Hyperbolic discount curves: a reply to Ainslie

Abstract

Ainslie (Theory and Decision, 73, 3–34, 2012) challenges our interpretation of the properties of hyperbolic discount curves in an iterated prisoners’ dilemma (IPD) model. In this reply, we discuss the emergence of hyperbolic discount functions in the behavioral economics literature and evaluate their properties. Furthermore, we present a summarized version of our IPD model and evaluate Ainslie’s points of contention.

Appendices

Appendix 1: finitely iterated prisoners’ dilemma

Consider the IPD model described in Sect. 4 and suppose that the game is repeated $T$ times in periods $1, 2,\ldots , T$ where $T\in \mathbb Z$ s.t. $1 \le T \ge \infty$. If this is common knowledge, we prove that the game has a unique SPE in which each firm plays Defect at all periods. We split the proof into 2 parts.

•   [Part 1: First, we determine the Nash equilibrium of the stage game using the best response function of firm $i\in N$:

• Definition $D-2$: In a 2 player game, the best response function of player $i$ is the function $R^{i}(s_{j})$ that for every given strategy $s_{j}$ of player $j$ assigns a strategy $s_{i}= R^{i}(s_{j})$ that maximizes player $i$’s payoff $\pi ^{i}(s_{i},s_{j})$ (Shy 1995, p. 21).

• From the model description in Sect. 4, the best response function of firm $i\in N$ is:

  $$\begin{aligned} R^{i}(s_{j}) ={\left\{ \begin{array}{ll} Defect \;\;if\;s_{j} \;=\;Cooperate &{}\\ Defect \;\; if\;s_{j} \;=\; Defect &{} \end{array}\right. } \end{aligned}$$

  (34)

• Definition $D-3$: An outcome $\hat{s} = (\hat{s}^{1}, \hat{s}^{2},\ldots , \hat{s}^{N})$ (where $\hat{s}^{i}\in S_{i}$ for every $i= 1,2,\ldots ,N$) is said to be a Nash equilibrium (NE) if for every player $i$, $\pi ^{i}(\hat{s}^{i}, \hat{s}^{\lnot i})\ge \pi ^{i}(s^{i}, \hat{s}^{\lnot i})$ for every $s^{i}\in S_{i}$ (Shy 1995, p.18).

  • For the outcome $s^{1}= (s_{i}^{1}, s_{i}^{1})$; $\pi ^{1}(s_{i}^{1}, s_{i}^{1})= C < \pi ^{1}(s_{i}^{2}, s_{i}^{1})= A$, contradicting $D-3\; \Rightarrow \; s^{1}$ is not NE.

  • For the outcome $s^{2}= (s_{i}^{1}, s_{i}^{2})$; $\pi ^{1}(s_{i}^{1}, s_{i}^{2})= Z < \pi ^{1}(s_{i}^{2}, s_{i}^{2})= D$, contradicting $D-3\; \Rightarrow \; s^{2}$ is not NE.

  • For the outcome $s^{3}= (s_{i}^{2}, s_{i}^{1})$; $\pi ^{2}(s_{i}^{2}, s_{i}^{1})= Z < \pi ^{2}(s_{i}^{2}, s_{i}^{2})= D$, contradicting $D-3\; \Rightarrow \; s^{3}$ is not NE.

  • For the outcome $s^{4}= (s_{i}^{2}, s_{i}^{2})$; $\pi ^{1}(s_{i}^{2}, s_{i}^{2})= D > \pi ^{1}(s_{i}^{1}, s_{i}^{2})= Z$ & $\pi ^{2}(s_{i}^{2}, s_{i}^{2})= D > \pi ^{2}(s_{i}^{2}, s_{i}^{1})= Z$ $\Rightarrow \; s^{4}$ is NE.

• The outcome $s^{4}= (Defect, Defect )$ thus constitutes an equilibrium in dominant strategies (EDS) and a unique NE for the stage IPD game.

•   [Part 2: Having established that (Defect, Defect) is an NE of the stage game, we suppose that both firms have played the IPD game in $T-1$ periods and they are ready to play for one last time in period $T$. At this point, the game is identical to the stage game and firm $i\in N$ plays its dominant strategy Defect (refer to the best response function of firm $i$ in Eq. 34). Therefore, the outcome of the game is the NE of the stage game (Defect, Defect). Now consider the game in period $T-1$. Both firms know that following this period, they will have one game to play and the outcome of the game involves both playing Defect. Again, at this period, both firms will play their dominant strategy resulting in the outcome (Defect, Defect). Using backward induction, we note that at each period $T-2, T-3, \ldots ,1$, the outcome where both firms play Defect will result hence SPE. $\square$

Appendix 2: cooperation under exponential discounting

Consider Case 1 and Case 2 defined in Sect. 4. We prove that the outcome (Cooperate, Cooperate) is SPE under exponential discounting if $\delta \ge \frac{A-C}{A-D}$.

• The sum of discounted payoffs under Case 1 is given by:

  $$\begin{aligned} C + \delta C + \delta ^{2} C + \ldots \end{aligned}$$

  (35)

  To find the sum of the series in Eq. 35, we exploit a property of the exponential discount function. Claim: The following sum, $1 + \delta + \delta ^{2} + \ldots$, converges to $\frac{1}{1-\delta }$ if $\delta < 1$.

Proof

Define the partial sums of the series as follows: $s_{1} = 1$, $s_{2} = 1+ \delta$, $s_{3} =1 + \delta + \delta ^{2}$,..., $s_{n} =1 + \delta + \cdots + \delta ^{n-1}$ where $s_{i}$ represents the $i$th partial sum $(i=1,2,\ldots n)$. Multiply $s_{n}$ by $\delta$ and obtain $\delta s_{n} = \delta + \delta ^{2}+ \cdots + \delta ^{n}$. Subtract $\delta s_{n}$ from $s_{n}$ and obtain: $s_{n} - \delta s_{n} = 1 - \delta ^{n}$. Solve for $s_{n}$: $s_{n} = \frac{1 - \delta ^{n}}{1 - \delta },\;\;(\delta \ne 1)$. Finally taking the value for $s_{n}$, note that if $\mid \delta \mid \!<\! 1$ then $\delta ^{n}\!\rightarrow \! 0$ as $n \!\rightarrow \! \infty$ and $s_{n}\!\rightarrow \! \frac{1}{1 -\delta }$. $\square$

• From this property, we establish that the sum in Eq. 35 is $C\left( \frac{1}{1-\delta }\right)$

• The sum of discounted payoffs under Case 2 is given by:

  $$\begin{aligned} A + \delta D + \delta ^{2} D + \cdots = A + D\left( \frac{\delta }{1-\delta }\right) \end{aligned}$$

  (36)

• For the outcome (Cooperate, Cooperate) to be an SPE, we require that:

  $$\begin{aligned} C\left( \frac{1}{1-\delta }\right) \ge A + D\left( \frac{\delta }{1-\delta }\right) \end{aligned}$$

  (37)
  $$\begin{aligned} \Leftrightarrow \frac{C}{1-\delta }\ge A+ \frac{\delta D}{1-\delta }\Leftrightarrow \frac{C-\delta D}{1-\delta }\ge A \Leftrightarrow C- \delta D \ge A - \delta A \end{aligned}$$
  $$\begin{aligned} \Leftrightarrow \delta (A-D)\ge A-C \Leftrightarrow \delta \ge \frac{A-C}{A-D}\square \end{aligned}$$

Appendix 3: cooperation under quasi-hyperbolic discounting

Consider Case 1 and Case 2 defined in Sect. 4. We prove that the outcome (Cooperate, Cooperate) is SPE under quasi-hyperbolic discounting if $\delta \ge \frac{A-C}{\beta (C-D)+A-C}$.

• The sum of discounted payoffs under Case 1 is given by:

  $$\begin{aligned} C +\beta \delta C + \beta \delta ^{2} C + \cdots \end{aligned}$$

  (38)

  Similarly, we exploit the convergence property of exponential discounting to determine this sum. The sum in Eq. 38 is thus:

  $$\begin{aligned} \beta C\left( \frac{\delta }{1-\delta }\right) + C \end{aligned}$$

• The sum of discounted payoffs under Case 2 is given by:

  $$\begin{aligned} A +\beta \delta D + \beta \delta ^{2} D + \cdots \end{aligned}$$

  (39)

  The sum in Eq. 39 is:

  $$\begin{aligned} \beta D\left( \frac{\delta }{1-\delta }\right) + A \end{aligned}$$

• For the outcome (Cooperate, Cooperate) to be an SPE, we require that:

  $$\begin{aligned} \beta C\left( \frac{\delta }{1-\delta }\right) + C \ge \beta D\left( \frac{\delta }{1-\delta }\right) + A \end{aligned}$$

  (40)
  $$\begin{aligned} \Leftrightarrow \frac{\delta (\beta C - \beta D)}{1-\delta }\ge A-C \Leftrightarrow \delta (\beta C - \beta D)\ge (A-C) (1-\delta ) \end{aligned}$$
  $$\begin{aligned} \Leftrightarrow \delta (\beta C- \beta D + A - C) \ge A- C \Leftrightarrow \delta \ge \frac{A-C}{\beta (C-D)+ A-C}\square \end{aligned}$$

Appendix 4: analysis of the break-even quasi-hyperbolic discount factor

We show that the following relation in Eq. 20 holds:

$$\begin{aligned} \frac{d}{d\beta }\; \delta ^{*}(\beta )<0 \end{aligned}$$

Define $(A-C)$ as $\alpha$ and $(C-D)$ as $\gamma$ in Eq. 19 and re-write $\delta ^{*}(\beta )$ as follows:

$$\begin{aligned} \delta ^{*}(\beta ) = \frac{A-C}{\beta (C-D) +A-C}= \frac{\alpha }{\beta \gamma + \alpha } \end{aligned}$$

Differentiating $\delta ^{*}(\beta )$ with respect to $\beta$, we obtain:

$$\begin{aligned} \frac{d}{d\beta }\left( \frac{\alpha }{\beta \gamma + \alpha }\right) = - \frac{\alpha \gamma }{(\beta \gamma + \alpha )^{2}} \end{aligned}$$

From the model description in Sect. 4, we have that $A>\;C>\;D$ implying $\alpha >0$ and $\gamma >0$:

$$\begin{aligned} \alpha = \underbrace{(A - C)}_{ +}\;\;\;\gamma = \underbrace{(C - D)}_{ +} \;\;\;\Rightarrow (A-C)(C-D)\;>\;0 \end{aligned}$$

Therefore, we establish the result in Eq. 20:

$$\begin{aligned} - \frac{\alpha \gamma }{(\beta \gamma + \alpha )^{2}}\;<\;0\;\Leftrightarrow - \frac{(A-C)(C-D)}{(\beta (C-D) + (A-C))^{2}}\;<\;0 \end{aligned}$$

Similarly, we show that the following relation holds for the second order derivative:

$$\begin{aligned} \frac{d^{2}}{d\beta ^{2}}\;\delta ^{*}(\beta )> 0 \end{aligned}$$

$$\begin{aligned} \frac{d^{2}}{d\beta ^{2}}\left( \frac{A - C}{\beta (C - D) + A - C}\right) = \frac{d}{d\beta }\;\left( \frac{-\alpha \gamma }{(\beta \gamma + \alpha )^{2}}\right) =-\alpha \gamma \;\frac{d}{d\beta }\left( \frac{1}{(\beta \gamma + \alpha )^{2}}\right) \end{aligned}$$

$$\begin{aligned} = -\alpha \gamma \;\cdot \;\frac{-\;2\gamma (\beta \gamma + \alpha )}{(\beta \gamma + \alpha )^{4}}= 2\alpha \gamma ^{2}\;\frac{1}{(\beta \gamma + \alpha )^{3}}\;>\;0 \end{aligned}$$

$$\begin{aligned} \Leftrightarrow \;\frac{2(A-C)(C-D)^{2}}{(\beta (C-D) + (A-C))^{3}}\;>\;0 \end{aligned}$$

$\square$