ≺ (ˆ y, ˆ s + σ ) for − ˆ t ≤ σ <
0, and (ˆ x, ˆ t + σ ) (cid:31) (ˆ y, ˆ s + σ ) for σ > Case 1.
Note that letting α = σ + ˆ t ≥ σ = ˆ s − ˆ t > x, α ) ∼ (ˆ y, ˆ σ + α ) for all α ≥ . Using the DU representation it follows that u (ˆ x ) u (ˆ y ) = D ( α + ˆ σ ) D ( α ) for all α ≥ . (8)Consider some t (cid:48) < s (cid:48) . Then (8) implies u (ˆ x ) u (ˆ y ) = D ( t (cid:48) + ˆ σ ) D ( t (cid:48) ) = D ( s (cid:48) + ˆ σ ) D ( s (cid:48) ) . Rearranging, D ( s (cid:48) ) D ( t (cid:48) ) = D ( s (cid:48) + ˆ σ ) D ( t (cid:48) + ˆ σ ) . (9)By continuity we can choose x (cid:48) < y (cid:48) such that D ( s (cid:48) ) D ( t (cid:48) ) = u ( x (cid:48) ) u ( y (cid:48) ) . (10)Hence, it follows from (9), (10) that( x (cid:48) , t (cid:48) ) ∼ ( y (cid:48) , s (cid:48) ) and ( x (cid:48) , t (cid:48) + ˆ σ ) ∼ ( y (cid:48) , s (cid:48) + ˆ σ ) . D ( s (cid:48) ) D ( t (cid:48) ) = D ( s (cid:48) + µ ) D ( t (cid:48) + µ ) for all µ > . Since t (cid:48) < s (cid:48) were arbitrary, it follows by Proposition 8 that (cid:60) exhibits constant impa-tience. Case 2.
Defining α, ˆ σ as for Case 1, we have(ˆ x, α ) (cid:31) (ˆ y, ˆ σ + α ) for 0 ≤ α < ˆ t, and(ˆ x, α ) ≺ (ˆ y, ˆ σ + α ) for σ > ˆ t. Therefore, using the DU representation u (ˆ x ) u (ˆ y ) > D (ˆ σ + α ) D ( α ) for 0 ≤ α < ˆ t, and u (ˆ x ) u (ˆ y ) < D (ˆ σ + α ) D ( α ) for α > ˆ t. Hence, D ( t (cid:48) + ˆ σ ) D ( t (cid:48) ) < u (ˆ x ) u (ˆ y ) < D ( s (cid:48) + ˆ σ ) D ( s (cid:48) ) for any t (cid:48) < ˆ t < s (cid:48) . Rearranging D ( s (cid:48) ) D ( t (cid:48) ) < D ( s (cid:48) + ˆ σ ) D ( t (cid:48) + ˆ σ ) for any t (cid:48) < ˆ t < s (cid:48) . (11)By continuity we can choose x (cid:48) < y (cid:48) such that D ( s (cid:48) ) D ( t (cid:48) ) = u ( x (cid:48) ) u ( y (cid:48) ) . (12)It follows from (11), (12) and the one-switch property that D ( s (cid:48) − µ ) D ( t (cid:48) − µ ) < D ( s (cid:48) ) D ( t (cid:48) ) < D ( s (cid:48) + µ ) D ( t (cid:48) + µ ) (13)for any µ > t (cid:48) < ˆ t < s (cid:48) .Consider some t (cid:48) < s (cid:48) . There are three possible sub-cases:(a) t (cid:48) < ˆ t < s (cid:48) ,(b) t (cid:48) < s (cid:48) ≤ ˆ t , and(c) ˆ t ≤ t (cid:48) < s (cid:48) . 13e will show that, in each of these three sub-cases, D ( s (cid:48) ) D ( t (cid:48) ) < D ( s (cid:48) + µ ) D ( t (cid:48) + µ ) for all µ > . (14)From Proposition 8 we may then conclude that the preferences exhibit strict DI.In sub-case (a), it follows directly from (13) that (14) holds.In sub-case (b) choose ε > s (cid:48) + ε < ˆ t < t (cid:48) + ε . Let t (cid:48)(cid:48) = t (cid:48) + ε and s (cid:48)(cid:48) = s (cid:48) + ε .It follows from (13) that D ( s (cid:48)(cid:48) − σ ) D ( t (cid:48)(cid:48) − σ ) < D ( s (cid:48)(cid:48) ) D ( t (cid:48)(cid:48) ) < D ( s (cid:48)(cid:48) + σ ) D ( t (cid:48)(cid:48) + σ ) (15)for all σ >
0. Let σ = ε . Then we have D ( s (cid:48) ) D ( t (cid:48) ) < D ( s (cid:48) + ε ) D ( t (cid:48) + ε ) . (16)By continuity we can choose x (cid:48) < y (cid:48) such that D ( s (cid:48) ) D ( t (cid:48) ) = u ( x (cid:48) ) u ( y (cid:48) ) . (17)Therefore, it follows from (16), (17) and the one-switch property that (14) holds.In sub-case (c) choose ε > t (cid:48) − ε < ˆ t < s (cid:48) − ε . Let t (cid:48)(cid:48) = t (cid:48) − ε and s (cid:48)(cid:48) = s (cid:48) − ε .Then it follows from (13) that D ( s (cid:48)(cid:48) − σ ) D ( t (cid:48)(cid:48) − σ ) < D ( s (cid:48)(cid:48) ) D ( t (cid:48)(cid:48) ) < D ( s (cid:48)(cid:48) + σ ) D ( t (cid:48)(cid:48) + σ ) (18)for all σ >
0. Let σ = ε . Then we have D ( s (cid:48) − ε ) D ( t (cid:48) − ε ) < D ( s (cid:48) ) D ( t (cid:48) ) . (19)Choose x (cid:48) < y (cid:48) such that D ( s (cid:48) ) D ( t (cid:48) ) = u ( x (cid:48) ) u ( y (cid:48) ) . (20)Therefore, (14) follows by (19), (20) and the one-switch property.Therefore, in Case 2 (cid:60) exhibits strictly DI.Finally, in Case 3 we may show that (cid:60) exhibits strictly II by a symmetric proof tothat for Case 2. “If”.
Suppose that there are some x, y, t, s with t ≤ s and some σ ∗ ≥ x, t + σ ∗ ) ∼ ( y, s + σ ∗ ). It suffices to show that either( x, t + σ ) ∼ ( y, s + σ ) for all σ, (21)14r ( x, t + σ ) (cid:31) ( y, s + σ ) for all σ < σ ∗ , and (22)( x, t + σ ) ≺ ( y, s + σ ) for all σ > σ ∗ . (23)If t = s then ( x, t + σ ∗ ) ∼ ( y, t + σ ∗ ). It follows by monotonicity that x = y . Hence, (cid:60) satisfies the one-switch property for dated outcomes.Assume that t < s and (cid:60) exhibits constant impatience. It follows that ( x, t + σ ∗ + σ (cid:48) ) ∼ ( y, s + σ ∗ + σ (cid:48) ) for all σ (cid:48) >
0, or ( x, t + σ ) ∼ ( y, s + σ ) for all σ > σ ∗ . To show that (cid:60) satisfiesthe one-switch property for dated outcomes we need to demonstrate that ( x, t + σ ) ∼ ( y, s + σ ) for all σ < σ ∗ such that t + σ ≥
0, or ( x, t + σ ∗ − σ (cid:48) ) ∼ ( y, s + σ ∗ − σ (cid:48) ) for all σ (cid:48) > σ ∗ such that t + σ ∗ − σ (cid:48) ≥
0. The proof is by contradiction. Suppose that there exists σ (cid:48)(cid:48) > σ ∗ such that, say, ( x, t + σ ∗ − σ (cid:48)(cid:48) ) (cid:31) ( y, s + σ ∗ − σ (cid:48)(cid:48) ) with t + σ ∗ − σ (cid:48)(cid:48) ≥
0. By continuityand impatience there exist t (cid:48) > t such that ( x, t (cid:48) + σ ∗ − σ (cid:48)(cid:48) ) ∼ ( y, s + σ ∗ − σ (cid:48)(cid:48) ). Then, since (cid:60) exhibits constant impatience it follows that ( x, t (cid:48) + σ ∗ − σ (cid:48)(cid:48) + γ ) ∼ ( y, s + σ ∗ − σ (cid:48)(cid:48) + γ )for all γ >
0. Let γ = σ (cid:48)(cid:48) >
0. Then we obtain ( x, t (cid:48) + σ ∗ ) ∼ ( y, s + σ ∗ ). Since t (cid:48) > t itimplies by impatience that ( x, t (cid:48) + σ ∗ ) ≺ ( x, t + σ ∗ ). Hence, ( x, t + σ ∗ ) (cid:31) ( y, s + σ ∗ ), acontradiction.Suppose that t < s and (cid:60) exhibits strictly DI. It follows that ( x, t + σ ∗ + α ) ≺ ( y, s + σ ∗ + α ) for all α >
0, or ( x, t + σ ) ≺ ( y, s + σ ) for all σ > σ ∗ . We now need toshow that ( x, t + σ ) (cid:31) ( y, s + σ ) for all σ < σ ∗ , or ( x, t + σ ∗ − σ (cid:48) ) (cid:31) ( y, s + σ ∗ − σ (cid:48) ) forall σ (cid:48) > σ ∗ such that t + σ ∗ − σ (cid:48) ≥
0. The proof is by contradiction. Suppose there exist σ (cid:48)(cid:48) > σ ∗ such that ( x, t + σ ∗ − σ (cid:48)(cid:48) ) (cid:52) ( y, s + σ ∗ − σ (cid:48)(cid:48) ) and t + σ ∗ − σ (cid:48)(cid:48) ≥ x, t + σ ∗ − σ (cid:48)(cid:48) ) ∼ ( y, s + σ ∗ − σ (cid:48)(cid:48) ) with σ (cid:48)(cid:48) > σ ∗ and t + σ ∗ − σ (cid:48)(cid:48) ≥ (cid:60) satisfy strictly DI it follows that ( x, t + σ ∗ − σ (cid:48)(cid:48) + γ ) ≺ ( y, s + σ ∗ − σ (cid:48)(cid:48) + γ )for all γ >
0. Let γ = σ (cid:48)(cid:48) >
0. Then we have ( x, t + σ ∗ ) ≺ ( y, s + σ ∗ ), which is acontradiction. Secondly, consider ( x, t + σ ∗ − σ (cid:48)(cid:48) ) ≺ ( y, s + σ ∗ − σ (cid:48)(cid:48) ) with σ (cid:48)(cid:48) > σ ∗ and t + σ ∗ − σ (cid:48)(cid:48) ≥
0. It follows by continuity and impatience that there exist s (cid:48) > s suchthat ( x, t + σ ∗ − σ (cid:48)(cid:48) ) ∼ ( y, s (cid:48) + σ ∗ − σ (cid:48)(cid:48) ). Hence, since (cid:60) exhibit strictly DI it implies that( x, t + σ ∗ − σ (cid:48)(cid:48) + γ ) ≺ ( y, s (cid:48) + σ ∗ − σ (cid:48)(cid:48) + γ ) for all γ > σ (cid:48)(cid:48) > σ ∗ and t + σ ∗ − σ (cid:48)(cid:48) ≥ γ = σ (cid:48)(cid:48) . Then we have ( x, t + σ ∗ ) ≺ ( y, s (cid:48) + σ ∗ ). Since s (cid:48) > s it follows by impatiencethat ( y, s (cid:48) + σ ∗ ) ≺ ( y, s + σ ∗ ), therefore, ( x, t + σ ∗ ) ≺ ( y, s + σ ∗ ), a contradiction.If we assume that (cid:60) exhibits strictly II, the proof is analogous.While the assumption of a DU representation is not necessary for the “If” part ofthe proof, it is essential for our proof of the “Only If” part. The necessity of a DUrepresentation for the “Only If” result of Lemma 11 remains an open question.It was demonstrated in the working paper [6], that for differentiable discount func-tions a strictly increasing time preference rate corresponds to strictly II, while a strictlydecreasing time preference rate corresponds to strictly DI. The proof is along the linesof [6, Lemma 11]. We use this result to prove the following proposition.
Proposition 12.
Suppose that (cid:60) has a DU representation ( u, D ) . The time preference rate , r ( t ), is defined as follows: r ( t ) = − D (cid:48) ( t ) D ( t ) . If D ( t ) = (1 + ct ) e − rt , where r ≥ c ≥ and r > , then (cid:60) exhibits strictly II when c > and (cid:60) exhibits stationarity when c = 0 . • If D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c > , a ≤ b/c + 1 , then (cid:60) exhibitsstrictly DI when a < , strictly II when < a ≤ b/c + 1 and stationarity when a = 1 .Proof. (a) Linear times exponential. Assume first that c >
0. Since D (cid:48) ( t ) = e − rt ( c − r − crt ), the time preference rate is: − D (cid:48) ( t ) D ( t ) = e − rt ( crt − c + r )(1 + ct ) e − rt = r (1 + ct ) − c ct = r − c ct . The derivative of time preference rate is c (1+ ct ) − >
0, therefore, linear times exponentialdiscount function exhibits strictly increasing impatience. Otherwise, if c = 0, then D ( t ) = e − rt and the preferences (cid:60) exhibit stationarity (see, for example, [17]). (b) Sums of exponentials. The time preference rate is − D (cid:48) ( t ) D ( t ) = abe − bt + (1 − a )( b + c ) e − ( b + c ) t ae − bt + (1 − a ) e − ( b + c ) t = e − bt [ ab + (1 − a )( b + c ) e − ct ] e − bt [ a + (1 − a ) e − ct ]= ab + (1 − a )( b + c ) e − ct a + (1 − a ) e − ct . The derivative of time preference rate is (cid:20) ab + (1 − a )( b + c ) e − ct a + (1 − a ) e − ct (cid:21) (cid:48) = − c (1 − a )( b + c ) e − ct [ a + (1 − a ) e − ct ] + [ ab + (1 − a )( b + c ) e − ct ] c (1 − a ) e − ct [ a + (1 − a ) e − ct ] . The sign of the derivative depends on the sign of the numerator of this expression: Q ( t ) = − c (1 − a )( b + c ) e − ct (cid:2) a + (1 − a ) e − ct (cid:3) + (cid:2) ab + (1 − a )( b + c ) e − ct (cid:3) c (1 − a ) e − ct . Simplifying Q ( t ): Q ( t ) = c (1 − a ) e − ct (cid:2) ab + (1 − a )( b + c ) e − ct − ( b + c )( a + (1 − a ) e − ct ) (cid:3) = c e − ct a ( a − . Recall that a > a ≤ b/c + 1. Therefore, Q ( t ) = 0 if a = 1, Q ( t ) is strictly negativeif 0 < a < Q ( t ) is strictly positive if 1 < a ≤ b/c + 1 and a (cid:54) = 1. Hence, the timepreference rate is constant if a = 1, strictly decreasingif 0 < a < < a ≤ b/c + 1. This in turn implies that (cid:60) exhibit stationarity if a = 1, strictly DI if0 < a < < a ≤ b/c + 1.A linear times exponential discount function and two sum of exponentials discountfunctions, with their associated rates of time preference, are illustrated in Figure 2.It is worth mentioning that Bell’s [9] definitions of the terms “decreasing impatience”and “increasing impatience” are different from the ones used here. Bell’s definitions aregiven below: 16 inear times exponentialSum of exponentialsSum of exponentials 2Exponential Linear times exponentialSum of exponentialsSum of exponentials 2Exponential Time T i m e p r e f e r e n ce r a t e Time D i s c o un t f un c t i o n . . . . . . . . . Figure 2: Linear times exponential discount function D ( t ) = (1+0 . t ) e − . t , exponentialdiscount function D ( t ) = e − . t , sum of exponentials D ( t ) = 0 . e . t + 0 . e . t , sum ofexponentials D ( t ) = 1 . e . t − . e . t and their associated rates of time preference Definition 13 ([9]) . Let (cid:60) on A have a DU representation with a discount function D .Then we say that preferences (cid:60) exhibit DI ∗ [II ∗ ] if D ( s + t ) > [ < ] D ( s ) D ( t ) for any s, t > . Note that Bell’s [9] DI ∗ (II ∗ ) corresponds to strict log-superadditivity (log-subadditivity) of D . Obviously, strict log-superadditivity (strict log-subadditivity) is aspecial case of strict log-convexity (strict log-concavity). Therefore, Bell’s definitions ofDI ∗ and II ∗ are weaker properties than our strictly DI and strictly II. Bell [9, Proposition8] specifies the parameter values for a sum of exponentials discount function such that (cid:60) exhibit DI ∗ /II ∗ : Proposition 14 ([9, Proposition 8]) . Let D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c > and a ≤ b/c . Then it is DI ∗ if a < and II ∗ if < a ≤ b/c . Comparing Proposition 14 to Proposition 11, it is easy to see that strictly II andII ∗ have the same implications for the parameters for a sum of exponentials discountfunction (and similarly for strictly DI and DI ∗ ). It is also straightforward to observe thatthe restrictions imposed by the properties of strictly II and II ∗ on the parameters for alinear times exponential discount function coincide. Bell [9, Proposition 8] uses a strict inequality a < b/c to guarantee that D (cid:48) ( t ) < t = 0.However, if D (cid:48) ( t ) = 0 when t = 0 and D (cid:48) ( t ) < t >
0, then D ( t ) is strictly decreasing on [0 , ∞ ).Therefore, we allow a ≤ b/c , since this weak inequality is consistent with the properties of a discountfunction. .3 Representation of preferences with the one-switchproperty We first observe that constant impatience is equivalent to the zero-switch property:
Definition 15.
We say that (cid:60) on A exhibit the zero-switch property if for every pair ( x , t ) , ( y , s ) ∈ A the ranking of ( x , t + σ ) and ( y , s + σ ) is independent of σ . It follows from this definition that if (cid:60) on A exhibit the zero-switch property, thenfor any ( x , t ) , ( y , s ) ∈ A , if there exists ˆ σ ≥ x , t + ˆ σ ) ∼ ( y , s + ˆ σ ), then( x , t + σ ) ∼ ( y , s + σ ) for any σ ≥ Proposition 16.
Let (cid:60) on A have a DU representation ( u, D ) .Then (cid:60) exhibit the one-switch property only if D ( t ) has one of the following forms: • D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , with a, b, c > and a ≤ b/c + 1 , • D ( t ) = (1 + ct ) e − rt , where r ≥ c ≥ and r > . We adapt Bell’s [9, 10] method of proof for the risk (expected utility) context to ourtime preference framework. The required adaptation is non-trivial. The main reason isthat probabilities sum up to one, whereas utilities of outcomes do not. In the originalBell proof [9] this property of probabilities was used to obtain a system of two equationsin two variables. The conditions under which the solutions of this system exist are well-known. In the time preference setting, we use Lemma 11 to obtain two sequences ofdated outcomes that are indifferent at two different points of delay. As a result of whichwe obtain a homogeneous second order linear difference equation. The solutions of thisequation extended to continuous time give us three types of functions. We further showthat only linear times exponential and sums of exponentials (with suitable parameterrestrictions) satisfy the one-switch property and the properties of a discount function(using Proposition 4). It is also worth mentioning that in the risk setting Bell [9, 10]obtains a third order difference equation, rather than the second order difference equationthat we obtain in the time preference framework.
Proof.
Fix some σ >
0. Suppose we can find u ( α ) , u ( β ) , u ( γ ) > u ( α ) + u ( β ) D (2 σ ) = u ( γ ) D ( σ ) and u ( α ) D ( σ ) + u ( β ) D (3 σ ) = u ( γ ) D (2 σ ) (24)Then, since (cid:60) has a DU representation, it implies that( x , t ) ∼ ( y , s ) and ( x , t + σ ) ∼ ( y , s + σ ) , where ( x , t ) = (( α, β ) , (0 , σ )), ( y , s ) = ( γ, σ ) and ( x , t + σ ) = (( α, β ) , ( σ, σ )), ( y , s + σ ) =( γ, σ ).We first show that for any D ( σ ) , D (2 σ ) , D (3 σ ) we can always find u ( α ) , u ( β ) , u ( γ ) > u ( α ) = u ( γ ) D ( σ ) − u ( β ) D (2 σ ), whichwe may substitute into the second equation as follows:( u ( γ ) D ( σ ) − u ( β ) D (2 σ )) D ( σ ) + u ( β ) D (3 σ ) = u ( γ ) D (2 σ ) . u ( γ )( D ( σ ) − D (2 σ )) + u ( β ) ( D (3 σ ) − D ( σ ) D (2 σ )) = 0 . (25)Since preferences (cid:60) satisfy the one-switch property they will also satisfy the one-switchproperty for dated outcomes, and hence it follows by Lemma 11 that (cid:60) exhibits strictly DI,strictly II or stationarity. Assume first that (cid:60) exhibits strictly DI. Then, by Proposition8 we have 1 D ( σ ) > D ( σ ) D (2 σ ) > D (2 σ ) D (3 σ ) > · · · Therefore, D ( σ ) − D (2 σ ) < D (3 σ ) − D ( σ ) D (2 σ ) >
0, and hence, by continuity itis always possible to find u ( β ) , u ( γ ) > u ( β ) u ( γ ) = D (2 σ ) − D ( σ ) D (3 σ ) − D ( σ ) D (2 σ ) , we have u ( β ) u ( γ ) − D ( σ ) D (2 σ ) = D (2 σ ) − D ( σ ) D (3 σ ) − D ( σ ) D (2 σ ) − D ( σ ) D (2 σ ) = D (2 σ ) − D ( σ ) D (3 σ ) D (2 σ ) ( D (3 σ ) − D ( σ ) D (2 σ )) < . Hence, u ( β ) D (2 σ ) − u ( γ ) D ( σ ) <
0, which implies that u ( α ) = u ( γ ) D ( σ ) − u ( β ) D (2 σ ) > . If (cid:60) exhibits strictly II the argument is analogous with the inequalities reversed.In the case when (cid:60) exhibits stationarity, we have D ( σ ) − D (2 σ ) = 0 and D (3 σ ) − D ( σ ) D (2 σ ) = 0, therefore (25) holds for any u ( β ) , u ( γ ). Hence, we can choose u ( β ) , u ( γ ) > u ( α ) = u ( γ ) D ( σ ) − u ( β ) D (2 σ ) > x , t ) , ( y , s ) ∈ A such that( x , t + t ) ∼ ( y , s + t ) , where t = 0 , σ. Then, since (cid:60) satisfies the one-switch property it must be true that( x , t + t ) ∼ ( y , s + t ) for any t ≥ . In particular, u ( α ) D ( t ) + u ( β ) D ( t + 2 σ ) = u ( γ ) D ( t + σ ).Since u ( γ ) (cid:54) = 0, we can write D ( t +2 σ )+ aD ( t + σ )+ bD ( t ) = 0, where a = − u ( γ ) /u ( β ), b = u ( α ) /u ( β ).For some σ > t ≥ D ( t ,σ ) n = D ( t + nσ ), n ∈ N . We then obtain ahomogeneous second order linear difference equation D ( t ,σ ) n +2 + aD ( t ,σ ) n +1 + bD ( t ,σ ) n = 0 . (26)It is well known (see, for example, [21]) that this equation has three types of solutions,which are derived from the characteristic equation: z + az + b = 0. These three solutionsare as follows: Solution 1. If a − b >
0, then there are two distinct real roots denoted as z and z . Inthis case the two linearly independent solutions to (26) are z n and z n , where n ∈ N . The general solution is D ( t ,σ ) n = c z n + c z n , where c , c = const .19 olution 2. If a − b = 0, then the roots coincide so z = z = z . In this case the twolinearly independent solutions to (26) are z n and nz n . The general solutionis D ( t ,σ ) n = ( c + c n ) z n , where c , c = const . Solution 3. If a − b <
0, the roots are complex z ± = x ± iy = r (cos θ ± i sin θ ) = re ± iθ , where y > r = (cid:112) x + y , cos θ = x/r and sin θ = y/r with θ ∈ (0 , π )(since y > Rez n + = r n cos nθ and Imz n + = r n sin nθ . The general solution is D ( t ,σ ) n = r n [ c cos nθ + c sin nθ ] , where c , c = const. Note that by letting C = (cid:112) c + c , cos ω = c C , sin ω = c C and ω = tan − ( c c ), Solution3 can be rewritten as follows: D ( t ,σ ) n = r n [ c cos nθ + c sin nθ ] = Cr n [cos ω cos nθ + sin ω sin nθ ] = Cr n cos( nθ − ω ) . Recall that θ ∈ (0 , π ). Therefore, Solution 3 can be excluded because it implies multiplechanges of sign (it does not satisfy monotonicity).Note that equation (26) holds for arbitrary σ > t ≥
0, though theroots z and z and the constants c and c may depend, in a continuous fashion, on t and σ . Bell [10, 9] argues that D must therefore satisfy the corresponding limit of one ofthese solutions, as σ → The solutions of (26) converge, respectively to
Solution 1. (Sum of exponentials): D ( t ) = c e r t + c e r t , where r (cid:54) = r , Solution 2. (Linear times exponential): D ( t ) = ( c + c t ) e r t .By Proposition 4 it follows that: Solution 1. D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , where a, b, c >
0, and a ≤ b/c + 1; Solution 2. D ( t ) = (1 + ct ) e − rt , where r ≥ c ≥ r >
0. Note that this solutionincludes the exponential discount function as a special case. That is, if c = 0, then D ( t ) = e − rt , where r > Romanian mathematician Rad´o [23] proved a more general result which was recently extended tomultidimensional case in [4]. Rad´o [23, Theorem 2] proves that the set of continuous functions f : R → R ,which satisfy the equation a ( σ ) f ( t ) + a ( σ ) f ( t + σ ) + . . . + a n ( σ ) f ( t + nσ ) = 0 with continuous functions a i ( σ ) : (0 , W ) → R , where W > i = 0 , . . . , n with a n ( σ ) (cid:54) = 0, coincides with the set of functions f : R → R which satisfy a linear differential equation A f + A f (cid:48) + . . . + A n f ( n ) = 0 for some realcoefficients A , . . . , A n . When n = 2 it is well-known that the solutions to such a differential equationcoincide with the limits of our three solution types. Corollary 17.
Let (cid:60) on A have a DU representation ( u, D ) . Then (cid:60) exhibits the one-switch property if and only if D ( t ) has one of the following forms: • D ( t ) = ae − bt + (1 − a ) e − ( b + c ) t , with a, b, c > and a ≤ b/c + 1 , • D ( t ) = (1 + ct ) e − rt , where r ≥ c ≥ and r > . The one-switch property is equivalent to the weak one-switch property in the risksetting when preferences over lotteries have an expected utility representation [9]. Thisequivalence follows from [9, Lemma 3], where mixture linearity and other properties ofexpected utility are used to show that if two lotteries are indifferent at two wealth levelsthen they should be indifferent for any wealth level. In the intertemporal framework of thispaper a direct adaptation of the proof of this equivalence is not possible, even if we assumethat preferences have a DU representation, because a DU representation is, in general,not mixture linear. However, we can adapt the Anscombe and Aumann (AA) setting [7]to preferences over streams of consumption lotteries. Working in this environment it ispossible to establish the equivalence of the weak one-switch property and the one-switchproperty for time preferences with a discounted expected utility representation. We firsthave to adapt the AA framework to continuous time for this purpose.Assume that X is a mixture set ([16]); that is, for every x, y ∈ X and every λ, µ ∈ [0 , xλy ∈ X satisfying: • x y = x , • xλy = y (1 − λ ) x , • ( xµy ) λy = x ( λµ ) y .We assume that X contains a “neutral” outcome, denoted by 0. We can think of X as a set of lotteries with monetary outcomes, and 0 corresponds to the lottery which pays0 with certainty.We next introduce a mixture operation for sequences of dated outcomes, analogous tothe AA mixing operation. To do so, we recall that the neutral outcome obtains at anydate not specified in the sequence. First, define( x , t ) λ ( y , t ) = ( x λ y , t ) for any ( x , t ) , ( y , t ) ∈ A and all λ ∈ [0 , , (27)where x λ y is defined the usual way (see [16]).21 imes s s s ⇒ timet t t timet s t s s t s l l l l l l l Figure 3: Concatenation of t = ( t , t , t ) and s = ( s , s , s , s )Let ( x , t ) ∈ A with t = ( t , t , . . . , t n ) and let s = ( s , s , . . . , s m ). Define t | s = l = ( l , l , . . . , l k ) , where l < l < . . . < l k and { l , l , . . . , l k } = { t , t , . . . , t n } ∪{ s , s , . . . , s m } . An example of concatenation procedure for time vectors t = ( t , t , t )and s = ( s , s , s , s ) is given in Figure 3.For any ( x , t ) , ( y , s ) ∈ A and any λ ∈ [0 ,
1] define the mixture operation as follows:( x , t ) λ ( y , s ) = ( z , t | s ) λ ( z (cid:48) , s | t ) , (28)where z is defined so that if l j = t i , then z j = x i , otherwise z j = 0, and z (cid:48) is defined sothat if l j = s i , then z (cid:48) j = y i , otherwise z (cid:48) j = 0. Note that ( x , t ) and ( z , t | s ) are identicalsequences of dated outcomes, and likewise ( y , s ) and ( z (cid:48) , s | t ) are identical sequences.Figure 4 illustrates the transformation of the sequence ( x , t ) = (( x , x , x ) , ( t , t , t ))to the sequence ( z , t | s ) and the sequence ( y , s ) = (( y , y , y , y ) , ( s , s , s , s )) to thesequence ( z (cid:48) , s | t ). Note that t | s = s | t .( x , t ) → ( z , t | s ) timet t t x x x s s s s l l l l l l l z z z z z z z ( y , s ) → ( z (cid:48) , s | t ) times s s s y y y y t t t l l l l l l l z (cid:48) z (cid:48) z (cid:48) z (cid:48) z (cid:48) z (cid:48) z (cid:48) Figure 4: Transformation of ( x , t ) = (( x , x , x ) , ( t , t , t )) to ( z , t | s ) and ( y , s ) =(( y , y , y , y ) , ( s , s , s , s )) to ( z (cid:48) , s | t )It is not hard to see that A is a mixture set.The utility function u : X → R is called mixture linear if for every x, y ∈ X we have u ( xλy ) = λu ( x ) + (1 − λ ) u ( y ).We say that preferences in this environment have a DEU (discounted expected utility)representation if they have a DU representation ( u, D ) in which u is mixture linear on X . We next show that the induced utility U on sequences of dated outcomes is mixturelinear on A . It follows from (28) that U (( x , t ) λ ( y , s )) = U (( z , t | s ) λ ( z (cid:48) , s | t )) = U (( z , l ) λ ( z (cid:48) , l )) = U ( z λ z (cid:48) , l ) , (29)22here l = t | s = s | t . Since u is mixture linear it follows that U ( z λ z (cid:48) , l ) = λU ( z , l ) + (1 − λ ) U ( z (cid:48) , l ) = λU ( z , t | s ) + (1 − λ ) U ( z (cid:48) , s | t )= λU ( x , t ) + (1 − λ ) U ( y , s ) . Hence, U is mixture linear on A .It is worth mentioning that the problem of finding axiomatic foundations for a DEUrepresentation remains an open question. It is natural to assume that Fishburn andRubinstein’ s axioms [17] should be satisfied for (cid:60) restricted to A with X being amixture set. However, it is beyond the scope of the present paper to address this issue.The following lemma is a part of [9, Lemma 3]. The proof is given for completeness.
Lemma 18 (cf., [9]) . Let preference (cid:60) on A have a DU representation. If (cid:60) on A exhibitthe weak one-switch property, then for any ( x , t ) , ( y , s ) ∈ A if there exist σ , σ such that σ < σ and ( x , t + σ ) ∼ ( y , s + σ ) , and ( x , t + σ ) ∼ ( y , s + σ ) , then ( x , t + σ ) ∼ ( y , s + σ ) for any σ ∈ ( σ , σ ) .Proof. Suppose ( x , t ) , ( y , s ) ∈ A and σ , σ are such that σ < σ with( x , t + σ ) ∼ ( y , s + σ ) , ( x , t + σ ) ∼ ( y , s + σ ) . We need to show that ( x , t + σ ) ∼ ( y , s + σ ) for any σ ∈ ( σ , σ ). The proof is bycontradiction. Assume that we can have ˆ σ ∈ ( σ , σ ) such that ( x , t + ˆ σ ) (cid:31) ( y , s + ˆ σ ).Consider y (cid:48) = y + ε , where ε > x , t + ˆ σ ) (cid:31) ( y (cid:48) , s + ˆ σ ) (bycontinuity, such ε can always be found). However, ( x , t + σ ) ≺ ( y (cid:48) , s + σ ) and ( x , t + σ ) ≺ ( y (cid:48) , s + σ ), which implies a double switch. We obtained the desired contradiction. Thecase ( x , t + ˆ σ ) ≺ ( y , s + ˆ σ ) is similar. Therefore, ( x , t + σ ) ∼ ( y , s + σ ) for any σ ∈ ( σ , σ ).Lemma 18 implies the following corollary: Corollary 19.
The weak one-switch property implies that { σ | ∆( σ ) = 0 } is a closed(possibly empty) interval. In the next proposition the mixtures of sequences of dated outcomes will be used.It adapts [9, Lemma 3] to the time preference setting. Proposition 20 is proved bycontradiction. We first need to find two sequences of dated outcomes such that theirDEU difference changes its sign as the delay increases. Using Corollary 19 we thenconsider two cases depending on whether the just obtained DEU difference equals zero ata unique point or on the interval. A double switch (contradiction to the weak one-switchproperty) can be obtained in each case by introducing suitable mixtures of sequences ofdated outcomes. The illustrations to support the proof are given in the Appendix. For example, Bleichrodt et al. [14] assume the existence of DU representation for (cid:60) on A , where X is not necessarily restricted to reals. Similarly, Rohde [24] applies the DU representation to (cid:60) on A requiring only that X is a connected topological space which contains a “neutral” outcome. roposition 20 (cf., [9, Lemma 3]) . Let preference (cid:60) on A have a DEU representation.If (cid:60) exhibits the weak one-switch property, then for any ( x , t ) , ( y , s ) ∈ A if ( x , t + σ ) ∼ ( y , s + σ ) and ( x , t + σ ) ∼ ( y , s + σ ) for some σ ≥ and some σ > σ , then ( x , t + σ ) ∼ ( y , s + σ ) for any σ ≥ .Proof. Suppose 0 ≤ σ < σ and( x , t + σ ) ∼ ( y , s + σ ) , ( x , t + σ ) ∼ ( y , s + σ ) . By Lemma 18 it follows that( x , t + σ ) ∼ ( y , s + σ ) for any σ ∈ ( σ , σ ) . (30)The weak one-switch property implies weak preference in one direction above σ and inthe other direction below σ . We assume that ( x , t + σ ) is weakly preferred to ( y , s + σ )above σ . The other case can be treated similarly.Suppose that (30) is satisfied but( x , t + σ (cid:48) ) (cid:31) ( y , s + σ (cid:48) ) for some σ (cid:48) > σ . (31)We will prove that a contradiction follows.Let ∆ ( σ ) = U ( x , t + σ ) − U ( y , s + σ ). Therefore,∆ ( σ ) = 0 , if σ ∈ [ σ , σ ] , and ∆ ( σ (cid:48) ) > . Given the DEU representation we can always find a, b ∈ X and q > p such that∆ ( σ ) = U ( a, q + σ ) − U ( b, p + σ ) < σ. Let ( z , t ) = ( a, q ) λ ( x , t ), where λ ∈ (0 , z , t ) = ( b, p ) λ ( y , s ).Consider the function∆ ( σ ) = U ( z , t + σ ) − U ( z , t + σ )= λ ( U ( a, q + σ ) − U ( b, p + σ )) + (1 − λ ) ( U ( x , t + σ ) − U ( y , s + σ ))= λ ∆ ( σ ) + (1 − λ )∆ ( σ ) . Then for λ sufficiently small we have (see Figure A.1 in the Appendix ):∆ ( σ ) < , ∆ ( σ ) < , and ∆ ( σ (cid:48) ) > . Therefore, by continuity, there must exist at least one σ ∗ ∈ ( σ , σ (cid:48) ) such that ∆ ( σ ∗ ) = 0.By Corollary 19 there are two possible cases: either σ ∗ is unique, or ∆ ( σ ) = 0 for σ ∈ [ σ ∗ , σ ∗ ] ⊂ ( σ , σ (cid:48) ) (see Figure A.2 in the Appendix). Note that for all figures in Appendix only sign is relevant in the vertical dimension. ase 1 . Assume first that σ ∗ is unique; i.e., (using the one-switch property) ∆ ( σ ) < σ < σ ∗ and ∆ ( σ ) > σ > σ ∗ . Consider the reflection of ∆ ( σ ) about the σ -axis;i.e., ˆ∆ ( σ ) = − ∆ ( σ ). Then ˆ∆ ( σ ) > σ < σ ∗ and ˆ∆ ( σ ) < σ > σ ∗ . Next, chooseˆ σ ∈ ( σ , σ ) and shift the function ˆ∆ ( σ ) to the left as follows˜∆ ( σ ) = ˆ∆ ( σ + ( σ ∗ − ˆ σ )) = − ∆ ( σ + ( σ ∗ − ˆ σ ))= U ( z , t + σ ∗ − ˆ σ + σ ) − U ( z , t + σ ∗ − ˆ σ + σ ) , so that for the shifted function (see Figure A.3 in the Appendix):˜∆ ( σ ) > , ˜∆ (ˆ σ ) = 0 , ˜∆ ( σ ) < , and ˜∆ ( σ (cid:48) ) < . Define the mixtures ( w , l ) = ( z , t + σ ∗ − ˆ σ ) λ ( x , t ) and ( w , l ) = ( z , t + σ ∗ − ˆ σ ) λ ( y , s ).Analyse the function∆ ( σ ) = U ( w , l + σ ) − U ( w , l + σ )= λ ( U ( z , t + σ ∗ − ˆ σ + σ ) − U ( z , t + σ ∗ − ˆ σ + σ ))+ (1 − λ ) ( U ( x , t + σ ) − U ( y , s + σ ))= λ ˜∆ ( σ ) + (1 − λ )∆ ( σ ) . Choosing λ sufficiently small we obtain (see Figure A.4 in the Appendix):∆ ( σ ) > , ∆ ( σ ) < , and ∆ ( σ (cid:48) ) > . Therefore, we have arrived at a contradiction to the one-switch property.
Case 2 . We next assume that there exist σ ∗ < σ ∗ such that ∆ ( σ ) = 0 if and onlyif σ ∈ [ σ ∗ , σ ∗ ] ⊂ ( σ , σ (cid:48) ). Also, by the one-switch property, we must have ∆ ( σ ) < σ < σ ∗ , and ∆ ( σ ) > σ > σ ∗ . We consider the reflection of ∆ ( σ ) about the σ -axisand shift it to the left by a small amount (cid:15) > ( σ ) as follows¯∆ ( σ ) = − ∆ ( σ + ε ) = U ( z , t + ε + σ ) − U ( z , t + ε + σ ) . Then (see Figure A.5 in the Appendix):¯∆ ( σ ) > σ < σ ∗ − ε, ¯∆ ( σ ) = 0 if σ ∈ [ σ ∗ − ε, σ ∗ − ε ] , and ¯∆ ( σ ) < σ > σ ∗ − ε. Define the mixtures ( v , k ) = ( z , t + ε ) λ ( z , t ) and ( v , k ) = ( z , t + ε ) λ ( z , t ).Analyse the function¯∆ ( σ ) = U ( v , k + σ ) − U ( v , k + σ )= λ ( U ( z , t + ε + σ ) − U ( z , t + ε + σ )) +(1 − λ ) ( U ( z , t + σ ) − U ( z , t + σ ))= λ ¯∆ ( σ ) + (1 − λ )∆ ( σ ) . Choosing λ sufficiently small we obtain (see Figure A.6 in the Appendix):¯∆ ( σ ) < , ¯∆ ( σ ∗ ) = 0 , and ¯∆ ( σ ∗ ) < . b, p ) λ ( v , k ) and ( a, q ) λ ( v , k ) we have∆ ( σ ) = λ ( U ( b, p + σ ) − U ( a, q + σ )) + (1 − λ ) ( U ( v , k + σ ) − U ( v , k + σ ))= − λ ∆ ( σ ) + (1 − λ ) ¯∆ ( σ ) . Recall that ∆ ( σ ) < σ . Letting λ be sufficiently small we obtain a double switch(see Figure A.7 in the Appendix):∆ ( σ ) < , ∆ ( σ ∗ ) > , and ∆ ( σ ∗ ) < , which is a contradiction.The implication of Proposition 20 is as follows: Corollary 21.
If preferences (cid:60) on A have a DEU representation, then the one-switchproperty is equivalent to the weak one-switch property. In this section we consider preferences (cid:60) on A and return to the initial assumptionthat X = [0 , ∞ ). Definition 22.
We say that (cid:60) exhibits the weak one-switch property for dated outcomes if (cid:60) exhibits the weak one-switch property on A . The following proposition gives a partial characterisation of the weak one-switch prop-erty for dated outcomes.
Proposition 23.
Let (cid:60) restricted to A has a DU representation. Then preferences (cid:60) exhibit the weak one-switch property for dated outcomes if (cid:60) exhibit DI or II.Proof. We show that if (cid:60) exhibits II or DI it must also exhibit the weak one-switchproperty for dated outcomes. The proof is by contradiction. Suppose that for some( x, t ) , ( y, s ) ∈ A and some σ, ε such that 0 < σ < ε we can have:( x, t ) (cid:31) ( y, s ) , (32)( x, t + σ ) ≺ ( y, s + σ ) , (33)( x, t + ε ) (cid:31) ( y, s + ε ) . (34)W.l.o.g. assume that t < s . Then we also must have that x < y , otherwise (33) contradictsimpatience and monotonicity. By continuity we can find s (cid:48) ∈ ( t, s ) such that ( x, t ) ∼ ( y, s (cid:48) ). Therefore, by II we have ( x, t + σ ) (cid:60) ( y, s (cid:48) + σ ). By impatience, ( y, s (cid:48) + σ ) (cid:31) ( y, s + σ ), hence, by transitivity, ( x, t + σ ) (cid:31) ( y, s + σ ), which is a contradiction to (33).Therefore, we have shown that if (cid:60) does not satisfy the one-switch property for datedoutcomes it also does not exhibit II. 26he proof for DI is analogous. Indeed, consider ( x, t + σ ) ≺ ( y, s + σ ), with t < s , x < y , as before. By continuity we can find s (cid:48)(cid:48) > s such that ( x, t + σ ) ∼ ( y, s (cid:48)(cid:48) + σ ).Since ε > σ , let ε = σ + γ for some γ >
0. From the equivalence ( x, t + σ ) ∼ ( y, s (cid:48)(cid:48) + σ ) itfollows by DI that ( x, t + σ + γ ) (cid:52) ( y, s (cid:48)(cid:48) + σ + γ ), or, equivalently, ( x, t + ε ) (cid:52) ( y, s (cid:48)(cid:48) + ε ).Since s (cid:48)(cid:48) > s , impatience implies that ( y, s (cid:48)(cid:48) + ε ) ≺ ( y, s + ε ). Therefore, we must have( x, t + ε ) ≺ ( y, s + ε ), which brings us to a contradiction to (34).We have not been able to establish the converse to Proposition 23. The arguments usedin Lemma 11 do not adapt straightforwardly to the present situation. However, Propo-sition 23, together with previous characterisation of the one-switch property for datedoutcomes (Lemma 11), already imply that the one-switch property for dated outcomesand the weak one-switch property for dated outcomes are not equivalent for intertemporalpreferences with a DU representation. This paper fills a gap in Bell’s original characterisation [9, Proposition 8] of discountfunctions compatible with the one-switch property for sequences of dated outcomes. Weshowed that functions of the linear times exponential form also have this property andthat such discount functions exhibit strictly increasing impatience. Although decreasingimpatience is commonly found in experiments [18], there is also much evidence for in-creasing impatience (see, for example, [8], [26]). To the best of our knowledge, the lineartimes exponential function has not been used in the literature on time preferences before.Therefore, we introduce a new type of a discount function that accommodates strictlyincreasing impatience and the one-switch property.With regard to sums of exponentials, there has recently been some interest in thistype of discount function. In their experiments, McClure et al. [20] used magnetic neuro-images of individuals’ brains to study intertemporal choice. They found that making adecision is a result of the interaction of two separate neural systems with different levelsof impatience. To describe the involvement of these two brain areas in discounting, theysuggested sum of exponential discount functions which they refer to as double exponential discounting. The recent empirical study by Cavagnaro et al. [15] demonstrates that doubleexponential discounting provides a better fit to actual time preferences than exponential,quasi-hyperbolic, proportional hyperbolic and generalized hyperbolic discounting.A second contribution of this paper is to clarify the definition of the original one-switchrule introduced by Bell [9], by considering two definitions: the weak one-switch propertyand the (strong) one-switch property. We demonstrate that if X is a mixture set and ifpreferences have a DEU representation, then the one-switch property is equivalent to theweak one-switch property.Third, we prove that if preferences have a DU representation, then preferences exhibitthe one-switch property for dated outcomes if and only if they exhibit either strictlyincreasing impatience, or strictly decreasing impatience, or constant impatience. A partialanalogue is obtained for the weak one-switch property for dated outcomes. That is,27he preferences exhibit the weak one-switch property for dated outcomes if they exhibitincreasing impatience or decreasing impatience. Acknowledgments
I thank my supervisor Matthew Ryan for valuable suggestions and detailed comments.I am grateful to participants of 6th Microeconomic Theory Workshop at Victoria Uni-versity of Wellington for helpful comments. Financial support from the University ofAuckland is gratefully acknowledged.
A Appendix
A.1 Illustrations for the proof of Proposition 20
In this appendix we provide the illustrations to accompany the proof of Proposition20. Note that for all the figures in the appendix only the sign (but not the value) of avertical coordinate of a point is relevant. 28 ∆ ( σ ) σ σ σ (cid:48) σ ∆ ( σ ) σ σ σ (cid:48) σ ∆ ( σ ) = λ ∆ ( σ ) + (1 − λ )∆ ( σ ) σ σ σ (cid:48) Figure A.1: The mixture of ∆ ( σ ) and ∆ ( σ ) (with small λ ) σ ∆ ( σ ) σ ∗ σ σ σ (cid:48) σ ∆ ( σ ) σ ∗ σ ∗ σ σ σ (cid:48) Figure A.2: Function ∆ ( σ ) equals zero at one point or at the interval29 ˆ∆ ( σ ) = − ∆ ( σ ) σ ∗ σ (cid:48) σ σ σ ˜∆ ( σ ) = ˆ∆ ( σ ∗ − σ (cid:48) + σ )ˆ σσ σ σ (cid:48) σ ˜∆ ( σ ) ˆ σ σ σ (cid:48) σ Figure A.3: Transformation of ∆ ( σ ) to ˜∆ ( σ ), and then with the new values at indicated σ , σ , σ (cid:48) ∆ ( σ ) σ σ σ (cid:48) σ ˜∆ ( σ ) ˆ σσ σ σ (cid:48) σ ∆ ( σ ) = λ ˜∆ ( σ ) + (1 − λ )∆ ( σ )ˆ σ σ σ σ (cid:48) Figure A.4: The mixture of ˜∆ ( σ ) and ∆ ( σ ) (with small λ ) produces a double switch31 ∆ ( σ ) σ ∗ σ ∗ σ σ σ (cid:48) σ − ∆ ( σ ) σ ∗ σ ∗ σ (cid:48) σ σ σ ¯∆ ( σ ) = − ∆ ( σ + ε ) σ (cid:48) σ σ σ ∗ σ ∗ σ ¯∆ ( σ ) σ (cid:48) σ σ σ ∗ σ ∗ Figure A.5: Transformation of ∆ ( σ ) to ¯∆ ( σ ) with the new values at σ , σ , σ ∗ , σ (cid:48) ∆ ( σ ) σ ∗ σ ∗ σ σ σ (cid:48) σ ¯∆ σ ∗ σ (cid:48) σ σ σ ∗ σ ¯∆ ( σ ) = λ ¯∆ ( σ ) + (1 − λ )∆ ( σ ) σ σ ∗ σ σ ∗ σ (cid:48) Figure A.6: The mixture of ¯∆ ( σ ) and ∆ ( σ ) (with small λ ) at the three points σ , σ ∗ , σ ∗ ¯∆ ( σ ) σ σ ∗ σ σ ∗ σ (cid:48) σ − ∆ ( σ ) σ σ σ ∗ σ ∗ σ (cid:48) σ ∆ ( σ ) = − λ ∆ ( σ ) + (1 − λ ) ¯∆ ( σ ) σ σ σ ∗ σ ∗ σ (cid:48) Figure A.7: The mixture of − ∆ ( σ ) and ¯∆ ( σ ) (with small λ ) produces a double switch.It is sufficient to consider the value of ∆ ( σ ) at the three points σ , σ ∗ , σ ∗ References [1] A. E. Abbas and D. E. Bell. One-switch independence for multiattribute utilityfunctions.
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