# Comparative Patience

Mark Whitmeyer Arizona State University<EMAIL_ADDRESS>I thank
Joseph Whitmeyer for his comments. Special thanks is due to Michelle
Whitmeyer, who has already taught me a lot about patience.

###### Abstract

We begin by formulating and characterizing a dominance criterion for prize
sequences: $x$ dominates $y$ if any impatient agent prefers $x$ to $y$. With
this in hand, we define a notion of comparative patience. Alice is more
patient than Bob if Alice’s normalized discounted utility gain by going from
any $y$ to any dominating $x$ is less than Bob’s discounted utility gain from
such an improvement. We provide a full characterization of this relation in
terms of the agents’ discount rules.

##  1 Introduction

Typical economic agents prefer that good things happen sooner rather than
later. That is, they are impatient. This paper seeks to clarify what exactly
it means for one person to be more patient than another.

One possible definition of comparative patience is the naïve one: Alice is
more patient than Bob if for any stream of payoffs, Alice’s discounted sum of
utilities is larger than Bob’s. This is the modal definition of comparative
patience in the literature–see for instance the literature on repeated games,
where “sufficiently patient” refers to an exponential discount factor that is
close enough to one.

Is this really what it means to be more patient, however? In particular,
suppose Alice’s discounted sum of utilities is higher than Bob’s for some
stream of payoffs, and it remains higher after we delay the arrival rate of
the awards, yet Alice suffers a loss in her value for the stream as a result
of the delay whereas Bob is unaffected. From this perspective, it seems like
Bob ought to be characterized as more patient than Alice, as he is unfazed by
the delay.

Our agenda in this paper, therefore, is to provide a behavioral
characterization of what it means for one agent to be more patient than
another. Here is how we define greater patience: Alice is more patient than
Bob if the normalized difference between the discounted sums of her payoffs
for any two payoff sequences with one worse (more delayed) than the other is
lower than Bob’s. Of course, this definition relies heavily on what we mean by
“worse” in the context of payoff sequences. Thus, our first task is to define
and characterize this concept.

We say that one prize sequence $x$ is superior to another sequence $y$ if
$\sum_{t=1}^{T}\beta_{t}x_{t}\geq\sum_{t=1}^{T}\beta_{t}y_{t}$ for any
positive decreasing sequence $\beta=\left\\{\beta_{t}\right\\}_{t=1}^{T}$,
where the time horizon $T$ is either finite or infinite. In words, $x$ is
superior to $y$ if any impatient agent (one who discounts the future) prefers
$x$ to $y$. Inferiority in this sense, then, is how we formalize one sequence
being worse than another. Our first result, Proposition 2.7, characterizes
this superiority in terms of the sequences $x$ and $y$ themselves.

We then turn our attention to comparative patience, defined as follows. Alice,
who discounts according to discounting sequence $\alpha$, is more patient than
Bob, who discounts according to $\beta$, if for any positive prize sequences
$x$ and $y$ with $x$ superior to $y$ and
$\sum_{t=1}^{T}x_{t}=\sum_{t=1}^{T}y_{t}$ (and such that the denominators of
the following are nonzero),

$\frac{\sum_{t=1}^{T}\alpha_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}y_{t}}\leq\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\beta_{t}y_{t}}\text{.}$

This is equivalent to the normalized difference in Alice’s discounted payoffs
from $x$ and $y$ being less than the difference of Bob’s discounted payoffs
from the two streams.

In the main result of the paper, Theorem 3.8, we show that Alice is more
patient than Bob if and only if

$\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\beta_{1}}{\alpha_{1}}$

for every $t$. This is an exceptionally strong condition; for one, it implies
that $\alpha_{t}/\beta_{t}$ is monotonically increasing in $t$. Notably, it is
only if the time horizon is two that it is satisfied for any two exponential
discounters. That is, if there are three or more periods Alice can have
exponential discount factor $a$ and Bob $b$, with $a>b$, yet Alice is not more
patient than Bob.

Economists have long been interested in the time preferences of agents (von
Böhm-Bawerk (1891), Fisher (1930)). Many–Koopmans (1960), Burness (1973),
Burness (1973), and Nachman (1975)–have explored patience (impatience) and how
agents evaluate consumption streams, or at least ought to. We ignore these
important issues. Instead, we merely specify that agents’ utility streams over
per-period rewards are additively separable over time, and that the agents are
impatient in the sense that they weigh these utilities according to a
decreasing discount function.

Two works are especially close to this one. The first is the working-paper
version Chambers and Echenique (2018). There, the authors provide a robust
ranking of reward streams, where one dominates another if it has a higher
discounted sum for all exponential discounters. The first part of this paper,
instead, provides a robust ranking of reward streams, where the superiority is
required to hold for all discounters rather than exponential ones.

The second similar work is Quah and Strulovici (2013), who explore the effect
of an agent’s discount rate on her behavior in stopping problems. They
formulate a “more patient than” partial order. In this conception, Alice is
more patient than Bob if $\alpha_{t}/\beta_{t}$ is increasing in $t$. They
show that this is equivalent to Bob preferring a later prize to an earlier one
implying Alice does too. In general, the patience order in this paper is
strictly stronger than Quah and Strulovici’s, though in Proposition 3.6, we
show that when there are just two periods the relations are equivalent. For
three or more periods; however, they are not, and, in fact, a more patient
Alice in Quah and Strulovici’s sense may suffer a greater normalized loss from
moving from a superior to an inferior stream than Bob.

##  2 Dominance of Sequences

Throughout the paper, we make the standing assumption that all infinite
sequences under consideration are absolutely summable. For
$T\in\mathbb{N}\cup\left\\{\infty\right\\}$, let
$x=\left\\{x_{t}\right\\}_{t=1}^{T}$ and $y=\left\\{y_{t}\right\\}_{t=1}^{T}$
be two prize sequences (in utils). We define

$\mathcal{T}\coloneqq\left\\{t\in\mathbb{N}\colon t\leq T\right\\}\text{.}$

A sequence, $\beta$, is positive if $\beta_{t}\geq 0$ for all
$t\in\mathcal{T}$ and decreasing if $\beta_{t}\geq\beta_{t+1}$ for all
$t\in\mathcal{T}$. Here are three notions of superiority of prize sequences.

######  Definition 2.1.

Prize sequence $x$ dominates $y$ pointwise if $x_{t}\geq y_{t}$ for all
$t\in\mathcal{T}$.

######  Definition 2.2.

Prize sequence $x$ dominates $y$ if
$\sum_{t=1}^{p}x_{t}\geq\sum_{t=1}^{p}y_{t}$ for all $p\in\mathcal{T}$. We
write this $x\trianglerighteq y$.

Naturally, if $x$ dominates $y$ pointwise, $x$ dominates $y$, while the
converse is false.

######  Definition 2.3.

Prize sequence $x$ is superior to $y$ if
$\sum_{t=1}^{T}\beta_{t}x_{t}\geq\sum_{t=1}^{T}\beta_{t}y_{t}$ for any
positive decreasing sequence $\beta=\left\\{\beta_{t}\right\\}_{t=1}^{T}$.

We note the following lemma. Its proof, as well as all proofs omitted from the
main text, may be found in the Appendix (A).

######  Lemma 2.4.

If $x\trianglerighteq y$, there exists a $\tilde{x}$ that $x$ dominates
pointwise, and that satisfies
$\sum_{t=1}^{T}\tilde{x}_{t}=\sum_{t=1}^{T}y_{t}$ and
$\sum_{t=1}^{p}\tilde{x}_{t}\geq\sum_{t=1}^{p}y_{t}$ for all
$p\in\mathcal{T}$.

This lemma is unsurprising. If the sum of the elements of $x$ is strictly
higher than the sum of the elements of $y$, we can just remove excess amounts
from particular elements of $x$ in order to obtain a balance (equal sums),
while preserving the dominance of the partial sums. Next we state the
following lemma, which is a discrete version of the integration by parts
formula.

######  Lemma 2.5 (Abel’s Lemma).

Let $a$ and $b$ be two sequences of real numbers. For all $k\in\mathcal{T}$,
we define $A_{k}=\sum_{i=1}^{k}a_{i}$. Then, for $T\in\mathbb{N}$,

$\sum_{t=1}^{T}a_{t}b_{t}=\sum_{t=1}^{T-1}A_{t}(b_{t}-b_{t+1})+A_{T}b_{t}\text{,}$

and for $T=\infty$,

$\sum_{t=1}^{\infty}a_{t}b_{t}=\sum_{t=1}^{\infty}A_{t}(b_{t}-b_{t+1})\text{.}$

Using Abel’s lemma, we obtain our next result. This is almost exactly the
characterization of first-order stochastic dominance–once we normalize the
sums so that they sum to one–with the single exception that elements of
$\tilde{x}$ and $y$ may take negative values.

######  Lemma 2.6.

Let $\tilde{x}$ and $y$ be two sequences that satisfy
$\sum_{t=1}^{T}\tilde{x}_{t}=\sum_{t=1}^{T}y_{t}$ and
$\sum_{t=1}^{p}\tilde{x}_{t}\geq\sum_{t=1}^{p}y_{t}$ for all
$p\in\mathcal{T}$. Then, for any positive decreasing sequence $\beta$,
$\sum_{t=1}^{T}\beta_{t}\tilde{x}_{t}\geq\sum_{t=1}^{T}\beta_{t}y_{t}$.

With these lemmas in hand, we equate superiority with dominance.

######  Proposition 2.7.

$x$ is superior to $y$ if and only if $x$ dominates $y$.

###### Proof.

$\left(\Leftarrow\right)$ Let $x$ dominate $y$. Then, for any positive
decreasing sequence $\beta$,

$\sum_{t=1}^{T}\beta_{t}x_{t}\geq\sum_{t=1}^{T}\beta_{t}\tilde{x}_{t}\geq\sum_{t=1}^{T}\beta_{t}y_{t}\text{,}$

where the first inequality follows from Lemma 2.4 and the second from Lemma
2.6.

$\left(\Rightarrow\right)$ Suppose for the sake of contraposition that $x$
does not dominate $y$. Then, there exists some $p\in\mathbb{N}$ such that

$\sum_{t=1}^{p}x_{t}<\sum_{t=1}^{p}y_{t}\text{.}$

Then, let $\beta_{t}=1$ for all $t\in\left\\{1,\dots,p\right\\}$ and
$\beta_{t}=0$ for all $t\in\left\\{p+1,\dots,T\right\\}$. Consequently,

$\sum_{t=1}^{T}\beta_{t}x_{t}=\sum_{t=1}^{p}x_{t}<\sum_{t=1}^{p}y_{t}=\sum_{t=1}^{T}\beta_{t}y_{t}\text{.}$

and so $x$ is not superior to $y$. ∎

This proposition is more-or-less first-order stochastic dominance, which makes
sense due to the mononicity of $\beta$.

##  3 Relative Patience

Now that we have provided a ranking of prize streams, we can explore relative
patience. Henceforth, we specify that any discounting sequence, $\beta$,
besides being absolutely summable and decreasing, is such that $\beta_{t}>0$
for any $t\in\mathcal{T}$, i.e., is strictly positive.

A first pass at defining what it means for one agent to be more patient than
another is the following.

######  Definition 3.1.

Alice, who discounts according to $\alpha$, is more serene than Bob, who
discounts according to $\beta$, if for any prize sequences $x$ and $y$ with
$x\trianglerighteq y$ and $\sum_{t=1}^{T}x_{t}=\sum_{t=1}^{T}y_{t}$,

$\sum_{t=1}^{T}\alpha_{t}x_{t}-\sum_{t=1}^{T}\alpha_{t}y_{t}\leq\sum_{t=1}^{T}\beta_{t}x_{t}-\sum_{t=1}^{T}\beta_{t}y_{t}\text{.}$

We can characterize relative serenity in terms of the agents’ discounting
behavior.

######  Proposition 3.2.

Alice is more serene than Bob if and only if $\beta-\alpha$ is a decreasing
sequence with $\beta_{t}\geq\alpha_{t}$ for all $t\in\mathcal{T}$.

###### Proof.

Obviously, for $x$ and $y$,

$\sum_{t=1}^{T}\alpha_{t}x_{t}-\sum_{t=1}^{T}\alpha_{t}y_{t}\leq\sum_{t=1}^{T}\beta_{t}x_{t}-\sum_{t=1}^{T}\beta_{t}y_{t}\
\Leftrightarrow\
\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)y_{t}\leq\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)x_{t}\text{.}$

$\left(\Leftarrow\right)$ Suppose $\beta-\alpha$ is a decreasing sequence with
$\beta_{t}\geq\alpha_{t}$ for all $t\in\mathcal{T}$. Then, as
$x\trianglerighteq y$,

$\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)x_{t}\geq\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)y_{t}\text{.}$

$\left(\Rightarrow\right)$ Suppose first for the sake of contraposition that
$\beta-\alpha$ is not a decreasing sequence. Then there exist
$k,k+1\in\mathcal{T}$ such that
$\beta_{k}-\alpha_{k}<\beta_{k+1}-\alpha_{k+1}$. Define $x_{t}=0$ for all
$t\neq k$ and $x_{k}=1$, and define $y_{t}=0$ for all $t\neq k+1$ and
$y_{k+1}=1$. By construction $x\trianglerighteq y$, but

$\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)x_{t}=\beta_{k}-\alpha_{k}<\beta_{k+1}-\alpha_{k+1}=\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)y_{t}\text{,}$

and so Alice is not more serene than Bob.

Finally, suppose for the sake of contraposition that $\beta-\alpha$ is a
decreasing sequence, but that there exists $k\in\mathcal{T}$ for which
$\beta_{k}-\alpha_{k}<0$. But then let $x_{t}=0$ for all $t\in\mathcal{T}$,
$y_{k}=-1$ and $y_{t}=0$ for all $t\in\mathcal{T}\setminus\left\\{k\right\\}$.
So,

$\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)x_{t}=0<\alpha_{k}-\beta_{k}=\sum_{t=1}^{T}\left(\beta_{t}-\alpha_{t}\right)y_{t}\text{,}$

and so Alice is not more serene than Bob. ∎

When Alice and Bob are both exponential discounters, i.e.,
$\alpha_{t}=a^{t-1}$ and $\beta_{t}=b^{t-1}$, $\beta_{2}\geq\alpha_{2}$ if and
only if $b\geq a$. However, $\beta_{1}-\alpha_{1}=0$, so $\beta-\alpha$ is
decreasing only if $a\geq b$. Accordingly,

######  Remark 3.3.

If Alice and Bob are exponential discounters, Alice is more serene than Bob if
and only if their discount factors are the same.

More generally,

######  Corollary 3.4.

Let $\alpha_{1}=\beta_{1}$. Then, Alice is more serene than Bob if and only if
$\alpha=\beta$.

This corollary is easy: we need $\beta-\alpha$ to be decreasing, but the
difference to also be weakly positive. As there is nowhere to go but down and
the difference is starting at $0$, it must remain at $0$. Proposition 3.2 and
Corollary 3.4 reveal that this definition is not terribly satisfactory. Can we
improve upon it?

Upon reflecting the definition of “more serene than,” we see that it could be
viewed as a bit of an “apples to oranges” comparison:
$\sum_{t=1}^{T}\alpha_{t}y_{t}$ and $\sum_{t=1}^{T}\beta_{t}y_{t}$ need not be
the same, so perhaps the proper definition should normalize things. In any
case, let us take a second pass at defining comparative patience.

If both $x$ and $y$ are positive sequences, $\sum_{t=1}^{T}\alpha_{t}y_{t}\neq
0$, and $\sum_{t=1}^{T}\beta_{t}y_{t}\neq 0$, we can normalize Alice’s payoff
difference by

$\frac{\sum_{t=1}^{T}\beta_{t}y_{t}}{\sum_{t=1}^{T}\alpha_{t}y_{t}}\text{.}$

Doing so, we note that

$\frac{\sum_{t=1}^{T}\beta_{t}y_{t}}{\sum_{t=1}^{T}\alpha_{t}y_{t}}\left(\sum_{t=1}^{T}\alpha_{t}x_{t}-\sum_{t=1}^{T}\alpha_{t}y_{t}\right)\leq\sum_{t=1}^{T}\beta_{t}x_{t}-\sum_{t=1}^{T}\beta_{t}y_{t}\quad\Leftrightarrow\quad\frac{\sum_{t=1}^{T}\alpha_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}y_{t}}\leq\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\beta_{t}y_{t}}\text{.}$

######  Definition 3.5.

Alice, who discounts according to $\alpha$, is more patient than Bob, who
discounts according to $\beta$, if for any positive prize sequences $x$ and
$y$ with $x\trianglerighteq y$ and $\sum_{t=1}^{T}x_{t}=\sum_{t=1}^{T}y_{t}$,
such that $\sum_{t=1}^{T}\alpha_{t}y_{t}\neq 0$ and
$\sum_{t=1}^{T}\beta_{t}y_{t}\neq 0$,

$\frac{\sum_{t=1}^{T}\alpha_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}y_{t}}\leq\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\beta_{t}y_{t}}\text{.}$

Of course, as $x\trianglerighteq y$, $\sum_{t=1}^{T}\alpha_{t}y_{t}\neq 0$ and
$\sum_{t=1}^{T}\beta_{t}y_{t}\neq 0$ imply $\sum_{t=1}^{T}\alpha_{t}x_{t}\neq
0$ and $\sum_{t=1}^{T}\beta_{t}x_{t}\neq 0$. Furthermore,

$\frac{\sum_{t=1}^{\infty}\alpha_{t}x_{t}}{\sum_{t=1}^{\infty}\alpha_{t}y_{t}}\leq\frac{\sum_{t=1}^{\infty}\beta_{t}x_{t}}{\sum_{t=1}^{\infty}\beta_{t}y_{t}}\quad\Leftrightarrow\quad\frac{\sum_{t=1}^{\infty}\beta_{t}y_{t}}{\sum_{t=1}^{\infty}\alpha_{t}y_{t}}\leq\frac{\sum_{t=1}^{\infty}\beta_{t}x_{t}}{\sum_{t=1}^{\infty}\alpha_{t}x_{t}}\text{.}$

When $T=2$, there is a nice characterization of patience in terms of the
discount factors.

######  Proposition 3.6.

If $T=2$, Alice is more patient than Bob if and only if
$\frac{\alpha_{1}}{\beta_{1}}\leq\frac{\alpha_{2}}{\beta_{2}}$.

As we noted in the introduction, this is precisely Quah and Strulovici
(2013)’s definition of comparative patience. This makes sense, their
definition of comparative patience–that if Bob prefers a later prize to an
earlier one, so too does a more patient Alice–inherently (as the discounting
is free) makes theirs a statement about two-period problems.

Alas, it is not the case that

$\frac{\alpha_{1}}{\beta_{1}}\leq\frac{\alpha_{2}}{\beta_{2}}\leq\frac{\alpha_{3}}{\beta_{3}}$

implies Alice is more patient than Bob when $T=3$. To see this, let

$\alpha_{1}=\frac{1}{2},\ \alpha_{2}=\frac{12}{25},\
\alpha_{3}=\frac{91}{250},\ \beta_{1}=1,\ \beta_{2}=\frac{2}{3},\text{ and
}\beta_{3}=\frac{1}{2}\text{.}$

We have

$\frac{1}{2}<\frac{18}{25}<\frac{91}{125}\text{,}$

so $\alpha/\beta$ is monotone. However, if

$x_{1}=y_{1}=1,\ x_{2}=y_{3}=\frac{3}{2},\text{ and }x_{3}=y_{2}=1\text{,}$
$\frac{\alpha_{1}x_{1}+\alpha_{2}x_{2}+\alpha_{3}x_{3}}{\alpha_{1}y_{1}+\alpha_{2}y_{2}+\alpha_{3}y_{3}}-\frac{\beta_{1}x_{1}+\beta_{2}x_{2}+\beta_{3}x_{3}}{\beta_{1}y_{1}+\beta_{2}y_{2}+\beta_{3}y_{3}}>.0035>0\text{,}$

so Alice is not more patient than Bob.

The following observation will be useful. For two positive sequences $x$ and
$z$, $z$ is a binary deterioration of $x$ if there exist
$t_{1},t_{2}\in\mathcal{T}$ (with $t_{1}<t_{2}$), and $\eta>0$ such that
$x_{t}=z_{t}$ for all $t\neq t_{1}$ and $t\neq t_{2}$,
$z_{t_{1}}=x_{t_{1}}-\eta$, and $z_{t_{2}}=x_{t_{2}}+\eta$.

######  Lemma 3.7.

Let $\sum_{t=1}^{T}x_{t}=\sum_{t=1}^{T}y_{t}$. $x\trianglerighteq y$ if and
only if there exists a sequence of binary deteriorations
$x,z^{1},\dots,z^{k},y$.

###### Proof.

$\left(\Leftarrow\right)$ This direction is immediate: by construction,

$x\trianglerighteq z^{1}\trianglerighteq\dots\trianglerighteq
z^{k}\trianglerighteq y\text{.}$

$\left(\Rightarrow\right)$ If $x$ and $y$ differ at only finitely many values
of $t$, the result follows from the discussion on pages 630 and 631 of
Fishburn (1982). It is extremely likely that this result is known in the
infinite case, but we prove it anyway, in Appendix A.4 ∎

We define

$\bar{\mathcal{T}}\coloneqq\begin{cases}t\in\mathbb{N}\colon t\leq
T-1,\quad&\text{if}\quad T\in\mathbb{N}\\\ \mathbb{N},\quad&\text{if}\quad
T=\infty\text{.}\end{cases}$

Here is the main characterization result.

######  Theorem 3.8.

Alice is more patient than Bob if and only if

$\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\beta_{1}}{\alpha_{1}}\text{.}$

This condition is strictly stronger than monotonicity of
$\alpha_{t}/\beta_{t}$, and, as the three-period example above reveals, is
equivalent to such monotonicity if and only if there are two periods.

######  Corollary 3.9.

If Alice is more patient than Bob, $\alpha_{t}/\beta_{t}$ is increasing in
$t$.

It turns out that if Alice is more patient than Bob, and Bob is more patient
than Michelle, Alice is more patient than Michelle. Viz.,

######  Proposition 3.10.

Patience is transitive.

The two most common discounting sequences are exponential discounting and
quasi-hyperbolic discounting (Laibson (1997)). Observe that if Alice’s and
Bob’s discounting rules are both in one of these classes,
$\alpha_{1}=\beta_{1}=1$. That inspires this corollary:

######  Corollary 3.11.

Let $\alpha_{1}=\beta_{1}$. Then, Alice is more patient than Bob if and only
if $\beta_{t}-\beta_{t+1}\geq\alpha_{t}-\alpha_{t+1}$ for all
$t\in\bar{\mathcal{T}}$.

Interestingly, we see that for $T\geq 3$, it could be that Alice and Bob are
both exponential discounters, with $\alpha_{t}=a^{t-1}$, $\beta_{t}=b^{t-1}$,
and $1>a>b>0$, yet Alice is not more patient than Bob. A similar
incomparibility is true when both are quasi-hyperbolic discounters.
Nevertheless, we observe

######  Corollary 3.12.

Let Alice and Bob be exponential discounters and $T\in\mathbb{N}$. For any
discounting rule for Bob, $\beta_{t}=b^{t-1}$ ($b\in\left(0,1\right)$), there
exists an $\eta>0$ such that if $\alpha_{t}=a^{t-1}$ for
$a\in\left(1-\eta,1\right)$, Alice is more patient than Bob.

###### Proof.

$b^{t-1}-b^{t}$ is strictly decreasing in $t$ and $\beta^{T-1}-\beta^{T}>0$.
Moreover $a^{t-1}-a^{t}$ is continuous in $a$ for all $t\in\bar{\mathcal{T}}$
and equals $0$ for all $t\in\bar{\mathcal{T}}$ when $a=1$. Thus, the result
follows from the intermediate-value theorem and Corollary 3.11. ∎

On the other hand, the infinite-horizon environment is troubling.

######  Corollary 3.13.

Let Alice and Bob be exponential discounters and $T=\infty$. Alice is more
patient than Bob if and only if they have the same discount factor.

###### Proof.

Let the discount factor for Alice $a\in\left(0,1\right)$ be strictly greater
than that for Bob, $b$. Observe that for any $t\in\mathbb{N}$,

$\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}=\left(\frac{b}{a}\right)^{t-1}\frac{1-b}{1-a}\text{.}$

This expression is strictly decreasing in $t$ and equals $0$ when
$t\to\infty$. Thus, there exists a $t^{\dagger}\in\mathbb{N}$ such that this
expression is strictly less than $1$ for all $t\geq t^{\dagger}$. We conclude
that Alice is not more patient than Bob. ∎

Finally, as the patience relation is antisymmetric when
$\beta_{1}=\alpha_{1}$,

######  Corollary 3.14.

If $\beta_{1}=\alpha_{1}$, patience is a partial order on a set of agents.

## References

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##  Appendix A Omitted Proofs

###  A.1 Lemma 2.4 Proof

###### Proof.

First, suppose $T=\infty$. The sequence
$\left\\{\sum_{i=1}^{t}\left(x_{i}-y_{i}\right)\right\\}_{t=1}^{\infty}$
converges and is, therefore, bounded. Consequently, for any $t\in\mathbb{N}$,

$s_{t}\coloneqq\inf_{k\geq t}\sum_{i=1}^{k}\left(x_{i}-y_{i}\right)$

is well-defined. Then, for any $t\in\mathbb{N}$, we define
$z_{t}=s_{t}-s_{t-1}$, where $s_{0}=0$. By construction $z_{t}\geq 0$ for all
$t\in\mathbb{N}$. We define $\tilde{x}_{t}=x_{t}-z_{t}$ for all
$t\in\mathbb{N}$ and observe that, by construction,

$\sum_{t=1}^{p}\tilde{x}_{t}=\sum_{t=1}^{p}\left(x_{t}-z_{t}\right)=\sum_{t=1}^{p}x_{t}-s_{p}=\sum_{t=1}^{p}x_{t}-\inf_{k\geq
p}\sum_{i=1}^{k}\left(x_{i}-y_{i}\right)\geq\sum_{t=1}^{p}y_{t}\text{,}$

for all $p\in\mathbb{N}$. Moreover,

$\sum_{t=1}^{\infty}\tilde{x}_{t}=\lim_{p\to\infty}\sum_{t=1}^{p}\tilde{x}_{t}=\sum_{t=1}^{\infty}x_{t}-\liminf_{p\to\infty}\sum_{i=1}^{p}\left(x_{i}-y_{i}\right)=\sum_{t=1}^{\infty}y_{t}\text{,}$

as desired.

If $T\in\mathbb{N}$ the proof is virtually identical, mutatis mutandis.∎

###  A.2 Lemma 2.6 Proof

###### Proof.

Suppose $T=\infty$. By Lemma 2.5,

$\sum_{t=1}^{\infty}\beta_{t}\tilde{x}_{t}-\sum_{t=1}^{\infty}\beta_{t}y_{t}=\sum_{t=1}^{\infty}\underbrace{\left(\beta_{t}-\beta_{t+1}\right)}_{\geq
0}\underbrace{\sum_{i=1}^{t}(\tilde{x}_{i}-y_{i})}_{\geq 0}\geq 0\text{.}$

Now suppose $T\in\mathbb{N}$. By Lemma 2.5,

$\sum_{t=1}^{T}\beta_{t}\tilde{x}_{t}-\sum_{t=1}^{T}\beta_{t}y_{t}=\sum_{t=1}^{T-1}\underbrace{\left(\beta_{t}-\beta_{t+1}\right)}_{\geq
0}\underbrace{\sum_{i=1}^{t}(\tilde{x}_{i}-y_{i})}_{\geq
0}+\beta_{T}\underbrace{\sum_{i=1}^{T}\left(\tilde{x}_{i}-y_{i}\right)}_{=0}\geq
0\text{,}$

where the last term equals $0$ by assumption.∎

###  A.3 Proposition 3.6 Proof

###### Proof.

$\left(\Leftarrow\right)$ Suppose
$\frac{\alpha_{1}}{\beta_{1}}\leq\frac{\alpha_{2}}{\beta_{2}}$. We have

$\frac{\beta_{1}x_{1}+\beta_{2}x_{2}}{\beta_{1}y_{1}+\beta_{2}y_{2}}=\frac{\beta_{1}x_{1}+\beta_{2}x_{2}}{\beta_{1}y_{1}+\beta_{2}(x_{2}+x_{1}-y_{1})}\text{.}$

The derivative of this with respect to $\beta_{1}$ simplifies to

$\frac{\beta_{2}\left(x_{1}+x_{2}\right)\left(x_{1}-y_{1}\right)}{\left(\beta_{1}y_{1}+\beta_{2}(x_{2}+x_{1}-y_{1})\right)^{2}}\geq
0\text{.}$

Consequently,

$\frac{\beta_{1}x_{1}+\beta_{2}x_{2}}{\beta_{1}y_{1}+\beta_{2}y_{2}}\geq\frac{x_{1}\frac{\beta_{2}}{\alpha_{2}}\alpha_{1}+\beta_{2}x_{2}}{y_{1}\frac{\beta_{2}}{\alpha_{2}}\alpha_{1}+\beta_{2}y_{2}}=\frac{\alpha_{1}x_{1}+\alpha_{2}x_{2}}{\alpha_{1}y_{1}+\alpha_{2}y_{2}}\text{.}$

$\left(\Rightarrow\right)$ Suppose for the sake of contraposition
$\frac{\alpha_{1}}{\beta_{1}}>\frac{\alpha_{2}}{\beta_{2}}$. Let
$x_{1}=y_{2}=1$ and $x_{2}=y_{1}=0$. Thus,

$\frac{\alpha_{1}x_{1}+\alpha_{2}x_{2}}{\alpha_{1}y_{1}+\alpha_{2}y_{2}}=\frac{\alpha_{1}}{\alpha_{2}}>\frac{\beta_{1}}{\beta_{2}}=\frac{\beta_{1}x_{1}+\beta_{2}x_{2}}{\beta_{1}y_{1}+\beta_{2}y_{2}}\text{,}$

so Alice is not more patient than Bob.∎

###  A.4 Lemma 3.7 Proof

###### Proof.

Suppose $x\trianglerighteq y$ and $x\neq y$ (if $x=y$, we are done). Let

$t_{1}\coloneqq\min\left\\{t\in\mathcal{T}\colon x_{t}\neq
y_{t}\right\\}\text{.}$

As $x\trianglerighteq y$, $x_{t_{1}}>y_{t_{1}}$. Now let

$t_{1}^{\prime}\coloneqq\min\left\\{t\in\mathcal{T}\colon
x_{t}<y_{t}\right\\}\text{.}$

As $x\trianglerighteq y$, $t_{1}^{\prime}>t_{1}$. There are two possibilities:
either $x_{t_{1}}-y_{t_{1}}\geq y_{t_{1}^{\prime}}-x_{t_{1}^{\prime}}$, or
$x_{t_{1}}-y_{t_{1}}<y_{t_{1}^{\prime}}-x_{t_{1}^{\prime}}$. In either case,
we define $z^{1}_{t}=x_{t}$ for all $t\neq t_{1},t_{1}^{\prime}$,
$z^{1}_{t_{1}}=x_{t_{1}}-\eta$ and
$z^{1}_{t_{1}^{\prime}}=x_{t_{1}^{\prime}}+\eta$; where
$\eta=y_{t_{1}^{\prime}}-x_{t_{1}^{\prime}}$ in the first case and
$\eta=x_{t_{1}}-y_{t_{1}}$ in the second.

By construction $z^{1}$ is a binary deterioration of $x$. We then construct
$z^{2}$ from $z^{1}$–and, subsequently, each $z^{i+1}$ from $z^{i}$ ($k-1\geq
i\geq 1$)–in the same manner. If $T\in\mathbb{N}$, we may conclude the result.
Let $T=\infty$. Suppose for the sake of contradiction that there exists
$\eta>0$ and a $t\in\mathbb{N}$ such that $z_{t}^{j}-y_{t}>\eta$ for all
$j\in\mathbb{N}$. By the construction of the sequence of $z$s, there exists
$k\in\mathbb{N}$ such that $z_{i}^{k}=y_{i}$ for all $i<k$. Moreover, by the
construction of the sequence of $z$s, it must be that
$z_{t}^{k}-y_{t}-\sum_{i=t+1}^{\infty}\left(y_{i}-z_{i}^{k}\right)>\eta>0$.
However, this contradicts the fact that
$\sum_{t=1}^{\infty}\left(z^{k}_{t}-y_{t}\right)=0$. ∎

###  A.5 Theorem 3.8 Proof

###### Proof.

$\left(\Rightarrow\right)$ Suppose for the sake of contraposition that

$\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}<\frac{\beta_{1}}{\alpha_{1}}\text{.}$

This implies that there exists a $k\in\bar{\mathcal{T}}$ such that

$\frac{\beta_{k}-\beta_{k+1}}{\alpha_{k}-\alpha_{k+1}}<\frac{\beta_{1}}{\alpha_{1}}\text{.}$

If $k=1$, this implies $\alpha_{1}\beta_{2}>\alpha_{2}\beta_{1}$, in which
case, following the proof of Proposition 3.6, we are done. Consequently,
without loss of generality we specify $k=2$, i.e.,

$\frac{\beta_{2}-\beta_{3}}{\alpha_{2}-\alpha_{3}}<\frac{\beta_{1}}{\alpha_{1}}\text{.}$

Moreover, if

$\frac{\beta_{2}-\beta_{3}}{\alpha_{2}-\alpha_{3}}<\frac{\beta_{2}}{\alpha_{2}}\text{,}$

again following the proof of Proposition 3.6 we are done. Accordingly, we
assume

$\frac{\beta_{2}}{\alpha_{2}}\leq\frac{\beta_{2}-\beta_{3}}{\alpha_{2}-\alpha_{3}}\text{,}$

which holds if and only if $\alpha_{3}\beta_{2}\geq\alpha_{2}\beta_{3}$. In
sum, we have

$\frac{\beta_{3}}{\alpha_{3}}\leq\frac{\beta_{2}}{\alpha_{2}}\leq\frac{\beta_{2}-\beta_{3}}{\alpha_{2}-\alpha_{3}}<\frac{\beta_{1}}{\alpha_{1}}\leq\frac{\beta_{1}-\beta_{2}}{\alpha_{1}-\alpha_{2}}\text{.}$

Let $x_{1}=y_{1}=1$, $x_{3}=y_{2}=0$, and $x_{2}=y_{3}=\eta>0$. Then,

$\begin{split}\frac{\alpha_{1}x_{1}+\alpha_{2}x_{2}+\alpha_{3}x_{3}}{\alpha_{1}y_{1}+\alpha_{2}y_{2}+\alpha_{3}y_{3}}-\frac{\beta_{1}x_{1}+\beta_{2}x_{2}+\beta_{3}x_{3}}{\beta_{1}y_{1}+\beta_{2}y_{2}+\beta_{3}y_{3}}&>0\quad\Leftrightarrow\\\
\frac{\alpha_{1}+\alpha_{2}\eta}{\alpha_{1}+\alpha_{3}\eta}-\frac{\beta_{1}+\beta_{2}\eta}{\beta_{1}+\beta_{3}\eta}&>0\quad\Leftrightarrow\\\
\eta\left(\beta_{1}\left(\alpha_{2}-\alpha_{3}\right)-\alpha_{1}\left(\beta_{2}-\beta_{3}\right)+\eta\left(\alpha_{2}\beta_{3}-\alpha_{3}\beta_{2}\right)\right)&>0\end{split}$

If $\alpha_{2}\beta_{3}=\alpha_{3}\beta_{2}$, the last inequality holds for
any strictly positive $\eta$. Suppose instead
$\alpha_{2}\beta_{3}<\alpha_{3}\beta_{2}$. Then, the last inequality holds for
all

$\eta\in\left(0,\frac{\beta_{1}\left(\alpha_{2}-\alpha_{3}\right)-\alpha_{1}\left(\beta_{2}-\beta_{3}\right)}{\alpha_{3}\beta_{2}-\alpha_{2}\beta_{3}}\right)\text{,}$

which interval is nonempty by assumption.

$\left(\Leftarrow\right)$ Now let

$\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\beta_{1}}{\alpha_{1}}\text{.}$

We want to show that for any positive prize sequences $x$ and $y$ with
$x\trianglerighteq y$ and $\sum_{t=1}^{T}x_{t}=\sum_{t=1}^{T}y_{t}$, such that
$\sum_{t=1}^{T}\alpha_{t}y_{t}\neq 0$ and $\sum_{t=1}^{T}\beta_{t}y_{t}\neq
0$,

$\frac{\sum_{t=1}^{T}\alpha_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}y_{t}}\leq\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\beta_{t}y_{t}}\text{.}$

Recall also that, by assumption, $\alpha_{t},\beta_{t}>0$ for
$t\in\mathcal{T}$.

Lemma 3.7 means that it suffices to show that for any $x$ with
$\sum_{t=1}^{T}\alpha_{t}x_{t}\neq 0$ and $\sum_{t=1}^{T}\beta_{t}x_{t}\neq 0$
and any binary deterioration of $x$, $z$, with
$\sum_{t=1}^{T}\alpha_{t}z_{t}\neq 0$ and $\sum_{t=1}^{T}\beta_{t}z_{t}\neq
0$,

$\frac{\sum_{t=1}^{T}\beta_{t}z_{t}}{\sum_{t=1}^{T}\alpha_{t}z_{t}}\leq\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}x_{t}}\text{.}$

Now take

$\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}x_{t}}\text{,}$

with $\sum_{t=1}^{T}\alpha_{t}x_{t}\neq 0$ and
$\sum_{t=1}^{T}\beta_{t}x_{t}\neq 0$ and an arbitrary binary deterioration of
$x$, $z$, with $\sum_{t=1}^{T}\alpha_{t}z_{t}\neq 0$ and
$\sum_{t=1}^{T}\beta_{t}z_{t}\neq 0$. That is, take some $x_{k}$ and subtract
$\eta$, add $\eta$ to some $x_{s}$ with $s,k\in\bar{\mathcal{T}}$ and $s>k$.
We have

$\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}x_{t}}-\frac{\sum_{t=1}^{T}\beta_{t}z_{t}}{\sum_{t=1}^{T}\alpha_{t}z_{t}}=\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}x_{t}}-\frac{\sum_{t=1}^{T}\beta_{t}x_{t}-\eta(\beta_{k}-\beta_{s})}{\sum_{t=1}^{T}\alpha_{t}x_{t}-\eta(\alpha_{k}-\alpha_{s})}\text{.}$

The right-hand side of this is positive if and only if

$\frac{\beta_{k}-\beta_{s}}{\alpha_{k}-\alpha_{s}}\geq\frac{\sum_{t=1}^{T}\beta_{t}x_{t}}{\sum_{t=1}^{T}\alpha_{t}x_{t}}\text{.}$

Observe that, appealing to the mediant inequality, it suffices to show

$\inf_{k,s\in\bar{\mathcal{T}},s>k}\frac{\beta_{k}-\beta_{s}}{\alpha_{k}-\alpha_{s}}\geq\sup_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}}{\alpha_{t}}\text{.}$

######  Claim A.1.

For any $k,s\in\bar{\mathcal{T}}$ with $s>k$

$\frac{\beta_{k}-\beta_{s}}{\alpha_{k}-\alpha_{s}}\geq\min\left\\{\frac{\beta_{k}-\beta_{k+1}}{\alpha_{k}-\alpha_{k+1}},\frac{\beta_{k+1}-\beta_{k+2}}{\alpha_{k+1}-\alpha_{k+2}},\dots,\frac{\beta_{s-1}-\beta_{s}}{\alpha_{s-1}-\alpha_{s}}\right\\}\text{.}$

###### Proof.

As

$\beta_{k}-\beta_{s}=\left(\beta_{k}-\beta_{k+1}\right)+\left(\beta_{k+1}-\beta_{k+2}\right)+\cdots+\left(\beta_{s-1}-\beta_{s}\right)\text{,}$

and

$\alpha_{k}-\alpha_{s}=\left(\alpha_{k}-\alpha_{k+1}\right)+\left(\alpha_{k+1}-\alpha_{k+2}\right)+\cdots+\left(\alpha_{s-1}-\alpha_{s}\right)\text{,}$

this follows from the mediant inequality. ∎

######  Claim A.2.

$\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\beta_{1}}{\alpha_{1}}\quad\Rightarrow\quad\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\sup_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}}{\alpha_{t}}$

###### Proof.

For any $k\in\bar{\mathcal{T}}$,

$\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\beta_{1}}{\alpha_{1}}\
\Rightarrow\
\frac{\beta_{1}-\beta_{k}}{\alpha_{1}-\alpha_{k}}\geq\frac{\beta_{1}}{\alpha_{1}}\
\Rightarrow\
\frac{\beta_{1}}{\alpha_{1}}\geq\frac{\beta_{k}}{\alpha_{k}}\text{,}$

where the first implication uses Claim A.1. ∎

From Claim A.1 we have

$\begin{split}\inf_{k,s\in\bar{\mathcal{T}},s>k}\frac{\beta_{k}-\beta_{s}}{\alpha_{k}-\alpha_{s}}&\geq\inf_{k,s\in\bar{\mathcal{T}},s>k}\min\left\\{\frac{\beta_{k}-\beta_{k+1}}{\alpha_{k}-\alpha_{k+1}},\frac{\beta_{k+1}-\beta_{k+2}}{\alpha_{k+1}-\alpha_{k+2}},\dots,\frac{\beta_{s-1}-\beta_{s}}{\alpha_{s-1}-\alpha_{s}}\right\\}\\\
&=\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\text{.}\end{split}$

Finally, combining this with Claim A.2, we have

$\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\beta_{1}}{\alpha_{1}}\quad\Rightarrow\quad\inf_{k,s\in\bar{\mathcal{T}},s>k}\frac{\beta_{k}-\beta_{s}}{\alpha_{k}-\alpha_{s}}\geq\sup_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}}{\alpha_{t}}\text{,}$

and we conclude the theorem.∎

###  A.6 Proposition 3.10 Proof

###### Proof.

Let Michelle discount according to $\gamma$, Alice and Bob according to
$\alpha$ and $\beta$. Let

$\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\beta_{1}}{\alpha_{1}}\quad\text{and}\quad\inf_{t\in\bar{\mathcal{T}}}\frac{\gamma_{t}-\gamma_{t+1}}{\beta_{t}-\beta_{t+1}}\geq\frac{\gamma_{1}}{\beta_{1}}\text{.}$

Then, suppose for the sake of contradiction that

$\frac{\gamma_{1}}{\alpha_{1}}>\inf_{t\in\bar{\mathcal{T}}}\frac{\gamma_{t}-\gamma_{t+1}}{\alpha_{t}-\alpha_{t+1}}\text{.}$

Consequently

$\inf_{t\in\bar{\mathcal{T}}}\frac{\gamma_{t}-\gamma_{t+1}}{\beta_{t}-\beta_{t+1}}\geq\frac{\gamma_{1}}{\beta_{1}}>\frac{\alpha_{1}}{\beta_{1}}\inf_{t\in\bar{\mathcal{T}}}\frac{\gamma_{t}-\gamma_{t+1}}{\alpha_{t}-\alpha_{t+1}}\geq\frac{\inf_{t\in\bar{\mathcal{T}}}\frac{\gamma_{t}-\gamma_{t+1}}{\alpha_{t}-\alpha_{t+1}}}{\inf_{t\in\bar{\mathcal{T}}}\frac{\beta_{t}-\beta_{t+1}}{\alpha_{t}-\alpha_{t+1}}}\text{,}$

a contradiction. ∎