Do It Today or Do It Tomorrow: Empirical Non-exponential Discounting Explained by Symmetry Ideas

At first glance, it seems to make sense to conclude that when a 1 dollar reward tomorrow is equivalent to a D < 1 dollar reward today, the day-after-tomorrow’s 1 dollar reward would be equivalent to D ·D = D dollars today, and, in general, a reward after time t is equivalent to D(t) = D dollars today. This exponential discounting function D(t) was indeed proposed by the economists, but it does not reflect the actual human behavior. Indeed, according to this formula, the effect of distant future events is negligible, and thus, it would be reasonable for a person to take on huge loans or get engaged in unhealthy behavior even when the long-term consequences will be disastrous. In real life, few people behave like that, since the actual empirical discounting function is different: it is hyperbolic D(t) = 1/(1 + k · t). In this paper, we use symmetry ideas to explain this empirical phenomenon. 1 Discounting: Theoretical Foundations, Empirical Data, and Related Challenge What is discounting. Future awards are less valuable than the same size awards given now. This phenomenon is known as discounting; see, e.g., [1, 3, 5–9, 11] for details. Procrastination is an inevitable consequence of discounting. Suppose that we have a task which is due by a certain deadline. This can be a task of submitting a grant proposal, or of submitting a paper to a conference. In this case, the reward is the same no matter when we finish this task – as long as we finish it before the deadline. Similarly, the overall negative effect caused by the need to do some boring stuff is the same no matter when we do it. But, due to discounting, if we perform this task later, today’s negative effect is smaller than if we perform this task today. The further in the future is this negative effect, the smaller is its influence on our today’s happiness. Thus, a natural way to maximize today’s happiness is to postpone this task as much as possible – which is exactly what people do; see, e.g., [2, 7]. 2 F. Zapata, O. Kosheleva, V. Kreinovich, and T. Dumrongpokaphan A simple theoretical model of discounting. How can we describe discounting in numerical terms? At first glance, providing numerical description for discounting is a straightforward idea. Indeed, let us assume that 1 dollar tomorrow is equivalent to D < 1 dollars today. This is true for every day: 1 dollar at the day t + 1 is equivalent to D dollars in day t. This means, in particular, that 1 dollar in day t0+2 is equivalent to D dollars at time t0+1. Since 1 dollar on day t0+1 is equivalent to D dollars at the initial moment of time t0, D dollars on day t0 + 1 are equivalent to D · D dollars on day t0. Thus, we can conclude that 1 dollar at day t0 + 2 is equivalent to D 2 dollars at moment t0. Similarly, 1 dollar at moment t0 + 3 is equivalent to D dollars at moment t0 + 2 and thus, to D ·D = D dollars at moment t0. In general, by induction over t, we can show that 1 dollar at moment t0 + t is equivalent to D(t) def = D dollars at the current moment t0. We can rewrite the above expression D(t) = D as D(t) = exp(−a · t), (1) where a def = − ln(D). Because of this form, this discounting is known as exponential. Practical problem with exponential discounting. At first glance, exponential discounting is a very reasonable idea. However, it has a problem: exponential functions decrease very fast, and for large t, the value exp(−a · t) becomes indistinguishable from 0. In practical terms, this means that a person looks for an immediate reward even if there is a significant negative downside in the distant future. Such behavior indeed happens: a young man takes many loans without taking into account that in the future, he will have to pay; a young person ruins his health by using drugs, not taking into account that in the future, this may lead to a premature death. A person commits a crime without taking into consideration that eventually, he will be caught and punished. Such behavior does happen, but such behavior is abnormal. Most people do not take an unrealistic amount of loans, most people do not ruin their health during their youth, most people do not commit crimes. This means that for most people, discounting decreases much slower than the exponential function. So how to we describe discounting: empirical data. Empirical data shows that discounting indeed decrease much slower than predicted by the exponential function: namely, 1 dollar at moment t0 + t is equivalent to D(t) = 1 1 + k · t (2) dollars at moment t0. This formula is known as hyperbolic discounting; see, e.g., [1, 3, 5–9, 11]. Empirical Non-Exponential Discounting Explained 3 Problem: how can we explain the empirical data. In principle, there exist many functions that decrease slower than the exponential function exp(−a · t). So why, out of all these functions, we observe the hyperbolic one? What we do in this paper. In this paper, we use symmetries to provide a theoretical explanation for the empirical discounting formula. To be more precise, our theoretical explanation leads to a family of functions of which hyperbolic discounting is one of the possibilities. 2 Analysis of the Problem The idea of a re-scaling. Let D(t) denote the discounting of a reward which is t moments into the future, i.e., the amount of money such that getting D(t) dollars now is equivalent to getting 1 dollar after time t. By definition, D(0) means getting 1 dollar with no delay, so D(0) = 1. It is also reasonable to require that as the time period time t increases, the value of the reward goes to 0, so that lim t→+∞ D(t) = 0. It is also reasonable to require that a small change in t should lead to small changes in D(t), i.e., that the function D(t) be differentiable (smooth). The further into the future we get the reward, the less valuable this reward is now, so the function D(t) is increasing as the time t increases. Thus, if we further delay all the rewards by some time s, then each value D(t) will be replaced by a smaller value D(t+s). We can describe this replacement as D(t+s) = Fs(D(t)), where the function Fs(x) re-scales the original discount value D(t) into the new discount value D(t+ s). For the exponential discounting (1), the re-scaling Fs(x) is linear: D(t+s) = C ·D(t), where C def = exp(−a · s), so we have Fs(x) = C · x. For the hyperbolic discounting (2), the corresponding re-scaling Fs(x) is not linear. Which re-scaling should we select? Which re-scalings are reasonable: formulating this question in precise mathematical terms. We want to select some reasonable re-scalings. What does “reasonable” mean? Of course, linear re-scalings should be reasonable. Also, intuitively, if a re-scaling is reasonable, then its inverse should also be reasonable. Similarly, if two re-scalings are reasonable, then applying them one after another should also lead to a reasonable re-scaling. In other words, a composition of two re-scalings should also be reasonable. In mathematical terms, we can conclude that the class of all reasonable re-scalings should be closed under inverse transformation and composition of two mappings. This means that with respect to the composition operation, such re-scalings must form a group. We want to be able to determine the transformation from this group based on finitely many experiments. In each experiment, we gain a finite number of values, so after a finite number of experiments, we can only determine a finite number of parameters. Thus, we should be able to select an elements of the desired transformation group based on the values of finitely many parameters. 4 F. Zapata, O. Kosheleva, V. Kreinovich, and T. Dumrongpokaphan In mathematical terms, this means that the corresponding transformation group should be finite-dimensional. Summarizing: we want all the transformations Fs(x) to belong to a finitedimensional transformation group of functions of one variable that contains all linear transformations. Which re-scalings are reasonable: answer to the question. It is known (see, e.g., [4, 10, 12]) that the only finite-dimensional transformation groups of functions of one variable that contain all linear transformations are the group of all linear transformations and the group of all fractional-linear transformations a+ b · x 1 + c · x . Thus, our informal requirement that each re-scaling is reasonable implies that each re-scaling should be fractionally linear: Fs(x) = a(s) + b(s) · x 1 + c(s) · x . So, we arrive at the following requirement. 3 Definition and the Main Result Definition 1.We say that a smooth decreasing function D(t) for which D(0) = 1 and lim t→∞ D(t) = 0 is a reasonable discounting function if for every s, there exist values a(s), b(s), and c(s) for which D(t+ s) = a(s) + b(s) ·D(t) 1 + c(s) ·D(t) . (3) Proposition 1. A function D(t) is a reasonable discounting function if and only if it has one of the following forms: D(t) = exp(−a · t), D(t) = 1 1 + k · t , D(t) = 1 + a 1 + a · exp(k · t) , or D(t) = a (a+ 1) · exp(k · t)− 1 , for some a > 0 and k > 0. Comment. The first discounting function corresponds to exponential discounting, the second to the hyperbolic discounting, the other two functions correspond to the more general case. Both exponential and hyperbolic discounting can be viewed as the limit case of the more general formulas. Indeed, in the limit a → ∞, both general expressions tend to the formula D(t) = exp(−k · t) corresponding to the exponential discounting. Empirical Non-Exponential Discounting Explained 5 On the other hand, if we tend k to 0, we get exp(k · t) ≈ 1 + k · t, so for a(k) = α · k, the second general formula takes the form D(t) = α · k (1 + α · k) · (1 + k · t)− 1 . The denominator of this expression has the form 1 + α · k + k · t+ α · k · t− 1 = α · k + (k + α · k) · t, so D(t) = α · k α · k + (k + α · k2) · t . Dividing both numerator and denominator of this formula by α · k, we get the hyperbolic discounting D(t) = 1 1 + k′ · t , with k′ = k+ 1 α , i.e., in the limit k → 0, with k′ = 1 α . Proof. 1◦. Let us first show that each of the four functions D(t) listed in the formulation of the Proposition is a reasonable discounting function in the sense of Definition 1. It is easy to see that all four functions are smooth and decreasing, and that for all of them, we have D(0) = 1 and lim t→∞ D(t) = 0. Let us show, one by one, that each of these four functions satisfies the property (3) for appropriate auxiliary functions a(s), b(s), c(s), and d(s). 1.1◦. For D(t) = exp(−a · t), we have exp(a−a ·(t+s)) = exp(−a ·s) ·exp(−a · t), i.e., D(t+s) = exp(−a ·s) ·D(t). Thus, the condition (3) is satisfied for a(s) = 0, b(s) = exp(−a · s), and c(s) = 0. 1.2◦. For the function D(t) = 1 1 + k · t , we have 1 + k · t = 1 D(t) hence 1 + k · (t+ s) = (1 + k · t) + k · s = 1 D(t) + k · s and thus, D(t+ s) = 1 1 + k · (t+ s) = 1 1 D(t) + k · s . Multiplying both numerator and denominator of the right-hand side by D(t), we conclude that D(t+ s) = D(t) 1 + k · s ·D(t) . Thus, the condition (3) is satisfied for a(s) = 0, b(s) = 1, and c(s) = k · s. 6 F. Zapata, O. Kosheleva, V. Kreinovich, and T. Dumrongpokaphan 1.3◦. For the function D(t) = 1 + a 1 + a · exp(k · t) , we have 1+a ·exp(k · t) = 1 + a D(t) . Thus, a · exp(k · t) = 1 + a D(t) − 1 = 1 + a−D(t) D(t) , hence a · exp(k · (t+ s)) = exp(k · s) · (a · exp(k · t)) = exp(k · s) · 1 + a−D(t) D(t) = exp(k · s) · (1 + a)− exp(k · s) ·D(t) D(t) .