Hyperbolic Discounting of the Far-Distant Future
HHyperbolic Discounting of the Far-Distant Future
Nina Anchugina , Matthew Ryan , and Arkadii Slinko Department of Mathematics, University of Auckland School of Economics, Auckland University of Technology [email protected], [email protected], [email protected]
February 2017
Abstract.
We prove an analogue of Weitzman’s [7] famous result that an exponentialdiscounter who is uncertain of the appropriate exponential discount rate should discountthe far-distant future using the lowest (i.e., most patient) of the possible discount rates.Our analogous result applies to a hyperbolic discounter who is uncertain about the appro-priate hyperbolic discount rate. In this case, the far-distant future should be discountedusing the probability-weighted harmonic mean of the possible hyperbolic discount rates.
Keywords:
Hyperbolic discounting, Uncertainty.
JEL Classification:
D71, D90. a r X i v : . [ q -f i n . E C ] F e b Introduction
Consider an individiual – or Social Planner – who ranks streams of outcomes over acontinuous and unbounded time horizon T = [0 , ∞ ) using a discounted utility criterionwith discount function D : T → (0 , D is differentiable,strictly decreasing and satisfies D (0) = 1. For example, D might have the exponential form D ( t ) = e − rt for some constant discount (or time preference) rate , r >
0. Such discounting may be mo-tivated by suitable preference axioms ([4]) or as a survival function with constant hazardrate, r ([6]). For an arbitrary (i.e., not necessarily exponential) discount function, Weitz-man ([7]) defines the local (or instantaneous) discount rate , r ( t ), using the relationship: D ( t ) = exp (cid:18) − (cid:90) t r ( τ ) dτ (cid:19) ⇔ r ( t ) = − D (cid:48) ( t ) D ( t ) (1)Note that r ( t ) is constant if (and only if) D has the exponential form.Weitzman ([7]) considers a scenario in which the decision-maker is uncertain about theappropriate discount function to use. She may, for example, be uncertain about the true(constant) hazard rate of her survival function, as in [6]. The decision-maker entertains n possible scenarios corresponding to n possible discount functions D i , i = 1 , , ..., n , withassociated local discount rate functions r i . Suppose that scenario i has probability p i > (cid:80) ni =1 p i = 1, and that the decision-maker discounts according to the expected (orcertainty equivalent) discount function D = n (cid:88) i =1 p i D i (2)(Such a discount function may also arise if the decision-maker is a utilitarian SocialPlanner for a population of n heterogeneous individuals, as in [5].)Let r be the local discount rate function associated with certainty equivalent discountfunction (2). Weitzman [7] studies the limit behaviour of r ( t ) as t → ∞ . He proves that ifeach r i ( t ) converges to a limit, then r ( t ) converges to the lowest of these limits. In otherwords, if lim t →∞ r i ( t ) = r ∗ i for each i , then lim t →∞ r ( t ) = min { r ∗ , . . . , r ∗ n } . (3)Moreover, if each r i is constant (i.e., each D i is exponential), then r ( t ) declines mono-tonically to this limit ([7]). Example 1.
Suppose each D i is exponential, so that r i ( t ) = r i is constant. Then theresults in [7] imply that r ( t ) declines monotonically with limit min i r i . Figure 1 illustratesfor the case n = 3 , r = 0 . , r = 0 . , r = 0 . and p = p = p = 1 / . In fact, this is true more generally – see [7, footnote 6]. D D D(t) r r r r(t) Time D i s c o un t R a t e Time E x p o n e n t i a l D i s c o un t F un c t i o n . . . . . . . . . Figure 1: Exponential Discount FunctionsWeitzman’s result may be interpreted as saying that the mixed discount function (2)behaves locally as an exponential discount function with discount rate (3) when discount-ing outcomes in the far distant future. This result is most salient if the the individual D i functions are themselves exponential, as in Example 1. However, many individuals do not discount exponentially ([2]). If the D i functions all fall within some non-exponentialclass, it is natural to characterise the local asymptotic behaviour of (2) using the sameclass of functions. The next section considers the hyperbolic class. In this section we assume that each D i has the (proportional) hyperbolic form D i ( t ) = 11 + h i t where h i > hyperbolic discount rate . We further assume that h > h > . . . > h n .In particular, D exhibits the most “patience” and D n the least – see [1] and Example 2.Note that r i ( t ) = − D (cid:48) i ( t ) D i ( t ) = h i h i t and hence r ∗ i = 0 for each i . In other words, the limiting local (exponential) discount rateis the same for each discount function, reflecting the fact that hyperbolic functions decline2ore slowly than exponentials for large t . Weitzman’s result is not very informative forthis scenario.Instead, we should like to have a local hyperbolic approximation to the mixed discountfunction (2), which may not itself have the proportional hyperbolic form. We thereforefollow Weitzman’s example and define the local (or instantaneous) hyperbolic discountrate , h ( t ), as follows: D ( t ) = 11 + h ( t ) t ⇔ h ( t ) = (cid:18) D ( t ) − (cid:19) t (4)Note that h ( t ) is constant if (and only if) D has the proportional hyperbolic form. How does h ( t ) behave as t → ∞ ? The following two results, which are proved in the Appendix, answer this question. Inorder to state the second result, we remind the reader that the weighted harmonic mean of non-negative values x , x , . . . , x n with non-negative weights a , a , . . . , a n satisfying a + . . . + a n = 1 is H ( x , a ; . . . ; x n , a n ) = (cid:32) n (cid:88) i =1 a i x i (cid:33) − . It is well-known that the weighted harmonic mean is smaller than the correspondingweighted arithmetic mean (i.e., expected value).
Theorem 1.
The local hyperbolic discount rate, h ( t ) , is strictly decreasing. Theorem 2.
The local hyperbolic discount rate of the certainty equivalent discount func-tion converges to the probability-weighted harmonic mean of the individual hyperbolic dis-count rates. That is h ( t ) → H ( h , p ; . . . ; h n , p n ) as t → ∞ . The following example illustrates both results.
Example 2.
Suppose n = 3 , h = 0 . , h = 0 . , h = 0 . and p = p = p = . Notethat h = 0 . corresponds to the arithmetic mean of h , h and h . Figure 2 displays themonotonic decline of h ( t ) towards the weighted harmonic mean H ( h , p ; h , p ; h , p ) ≈ . . With exponential discounting, uncertainty about the (exponential) discount rate im-plies that the far-distant future is discounted according to the most “patient” of thepossible discount functions. If discounting is hyperbolic, with uncertainty about the (hy-perbolic) discount rate, all possible discount functions matter for the discounting of thefar-distant future. The asymptotic local hyperbolic discount rate is, however, below theaverage (i.e., arithmetic mean) of the possible rates. See, in particular, the important reformulation of Weitzman’s result by Gollier and Weitzman ([3]),which resolves the so-called “Weitzman-Gollier puzzle”. D D D(t) h h h h(t)H(h , p ; h , p ; h , p ) Time H y p e r b o li c D i s c o un t R a t e Time H y p e r b o li c D i s c o un t F un c t i o n . . . . . . . . . Figure 2: Hyperbolic Discount Functions
Acknowledgments
Nina Anchugina gratefully acknowledges financial support from the University of Auck-land. Arkadii Slinko was supported by the Royal Society of New Zealand Marsden Fund3706352.
A Appendix
A.1 Proof of Theorem 1
We prove this statement by induction on n . First we need to prove that the statementholds for n = 2. In this case: h ( t ) = (cid:20) p (1 + h t ) − + p (1 + h t ) − − (cid:21) t for each t >
0. Rearranging: h ( t ) = (cid:20) (1 + h t )(1 + h t ) p (1 + h t ) + p (1 + h t ) − (cid:21) t = (cid:20) h + h ) t + h h t p + p + ( p h + p h ) t − (cid:21) t . Since p + p = 1 we obtain: h ( t ) = (cid:20) h + h ) t + h h t p h + p h ) t − (cid:21) t = p h + p h + h h t p h + p h ) t .
4y differentiating h ( t ): h (cid:48) ( t ) = h h (1 + ( p h + p h ) t ) − ( p h + p h + h h t ) ( p h + p h )[1 + ( p h + p h ) t ] (5)We need to show that h (cid:48) ( t ) <
0. Since the denominator of (5) is positive, the sign of h (cid:48) ( t )depends on the sign of the numerator. Therefore, we denote the numerator of (5) by Q and analyse it separately: Q ( t ) = h h [1 + ( p h + p h ) t ] − ( p h + p h + h h t ) ( p h + p h )= h h − ( p h + p h ) ( p h + p h ) . By expanding the brackets and using the fact that p + p = 1 implies 1 − p − p = 2 p p expression Q can be simplified further: Q ( t ) = h h − p h h − p p h − p p h − p h h = h h (1 − p − p ) − p p ( h + h )= − p p ( h − h ) . Therefore, since h (cid:54) = h we have Q <
0. Hence it follows that h (cid:48) ( t ) < h ( t ) isstrictly decreasing.Suppose that the proposition holds for n = k . We need to show that it also holds for n = k + 1. When n = k + 1 we have: D = k +1 (cid:88) i =1 p i D i = (1 − p k +1 ) (cid:32) k (cid:88) i =1 p i − p k +1 D i (cid:33) + p k +1 D k +1 . Since k (cid:88) i =1 p i − p k +1 = 1 , we may write D = (1 − p k +1 ) D ( k ) + p k +1 D k +1 . where D ( k ) = k (cid:88) i =1 p i − p k +1 D i . By the induction hypothesis it follows that D ( k ) = 11 + h ( k ) ( t ) t , where h ( k ) is strictly decreasing. Therefore, h ( t ) = (cid:20) − p k +1 ) D ( k ) + p k +1 D k +1 − (cid:21) t = (cid:34) − p k +1 ) (1 + h ( k ) ( t ) t ) − + p k +1 (1 + h k +1 t ) − − (cid:35) t . p = 1 − p k +1 , ˆ p = p k +1 , ˆ h ( t ) = h ( k ) ( t ) and ˆ h = h k +1 . Then we have h ( t ) = (cid:34) p (1 + ˆ h ( t ) t ) − + ˆ p (1 + ˆ h t ) − − (cid:35) t . Analogously to the case n = 2, this expression can be rearranged to give: h ( t ) = ˆ p ˆ h ( t ) + ˆ p ˆ h + ˆ h ( t ) ˆ h t p ˆ h t + ˆ p ˆ h ( t ) t . However, in contrast to the case n = 2, ˆ h is now a function of t . Thus: h (cid:48) ( t ) = N ( t ) (cid:104) p ˆ h t + ˆ p ˆ h ( t ) t (cid:105) . (6)where N ( t ) = (cid:16) ˆ p ˆ h (cid:48) ( t ) + ˆ h ( t ) ˆ h + ˆ h (cid:48) ( t ) ˆ h t (cid:17) (cid:16) p ˆ h t + ˆ p ˆ h ( t ) t (cid:17) − (cid:16) ˆ p ˆ h ( t ) + ˆ p ˆ h + ˆ h ( t ) ˆ h t (cid:17) (cid:16) ˆ p ˆ h + ˆ p ˆ h ( t ) + ˆ p ˆ h (cid:48) ( t ) t (cid:17) . The denominator of (6) is strictly positive, so the sign of the derivative is the same asthat of N ( t ). Note that N ( t ) = ˆ Q ( t )+ˆ h (cid:48) ( t ) (cid:104)(cid:16) ˆ p + ˆ h t (cid:17) (cid:16) p ˆ h t + ˆ p ˆ h ( t ) t (cid:17) − ˆ p t (cid:16) ˆ p ˆ h ( t ) + ˆ p ˆ h + ˆ h ( t )ˆ h t (cid:17)(cid:105) where ˆ Q ( t ) is defined as above, but with h = ˆ h ( t ) and h = ˆ h . Since ˆ Q ( t ) ≤ h ( t ) = h ) and ˆ h (cid:48) <
0, it suffices to show (cid:16) ˆ p + ˆ h t (cid:17) (cid:16) p ˆ h t + ˆ p ˆ h ( t ) t (cid:17) − ˆ p t (cid:16) ˆ p ˆ h ( t ) + ˆ p ˆ h + ˆ h ( t )ˆ h t (cid:17) > p (cid:16) p ˆ h t (cid:17) + ˆ h t (cid:16) p ˆ h t (cid:17) − (ˆ p ) ˆ h t. We now use the fact that (ˆ p ) = (1 − ˆ p ) = 1 − p + (ˆ p ) to getˆ p (cid:16) p ˆ h t (cid:17) + ˆ h t (cid:16) p ˆ h t (cid:17) − (cid:2) − p + (ˆ p ) (cid:3) ˆ h t = ˆ p + (cid:16) t ˆ h (cid:17) ˆ p + 2ˆ p ˆ h t, which is strictly positive. This establishes the required inequality (7) and completes theproof. 6 .2 Proof of Theorem 2 We note that p i h i t = p i h i t + (cid:15) i ( t ) , where (cid:15) i ( t ) /t → t → ∞ . Let (cid:15) ( t ) = (cid:15) ( t ) + . . . + (cid:15) n ( t ). Hence it follows that:11 + h ( t ) t = n (cid:88) i =1 p i D i ( t ) = p h t + . . . + p n h n t = p h t + . . . + p n h n t + (cid:15) ( t )= (cid:18) p h + . . . + p n h n (cid:19) t + (cid:15) ( t )= 1 H ( h , p ; . . . ; h n , p n ) t + (cid:15) ( t )= 11 + H ( h , p ; . . . ; h n , p n ) t + ˆ (cid:15) ( t ) , where ˆ (cid:15) ( t ) /t → t → ∞ . This implies that h ( t ) → H ( h , p ; . . . ; h n , p n ) as t → ∞ . References [1] N. Anchugina, M.J. Ryan, and A. Slinko. Aggregating time preferences with decreasingimpatience. arXiv preprint arXiv:1604.01819 , April 2016.[2] S. Frederick, G. Loewenstein, and T. O’Donoghue. Time discounting and time pref-erence: A critical review.
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