Optimal consumption sequences under habit formation and satiation

Optimal consumption sequences under habit formation and satiation

Accepted Manuscript Optimal consumption sequences under habit formation and satiation Juan Dubra, Martín Egozcue, Luis Fuentes García PII: DOI: Refer...

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Accepted Manuscript Optimal consumption sequences under habit formation and satiation Juan Dubra, Martín Egozcue, Luis Fuentes García

PII: DOI: Reference:

S0304-4068(18)30134-4 https://doi.org/10.1016/j.jmateco.2018.11.004 MATECO 2284

To appear in:

Journal of Mathematical Economics

Received date : 21 May 2018 Revised date : 15 November 2018 Accepted date : 30 November 2018 Please cite this article as: J. Dubra, M. Egozcue and L.F. García, Optimal consumption sequences under habit formation and satiation. Journal of Mathematical Economics (2018), https://doi.org/10.1016/j.jmateco.2018.11.004 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

Optimal Consumption Sequences under Habit Formation and Satiation.∗ Juan Dubra

Mart´ın Egozcue

Universidad de Montevideo

Universidad de Montevideo and ANII

[email protected]

[email protected] Luis Fuentes Garc´ıa

Departamento de Matem´aticas Universidade da Coru˜ na [email protected] Revised August, 2018.

Abstract In this note, we contribute to the literature of behavioral economics by extending a result in Baucells and Sarin (2010). We find the optimal consumption sequence for many goods and many periods in an intertemporal choice framework where preferences can exhibit at the same time habit formation and satiation. For the case in which habit formation matters more than satiation, we find that if consumers must consume a good in every period, for certain types of S-shaped value function, consuming the same good in all the periods is optimal. However, if consumers can choose not to consume in some periods, then this behavior is optimal only for certain kind of piece-wise linear value functions. We also discuss briefly the case when satiation dominates habituation. Keywords: Habit formation; satiation; S-shaped value functions. Journal of Economic Literature Classification Numbers: D11, D15, D90, D91. AMS 2010 subject classification: 62P05 ∗

We thank the editor and the anonymous referees for several suggestions which have improved the tech-

nical side, and the exposition, of this note.

1

A large literature in psychology has established that animals and humans exhibit habit formation and satiation. For example, Read et al. (1999, p. 183) argue that “The most important taste change effects are habit formation and satiation.” Also, a sizeable literature in economics has studied habit formation and satiation often separately, while Baucells and Sarin (2010) (henceforth BS) put forward a model for both satiation and habit formation in an intertemporal choice framework (see references therein for the psychology and economics literatures of habit formation, and satiation). In their model satiation and habituation, together with the shape of the value function, determine the trade-off between seeking variety and maintaining acquired habits. In a discrete choice context, BS show that if habituation dominates satiation, individuals will consume the same good in each period in a two goods and two periods framework. They ask whether this proposition also holds in a setting with more than two goods and two periods (see, BS p. 294). The aim of this note is to characterize the conditions under which it is true more generally.1 Analyzing the effects of joint habit formation and satiation beyond a two good, two period example is important. As has been established in Fuhrer (2000), and the literature that follows, one can confidently reject the hypothesis of no habit formation with macro data, as habituation helps explain certain patterns of consumption and asset prices. Still, models that only have habit formation can also be rejected by the data (see Cochrane, 2013). One feature that traditional real business cycle models fail to account for is the large volatility in employment (for an early review, see Hansen and Wright, 1992). An avenue that helps in matching employment volatility is assuming that labor is indivisible (you either work 8 hours, or you don’t work at all; see Hansen, 1985), which produces the same patterns as when satiation is present: people fluctuate from one state to the other more often than one would expect with traditional preferences. Also, satiation has been prominent in marketing, and it has often been studied in conjunction with habituation, as in Voss et al. (2010), “Do Satisfied Customers Always Buy More? The Roles of Satiation and Habituation in Customer Repurchase”. More generally, marketing science has focused on the purchasing behavior of narrow categories of goods (e.g. vacations in Cancun vs. vacations in Bermuda) where satiation plays a prominent role, while economics has focused in broader categories (e.g. beach holidays) where habituation may be more relevant.2 1 2

The problem was explicitly raised to the second author by Manel Baucells in private correspondence. See for example Hasegawa et al. (2012), McAlister (1982) or the seminal Gr¨ onroos (1984).

2

Another area where the results of this note might be of interest is in data for which inattention (rational, as in Sims (2003) or the literature that followed, or just “given” as a constraint as in Calvo (1983)) has sometimes been used to justify stickiness in choices. If there is some variable that decision makers don’t change as often as they should according to some traditional metric, that could be because they are not paying attention to external conditions, or because they have gotten used or habituated to the variable being at that value (and switching has a cost). In Section 3 below we give a simple example where both inattention and the model we study can be used to explain data that looks irrational. This brief account points to the fact that understanding models where both habit formation and satiation are present, beyond the two period-two good example, is of interest. More generally, there is the feeling among many economists that while behavioral economics is good at describing how individual behavior departs from rationality or traditional models, its impact on explaining market behavior or data is more limited. For example, about behavioral finance, Fama (2010) stated “I’ve always said they are very good at describing how individual behavior departs from rationality. That branch of it has been incredibly useful. It’s the leap from there to what it implies about market pricing where the claims are not so well-documented in terms of empirical evidence.” Part of the problem is that the adoption of well micro-founded (or behaviorally sound) models of preferences (such as BS) have been slow in their adoption in applied work, where more than two goods and periods are needed. This note provides one step in that direction, but more work is needed. We find that if consumers have to choose one good in each period, and habituation dominates satiation, the conjecture that consuming the same good in every period is optimal holds for certain types of S-shaped value functions. However, for general S-shaped value functions, the conjecture does not hold. Moreover, if subjects are not forced to consume one good in each period, the conjecture is true only for certain types of piece-wise linear functions.

1

Model setup

In this section, we present the BS model setup and mathematical notation. Assume T periods of time. A consumer must choose in each (discrete) period whether to consume or not one unit out of K goods. Let x1 = (x11 , x12 , ..., x1T ) be a vector that denotes the

3

consumption sequence of the first good, where x1i = 1 indicates that good 1 is consumed in period i (positive consumption) and x1i = 0 means that good 1 is not consumed in period i (zero consumption). Hereafter, to ease notation, the goods are also denoted with letters by a simple cipher substitution. In this case, since it is the first good, represented by an A. Therefore, for instance, consider T = 4, K = 3, and the following vectors: x1 = (1, 0, 0, 0), x2 = (0, 1, 0, 0, ), and x3 = (0, 0, 1, 0), then the overall consumption sequence denoted by c = {x1 , x2 , x3 } represents the following consumption stream (A, B, C, −), where B and C

stands for the consumption of good 2 and 3 and the dash “−” implies zero consumption in the last period (no consumption of goods is possible in the BS model). In each period, the consumer evaluates, for each good, the following expression v(x + y − r) − v(y), where v is a value function, x is the quantity consumed of the good, y the satiation level and r a reference point level. In other words, in each period the consumer evaluates the increment from current satiation with respect to a reference point level. Therefore, the discrete unit choice problem of the consumer is to maximize t=T K X X    v xkt + ytk − rtk − v ytk , U (c) =

(1)

k=1 t=1

 k k k k subject to the following recursion equations: ytk = γ xkt−1 + yt−1 − rt−1 , rt = rt−1 +  P K k α xkt−1 − rt−1 with xkt ∈ {0, 1}, y0k = xk0 = r0k = 0, k=1 xkt ≤ 1, for all t and the parameter γ ∈ (0, 1) denotes the speed of satiation while α ∈ (0, 1) denotes the speed of habituation.

BS assume symmetric goods, which means that the consumer evaluates each good by using the same v, γ and α. We maintain this assumption in this paper, but we discuss some extensions in Section 3. Given this assumption, if consuming only one good is optimal, we will assume without loss of generality that it is good A. Kahneman and Tversky (1979) introduce S-shaped value functions as one of the core features of prospect theory. One of its characteristics, is that the value function is concave in the positive domain and convex in the negative domain. It also presents what is coined loss aversion, that is: the value of losses loom larger than the value of gains. Nevertheless, there are many ways to define loss aversion, see K¨obberling and Wakker (2005) and the references therein. For this reason, we shall restrict our analysis to the next definitions of S-shaped value functions with loss aversion. 4

Definition 1 A function v : R → R is S−shaped if it is continuous, increasing, with v(0) =

0, concave for x ≥ 0 and convex for x < 0. Moreover, we say that v satisfies loss aversion whenever −v(−x) ≥ v(x) for all x > 0.

The next definition introduces three of the most common examples of S-shaped value function with loss aversion. Definition 2 The following value function v : R → R is the power value function:   xβ if x ≥ 0, v(x) =  −λ(−x)β if x ≤ 0, with 0 < β ≤ 1 and λ ≥ 1.

BS also use an S-shaped piecewise-linear value function. A function v : R → R is

canonical if

v(x) = with λ > 1, a > 0.

 

min(a, x)

for x ≥ 0,

 max(−λa, λx) for x ≤ 0,

K¨ obberling and Wakker (2005) introduce the following exponential value function   1−e−µx if x ≥ 0, µ v(x) =  λ eβx −1 if x < 0, β

with 0 < β < µ and λ ≥ 1.

The interested reader is referred to Dhami (2016, ch. 2), K¨obberling and Wakker (2005), Wakker (2010, ch. 3) and the references therein for further insights and features of S-shaped value functions. One of the novelties of the BS model is that there are two forces that affect individual consumption in each period: satiation and habituation. Satiation implies that individuals prefer to diversify their consumption, while habituation pushes individuals to continue consuming the same product in the future. BS prove the following result that finds the optimal consumption stream for two goods A and B in two periods: Proposition 1 Assume v is an S-shaped value function and satisfies loss aversion. If α ≥ γ then the consumption sequence (A, A) is optimal when T = 2. 5

In other words, if the speed of habituation is not smaller than the speed of satiation then the optimal consumption sequence is to consume the same product in both periods. BS also prove that if α < γ alternation of consumption is optimal, i.e., (A, B) is optimal. We shall discuss this case in Section 3. As BS note, the extension of this results for more than two periods and two goods is hard. Indeed, to gain some insight into this problem, they provide simulations using the canonical value function for three goods and twelve periods. Their analysis seems to suggest that the extension of Proposition 1 to any number of periods and goods must also be true (BS p. 294). In the next section, we tackle this problem and give the conditions under which the extension of Proposition 1 holds for many goods and periods.

2

Main results

Formally, the statement of the open problem can be written as follows: Conjecture 1 Let v be an S-shaped value function with loss aversion. If γ ≤ α then the op-

timal consumption sequence is c = {x1 , x2 , ..., xK } with x1 = (1, 1, ..., 1) and xk = (0, 0, ..., 0) for all k > 1, or equivalently, (A, A, ..., A) is optimal.

We shall solve this conjecture with a two step strategy. Step 1.a. First, we will show that, for instance, for some special S-shaped value functions with loss aversion: (A, B, C, D, B, A, D) is improved by (A, A, A, A, A, A, A) or (A, B, −, D, −, A, D) is improved by (A, A, −, A, −, A, A).

Step 1.b. Next, we will establish that, in general, (A, A, A, A, A, A, A) is not better than (A, A, −, A, −, A, A).

Step 2. Finally, we will show that for certain types of canonical value functions consuming A in each period is optimal. The next theorem, which establishes Step 1.a., shows that one can improve a consumption sequence by consuming the same good in all periods of positive consumption. It introduces the condition that w(x) = v(γ −1 x) − v(x) is subadditive in R+ , superadditive in R− and

exhibits loss aversion. To understand the role played by this condition, note that the key to

being able to prove our results is transforming the three variable per period value into a one 6

variable problem: for each good, given y1 = 0, we have t=T X t=1

[v (xt + yt − rt ) − v (yt )] =

T X    v γ −1 yt+1 − v (yt ) t=1

= v(γ

−1

yT +1 ) +

T X

w(yt ).

(2)

t=2

Then, the conditions on w ensure that for each good it is optimal to reduce movements in y : for example, consuming a good now increases y; ceasing to consume it later decreases y; but this rise and posterior decrease reduces utility because of loss aversion. This force pushes towards not switching between goods. Theorem 1 Assume v is an S-shaped value function with loss aversion, α ≥ γ, and that w(x) = v(γ −1 x) − v(x) is subadditive in R+ , superadditive in R− and features loss aversion.

Let c = {x1 , x2 , ..., xK } and ˆ c = {ˆ x1 , x ˆ2 , ..., x ˆK } be two consumption sequences. We have U (c) ≥ U (ˆ c), whenever x1t

=

K X k=1

xˆkt for all 1 ≤ t ≤ T and xk = (0, 0, ..., 0) for all k > 1.

We present all proofs in the appendix. It is worth noting that Theorem 1 does not say that, for instance, (A, B, C, −, D, −, A) is improved by (A, A, A, A, A, A, A); it says only

that it is improved by (A, A, A, −, A, −, A). That is, the optimal consumption sequence may

maintain the same periods of zero consumption.

The following example shows that if zero consumption is possible, then consuming a single good in all periods might not be optimal, establishing Step 1.b. Example 1. Let v(x) =

 

x0.9

 −2 (−x)

0.9

if x ≥ 0, if x ≤ 0.

Table 1 lists the five consumption sequences with highest utility for this value function and

7

its corresponding objective function value for α = γ = 0.7, K = 2 and T = 10. Table 1: Optimal Consumption Sequence x

U (x)

(A, −, A, A, A, A, A, A, A, A)

1.4030

(−, A, −, A, A, A, A, A, A, A)

(−, −, A, −, A, A, A, A, A, A)

(−, −, −, A, −, A, A, A, A, A) (A, A, A, A, A, A, A, A, A, A)

1.4028 1.4022 1.4002 1.3952

The single good consumption vector is in the fifth place of the overall ranking.  Trivially, if zero consumption is not allowed in any period, then we can state the following version of Conjecture 1, which is an immediate consequence of Theorem 1. Corollary 1 Assume α ≥ γ. Let v be an S-shaped power value function with loss aver-

sion, the canonical value function, or the exponential value function with µ ≤

2γlogγ . (1−γ)

If zero

consumption is not allowed in any period then (A, A, ..., A) is optimal.

Nevertheless, if we restrict our analysis only to the canonical value functions with a ≥ 1,

the single good consumption vector is globally optimal, even if we allow some periods to have zero consumption. This is the content of the next theorem and the last step (Step 2) of our proof strategy.

Theorem 2 For the canonical value function with a ≥ 1 and α ≥ γ, (A, A, ..., A) is optimal. The novel conditions in our previous results are those related to the function w(x) = v(γ

−1

x) − v(x). The following example shows that the assumptions imposed on w can not

be dispensed with: one cannot extend Corollary 1 and Theorem 2 to any type of S-shaped value functions even with loss aversion. Example 2. Consider the following piece-wise linear value function:   min(1/2, x) if x ≥ 0, v(x) =  max(−1/2, 2x) if x ≤ 0.

8

Consider, for instance, K = 2, T = 8, α = 0.96 and γ = 0.3 the following table depicts the top ranked consumption sequences. Table 2: Utility of some Consumption Sequences x

U (x)

(A, A, B, A, B, A, B, A)

1.49106

(A, A, A, −, A, A, −, A)

1.47716

(A, A, −, A, A, A−, A)

1.45536

(A, A, −, A, A, −, A, A)

1.44108

(A, A, A, A, A, A, A, A)

0.54166

...

As Table 2 shows, consuming the same good in all the periods is far below the first ranked consumption sequences. Constant consumption is not optimal because what fails in this example is that w does not feature loss aversion.  In sum, our analysis shows that if zero consumption is not allowed Conjecture 1 is true for certain types of S -shaped value functions. Otherwise, the conjecture is true only for some special cases of the canonical value function.

3

Extensions, satiation dominating habituation, and an application.

In order to follow the analysis of BS and simplify notation, we have assumed symmetry of goods, but the results in the previous section can be generalized somewhat for asymmetric goods, in the case in which habituation strictly dominates satiation, that is α > γ. To see how the extensions work, consider the next theorem. Theorem 3 Assume v is an S-shaped value function with strict loss aversion, i.e., −v(−x) > v(x) for all x > 0, α > γ, and suppose that w(x) = v(γ −1 x) − v(x) is subadditive in

R+ , superadditive in R− and exhibits strict loss aversion. Let c = {x1 , x2 , ..., xK } and

ˆ c = {ˆ x1 , x ˆ2 , ..., x ˆK } be two consumption sequences. We have U (c) > U (ˆ c) whenever x1t

=

K X k=1

xˆkt for all 1 ≤ t ≤ T and xk = (0, 0, ..., 0) for all k > 1, 9

and xˆkt00 = 1, xˆkt11 = 1 for some t0 < t1 and k0 6= k1 . In this case, consuming only one good is strictly better than any other consumption sequence; then, if each good k has a different αk and γ k , there exists an open set of (α, γ) ∈

R2K in the neighborhood of ((α, ..., α) , (γ, ..., γ)) ∈ R2K such that consuming a single good

is still optimal.

This extension is simple enough, and brings our results one step closer to applicability, as not all goods generate the same dependence (habit), or satiation. The size of the open set will depend on the application one has in mind, but in principle this extension allows for some asymmetry at least if it is not too large. In order to see how our results could be useful in applications, consider the following example. Example 3. We observe a person who has $2. We know the price of good A is $1 at date t = 0, and will remain fixed for t = 1, 2, 3. The price of good B is $1 today and will become $ 21 with probability p or $2 with probability 1 − p in period t = 1, and remain fixed at

its new value for t = 2, 3. We also know that the subject values both goods equally in  some informal sense. We observe (ex post) prices of B PB = 1, 21 , 21 , but that the subject

nevertheless consumes one unit of A in period 1, and another between periods 2 and 3, but not the “natural” alternative (A, B, B) . In economics inattention (rational as in Sims (2003) or just “given” as in Calvo, 1983) has sometimes been used to justify stickiness in choices (the subject consumed A after period t = 1, instead of switching to B). To illustrate, a

Calvo-style model would be one in which the subject values his consumption as the sum of goods consumed in all periods (perfect substitution among goods and periods), but is only allowed to re-optimize with probability q in each period after t = 1; if in a given period he is not allowed to re optimize, he consumes the same basket he chose in the previous period, if it is affordable, and nothing otherwise. The initial choice of good only matters if the individual is not allowed to re-optimize, in which case his consumption streams would be (A, A, −) ∼ (A, −, A) or (B, B, B) with

probability p or (B, −, −) with probability 1 − p. So (A, A, −) is optimal iff the value of

consuming (A, A, −) is greater than the expected value of consuming (B, B, B) or (B, −, −) 1 2 ≥ p3 + (1 − p) 1 ⇔ p ≤ . 2 So if p ≤

1 2

the subject will choose to consume A, and if prices of B rise, or they fall and the 10

individual cannot re-optimize, we will observe stickiness in choices as the observed stream will be (as observed) A in t = 1, and one more unit of A between periods 2 and 3. Another behaviorally sound alternative to the previous model is to assume that people have BS preferences, and habituation larger than satiation. More concretely, with a power value function as in Definition 2 with β =

4 5

and λ = 2, (A, −, B) is better than (B, B, B) with

probability p and (B, −, −) with probability 1 − p for p ≤

2 5

and α = 12 , γ = 52 . Since we know

by Theorem 1 that (A, −, A) dominates (A, −, B) , we then know that (A, −, A) dominates

(B, B, B) with probability p and (B, −, −) with probability 1−p; it also dominates (A, B, B)

(for similar parameters, (A, A, −) also dominates the available alternatives). Therefore, we

will observe that the subject will choose to consume A in period 1, and an additional unit of A between periods 2 and 3, instead of (A, B, B) , which is what needed an explanation.

Of course, three periods are not necessary to establish a result of this kind, but the point of the example is that results beyond two goods and periods are needed if the BS theory is to be applied more generally.  Of course, a natural question is what happens when γ > α, satiation dominating habit formation. The mathematical difficulties that arise in that case are different from the ones tackled in this note, and therefore exceed the scope of this work. To illustrate that the problem is non-trivial, and that even canonical value functions will not imply strict alternation, as one would expect, consider the examples presented in Table 3, which is based on a canonical value function, a = 1, λ = 2, T = 5 and only K = 2 goods. Table 3: Optimality when Satiation Dominates α

γ

Optimal Sequence

1 2 1 2 17 20

9 10 3 5 9 10

(−, B, A, A, B)

.

(B, B, B, B, B) (B, −, B, B, B)

Also, changing parameters has non-trivial consequences. For instance, letting T = 6 for the case in the first line of Table 3 makes (B, A, −, B, A, B) optimal. Again, the point of

these examples is only to show that the problem when satiation dominates is non-trivial, and there is no obvious conjecture to be proved.

11

4

Concluding remarks

In this note, we provide a proof of an open problem raised by Baucells and Sarin (2010). They show that consuming a single good in two periods is optimal if habit formation dominates satiation. They ask whether this result still holds in a more general setting. We find that for certain kinds of S-shaped value functions, the general result holds if the consumers must consume one good in every period. If zero consumption is allowed, then the general result is true only for certain types of piece-wise linear functions. Allowing for general asymmetry, and studying the case where satiation dominates habituation seem interesting topics for future research.

12

5

Appendix: Proofs.

We start proving some lemmas that we shall use in the sequel. First, we show some inequalities that hold for each good (to ease notation we suppress the upper index). Lemma 1 The following inequalities for the recursion equations, of each good, hold: 1. 0 ≤ rt < 1. Also if α ≥ γ then: 2. 0 ≤ rt − yt < 1. 3. −γ < yt ≤ γ and 4. if xt = 0 then yt+1 ≤ 0 and if xt = 1 then yt+1 > 0. Proof. The proof of the first assertion follows by induction, since 0 ≤ r0 < 1 and

assuming that 0 ≤ rt−1 < 1 then rt = αxt−1 + (1 − α)rt−1 < α + 1 − α = 1. Also all the

terms of this last equation are positive, therefore rt ≥ 0. Now, suppose that α ≥ γ. For the second assertion, using the recursion equations defined in (1), we have

rt − yt = (α − γ)xt−1 + (1 − α)rt−1 + γ(rt−1 − yt−1 ).

(3)

Since r0 − y0 = 0 ≥ 0 and assuming that rt−1 − yt−1 ≥ 0, using the first part of this lemma,

the assumption that α ≥ γ and xt−1 ≥ 0 implies that rt −yt ≥ 0. The second inequality is also

derived by induction, since r0 − y0 = 0 < 1 and rt − yt = (α − γ)xt−1 + (1 − α)rt−1 + γ(rt−1 −

yt−1 ) < α − γ + 1 − α + γ = 1. For the third assertion, we have seen in the previous result that 0 ≤ rt−1 − yt−1 < 1. So −1 < xt−1 − (rt−1 − yt−1 ) ≤ 1 and since yt = γ(xt−1 + yt−1 − rt−1 )

then −γ < yt ≤ γ. Finally, if xt = 0 then yt+1 = γ(0 − (rt − yt )) ≤ 0. And, if xt = 1 then yt+1 = γ(1 − (rt − yt )) > γ(1 − 1) = 0.

Next, we show the following fundamental lemma. If we have two consumption sequences

where the zero consumption occurs in the same periods then the summation of the satiation levels and the reference levels must coincide. Lemma 2 Let c = {x1 , x2 , ..., xK } and ˆ c = {ˆ x1 , x ˆ2 , ..., x ˆK } be two consumption sequences.

We have

K X k=1

xkt

=

K X k=1

xˆkt

for all t ⇒

K X

rtk

k=1

13

=

K X k=1

rˆtk

and

K X k=1

ytk

=

K X k=1

yˆtk .

Proof. The proof follows from the recursion equations and by an inductive argument as follows. For the first equation, we have ! K K X X k rt+1 = rtk + α k=1

k=1 K X

=

rˆtk

k=1

!

K X

k=1 K X



xkt xˆkt

k=1

!

!

− −

K X k=1 K X

rtk rˆtk

k=1

!!

!!

=

K X

k rˆt+1

k=1

Now, the second equation, K X

K X

k =γ yt+1

k=1

k=1 K X



xkt xˆkt

k=1

!

!

+ +

K X

k=1 K X

ytk yˆtk

k=1

!

!

− −

K X

k=1 K X k=1

rtk rˆtk

!!

!!

=

K X

k yˆt+1

k=1

We next present the following fundamental lemma. Lemma 3 Let f : R −→ R be a function subadditive for x ≥ 0 and satisfying the loss

aversion condition f (x) ≤ −f (−x) for all x ≥ 0. If x + y ≥ 0 and y ≤ 0, then f (x + y) ≥

f (x) + f (y).

Proof. By loss aversion: f (x + y) − f (y) ≥ f (x + y) + f (−y) and by subadditivity: f (x + y) + f (−y) ≥ f ((x + y) − y) = f (x).

We shall also use the following well known inequality (see, Kuczma, 2009, p. 217) Lemma 4 An S-shaped function is superadditive in R− and subadditive in R+ . Now, the proof of Theorem 1: Proof of Theorem 1.

Note that for each good, since y1 = 0, equation (2) above

implies U (c) =

K X

v(γ −1 yTk +1 )

+

T X K X t=2 k=1

k=1

14

w(ytk ),

and in the same manner, we have K X

U (ˆ c) =

T X K X

v(γ −1 yˆTk +1 ) +

w(ˆ ytk ).

t=2 k=1

k=1

PK

yˆtk = yt1 , for all t. Hence, ! ! T K T K X X X X U (c) = v(γ −1 yT1 +1 ) + w(yt1 ) = v γ −1 yˆTk +1 + w yˆtk .

Notice that by Lemma 2, we have

k=1

t=2

t=2

k=1

k=1

Since γ −1 > 0 it suffices to prove that for t = 2, 3, . . . , T + 1 the following inequalities hold ! K K X X w yˆtk ≥ w(ˆ ytk ), k=1

v

K X

γ −1 yˆTk +1

k=1

!

k=1



K X

v(γ −1 yˆTk +1 ).

k=1

The functions v and w verify the same conditions of subadditivity and superadditivity in the required domains then, by Lemma 4, it suffices to prove the inequality for w. Now, if x1t−1 = 0 then the consumption of all goods of the other sequence xˆkt−1 must be equal to 0, and by Lemma 1 then yˆtk ≤ 0 for all k > 1. By superadditivity of w in R− , P ˆtk > 0 and for only one good k, we get the desired inequality. If x1t−1 = 1 then yt1 = K k=1 y without loss of generality, say xˆ1t−1 = 1, while for the rest of the goods we must have zero

consumption. Then, using again Lemma 1, we know that yˆt1 > 0 and yˆtk ≤ 0 for all k > 1. Now, we simply apply Lemma 3 as follows:

K X

x = yˆt1 > 0,

k=2

such that: x+y =

K X

yˆtk ≤ 0,

yˆtk = yt1 > 0.

k=1

Therefore,

w(x + y) ≥ w(x) + w(y). Thus, we deduce the following inequality K K X X 1 k w( yˆt ) ≥ w(ˆ yt ) + w( yˆtk ). k=1

k=2

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Finally, since all yˆtk ≤ 0 for k ≥ 2, we apply again the superadditivity of w in R− and get: ! K K X X w yˆtk ≥ w(ˆ yt1 ) + w(ˆ ytk ). k=1

k=2

Proof of Corollary 1. For the power value function the proof is a direct consequence of Theorem 1, noticing that: w(x) = v(γ −1 x) − v(x) = (γ −β − 1)v(x), then w preserves the same conditions of convexity, concavity and loss aversion of v that are sufficient to prove Theorem 1. The proof for the canonical value function, is immediate, because w is subadditive in the positive domain, superadditive in the negative domain and has loss aversion. For the exponential value function, we show w00 (x) ≤ 0 for x positive and w00 (x) ≥ 0 for x ≤ 0, to

deduce that w is concave in the positive domain and convex in the negative domain and hence subadditive in R+ and superadditive in R− . In fact, as a consequence of Lemma 1, it

is sufficient to have the condition for x ∈ [−1, 1]. We see that for x positive w00 (x) ≤ 0 when

x ≤ x0 where x0 =

2γlogγ (1−γ)µ

satisfies w00 (x0 ) = 0. Then, because µ ≤

2γlogγ (1−γ)

00

we obtain x0 ≥ 1.

Also, since 0 < β < µ the same bound on µ guarantees w (x) ≥ 0 for x negative; moreover λ ≥ 1, and we have loss aversion, i.e. −w(−x) ≥ w(x) for x ≥ 0. Proof Theorem 2.

Let p be a natural number 1 ≤ p < T. Suppose we have two

consumption sequences c = (x, x2 , ..., xK ) and ˆ c = (ˆ x, x2 , ..., xK ), such that xk = (0, 0, ..., 0) xt }Tt=p = for all k > 1, xt = xˆt for all t = 0, . . . , p − 1 and {xt }Tt=p = (1, 1, 1, . . . , 1), {ˆ

(0, 1, 1, . . . , 1). We shall prove that U (c) ≥ U (ˆ c). Since by Lemma 1, if α ≥ γ then |yt | ≤

γ ≤ 1, then the value function is only evaluated at [−a, a]. We have:

yˆp+1 = γ(ˆ xp + yˆp − rˆp ) = γ(xp − 1 + yp − rp ) = yp+1 − γ, rˆp+1 = (1 − α)ˆ rp + αˆ xp = (1 − α)rp + αxp − α = rp+1 − α. Let a1 = −γ. For s ≥ 1 following an inductive argument: yˆp+s = yp+s + as , rˆp+s = rp+s − α(1 − α)s−1 .

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for as+1 = γ(as + α(1 − α)s−1 ). We have seen that the equations work for s = 1. Now assume, they are true for s, we will show that they are also true for s + 1: yˆp+s+1 = γ(ˆ xp+s + yˆp+s − rˆp+s ) = γ(xp+s + yp+s − rp+s + as + α(1 − α)s−1 ) = yp+s+1 + γ(as + α(1 − α)s−1 ) = yp+s+1 + as+1 . And for the second recursion rˆp+s+1 = (1 − α)ˆ rp+s + αˆ xp+s = (1 − α)(rp+s − α(1 − α)s−1 ) + αxp+s = rp+s+1 − α(1 − α)s . Let m = T − p. Since the elements of both vectors x and x ˆ are the same up to the period p − 1, we only need to prove the following inequality: p+m

U (c) − U (ˆ c) =

X  v(γ −1 yk+1 ) − v(yk ) − v(γ −1 yˆk+1 ) + v(ˆ yk ) ≥ 0. k=p

For k = p we have yk = yˆk . Moreover, since xp = 1 and xˆp = 0 by Lemma 1, yp+1 ≥ 0 and

yˆp+1 ≤ 0. We have shown that yˆp+1 = yp+1 − γ which implies that 0 ≤ yp+1 ≤ γ. Therefore, the first term of the summation is:

v(γ −1 yp+1 ) − v(γ −1 yp+1 ) = v(γ −1 yp+1 ) − v(γ −1 (yp+1 − γ)) = γ −1 yp+1 − λγ −1 yp+1 + λ. In addition, since i > p, xi = xˆi = 1 implies that yp+1 , yˆp+1 ≥ 0. For, k = p + 1 the summation is equal to:

v(γ −1 yp+2 ) − v(yp+1 ) − v(γ −1 yˆp+2 ) + v(ˆ yp+1 ) = −a2 γ −1 − yp+1 + λyp+1 − λγ, where a2 = γ(a1 + α) = γ(α − γ).

For k > p + 1, owing that yˆp+s = yp+s + as and as+1 = γ(as + α(1 − α)s−1 ), we get

m X  s=2

m m X  X   v(γ −1 yp+s+1 ) − v(yp+s ) − v(γ −1 yˆp+s+1 ) + v(ˆ yp+s ) = −as+1 γ −1 + as = −α (1−α)s−1 . s=2

Since this last term is a geometric series, we obtain: −α ·

(1 − α)m − (1 − α) = (1 − α)m − (1 − α). 1−α−1 17

s=2

Therefore, collecting terms, the summation is equal to S = γ −1 yp+1 − λγ −1 yp+1 + λ − (α − γ) − yp+1 + λyp+1 − λγ + (1 − α)m − (1 − α) = yp+1 (γ −1 − λγ −1 − 1 + λ) + λ + γ − λγ + (1 − α)m − 1 = yp+1 (γ −1 − 1)(1 − λ) + λ + γ(1 − λ) + (1 − α)m − 1 = yp+1 (γ −1 − 1)(1 − λ) + (γ − 1)(1 − λ) + (1 − α)m . Since 1 − λ ≤ 0, γ −1 − 1 ≥ 0 and yp+1 ≤ γ, we have the following inequalities: S ≥ γ(γ −1 − 1)(1 − λ) + (γ − 1)(1 − λ) + (1 − α)m = (1 − γ)(1 − λ) + (γ − 1)(1 − λ) + (1 − α)m = (1 − α)m > 0.

Sketch of Proof of Theorem 3.

The proof follows the same argument as the one

in Theorem 1 with an additional fact. By hypothesis, xˆkt11 = 1. Without loss of generality suppose k1 = 1 so xˆkt1 = 0 for all k > 1. Adapting Lemma 1 for this case, with α > γ, we know that yˆt11 +1 > 0, yˆtk1 +1 ≤ 0 and noticing that: since t1 + 1 > t1 > t0 then yˆtk10+1 < 0. Now,

modifying Lemma 3 for the case of strict loss aversion, we obtain: x = yˆt1 > 0, such that: x+y = Therefore,

y=

Pk=K k=1

Pk=K k=1

yˆtk < 0,

yˆtk = yt1 > 0.

w(x + y) > w(x) + w(y). And the strict inequality follows.

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