Temporal dominance and relative patience in intertemporal choice

Abstract

A stream of intertemporal payoffs X first-order (second-order) temporarily dominates another stream Y if (the area below) the cumulative payoff function of X up till any moment of time is always at least as high as that of Y; or, equivalently, any rank-dependent discounted utility maximizer with a strictly increasing (concave) utility function and a strictly decreasing (convex) discount function prefers X over Y. An individual A is relatively more patient than an individual B if, whenever B is willing to undertake some investment project, A is moreover so; or, equivalently, for any stream of payoffs, A’s present-day equivalent is at least as high as that of B; or, equivalently, A’s discount factor is at least as high as that of B and B’s utility function is “more concave” than that of A. Related concepts of absolute patience and comparative probabilistic patience are briefly discussed.

Appendix

Proof of proposition 1

(1) Consider the case when x first-order temporarily dominates y. Then x can be obtained from y through a series of increased and/or advanced payments as follows. If $y_{T}\le x_{T}$, then we add $x_{T}-y_{T} \ge 0$ to the payoff of y at the last moment T. If $y_{T}>x_{T}$, then we advance the difference $y_{T}-x_{T}$ from the payoff of y at the last moment T to the payoff of y at the before-last moment $T-1$. As a result, we obtain a new stream ${{\varvec{y}}}^{*}$ with payoff $x_{T}$ at the last moment T.

Next, we transform stream ${{\varvec{y}}}^{*}$ in the same way as we already transformed stream y only with respect to payoffs at the before-last moment of time $T-1$ (rather than the last moment of time T). Continuing this series of increased and/or advanced payments, we eventually reach the present moment $t=0$. At this stage, the transformed stream has the same payoffs as stream x in all but the present moment of time.

If the transformed stream has payoff $y_{0}$ at the present moment of time, then we must have $y_{0}\le x_{0}$ (because x first-order temporarily dominates y). Then we can add $x_{0}-y_{0} \ge 0$ to the payoff of stream y at the present moment of time to obtain stream x.

Otherwise, the payoff of the transformed stream at the present moment of time can be written as $y_0 +\mathop \sum \nolimits _{s=1}^\tau \left( {y_s -x_s } \right)$ where $\tau \in \{1,{\ldots }, T\}$ denotes the most distant moment of time from which onward we performed only advanced payments (i.e., $\tau \in \{1,{\ldots }, T\}$ is the highest number such that $\mathop \sum \nolimits _{s=t}^\tau \left( {y_s -x_s } \right) >0$ for all $t\in \{1,{\ldots }, \tau \}$. Since x first-order temporarily dominates y, we must have $F_{x}(\tau )\ge F_{y}(\tau )$ so that $y_0 +\mathop \sum \nolimits _{s=1}^\tau \left( {y_s -x_s } \right) \le x_0$. Then we can add $\mathop \sum \nolimits _{s=0}^\tau \left( {x_s -y_s } \right) \ge 0$ to the payoff of stream y at the present moment to obtain stream x.

(2) Consider the case when stream x is obtained from stream y through a series of increased and/or advanced payments. During any increased payment the cumulative payoff function of a new stream at any moment of time cannot be less than the cumulative payoff function of an old stream at the same moment of time. Similarly, during any advanced payment the cumulative payoff function of a new stream at any moment of time cannot be less than the cumulative payoff function of an old stream at the same moment of time. Hence, if x is obtained from y through a series of increased and/or advanced payments, then $F_{x}(t) \ge F_{y}(t)$ for all moments of time $t\in \{0, 1,{\ldots }, T\}$. If this inequality holds as equality at all moments of time $t\in \{0, 1,{\ldots }, T\}$ then streams x and y must be identical. Since we consider distinct streams x and y, there must be at least one moment $s\in \{0, 1,{\ldots }, T\}$ such that $F_{x}(s)> F_{y}(s)$. In other words, x first-order temporarily dominates y. $\square$

Proof of proposition 2

(1) Consider the case when stream x first-order temporarily dominates stream y. For any discount function $D: \{0, 1,{\ldots }, T\} \rightarrow (0,1]$, we can rewrite the present discounted value of stream x as $\mathop \sum \nolimits _{t=0}^T D\left( t \right) x_t =\mathop \sum \nolimits _{t=0}^{T-1} \left[ {D\left( t \right) -D\left( {t+1} \right) } \right] F_x \left( t \right) +D\left( T \right) F_x \left( T \right)$. If discount function D(.) is strictly decreasing, then $D(t)-D(t+1)>0$ for all $t\in \{0, 1,{\ldots }, T-1\}$. By definition, if x first-order temporarily dominates y, then $F_{x}(t)\ge F_{y}(t)$ for all $t\in \{0, 1,{\ldots }, T\}$ and there is at least one $s\in \{0, 1,{\ldots }, T\}$ such that $F_{x}(s)> F_{y}(s)$. Therefore, we must have $\mathop \sum \nolimits _{t=0}^{T-1} \left[ {D\left( t \right) -D\left( {t+1} \right) } \right] F_x \left( t \right) +D\left( T \right) F_x \left( T \right) >\mathop \sum \nolimits _{t=0}^{T-1} \left[ {D\left( t \right) -D\left( {t+1} \right) } \right] F_y \left( t \right) +D\left( T \right) F_y \left( T \right)$. Simplifying this inequality yields $\mathop \sum \nolimits _{t=0}^T D\left( t \right) x_t >\mathop \sum \nolimits _{t=0}^T D\left( t \right) y_t$.

(2) Consider the case when $\mathop \sum \nolimits _{t=0}^T D\left( t \right) x_t >\mathop \sum \nolimits _{t=0}^T D\left( t \right) y_t$ for any strictly decreasing discount function $D:\{0, 1,{\ldots }, T\}\rightarrow (0,1]$. Proof by contradiction: suppose that x does not first-order temporarily dominates y. Two cases are possible. First, if x is identical to y then inequality $\mathop \sum \nolimits _{t=0}^T D\left( t \right) x_t >\mathop \sum \nolimits _{t=0}^T D\left( t \right) y_t$ cannot hold for any discount function D(.).

Second, if x is not identical to y then there must be at least one $s\in \{0, 1,{\ldots }, T\}$ such that $F_{x}(s)<F_{y}(s)$. In this case, let us consider a discount function ${\varPsi } (t)=\beta ^{t}$, for all $t\in \{0,{\ldots }, s\}$, and ${\varPsi } (t)=\alpha ^{t}$, for all (if any) $t> s$, where $\alpha ,\beta \in (0,1)$ and $\beta > \alpha$. In the limit, as $\beta$ converges to one and $\alpha$ converges to zero, present discounted value $\mathop \sum \nolimits _{t=0}^T {\varPsi } \left( t \right) x_t$ converges to $F_{x}(s)$ and present discounted value $\mathop \sum \nolimits _{t=0}^T {\varPsi } \left( t \right) y_t$ converges to $F_{y}(s)> F_{x}(s)$. Thus, we can find $\beta$ sufficiently close to one and $\alpha$ sufficiently close to zero such that $\mathop \sum \nolimits _{t=0}^T {\varPsi } \left( t \right) y_t >\mathop \sum \nolimits _{t=0}^T {\varPsi } \left( t \right) x_t$. This contradicts our starting premise that $\mathop \sum \nolimits _{t=0}^T D\left( t \right) x_t >\mathop \sum \nolimits _{t=0}^T D\left( t \right) y_t$ for any strictly decreasing discount function D(.). $\square$

Proof of proposition 3

(1) Consider the case when stream x first-order temporarily dominates stream y. The rank-dependent discounted utility of x is $\mathop \sum \nolimits _{t=0}^{T-1} \left[ D\left( t \right) -D\left( {t+1} \right) \right] u\circ F_x \left( t \right) +D\left( T \right) u\circ F_x \left( T \right)$. If discount function D(.) is strictly decreasing, then $D(t)-D(t+1)>0$ for all $t\in \{0, 1,{\ldots }, T-1\}$. By definition, if x first-order temporarily dominates y, then $F_{x}(t)\ge F_{y}(t)$ for all $t\in \{0, 1,{\ldots }, T\}$ and there is at least one $s\in \{0, 1, {\ldots }, T\}$ such that $F_{x}(s)> F_{y}(s)$. If utility function u(.) is strictly increasing, then we also have $u\circ F_{x}(t)\ge u \circ F_{y}(t)$ for all $t\in \{0, 1,{\ldots },T\}$ and there is at least one $s\in \{0, 1,{\ldots }, T\}$ such that $u \circ F_{x}(s)> u \circ F_{y}(s)$. Therefore, we must have $\mathop \sum \nolimits _{t=0}^{T-1} \left[ {D\left( t \right) -D\left( {t+1} \right) } \right] u\circ F_x \left( t \right) +D\left( T \right) u\circ F_x \left( T \right) >\mathop \sum \nolimits _{t=0}^{T-1} \left[ {D\left( t \right) -D\left( {t+1} \right) } \right] u\circ F_y \left( t \right) +D\left( T \right) u\circ F_y \left( T \right)$.

(2) Consider the case when the rank-dependent discounted utility of x is greater than the rank-dependent discounted utility of y for any strictly increasing utility function and any strictly decreasing discount function. Proof by contradiction: suppose that x does not first-order temporarily dominates y. Two cases are possible. First, if x is identical to y then the rank-dependent discounted utility of x cannot be greater than the rank-dependent discounted utility of y for any utility/discount function.

Second, if x is not identical to y then there must be at least one $s\in \{0, 1,{\ldots }, T\}$ such that $F_{x}(s)<F_{y}(s)$. In this case, let us consider a discount function ${\varPsi } (t)=\beta ^{t}$, for all $t\in \{0,{\ldots }, s\}$, and ${\varPsi } (t)=\alpha ^{t}$, for all (if any) $t> s$, where $\alpha ,\beta \in (0,1)$ and $\beta > \alpha$. In the limit, as $\beta$ converges to one and $\alpha$ converges to zero, the rank-dependent discounted utility of x converges to $u\circ F_{x}(s)$ and the rank-dependent discounted utility of y converges to $u \circ F_{y}(s)> u \circ F_{x}(s)$. Thus, we can find $\beta$ sufficiently close to one and $\alpha$ sufficiently close to zero such that we arrive at the contradiction: the rank-dependent discounted utility of y is greater than the rank-dependent discounted utility of x. $\square$

Proof of proposition 4

(1) Consider the case when x second-order temporarily dominates y. For the subsequent argument it is convenient to rewrite the rank-dependent discounted utility of x as follows: $\mathop \sum \nolimits _{t=0}^{T-2} \left[ {D\left( t \right) +D\left( {t+2} \right) -2D\left( {t+1} \right) } \right] \mathop \sum \nolimits _{s=0}^t u\circ F_x \left( s \right) +\left[ {D\left( {T-1} \right) -D\left( T \right) } \right] \mathop \sum \nolimits _{s=0}^{T-1} u\circ F_x \left( s \right) +D\left( T \right) u\circ F_x \left( T \right)$. If discount function D(.) is strictly decreasing, then $D\left( {T-1} \right) -D\left( T \right) >0$. If discount function D(.) is convex then $D\left( t \right) +D\left( {t+2} \right) -2D\left( {t+1} \right) >0$ for all $t\in \{0,{\ldots }, T-2\}$.

By definition, if x second-order temporarily dominates y, then $F_{x}(T)\ge F_{y}(T)$, $\mathop \sum \nolimits _{s=0}^t F_x \left( s \right) \ge \mathop \sum \nolimits _{s=0}^t F_y \left( s \right)$ for all moments of time $t\in \{0, 1,{\ldots }, T-1\}$ and there is at least one moment of time $\tau \in \{0, 1,{\ldots }, T\}$ such that $\mathop \sum \nolimits _{s=0}^\tau F_x \left( s \right) >\mathop \sum \nolimits _{s=0}^\tau F_y \left( s \right)$. If utility function u(.) is strictly increasing, then we also have $u \circ F_{x}(T)\ge u \circ F_{y}(T)$. If utility function u(.) is concave, then we also have $\mathop \sum \nolimits _{s=0}^t u\circ F_x \left( s \right) \ge \mathop \sum \nolimits _{s=0}^t u\circ F_y \left( s \right)$ for all moments of time $t\in \{0, 1,{\ldots }, T-1\}$ and there is at least one moment of time $\tau \in \{0, 1, {\ldots }, T\}$ such that $\mathop \sum \nolimits _{s=0}^\tau u\circ F_x \left( s \right) >\mathop \sum \nolimits _{s=0}^\tau u\circ F_y \left( s \right)$. Therefore, we must have $\mathop \sum \nolimits _{t=0}^{T-2} \left[ {D\left( t \right) +D\left( {t+2} \right) -2D\left( {t+1} \right) } \right] \mathop \sum \nolimits _{s=0}^t u\circ F_x \left( s \right) +\left[ {D\left( {T-1} \right) -D\left( T \right) } \right] \mathop \sum \nolimits _{s=0}^{T-1} u\circ F_x \left( s \right) +D\left( T \right) u\circ F_x \left( T \right) >\mathop \sum \nolimits _{t=0}^{T-2} \left[ D\left( t \right) +D\left( {t+2} \right) -2D\left( {t+1} \right) \right] \mathop \sum \nolimits _{s=0}^t u\circ F_y \left( s \right) +\left[ D\left( {T-1} \right) -D\left( T \right) \right] \mathop \sum \nolimits _{s=0}^{T-1} u\circ F_y \left( s \right) +D\left( T \right) u\circ F_y \left( T \right)$. Simplifying this inequality yields the result $\mathop \sum \nolimits _{t=0}^{T-1} \left[ {D\left( t \right) -D\left( {t+1} \right) } \right] u\circ F_x \left( t \right) +D\left( T \right) u\circ F_x \left( T \right) >\mathop \sum \nolimits _{t=0}^{T-1} \left[ {D\left( t \right) -D\left( {t+1} \right) } \right] u\circ F_y \left( t \right) +D\left( T \right) u\circ F_y \left( T \right)$.

(2) Consider the case when the rank-dependent discounted utility of x is greater than the rank-dependent discounted utility of y for any strictly increasing concave utility function and any strictly decreasing convex discount function. Proof by contradiction: suppose that x does not second-order temporarily dominates y. Three cases are possible. First, if x is identical to y then the rank-dependent discounted utility of x cannot be greater than the rank-dependent discounted utility of y for any utility/discount function.

Second, if $F_{x}(T)< F_{y}(T)$, then let us consider a discount function ${\varPsi } (t)=\beta ^{t}$, for all $t\in \{0,{\ldots }, T\}$ and any strictly increasing concave utility function u(.). In the limit, as $\beta$ converges to one, the rank-dependent discounted utility of x converges to $u \circ F_{x}(T)$ and the rank-dependent discounted utility of y converges to $u \circ F_{y}(T)$. If utility function u(.) is strictly increasing that we must have $u \circ F_{y}(T)> u \circ F_{x}(T)$. Thus, we can find $\beta$ sufficiently close to one such that we arrive at the contradiction: the rank-dependent discounted utility of y is greater than the rank-dependent discounted utility of x for a strictly decreasing convex discount function ${\varPsi } (t)=\beta ^{t}$.

Third, if there is a moments of time $t\in \{0, 1,{\ldots }, T-1\}$ such that $\mathop \sum \nolimits _{s=0}^t F_x \left( s \right) <\mathop \sum \nolimits _{s=0}^t F_y \left( s \right)$ then let us consider rank-dependent discounted utility with a linear utility function and the following discount function:

$$\begin{aligned} {\varPsi } \left( \tau \right) =\left\{ \begin{array}{ll} {c-\root \of {\left( {c-1} \right) ^{2}+\tau \left( {t+1-\tau } \right) + \frac{\tau }{t+1}\left[ {2c\left( {1-\alpha ^{t+1}} \right) -1+\alpha ^{2t+2}} \right] },}&{} {\tau \le t+1} \\ {\alpha ^{\tau },}&{} {\tau \ge t+1} \\ \end{array} \right. \end{aligned}$$

for constant $c>0.5(t+1)^{2}+0.5$ and $\alpha \in (0,1)$. In the limit, as parameter c converges to infinity and parameter $\alpha$ converges to zero, this discount function converges to a piece-wise linear discount function with a slope ${\varPsi }( \tau )-{\varPsi } (\tau +1)=1/(t+1)$ for all $\tau \in \{0, 1,{\ldots }, t\}$ and a zero slope ${\varPsi } ( \tau )-{\varPsi } (\tau +1)=0$ for all (if any) $\tau >t+1$. Thus, in the limit as c converges to infinity and $\alpha$ converges to zero, the rank-dependent discounted utility of x converges to $\frac{1}{t+1}\mathop \sum \nolimits _{s=0}^t F_x \left( s \right)$ and the rank-dependent discounted utility of y converges to $\frac{1}{t+1}\mathop \sum \nolimits _{s=0}^t F_y \left( s \right)$. In other words, we can find c sufficiently large and $\alpha$ sufficiently close to zero such that we arrive at the contradiction: the rank-dependent discounted utility of y is greater than the rank-dependent discounted utility of x. $\square$

Proof of proposition 5

(1) We prove that decision maker A is more patient than decision maker B in the sense of definition 3 if B is more impatient than A in the sense of definition 4. Consider any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{A}}{{\varvec{x}}}$. If it happens so that ${{\varvec{x}}}\succcurlyeq _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$, then, according to definition 4, we must have ${{\varvec{x}}}\succcurlyeq _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$, which contradicts ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{A}}{{\varvec{x}}}$. Thus, for any x and $y_{0}$ such that ${{\varvec{y}}}_{\mathbf{0}}\succ _{\mathrm{A}}{{\varvec{x}}}$ we must have ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$, i.e., part a) of definition 3 is satisfied.

Next, let us consider any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{y}}}_{\mathbf{0} }\sim _{\mathrm{A}}{{\varvec{x}}}$. If it happens so that ${{\varvec{x}}}\succ _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$, then, according to definition 4, we must have ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$, which contradicts ${{\varvec{y}}}_{\mathbf{0} }\sim _{\mathrm{A}}{{\varvec{x}}}$. Thus, for any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{y}}}_{\mathbf{0} }\sim _{\mathrm{A}}{{\varvec{x}}}$ we must have ${{\varvec{y}}}_{\mathbf{0}} \succcurlyeq _{\mathrm{B}}{{\varvec{x}}}$, i.e., part b) of definition 3 is satisfied.

Finally, if B is more impatient than A in the sense of definition 4 then there exist a stream of outcomes x and an outcome $y_{0}$ such that either ${{\varvec{x}}}\sim _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ or ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$ and ${{\varvec{x}}}\succcurlyeq _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$. In the former case, we immediately have the existence of a stream of outcomes x and an outcome $y_{0}$ such that ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{y}}}_{\mathbf{0} }\succcurlyeq _{\mathrm{B}}{{\varvec{x}}}$. In the latter case, when ${{\varvec{y}}}_{\mathbf{0}}\succ _{\mathrm{B}}{{\varvec{x}}}$ and ${{\varvec{x}}}\succcurlyeq _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$, we have the existence of a stream of outcomes x and an outcome $y_{0}$ such that either ${{\varvec{y}}}_{\mathbf{0} }\sim _{\mathrm{A}}{{\varvec{x}}}$ and ${{\varvec{y}}}_{\mathbf{0}}\succ _{\mathrm{B}}{{\varvec{x}}}$ or ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$. In other words, part c) of definition 3 is satisfied as well.

(2) We prove that B is more impatient than A in the sense of definition 4 if A is more patient than B in the sense of definition 3. Consider any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{x}}}\succ _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$. If it happens so that ${{\varvec{y}}}_{\mathbf{0}}\succcurlyeq _{\mathrm{A}}{{\varvec{x}}}$, then, according to definition 3, we must have ${{\varvec{y}}}_{\mathbf{0} }\succcurlyeq _{\mathrm{B}}{{\varvec{x}}}$, which contradicts ${{\varvec{x}}}\succ _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$. Thus, for any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{x}}}\succ _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$ we must have ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$, i.e., part a) of definition 4 is satisfied.

Next, let us consider any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{x}}}\sim _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$. If it happens so that ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{A}}{{\varvec{x}}}$ then, according to definition 3, we must have ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$, which contradicts ${{\varvec{x}}}\sim _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0}}$. Thus, for any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{x}}}\sim _{\mathrm{B}}{{\varvec{y}}}_{\mathbf{0} }$ we must have ${{\varvec{x}}}\succcurlyeq _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$, i.e., part b) of definition 4 is satisfied as well.

Finally, if A is more patient than B in the sense of definition 3, then there exist a stream of outcomes x and an outcome $y_{0}$ such that either ${{\varvec{y}}}_{\mathbf{0} }\sim _{\mathrm{A}}{{\varvec{x}}}$ and ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$ or ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{y}}}_{\mathbf{0}}\succcurlyeq _{\mathrm{B}} {{\varvec{x}}}$. In the former case, we immediately have the existence of a stream of outcomes x and an outcome $y_{0}$ such that ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$ and ${{\varvec{x}}} \succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$. In the latter case, when ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{y}}}_{\mathbf{0} }\succcurlyeq _{\mathrm{B}}{{\varvec{x}}}$, we have the existence of a stream of outcomes x and an outcome $y_{0}$ such that either ${{\varvec{x}}}\sim _{\mathrm{B}} {{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ or ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$ and ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$. In other words, part c) of definition 4 is satisfied as well. $\square$

Proof of proposition 6

(1) We prove that decision maker A is more patient than decision maker B if ${ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ for all streams of outcomes x, and there exists at least one stream of outcomes y such that ${ PDE}_{\mathrm{A}}({{\varvec{y}}})>{ PDE}_{\mathrm{B}}({{\varvec{y}}})$.

Let us consider any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{A}}{{\varvec{x}}}$. By definition of the present-day equivalent, decision maker A is indifferent between x and a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{x}}})$ at the present moment of time and nothing in all subsequent moments of time. Thus, if A’s preferences are transitive, then A strictly prefers ${{\varvec{y}}}_{\mathbf{0}}$ over a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{x}}})$ at the present moment. This implies $y_{0}>{ PDE}_{\mathrm{A}}({{\varvec{x}}})$ given that A’s preferences are monotone. Thus, we have $y_{0}>{ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$. If B’s preferences are monotone, then B strictly prefers ${{\varvec{y}}}_{\mathbf{0}}$ over a degenerate stream that yields ${ PDE}_{\mathrm{B}}$(x) at the present moment. By definition, B is indifferent between the later and x. Thus, if B’s preferences are transitive, then we have ${{\varvec{y}}}_{\mathbf{0} }\succ _{\mathrm{B}}{{\varvec{x}}}$, i.e., part a) of definition 3 is satisfied.

Let us consider any stream of outcomes x and any outcome $y_{0}$ such that ${{\varvec{y}}}_{\mathbf{0} }\sim _{\mathrm{A}}{{\varvec{x}}}$, i.e., when $y_{0}={ PDE}_{\mathrm{A}}({{\varvec{x}}})$. In this case, we must have $y_{0}={ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$. If B’s preferences are monotone, then B weakly prefers ${{\varvec{y}}}_{\mathbf{0}}$ over a degenerate stream that yields ${ PDE}_{\mathrm{B}}$(x) at the present moment. By definition, B is indifferent between the later and x. Thus, if B’s preferences are transitive, then we have ${{\varvec{y}}}_{\mathbf{0} }\succcurlyeq _{\mathrm{B}}{{\varvec{x}}}$, i.e., part b) of definition 3 is satisfied.

Finally, if there exists at least one stream of outcomes ${{\varvec{y}}}$ such that ${ PDE}_{\mathrm{A}}({{\varvec{y}}})>{ PDE}_{\mathrm{B}}({{\varvec{y}}})$, we have a stream of outcomes y and an outcome ${ PDE}_{\mathrm{A}}({{\varvec{y}}})$ such that A is indifferent between y and a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{y}}})$ at the present moment. At the same time, given that ${ PDE}_{\mathrm{A}}({{\varvec{y}}})>{ PDE}_{\mathrm{B}}({{\varvec{y}}})$ and B’s preferences are monotone, B strictly prefers a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{y}}})$ at the present moment over a degenerate stream that yields ${ PDE}_{\mathrm{B}}({{\varvec{y}}})$ at the present moment. B is also indifferent between the later and y. Thus, if B’s preferences are transitive, B strictly prefers a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{y}}})$ at the present moment over y, i.e., part c) of definition 3 is satisfied.

(2) We prove that ${ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ for all streams of outcomes x and there exists at least one stream of outcomes y such that ${ PDE}_{\mathrm{A}}({{\varvec{y}}})>{ PDE}_{\mathrm{B}}({{\varvec{y}}})$ if decision maker A is more patient than decision maker B. By definition of the present-day equivalent, decision maker A is indifferent between x and a degenerate stream that yields ${ PDE}_{\mathrm{A}}$(x) at the present moment of time and nothing in all subsequent moments of time. According to part b) of definition 3, decision maker B then weakly prefers this degenerate stream over x. Thus, if B’s preferences are transitive, B weakly prefers a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{x}}})$ at the present moment over a degenerate stream that yields ${ PDE}_{\mathrm{B}}({{\varvec{x}}})$ at the present moment. This implies ${ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ if B’s preferences are monotone.

According to part c) of definition 3, there exist a stream of outcomes x and an outcome $y_{0}$ such that either ${{\varvec{y}}}_{\mathbf{0}}\sim _{\mathrm{A}}{{\varvec{x}}}$ and ${{\varvec{y}}}_{\mathbf{0}}\succ _{\mathrm{B}}{{\varvec{x}}}$ or ${{\varvec{x}}}\succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{y}}}_{\mathbf{0} } \succcurlyeq _{\mathrm{B}}{{\varvec{x}}}$. In the former case, $y_{0}={ PDE}_{\mathrm{A}}({{\varvec{x}}})$ and ${{\varvec{y}}}_{\mathbf{0}}\succ _{\mathrm{B}}{{\varvec{x}}}$ implies that decision maker B (with transitive preferences) strictly prefers a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{x}}})$ at the present moment over a degenerate stream that yields ${ PDE}_{\mathrm{B}}({{\varvec{x}}})$ at the present moment. This implies ${ PDE}_{\mathrm{A}}({{\varvec{x}}})>{ PDE}_{\mathrm{B}}({{\varvec{x}}})$ if B’s preferences are monotone. In the latter case when ${{\varvec{x}}} \succ _{\mathrm{A}}{{\varvec{y}}}_{\mathbf{0}}$ and ${{\varvec{y}}}_{\mathbf{0} }\succcurlyeq _{\mathrm{B}}{{\varvec{x}}}$, decision maker A (with transitive preferences) strictly prefers a degenerate stream that yields ${ PDE}_{\mathrm{A}}({{\varvec{x}}})$ at the present moment over ${{\varvec{y}}}_{\mathbf{0}}$. This implies ${ PDE}_{\mathrm{A}}({{\varvec{x}}})>y_{0}$ if A’s preferences are monotone. At the same time, decision maker B (with transitive preferences) weakly prefers ${{\varvec{y}}}_{\mathbf{0}}$ over a degenerate stream that yields ${ PDE}_{\mathrm{B}}({{\varvec{x}}})$ at the present moment. This implies $y_{0}\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ if B’s preferences are monotone. Thus, we must have ${ PDE}_{\mathrm{A}}({{\varvec{x}}})>y_{0}\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$. $\square$

Proof of proposition 7

(1) We prove that rank-dependent discounted utility maximizer A is more patient than rank-dependent discounted utility maximizer B if inequalities (3–4) are satisfied.

We prove by mathematical induction that ${ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ for all streams x. First, for any degenerate stream ${{\varvec{x}}}_{\mathbf{0}}$ A’s present-day equivalent is $x_{{ 0}}$, which is B’s present-day equivalent as well. Hence, inequality ${ PDE}_{\mathrm{A}}({{\varvec{x}}}_{\mathbf{0}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}}_{\mathbf{0}})$ holds trivially for all degenerate streams ${{\varvec{x}}}_{\mathbf{0}}$.

Second, assuming that ${ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ holds for all streams with nonzero outcomes up to the moment of time $t\in \{0, 1,{\ldots }, T-1\}$, we prove that it also holds for streams with nonzero outcomes up to the moment of time $t+1$. Let x be any stream with nonzero outcomes up to the moment of time $t+1$ ($x_{t+1}>0$). Let $d\in (x_{t}, x_{t}+x_{t+1})$ denote an outcome such that ${\varvec{\updelta }}\mathop {=}\limits ^{\mathrm{def}} (x_{0}, x_{1},{\ldots }, x_{t-1},d,0,{\ldots },0)\sim _{\mathrm{A}}{{\varvec{x}}}$ so that we have

$$\begin{aligned} \frac{u_A \left( b \right) -u_A \left( a \right) }{u_A \left( c \right) -u_A \left( a \right) }=\frac{D_A \left( {t+1} \right) }{D_A \left( t \right) } \end{aligned}$$

where $a=\mathop \sum \nolimits _{s=0}^t x_s$, $b=a-x_{t}+d$ and $c=a+x_{t+1}$. Using inequalities (3) and (4), we obtain

$$\begin{aligned} \frac{u_B \left( b \right) -u_B \left( a \right) }{u_B \left( c \right) -u_B \left( a \right) }\ge \frac{u_A \left( b \right) -u_A \left( a \right) }{u_A \left( c \right) -u_A \left( a \right) }=\frac{D_A \left( {t+1} \right) }{D_A \left( t \right) }\ge \frac{D_B \left( {t+1} \right) }{D_B \left( t \right) } \end{aligned}$$

which can be rearranged as ${ RDDU}_{\mathrm{B}}({\varvec{\updelta }})\ge { RDDU}_{\mathrm{B}}({{\varvec{x}}})$.¹¹ By assumption of mathematical induction ${ PDE}_{\mathrm{A}}(\varvec{\delta })\ge { PDE}_{\mathrm{B}}(\varvec{\delta })$ so that $u_{\mathrm{B}}\circ { PDE}_{\mathrm{A}}(\varvec{\delta })\ge u_{\mathrm{B}}\circ { PDE}_{\mathrm{B}}(\varvec{\delta })={ RDDU}_{\mathrm{B}}(\varvec{\delta })\ge { RDDU}_{\mathrm{B}}({{\varvec{x}}})=u_{\mathrm{B}}\circ { PDE}_{\mathrm{B}}({{\varvec{x}}})$. Since utility function $u_{\mathrm{B}}$(.) is strictly increasing, we must have ${ PDE}_{\mathrm{A}}(\varvec{\delta })\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$. If A is indifferent between a degenerate stream that yields ${ PDE}_{\mathrm{A}}(\varvec{\delta })$ at the present moment of time and $\varvec{\delta }$, and $\varvec{\delta }\sim _{\mathrm{A}}{{\varvec{x}}}$, then ${ PDE}_{\mathrm{A}}(\varvec{\delta })={ PDE}_{\mathrm{A}}({{\varvec{x}}})$. Thus, we have ${ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ for any stream x with strictly positive outcomes up to the moment of time $t+1$. By mathematical induction, we then must have ${ PDE}_{\mathrm{A}}({{\varvec{x}}})\ge { PDE}_{\mathrm{B}}({{\varvec{x}}})$ for all streams x.

Finally, there exist either a moment of time $t\in \{0, 1,{\ldots }, T-1\}$ or outcomes $a, b, c \in \mathbb {R}$, $a< b< c$, such that one of inequalities (3–4) holds as a strict inequality. Using either this moment of time or these outcomes we can construct a stream y such that ${ PDE}_{\mathrm{A}}({{\varvec{y}}})>{ PDE}_{\mathrm{B}}({{\varvec{y}}})$. By proposition 6, rank-dependent discounted utility maximizer A is then more patient than rank-dependent discounted utility maximizer B.

(2) We prove that inequalities (3–4) must hold if rank-dependent discounted utility maximizer A is more patient than rank-dependent discounted utility maximizer B.

Suppose that there is a moment of time $t\in \{0, 1,{\ldots }, T-1\}$ for which inequality (3) is violated. For any two outcomes $x_{t}$, $x_{t+1} >0$, we can construct a stream x that yields $x_{t}$ at the moment of time t, $x_{t+1}$ at the moment of time $t+1$ and nothing at all other moments of time. By definition of the present-day equivalent, we must have

$$\begin{aligned} u_A \circ PDE_A \left( {{\varvec{x}}} \right)= & {} D_A \left( t \right) \left[ {u_A \left( {x_t } \right) -u_A \left( 0 \right) } \right] \\&+D_A \left( {t+1} \right) \left[ {u_A \left( {x_t +x_{t+1} } \right) -u_A \left( {x_t } \right) } \right] \end{aligned}$$

Without the loss of generality, we can normalize utility functions of the two decision makers so that $u_{\mathrm{A}}\circ { PDE}_{\mathrm{A}}({{\varvec{x}}})=u_{\mathrm{B}}\circ { PDE}_{\mathrm{A}}({{\varvec{x}}})$ and $D_{\mathrm{A}}(t)[u_{\mathrm{A}}(x_{t})-u_{\mathrm{A}}(0)]=D_{\mathrm{B}}(t)[u_{\mathrm{B}}(x_{t})-u_{\mathrm{B}}(0)]$. In this case we have

$$\begin{aligned} u_B \circ PDE_A \left( {{\varvec{x}}} \right)= & {} u_A \circ PDE_A \left( {{\varvec{x}}} \right) \\= & {} D_B \left( t \right) \left[ {u_B \left( {x_t } \right) -u_B \left( 0 \right) } \right] +\frac{D_A \left( {t+1} \right) }{D_A \left( t \right) }\\&\left( {c\left[ {u_A \left( {x_t +x_{t+1} } \right) -u_A \left( 0 \right) } \right] -D_B \left( t \right) \left[ {u_B \left( {x_t } \right) -u_B \left( 0 \right) } \right] } \right) \\< & {} D_B \left( t \right) \left[ {u_B \left( {x_t } \right) -u_B \left( 0 \right) } \right] +\frac{D_B \left( {t+1} \right) }{D_B \left( t \right) }\\&\left( {D_A \left( t \right) \left[ {u_A \left( {x_t +x_{t+1} } \right) -u_A \left( 0 \right) } \right] -D_B \left( t \right) \left[ {u_B \left( {x_t } \right) -u_B \left( 0 \right) } \right] } \right) \end{aligned}$$

In the limit as $x_{t+1}$ converges to zero, $D_{\mathrm{A}}(t)[u_{\mathrm{A}}(x_{t}+x_{t+1})-u_{\mathrm{A}}(0)]$ converges to $D_{\mathrm{A}}(t)[u_{\mathrm{A}}(x_{t})-u_{\mathrm{A}}(0)]$. Since we normalized utility functions so that $D_{\mathrm{A}}(t)[u_{\mathrm{A}}(x_{t})-u_{\mathrm{A}}(0)]=D_{\mathrm{B}}(t)[u_{\mathrm{B}}(x_{t})-u_{\mathrm{B}}(0)]$ and utility functions are continuous, we can find outcome $x_{t+1}$ sufficiently close to zero such that $D_{\mathrm{B}}(t)[u_{\mathrm{B}}(x_{t}+x_{t+1})-u_{\mathrm{B}}(0)]$ is arbitrary close to $D_{\mathrm{A}}(t)[u_{\mathrm{A}}(x_{t}+x_{t+1})-u_{\mathrm{A}}(0)]$. In particular, we can find outcome $x_{t+1}$ sufficiently close to zero such that

$$\begin{aligned} u_B \circ PDE_A \left( {{\varvec{x}}} \right)< & {} D_B \left( t \right) \left[ {u_B \left( {x_t } \right) -u_B \left( 0 \right) } \right] +\frac{D_B \left( {t+1} \right) }{D_B \left( t \right) }\\&\left( {D_B \left( t \right) \left[ {u_B \left( {x_t +x_{t+1} } \right) -u_B \left( 0 \right) } \right] -D_B \left( t \right) \left[ {u_B \left( {x_t } \right) -u_B \left( 0 \right) } \right] } \right) \end{aligned}$$

This inequality can be rearranged as $u_{\mathrm{B}}\circ { PDE}_{\mathrm{A}}({{\varvec{x}}})<{ RDDU}_{\mathrm{B}}({{\varvec{x}}})=u_{\mathrm{B}}\circ { PDE}_{\mathrm{B}}({{\varvec{x}}})$. Since utility function $u_{\mathrm{B}}$(.) is strictly increasing, we must then have ${ PDE}_{\mathrm{A}}({{\varvec{x}}})<{ PDE}_{\mathrm{B}}({{\varvec{x}}})$. This contradicts the fact that A is more patient than B (cf. proposition 6).

Suppose now that there are outcomes $a,b,c\in \mathbb {R}$ such that $a< b< c$ and inequality (4) is violated. Let us pick an arbitrary moment of time t. Let z denote a stream that yields a at the present moment of time and $c-a$ at moment t.

If decision maker A weakly prefers a degenerate stream that yields b over stream z, then there must be an outcome $b'\in [b,c)$ such that decision maker A is indifferent between a degenerate stream that yields b’ and stream z. On the other hand, if A weakly prefers z over a degenerate stream that yields b, then there must be an outcome $b'\in (a,b]$ such that decision maker A is indifferent between a degenerate stream that yields $b'$ and stream z. In both cases, we end up with equation

$$\begin{aligned} \frac{u_A \left( {b^{\prime }} \right) -u_A \left( a \right) }{u_A \left( c \right) -u_A \left( a \right) }=D_A \left( t \right) \end{aligned}$$

If it happens so that

$$\begin{aligned} \frac{u_A \left( {b^{\prime }} \right) -u_A \left( a \right) }{u_A \left( c \right) -u_A \left( a \right) }>\frac{u_B \left( {b^{\prime }} \right) -u_B \left( a \right) }{u_B \left( c \right) -u_B \left( a \right) } \end{aligned}$$

then we must have

$$\begin{aligned} D_A \left( t \right) >\frac{u_B \left( {b^{\prime }} \right) -u_B \left( a \right) }{u_B \left( c \right) -u_B \left( a \right) } \end{aligned}$$

If we choose moment of time t sufficiently close to the present moment $t=0$, then $D_{\mathrm{B}}(t)$ is arbitrary close to $D_{\mathrm{A}}(t)$ so that we also have

$$\begin{aligned} D_B \left( t \right) >\frac{u_B \left( {b^{\prime }} \right) -u_B \left( a \right) }{u_B \left( c \right) -u_B \left( a \right) } \end{aligned}$$

which implies that decision maker B strictly prefers z over a degenerate stream that yields $b'$. This contradicts part b) of definition 3.

If it happens so that

$$\begin{aligned} \frac{u_A \left( {b^{\prime }} \right) -u_A \left( a \right) }{u_A \left( c \right) -u_A \left( a \right) }\le \frac{u_B \left( {b^{\prime }} \right) -u_B \left( a \right) }{u_B \left( c \right) -u_B \left( a \right) } \end{aligned}$$

then between outcomes b and b’ there must be an outcome d such that

$$\begin{aligned} \frac{u_A \left( d \right) -u_A \left( a \right) }{u_A \left( c \right) -u_A \left( a \right) }=\frac{u_B \left( d \right) -u_B \left( a \right) }{u_B \left( c \right) -u_B \left( a \right) } \end{aligned}$$

We can use this equality and the fact that inequality (4) is violated to obtain

$$\begin{aligned} \frac{u_A \left( b \right) -u_A \left( a \right) }{u_A \left( d \right) -u_A \left( a \right) }>\frac{u_B \left( b \right) -u_B \left( a \right) }{u_B \left( d \right) -u_B \left( a \right) } \end{aligned}$$

In other words, if $d> b$, then inequality (4) is also violated for outcomes $a,b,d\in \mathbb {R}$ such that $a< b< d$. Therefore, we can repeat the same argument as above with outcome c being replaced by outcome d (and a new outcome $b''\in (a,d)$ such that A is indifferent between a degenerate stream that yields $b''$ and a stream that yields a at the present moment of time and $d-a$ at moment t).

At the same time, the last inequality can be rearranged as

$$\begin{aligned} \frac{u_A \left( b \right) -u_A \left( a \right) }{u_A \left( d \right) -u_A \left( a \right) }\!=\!\frac{u_A \left( b \right) -u_A \left( d \right) }{u_A \left( d \right) -u_A \left( a \right) }+1>\frac{u_B \left( b \right) -u_B \left( d \right) }{u_B \left( d \right) -u_B \left( a \right) }+1 \!=\!\frac{u_B \left( b \right) -u_B \left( a \right) }{u_B \left( d \right) -u_B \left( a \right) } \end{aligned}$$

so that we have

$$\begin{aligned} \frac{u_A \left( b \right) -u_A \left( d \right) }{u_A \left( d \right) -u_A \left( a \right) }>\frac{u_B \left( b \right) -u_B \left( d \right) }{u_B \left( d \right) -u_B \left( a \right) } \end{aligned}$$

Moreover, we can rearrange

$$\begin{aligned} \frac{u_A \left( c \right) -u_A \left( a \right) }{u_A \left( d \right) -u_A \left( a \right) }=\frac{u_A \left( c \right) -u_A \left( d \right) }{u_A \left( d \right) -u_A \left( a \right) }+1=\frac{u_B \left( c \right) -u_B \left( d \right) }{u_B \left( d \right) -u_B \left( a \right) }\!+\!1\!=\!\frac{u_B \left( c \right) -u_B \left( a \right) }{u_B \left( d \right) -u_B \left( a \right) } \end{aligned}$$

so that we also have

$$\begin{aligned} \frac{u_A \left( c \right) -u_A \left( d \right) }{u_A \left( d \right) -u_A \left( a \right) }=\frac{u_B \left( c \right) -u_B \left( d \right) }{u_B \left( d \right) -u_B \left( a \right) } \end{aligned}$$

Using this equality and the last inequality, we obtain

$$\begin{aligned} \frac{u_A \left( b \right) -u_A \left( d \right) }{u_A \left( c \right) -u_A \left( d \right) }>\frac{u_B \left( b \right) -u_B \left( d \right) }{u_B \left( c \right) -u_B \left( d \right) } \end{aligned}$$

In other words, if $d< b$, then inequality (4) is also violated for outcomes $d,b,c\in \mathbb {R}_{+}$ such that $d< b< c$. Therefore, we can repeat the same argument as above with outcome a being replaced by outcome d (and a new outcome $b''\in (d,c)$ such that A is indifferent between a degenerate stream that yields $b''$ and a stream that yields d at the present moment of time and $c-d$ at moment t).

Iterating this argument sufficiently many times, we inevitably end up in a contradiction as there cannot be a sequence of triples of outcomes such that either the lowest or the highest outcome converges to outcome b and inequality (4) is violated for each element of this sequence but satisfied in the limit.

Finally, if there is neither a moment of time $t\in \{0, 1,{\ldots }, T-1\}$ nor outcomes $a,b,c \in \mathbb {R},a< b< c$, such that one of inequalities (3–4) holds as a strict inequality, then there cannot be any stream y such that ${ PDE}_{\mathrm{A}}({{\varvec{y}}})>{ PDE}_{\mathrm{B}}({{\varvec{y}}})$. This contradicts the fact that A is more patient than B. $\square$