A simple framework for the axiomatization of exponential and quasi-hyperbolic discounting

Abstract

The main goal of this paper is to investigate which normative requirements, or axioms, lead to exponential and quasi-hyperbolic forms of discounting. Exponential discounting has a well-established axiomatic foundation originally developed by Koopmans (Econometrica 28(2):287–309, 1960, 1972) and Koopmans et al. (Econometrica 32(1/2):82–100, 1964) with subsequent contributions by several other authors, including Bleichrodt et al. (J Math Psychol 52(6):341–347, 2008). The papers by Hayashi (J Econ Theory 112(2):343–352, 2003) and Olea and Strzalecki (Q J Econ 129(3):1449–1499, 2014) axiomatize quasi-hyperbolic discounting. The main contribution of this paper is to provide an alternative foundation for exponential and quasi-hyperbolic discounting, with simple, transparent axioms and relatively straightforward proofs. Using techniques by Fishburn (The foundations of expected utility. Reidel Publishing Co, Dordrecht, 1982) and Harvey (Manag Sci 32(9):1123–1139, 1986), we show that Anscombe and Aumann’s (Ann Math Stat 34(1):199–205, 1963) version of Subjective Expected Utility theory can be readily adapted to axiomatize the aforementioned types of discounting, in both finite and infinite horizon settings.

Appendix: Proof of Theorem 2

Proof

Necessity of the axioms is straightforward to verify. Therefore, we will focus on the proof of sufficiency.

Step 1. Applying Theorem 1 of Fishburn (1982) to the mixture set X, it follows from Axioms I1, I3, I4 that there exists a mixture linear utility function u preserving the order on X (unique up to positive affine transformations). Normalize u so that $u(x_0)=0$. Note that by non-triviality $u(x_0)$ is in the interior of the non-degenerate interval u(X).

Convert streams into their utility vectors by replacing the outcomes in each period by their utility values. Define the following order: $(v_1, v_2, \ldots ) \succcurlyeq ^* (u_1, u_2, \ldots ) \Leftrightarrow$ there exist $\mathbf{x}, \mathbf{y} \in X^{\infty }$ such that $\mathbf{x} \succcurlyeq \mathbf{y}$ and $u(x_t)=v_t$ and $u(y_t)=u_t$ for every t. This order is unambiguously defined because of the monotonicity assumption, i.e., if $x_i \sim x^{\prime }_i$ then $(x_1, \ldots , x_i, \ldots ) \sim (x_1, \ldots , x^{\prime }_i, \ldots )$.

The preference order $\succcurlyeq ^*$ inherits the properties of weak order, mixture independence and mixture continuity from $\succcurlyeq$. Note that $u(X)^{\infty }$ is a mixture set under the standard operation of taking convex combinations: if $\mathbf{v}, \mathbf{u} \in u(X)^{\infty }$ then

$$\begin{aligned} \mathbf{v}\lambda \mathbf{u}=\lambda \mathbf{v} + (1-\lambda ) \mathbf{u} \ \text {for every} \ \lambda \in (0,1). \end{aligned}$$

Therefore, by Theorem 1 of Fishburn (1982) we obtain a mixture linear representation $U :$ $u(X)^{\infty } \rightarrow \mathbb {R}$, where U is unique up to positive affine transformations.

Hence, $\mathbf{v} \succcurlyeq ^* \mathbf{u}$ if and only if $U(\mathbf{v}) \ge U(\mathbf{u})$.

Step 2. Normalize U so that $U(0, 0, \ldots )=U(\mathbf{0})=0$. Since 0 is in the interior of u(X), and since $U(\mathbf{v} \lambda \mathbf{0})=\lambda U(\mathbf{v})$ for any $\mathbf{v}\in \mathbb {R}^\infty$ and for every $\lambda \in (0, 1)$, we can assume that U is defined on $\mathbb {R}^{\infty }$.

Mixture linearity of U implies standard linearity of U on $\mathbb {R}^{\infty }$. To prove this, we need to show that $U(k \mathbf{v})= kU(\mathbf{v})$ for any k and $U(\mathbf{v}+\mathbf{u})=U(\mathbf{v})+U(\mathbf{u})$ for any $\mathbf{u}, \mathbf{v} \in \mathbb {R}^{\infty }$.

As $u(X)^{\infty }$ is a mixture set under the operation of taking convex combinations, $U(\mathbf{v}k\mathbf{0})=U(k\mathbf{v}+(1-k)\mathbf{0})=U(k \mathbf{v})=k U(\mathbf{v})$ for any $k \in (0, 1)$. If $k>1$ then $U(\mathbf{v})=U(\frac{k}{k}{} \mathbf{v})=\frac{1}{k}U(k\mathbf{v})$. Multiplying both parts of this equation by k, we obtain $U(k \mathbf{v})=k U(\mathbf{v})$ for all $k>1$. Therefore, $U(k \mathbf{v})= kU(\mathbf{v})$ for any $k>0$.

To prove that $U(\mathbf{v}+\mathbf{u})=U(\mathbf{v})+U(\mathbf{u})$, consider the mixture $\mathbf{v}\frac{1}{2}{} \mathbf{u}$. By mixture linearity of U we have:

$$\begin{aligned} U\left( \mathbf{v}\frac{1}{2}{} \mathbf{u}\right) =\frac{1}{2}U(\mathbf{v})+\frac{1}{2}U(\mathbf{u})=\frac{1}{2}\left( U(\mathbf{v})+U(\mathbf{u})\right) . \end{aligned}$$

(7)

On the other hand, $\mathbf{v}\frac{1}{2}{} \mathbf{u}=\frac{1}{2}\mathbf{v}+\frac{1}{2}{} \mathbf{u}=\frac{1}{2}(\mathbf{v}+\mathbf{u})$. Therefore,

$$\begin{aligned} U\left( \mathbf{v}\frac{1}{2}{} \mathbf{u}\right) =U\left( \frac{1}{2}(\mathbf{v}+\mathbf{u})\right) = \frac{1}{2} U(\mathbf{v}+\mathbf{u}) \end{aligned}$$

(8)

Comparing (7) and (8), we conclude that $U(\mathbf{v}+\mathbf{u})=U(\mathbf{v})+U(\mathbf{u})$.

Finally, note that

$$\begin{aligned} U(\mathbf{0})=U\left( \mathbf{v}+(\mathbf{-v})\right) =U(\mathbf{v})+U(\mathbf{-v})=0, \end{aligned}$$

hence $U(\mathbf{-v})=-U(\mathbf{v})$. Therefore, if $k<0$, then $U(k \mathbf{v})= -kU(\mathbf{-v})=kU(\mathbf{v})$.

For each T, consider the function $f :\mathbb {R}^T \rightarrow \mathbb {R}$ defined as follows:

$$\begin{aligned} f(v_1, \ldots , v_T)=U(v_1, \ldots , v_T, 0, 0, \ldots ). \end{aligned}$$

This function is linear on $\mathbb {R}^T$ and it satisfies $f(\mathbf{0})=0$; therefore,

$$\begin{aligned} f(v_1, \ldots , v_T)=\displaystyle \sum _{t=1}^T w_t^T v_t, \end{aligned}$$

where $\mathbf{w}^T=(w^T_1, \ldots , w^T_T)$. By monotonicity $w^T_t \ge 0$ for all $t \le T$.

Note that $w^T_t=U([1]_t)$, where $[1]_t$ is the vector with 1 in period t and 0 elsewhere. It follows that $w^{T}_t=w^{T^{\prime }}_t$ for any T and $T^{\prime }$. Hence, there is a vector $\mathbf{w} \in \mathbb {R}^\infty$ such that $U(v_1, \ldots , v_T, 0, 0, \ldots )=\sum \nolimits _{t=1}^{\infty } w_t v_t$ for any $(v_1, \ldots , v_T)\in \mathbb {R}^T$.

Recalling that $v_t=u(x_t)$, we obtain

$$\begin{aligned} U(u(x_1), \ldots , u(x_T), 0, 0, \ldots )=\sum _{t=1}^T w_t u(x_t)\quad \text { for all } \mathbf{x} \in X^{*}. \end{aligned}$$

Therefore, for every $\mathbf{x}, \mathbf{y} \in X^*$ we have $\mathbf{x} \succcurlyeq \mathbf{y}$ if and only if

$$\begin{aligned} \sum _{t=1}^T w_t u(x_t) \ge \sum _{t=1}^T w_t u(y_t). \end{aligned}$$

By slightly abusing the notation, re-define U so that:

$$\begin{aligned} U(x_1, \ldots , x_T, x_0, x_0, \ldots )=\sum _{t=1}^T w_t u(x_t) \quad \text { for all } \mathbf{x} \in X^{*}. \end{aligned}$$

Hence, $U(\mathbf{x})=\sum \nolimits _{t=1}^{\infty } w_t u(x_t)$ represents preferences on $X^*$.

Step 3. Next, we show that $U(x_1, x_2, \ldots )$ converges for any $(x_1, x_2, \ldots )$. Define $U_T:X^\infty \rightarrow {\mathbb R}$ as follows: $U_T(\mathbf{x}) = \sum \nolimits _{t=1}^T u_t(x_t)$, where $u_t(x_t)=w_t u(x_t)$. Consider the sequence of functions $U_1, U_2, \ldots , U_T, \ldots$ According to the Cauchy Criterion, a sequence of functions $U_T(\mathbf{x})$ defined on $X^\infty$ converges on $X^\infty$ if and only if for any $\varepsilon >0$ and any $\mathbf{x} \in X^\infty$ there exists $T \in \mathbb {N}$ such that $|U_N(\mathbf{x}) - U_M(\mathbf{x}) |< \varepsilon$ for any $N, M \ge T$.

Fix some $\mathbf{x}\in X^\infty$ and $\varepsilon >0$. Suppose that for some k it is possible to choose $x^+, x^-$ such that $[x^+]_k \succ [x_k]_k \succ [x^-]_k$. By Step 2 the preference $[x^+]_k \succ [x_k]_k \succ [x^-]_k$ implies that $w_k>0$. Therefore, as u is a continuous function, it is without loss of generality to assume that

$$\begin{aligned} u_k(x^+)-u_k(x_k)<\varepsilon /2 \text { and } u_k(x_k)-u_k(x^-)<\varepsilon /2. \end{aligned}$$

It follows that $u_k(x^+)-u_k(x^-)<\varepsilon ,$ or $u_k(x^-)-u_k(x^+)>-\varepsilon .$ By Axiom I6 there exist $T^+$ and $T^-$ satisfying $k\le \min \{T^-, T^+\}$ such that

$$\begin{aligned} \mathbf{x}_{k, N}^+ \succcurlyeq \mathbf{x} \succcurlyeq \mathbf{x}_{k, M}^-, \ \text {for all} \ N\ge T^+, \ M\ge T^-. \end{aligned}$$

Let $T^*=\max \{T^-, T^+\}$. It is necessary to demonstrate that $|U_N(\mathbf{x}) - U_M(\mathbf{x}) |< \varepsilon$ for any $N, M \ge T^*$. If $N=M$ the result is obviously true. If $N\ne M$ then it is without loss of generality to assume that $N>M$. By the additive representation:

$$\begin{aligned} U(\mathbf{x}_{k,N}^+) \ge U(\mathbf{x}_{k,M}^-). \end{aligned}$$

Expanding

$$\begin{aligned} u_k(x^+)+\sum _{t=1, t \ne k}^N u_t(x_t) \ge u_k(x^-)+\sum _{t=1, t \ne k}^M u_t(x_t). \end{aligned}$$

By rearranging this inequality

$$\begin{aligned} \sum _{t=M+1}^N u_t(x_t) \ge u_k(x^-)-u_k(x^+) > -\varepsilon . \end{aligned}$$

As $N>M \ge T^*$ it is also true that $U(\mathbf{x}_{k, M}^+) \ge U(\mathbf{x}_{k, N}^-$), hence

$$\begin{aligned} \sum _{t=M+1}^N u_t(x_t) \le u_k(x^+) - u_k(x^-) < \varepsilon . \end{aligned}$$

Note that

$$\begin{aligned} \sum _{t=M+1}^N u_t(x_t) = U_N(\mathbf{x})-U_M(\mathbf{x}). \end{aligned}$$

Hence, $|U_N(\mathbf{x}) - U_M(\mathbf{x}) |< \varepsilon$ and it follows that $U(\mathbf{x})$ converges by the Cauchy criterion.

Suppose now that it is not possible to find such k that $[x^+]_k \succ [x_k]_k \succ [x^-]_k$ for some $x^+, x^- \in X$. If $w_t=0$ for all t then the result is trivial. Suppose that $w_t>0$ for some t. Then for every period t for which $w_t>0$ we have

$$\begin{aligned} x_t \in X^e \equiv \{z \in X \ :\ z\succcurlyeq z' \ \text {for all} \ z'\in X \, \text {or} \ z'\succcurlyeq z \ \text {for all} \ z'\in X \}. \end{aligned}$$

For some $\lambda \in (0,1)$ replace $x_t$ with the mixture $x_t \lambda x_0$ for each t. Call the resulting stream $\mathbf{x}^*$. Then

$$\begin{aligned} U_T(\mathbf{x})-U_T(\mathbf{x}^*)=\sum _{t=1}^T u_t(x_t)-\sum _{t=1}^T u_t(x_t \lambda x_0)=(1-\lambda )\sum _{t=1}^T u_t(x_t)=(1-\lambda )U_T(\mathbf{x}). \end{aligned}$$

By rearranging this equation it follows that $U_T(\mathbf{x}^*)=\lambda U_T(\mathbf{x})$. By the previous argument $U_T(\mathbf{x}^*)$ converges; therefore, $U_T(\mathbf{x})$ converges.

Step 4. Show that $U(\mathbf{x})$ represents the order on $X^\infty$. Suppose that $\mathbf{x} \succcurlyeq \mathbf{y}$, where $\mathbf{x}, \mathbf{y} \in X^\infty$. If for some k, j it is possible to find $x^+, y^-$ such that $[x^+]_k \succ [x_k]_k$ and $[y^-]_j \prec [y_j]_j$, then $[x^+ \lambda x_k]_k \succ [x_k]_k$ for every $\lambda \in (0, 1)$ and $[y^- \mu y_j]_j \prec [y_j]_j$ for every $\mu \in (0,1)$. Let $x^*=x^+ \lambda x_k$ and $y^*=y^- \mu y_j$ for some $\lambda , \mu \in (0, 1)$. Denote

$$\begin{aligned} \mathbf{x}_{k,N}^*=(x_1, \ldots , x_{k-1}, x^*, x_{k+1}, \ldots , x_N, x_0, x_0, \ldots ), \end{aligned}$$

and

$$\begin{aligned} \mathbf{y}_{j,M}^*=(y_1, \ldots , y_{j-1}, y^*, y_{j+1}, \ldots , y_M, x_0, x_0, \ldots ). \end{aligned}$$

Then by Axiom I6, there exist $T^-, T^+$ such that

$$\begin{aligned} \mathbf{x}_{k,N}^*\succcurlyeq \mathbf{x}\succcurlyeq \mathbf{y} \succcurlyeq \mathbf{y}_{j,M}^*\ \end{aligned}$$

for all $N\ge T^+$ and for all $M\ge T^{-}$. Since $\mathbf{x}_{k,N}^* \succcurlyeq \mathbf{y}_{j,M}^*$ and U represents $\succcurlyeq$ on $X^*$ we have:

$$\begin{aligned} U(\mathbf{x}_{k,N}^*)\ge U(\mathbf{y}_{j,M}^*). \end{aligned}$$

By Step 3 we know that $U(x_1, \ldots , x_{k-1}, x^*, x_{k+1} \ldots )$ and $U(y_1, \ldots , y_{j-1}, y^*, y_{j+1}, \ldots )$ converge, so

$$\begin{aligned} U(x_1, \ldots , x_{k-1}, x^*, x_{k+1}, \ldots ) \ge U(y_1, \ldots , y_{j-1}, y^*, y_{j+1}, \ldots ). \end{aligned}$$

Recall that $x^*=x^+ \lambda x_k$ and $y^*=y^- \mu y_j$ for some $\lambda \in (0,1)$ and some $\mu \in (0,1)$. Since $\lambda$ and $\mu$ are arbitrary, it follows that $U(\mathbf{x}) \ge U(\mathbf{y})$.

If it is not possible to find $x^+$, $y^-$ such that $[x^+]_k \succ [x_k]_k$ and $[y^-]_j \prec [y_j]_j$, then either $w_t=0$ for all t, in which case $U(\mathbf{x})=U(\mathbf{y})$; or $x_t \succcurlyeq z'$ for all $z'\in X$ and all t with $w_t>0$, in which case $U(\mathbf{x}) \ge U(\mathbf{y})$; or $z' \succcurlyeq y_t$ for all $z'\in X$ and all t with $w_t>0$ in which case $U(\mathbf{x}) \ge U(\mathbf{y})$.

It is worth noting that as $\mathbf{x} \succcurlyeq \mathbf{y}$ implies $U(\mathbf{x}) \ge U(\mathbf{y})$, then by Axiom I2 it follows that $w_t>0$ for at least one t. Therefore, $\sum _{t=1}^\infty w_t>0$. Normalizing by $1/\sum _{t=1}^\infty w_t$, we can assume that $\sum _{t=1}^\infty w_t=1$.

Next, assume that $U(\mathbf{x}) \ge U(\mathbf{y})$. Suppose that it is possible to find k and $x^+, x^- \in X$ such that $\mathbf{x}_{k, N}^+\succcurlyeq \mathbf{x}\succcurlyeq \mathbf{x}_{k, N}^-$ for some fixed N. By mixture continuity, the set $\{ \alpha :\mathbf{x}_{k,N}^+ \alpha \mathbf{x}_{k, N}^- \succcurlyeq \mathbf {x}\}$ is closed. By assumption $\mathbf{x}_{k, N}^+\succcurlyeq \mathbf{x}$, so it follows that $\alpha = 1$ is included into the set. Analogously, the set $\{ \beta :\mathbf{x} \succcurlyeq \mathbf{x}_{k,N}^+ \beta \mathbf{x}_{k,N}^-\}$ is closed. In fact, $\beta = 0$ belongs to the set, as $\mathbf{x} \succcurlyeq \mathbf{x}_{k, N}^-$. Therefore, as both sets are closed, nonempty and form the unit interval, their intersection is nonempty. Hence, there exists $\lambda$ such that $\mathbf{x} \sim \mathbf{x}_{k, N}^+ \lambda \mathbf{x}_{k, N}^-$. Note that $\mathbf{x}_{k, N}^+ \lambda \mathbf{x}_{k,N}^- = (x_1, \ldots , x_{k-1}, x^+ \lambda x^-, x_{k+1} \ldots , x_N, x_0, x_0, \ldots )$. Let $x^+ \lambda x^- = x^*$. Define $\mathbf{x}_{k, N}^*= (x_1, \ldots ,x_{k-1}, x^*, x_{k+1}, \ldots , x_N, x_0, x_0, \ldots )$. Therefore, if there exist periods k, j and outcomes $x^+, x^-, y^+, y^-\in X$ such that $\mathbf{x}_{k, N}^+ \succcurlyeq \mathbf{x} \succcurlyeq \mathbf{x}_{k, N}^-$ and $\mathbf{y}_{j, M}^+ \succcurlyeq \mathbf{y} \succcurlyeq \mathbf{y}_{j, M}^-$ for some N and some M, we can find $\lambda , \mu \in [0, 1]$ such that $\mathbf{x} \sim \mathbf{x}_{k, N}^*$ and

$$\begin{aligned} \mathbf{y} \sim \mathbf{y}_{j, M}^* = (y_1, \ldots , y_{j-1}, y^*,y_{j+1}, \ldots , y_M, x_0, x_0, \ldots ), \end{aligned}$$

where $y^*=y^+\mu y^-$. We have already shown that if $\mathbf{x}\succcurlyeq \mathbf{y}$ then $U(\mathbf{x}) \ge U(\mathbf{y})$. From $\mathbf{x} \sim \mathbf{x}_{k, N}^*$ and $\mathbf{y} \sim \mathbf{y}_{j, M}^*$ it, therefore, follows that:

$$\begin{aligned} U(\mathbf{x}_{k, N}^*)= U(\mathbf{x}) \text { and } U(\mathbf{y}_{j, M}^*)= U(\mathbf{y}). \end{aligned}$$

Hence, from the assumption $U(\mathbf{x}) \ge U(\mathbf{y})$ we obtain:

$$\begin{aligned} U(\mathbf{x}_{k, N}^*) \ge U(\mathbf{y}_{j, M}^*). \end{aligned}$$

Recall that U is an order-preserving function on $X^*$. Thus, $\mathbf{x}_{k, N}^* \succcurlyeq \mathbf{y}_{j, M}^*$. Since $\mathbf{x} \sim \mathbf{x}_{k, N}^*$ and $\mathbf{y} \sim \mathbf{y}_{j, M}^*$, we obtain $\mathbf{x} \succcurlyeq \mathbf{y}$.

Suppose now that there is no such k, j or outcomes $x^+, x^-, y^+, y^-$ such that $\mathbf{x}_{k, N}^+\succcurlyeq \mathbf{x}\succcurlyeq \mathbf{x}_{k, N}^-$ and $\mathbf{y}_{j, M}^+\succcurlyeq \mathbf{y}\succcurlyeq \mathbf{y}_{j, M}^-$ for some N and some M. Then, using Axiom I6, we can conclude that either $x_t \in X^e$ for every t with $w_t>0$ or $y_t\in X^e$ for every t with $w_t>0$. Assume that there is only an upper bound to preferences; i.e., $X^e \equiv \{z \in X \ :\ z \succcurlyeq z' \ \text {for every} \ z'\in X\}$. Then $U(\mathbf{x}) \ge U(\mathbf{y})$ means that $x_t\in X^e$ whenever $w_t>0$. Therefore, $U(\mathbf{x})=U({\overline{\mathbf{x}}})$, where ${\overline{\mathbf{x}}}=(\overline{x}, \overline{x}, \ldots )$ and $\overline{x} \in X^e$. Hence, it follows by monotonicity that $\mathbf{x}\succcurlyeq \mathbf{y}$. In the case when there is only a lower bound, i.e., $\underline{x} \in X^e \equiv \{z \in X \ :\ z' \succcurlyeq z \ \text {for every} \ z'\in X\}$, the argument is similar.

Next, suppose that X is preference bounded above and below, i.e., there exist $\underline{x},\overline{x} \in X^e$ with $\overline{x}\succcurlyeq x \succcurlyeq \underline{x}$ for every $x\in X$. Assume that $U(\mathbf{x}) \ge U(\mathbf{y})$. We need to demonstrate that $\mathbf{x} \succcurlyeq \mathbf{y}$. By monotonicity and continuity there exist $\lambda , \mu \in [0, 1]$ such that $\mathbf{x} \sim {\overline{\mathbf{x}} }\lambda {\underline{\mathbf{x}}}$ and $\mathbf{y} \sim {\overline{\mathbf{x}}}\mu {\underline{\mathbf{x}}}$. Since by assumption $U(\mathbf{x}) \ge U(\mathbf{y})$ and U represents the preference order on constant streams, we have $U({\overline{\mathbf{x}} }\lambda { \underline{\mathbf{x}}}) \ge U({\overline{\mathbf{x}} }\mu {\underline{\mathbf{x}}})$. By rearranging this inequality $(\lambda -\mu )(U({ \overline{\mathbf{x}}})-U({\underline{\mathbf{x}}}))$, and using $U({ \overline{\mathbf{x}}}) > U({\underline{\mathbf{x}}})$ it follows that $\lambda \ge \mu$. Therefore, as $\mathbf{x} \sim {\overline{\mathbf{x}}}\lambda {\underline{\mathbf{x}}}$ and $\mathbf{y} \sim {\overline{\mathbf{x}} }\mu { \underline{\mathbf{x}}}$ and $\lambda \ge \mu$, we conclude that $\mathbf{x} \succcurlyeq \mathbf{y}$.

Thus, $(\mathbf{w}, u)$ is an AA representation for $\succcurlyeq$.

Step 5. Uniqueness of $w_t$. Assume that $(\mathbf{w}^\prime , u^\prime )$ is another AA representation. Then, for any t we have $w_t>0$ if and only if $w^\prime _t>0$. Consider the set of all constant programs $\{\mathbf{x}\in X^\infty : \mathbf{x} = (a, a, \ldots ), \text { where } a\in X\}$, which is a mixture set. Applying $(\mathbf{w}^\prime , u^\prime )$ and $(\mathbf{w}, u)$ to this set we conclude that $u(a)>u(b)$ if and only if $u^\prime (a)>u^\prime (b)$ for every $a, b \in X$. By Theorem 1 Fishburn (1982) it implies that $u=A u^\prime +B$ for some $A>0$ and some B. Hence,

$$\begin{aligned} \sum _{t=1}^ \infty w_t u(x_t) \ge \sum _{t=1}^ \infty w_t u(y_t) \ \text {if and only if} \ \sum _{t=1}^ \infty w^\prime _t u(x_t) \ge \sum _{t=1}^ \infty w^\prime _t u(y_t). \end{aligned}$$

For any t, s with $t \ne s$ and any $x', x'' \in X$, let $[x',x'']_{t,s}$ denote the stream with $x'$ in the t^th position, $x''$ in the s^th position and $x_0$ elsewhere. Fix t, s with $w_t>0$ and $w_s>0$. Using non-triviality, choose some $x^+, x^- \in X$ such that $x^+ \succ x^-$. Define $\mathbf{x}=[x^+, x^+]_{t,s}, \ \mathbf{y}= [x^+, x^-]_{t,s}, \ \mathbf{z}= [x^-, x^-]_{t,s}$. From the AA representation it follows that $\mathbf{x} \succ \mathbf{y} \succ \mathbf{z}$. By continuity of the AA representation there exists $\lambda \in (0, 1)$ such that $\mathbf{y} \sim \mathbf{x} \lambda \mathbf{z}$. Applying the AA representation to $\mathbf{y} \sim \mathbf{x} \lambda \mathbf{z}$ we obtain

$$\begin{aligned} w_t u(x^+)+w_s u(x^-) =\lambda (w_t+w_s) u(x^+)+(1-\lambda ) (w_t+w_s) u(x^-). \end{aligned}$$

It follows that $(1-\lambda )w_t = \lambda w_s$. Similarly, $(1-\lambda )w^\prime _t = \lambda w^\prime _s$. Therefore, $w_t/w_s =w^\prime _t/w^\prime _s$. As this is true for any t, s, we obtain that $\mathbf{w}=C {\mathbf{w}}^\prime$ for some $C>0$. The sufficiency of the uniqueness conditions follows by routine arguments. $\square$