A multiattribute decision time theory

Abstract

In this study, we analyze choice in the presence of some conflict that affects the decision time (response time), a subject that has been documented in the literature. We axiomatize a multiattribute decision time (MDT) representation, which is a dynamic extension of the classic multiattribute expected utility theory that allows potentially incomplete preferences. Under this framework, one alternative is preferred to another in a certain period if and only if the weighted sum of the attribute-dependent expected utility induced by the former alternative is larger than that induced by the latter for all attribute weights in a closed and convex set. MDT uniquely determines the decision time as the earliest period at which the ranking between alternatives becomes decisive. The comparative statics result indicates that the decision time provides useful information to locate indifference curves in a specific setting. MDT also explains various empirical findings in economics and other relevant fields.

Appendix

1.1 Proof of Theorem 1

The sufficiency part is straightforward. To show the necessity, we first fix $\tau$ and note that $X_i$ is separable for all $i \in I$ because it is compact. Because each $X_i$ is separable and metric, the partial order, continuity, independence, and single-attribute regularity axioms imply that there is a continuous von Neumann–Morgenstern utility function $u_i$ for each attribute i that represents the preference over the alternatives that differ with respect to only one attribute (using an argument similar to that of Fishburn 1965). Consistency implies that $u_i$ is independent of $\tau$; that is, for all p, $q \in {\mathcal {P}}(X)$ such that $p|_{-i} = q|_{-i}$, $p \succsim ^\tau q$ if and only if $E_{p|_i}[u_i(x_i)] \ge E_{q|_i}[u_i(x_i)]$. ($E_{(\cdot )|_i}[u_i(x_i)]$ denotes the expected utility in attribute i given alternative $(\cdot )$’s marginal probability distribution of $X_i$.) Because the $X_i$ are compact, we can normalize each utility function so that $\max _{x_i \in X_i} u_i(x_i) = 1$ and $\min _{x_i \in X_i} u_i(x_i) = 0$, which also derives the uniqueness of $u_i$. Define $u \equiv (u_1,\ldots , u_n)$.

Next, let $J_i = \{\eta \in \mathfrak {R}: \exists p \in {\mathcal {P}}(X), \eta = E_{p|_i}[u_i(x_i)]\} = [0, 1]$, and $J = \times ^n_{i=1} J_i = [0, 1]^n \subseteq \mathfrak {R}^n$. For all $p \in {\mathcal {P}}(X)$, we define $E_p[u(x)] \equiv (E_{p|_1}[u_1(x_1)],\ldots , E_{p|_n}[u_n(x_n)]) \in J$. We also define a binary relation $\succsim ^*$ on J such that $p \succsim ^\tau q$ if and only if $E_p[u(x)] \succsim ^* E_q[u(x)]$. From the monotonicity axiom, and by construction, it follows that $\succsim ^*$ is well-defined (i.e., $p \sim ^\tau q$ if $E_{p|_i}[u_i(x_i)] = E_{q|_i}[u_i(x_i)]$ for all $i \in I$) and satisfies the independence and continuity axioms.

Now, define a cone $K = \{\mu (s-t): \mu \ge 0, s, t \in J, s \succsim ^* t\} \subseteq \mathfrak {R}^n$. The following lemmas specify the characteristics of K.

Lemma 1

For all s, $t \in J$, $s \succsim ^* t$ if and only if $s-t \in K$.

Proof

We prove this by showing that for all s, $t \in J$, $s \succsim ^* t$ if and only if there are $s'$, $t' \in J$ and $\mu > 0$ such that $s' \succsim ^* t'$ and $s-t = \mu (s'-t')$. Assume the latter statement. Then, it follows from $s-t = \mu (s'-t')$ that

$$\begin{aligned} \frac{1}{1+\mu } s + \frac{\mu }{1+\mu } t'&=\, \frac{1}{1+\mu } t + \frac{\mu }{1+\mu } s'\\&\succsim ^*\, \frac{1}{1+\mu } t + \frac{\mu }{1+\mu } t', \end{aligned}$$

where the last preference relation holds because of the independence axiom. Now, $\frac{1}{1+\mu } s + \frac{\mu }{1+\mu } t' \succsim ^* \frac{1}{1+\mu } t + \frac{\mu }{1+\mu } t'$, which implies that $s \succsim ^* t$ by independence. The converse is straightforward. $\square$

Lemma 2

K is convex.

Proof

Suppose that s, t, $s'$, $t' \in J$ are such that $s-t$, $s'-t' \in K$. Lemma 1 implies that $s \succsim ^* t$ and $s' \succsim ^* t'$. For some $\mu \in (0, 1)$, independence implies that $\mu s + (1-\mu ) s' \succsim ^* \mu t + (1-\mu ) s'$ and $\mu t + (1-\mu ) s' \succsim ^* \mu t + (1-\mu ) t'$, from which it follows that $\mu s + (1-\mu ) s' \succsim ^* \mu t + (1-\mu ) t'$. Again, Lemma 1 implies that $\mu (s-t) + (1-\mu ) (s'-t') \in K$. $\square$

Lemma 3

K is closed.

Proof

The proof is an adaptation of that given by Ok (2007, Proposition H1, Claim 2). Let $\{\mu _m\}_{m=1}^\infty$, $\{s^m\}_{m=1}^\infty$, and $\{t^m\}_{m=1}^\infty$ be such that $\mu _m \ge 0$, $s^m$, $t^m \in J$, $s^m \succsim ^* t^m$ (i.e., $\mu _m(s^m-t^m) \in K$) for all m, and $\lim _{m \rightarrow \infty } \mu _m(s^m-t^m) = v \in \mathfrak {R}^n$. We show that $v \in K$. If $s^m = t^m$ for infinitely many m, it is trivial that $v = 0 \in K$. Accordingly, we can assume that $s^m \ne t^m$ for all m without loss of generality.

Because $J \subseteq \mathfrak {R}^n$, we apply the Euclidean distance $d(s, t) = (\sum _{i=1}^n |s_i-t_i|^2)^{1/2}$ to all $s = (s_1,\ldots , s_n)$, $t = (t_1,\ldots , t_n) \in J$. Fix an interior point $\hat{s}$ of J (note that the interior of J is nonempty because $J = [0, 1]^n$). Then, there is some $\epsilon > 0$ such that $N_{\epsilon , \mathfrak {R}^n}(\hat{s}) \equiv \{s \in \mathfrak {R}^n: d(s, \hat{s}) < \epsilon \} \subseteq J$. For some $\delta < \epsilon$, define $U = \{r \in J: r \succsim ^* \hat{s}, d(r, \hat{s}) \ge \delta \}$, which is closed because $\succsim ^*$ is continuous, and is compact because J is also compact.

Let $d_m = d(s^m, t^m)$. Then, because

$$\begin{aligned} \mu _m(s^m-t^m) = \frac{d_m \mu _m}{\delta } \cdot \frac{\delta }{d_m}(s^m-t^m) \end{aligned}$$

for all m, we obtain $\mu _m(s^m-t^m) = \mu '_m(r^m-\hat{s})$ by defining $\mu '_m = d_m \mu _m/\delta$ and $r^m = \hat{s} + \delta (s^m-t^m)/d_m$. Because $d(r^m, \hat{s}) = \delta < \epsilon$, $r^m \in N_{\epsilon , \mathfrak {R}^n}(\hat{s}) \subseteq J$ for all m. Moreover, because K is a cone, $s^m-t^m \in K$ implies $r^m-\hat{s} \in K$, thus, it follows from Lemma 1 that $r^m \succsim ^* \hat{s}$ for all m. Accordingly, $r^m \in U$ for all m, which implies that, from the compactness of U, there is a subsequence of $\{r^m\}_{m=1}^\infty$ that converges to some $r \in J$ such that $r \succsim ^* \hat{s}$. Additionally, because $\lim _{m \rightarrow \infty } \mu _m(s^m-t^m) = \lim _{m \rightarrow \infty } \mu '_m(r^m-\hat{s}) = v$ and $d(r^m, \hat{s}) = \delta$, $\{\mu '_m\}_{m=1}^\infty$ is a bounded sequence. Thus, there is a subsequence of $\{\mu '_m\}_{m=1}^\infty$ that converges to some $\mu ' \ge 0$, which implies that $v = \mu '(r-\hat{s})$. The last condition implies that $v \in K$. $\square$

Because K is convex and closed, it can be denoted by the intersection of closed half-spaces induced by the hyperplanes supporting it. Moreover, because K is a cone, each of the closed half-spaces is denoted by the set $\{\gamma \in \mathfrak {R}^n: \lambda \cdot \gamma \ge 0\}$, where $\lambda \in \mathfrak {R}^n$ satisfies $\lambda \cdot (s-t) = 0$ for some s, $t \in J$ such that $s-t \in K$. Eventually, by defining a closed and convex set $C \equiv \{\lambda \in \mathfrak {R}^n : \lambda \cdot (s-t) \ge 0 \mathrm{\ for \ all \ } s, t \in J \mathrm{\ such \ that \ } s-t \in K \}$, $K = \{s-t: s, t \in J, \lambda \cdot (s-t) \ge 0 \mathrm{\ for \ all \ \lambda \in C}\}$. The following lemma indicates that C only consists of non-negative vectors.

Lemma 4

$C \subseteq \mathfrak {R}^n_+ \equiv \{\lambda = (\lambda _1,\ldots , \lambda _n) \in \mathfrak {R}^n : \lambda _i \ge 0 \ \text {for all} \ i \in I \}$.

Proof

Suppose that there exists $\lambda = (\lambda _1,\ldots , \lambda _n) \in C$ such that $\lambda _j < 0$ for some $j \in I$. Then, by defining $s = (s_1,\ldots , s_n)$, $t = (t_1,\ldots , t_n) \in J$ such that $s_j < t_j$ and $s_i = t_i$ for all $i \ne j$, $s-t \in \mathfrak {R}^n_- \equiv \{r = (r_1,\ldots , r_n) \in \mathfrak {R}^n : r_i \le 0 \ \text {for all} \ i \in I \}$ and $\lambda \cdot (s-t) > 0$. This also implies that $t-s \in \mathfrak {R}^n_+$ and $\lambda \cdot (t-s) < 0$. It follows from the monotonicity axiom and $t-s \in \mathfrak {R}^n_+$ that $t \succ ^* s$. However, $\lambda \cdot (t-s) < 0$ implies $t-s \not \in K$ by the construction of C and K, which contradicts Lemma 1. $\square$

Now, we define $\varLambda ^\tau \equiv C \cap \varDelta ^{n-1}$, which is closed and convex because both C and $\varDelta ^{n-1}$ are closed and convex. It follows from Lemma 4 and the construction of K and C that for all p, $q \in {\mathcal {P}}(X)$, $p \succsim ^\tau q$ if and only if $\lambda \cdot E_p[u(x)] \ge \lambda \cdot E_q[u(x)]$ for all $\lambda \in \varLambda ^\tau$. To prove the uniqueness of $\varLambda ^\tau$, suppose that both $\varLambda ^\tau$ and $\varLambda '$ (along with u) represent $\succsim ^\tau$. Without loss of generality, we assume $\varLambda ' {\setminus } \varLambda ^\tau \ne \phi$. Then, because $\varLambda ^\tau$ is closed and convex, it follows from one version of the separating hyperplane theorem (Dunford and Schwartz 1957, V2.12) that there exist p, $q \in {\mathcal {P}}(X)$ such that $\lambda \cdot (E_p[u(x)] - E_q[u(x)]) > 0$ for all $\lambda \in \varLambda ^\tau$, but $\lambda ' \cdot (E_p[u(x)]-E_q[u(x)]) < 0$ for some $\lambda ' \in \varLambda ' {\setminus } \varLambda ^\tau$. Thus, the preference represented by $\varLambda ^\tau$ implies $p \succsim ^\tau q$, whereas that represented by $\varLambda '$ implies $p \bowtie ^\tau q$, which is a contradiction.

Finally, the following lemma characterizes the relationship among the $\varLambda ^\tau$ for different $\tau$.

Lemma 5

For all $\tau ,$ $\tau ' \in \mathbb {N} \cup \{0\},$ if $p \succsim ^\tau q$ implies $p \succsim ^{\tau '} q$ for all p,  $q \in {\mathcal {P}}(X),$ then $\varLambda ^\tau \supseteq \varLambda ^{\tau '}$.

Proof

Suppose, on the contrary, that $\varLambda ^{\tau '} {\setminus } \varLambda ^\tau \ne 0$. Then, because $\varLambda ^\tau$ is closed and convex, one version of the separating hyperplane theorem (Dunford and Schwartz 1957, V2.12) implies that there exist p, $q \in {\mathcal {P}}(X)$ such that $\lambda \cdot (E_p[u(x)]-E_q[u(x)]) > 0$ for all $\lambda \in \varLambda ^\tau$ and $\lambda ' \cdot (E_p[u(x)]-E_q[u(x)]) < 0$ for some $\lambda ' \in \varLambda ^{\tau '} {\setminus } \varLambda ^\tau$. From this, it follows that $p \succsim ^\tau q$ and $p \bowtie ^{\tau '} q$, which is a contradiction. $\square$

By the consistency axiom, Lemma 5 implies that $\varLambda ^\tau \supseteq \varLambda ^{\tau '}$ if $\tau < \tau '$, which concludes the proof. $\square$

1.2 Proof of Theorem 2

First, we show that statement (a) implies (b). Assume that for all $\tau \in \mathbb {N} \cup \{0\}$ and p, $q \in {\mathcal {P}}(X)$, $p \succsim ^\tau _2 q$ implies $p \succsim ^\tau _1 q$. Thus, an argument similar to that applied for Lemma 5 implies that $\varLambda _1^\tau \subseteq \varLambda _2^\tau$ for all $\tau \in \mathbb {N} \cup \{0\}$. Conversely, assume that statement (b) holds, that is, $\varLambda _1^\tau \subseteq \varLambda _2^\tau$ for all $\tau \in \mathbb {N} \cup \{0\}$. It follows from the definitions of $\varLambda _1^\tau$ and $\varLambda _2^\tau$ that, for all $\tau \in \mathbb {N} \cup \{0\}$ and p, $q \in {\mathcal {P}}(X)$, $p \succsim ^\tau _2 q$ implies $p \succsim ^\tau _1 q$, which implies statement (a). $\square$

1.3 Proof of Theorem 3

First, because alternative pairs (p, q) and $(p', q')$ are congruent with respect to the hyperplane $h^\infty$, which has normal vector $\lambda ^\infty \in \cap ^{\infty }_{\tau = 0} \varLambda ^\tau$, there exist $s \in \mathfrak {R}^n$ and $\psi \in \{-1, 1\}$ such that $\lambda ^\infty \cdot s = 0$, $E_{p-q}[u(x)] = \mu _1 s + \mu _2 \psi \lambda ^\infty$, and $E_{p'-q'}[u(x)] = \mu '_1 s + \mu '_2 \psi \lambda ^\infty$ for some $\mu _1$, $\mu _2$, $\mu '_1$, $\mu '_2 \ge 0$. Without loss of generality, we assume that $\psi = 1$ in the following. (The case $\psi = -1$ can be proved similarly.)

Now, we prove statement (a). Assume that $\tau ^*(p, q) < \tau ^*(p', q')$, which implies $\tau ^*(p, q) < \infty$. Because $\psi = 1$ implies that $\lambda ^\infty \cdot E_{p-q}[u(x)] \ge 0$ and $\lambda ^\infty \cdot E_{p'-q'}[u(x)] \ge 0$, there exists $\hat{\lambda } \in \varLambda ^{\tau ^*(p, q)} {\setminus } \varLambda ^{\tau ^*(p, q)+1}$ such that

$$\begin{aligned} \hat{\lambda } \cdot E_{p-q}[u(x)] \ge 0 > \hat{\lambda } \cdot E_{p'-q'}[u(x)]; \end{aligned}$$

(A.1)

that is, a hyperplane $\hat{h}$ with normal vector $\hat{\lambda }$ separates vectors $E_{p-q}[u(x)]$ and $E_{p'-q'}[u(x)]$. We also have $\mu '_1 > 0$ to satisfy (A.1); otherwise, $\mu '_1 = 0$ and $\hat{\lambda } \cdot E_{p'-q'}[u(x)] = \hat{\lambda } \cdot (\mu '_2 \lambda ^\infty ) \ge 0$, which is a contradiction. (Note that $\hat{\lambda } \cdot \lambda ^\infty \ge 0$ because $\hat{\lambda }$, $\lambda ^\infty \in \varLambda ^{\tau ^*(p, q)} \subseteq \varDelta ^{n-1}$.) It follows from $\psi = 1$ that there exists $r \in \mathfrak {R}$ such that r is parallel to $\hat{h}$ (i.e., $\hat{\lambda } \cdot r = 0$) and $r = \hat{\mu }_1 s + \hat{\mu }_2 \lambda ^\infty$ for some $\hat{\mu }_1$, $\hat{\mu }_2 \ge 0$. Now, we consider the following cases.

Case 1: $\hat{\mu }_1 = 0$

It follows from $r = \hat{\mu }_2 \lambda ^\infty$ and $\hat{\lambda } \cdot r = 0$ that $\hat{\lambda } \cdot (\hat{\mu }_2 \lambda ^\infty ) = 0$. Because $\hat{\mu }_2 > 0$ (otherwise, we must have $r = 0$, which is a contradiction), $\hat{\lambda } \cdot \lambda ^\infty = 0$. Thus, (A.1) implies that $\hat{\lambda } \cdot (\mu _1 s) \ge 0 > \hat{\lambda } \cdot (\mu '_1 s)$. Because we already showed that $\mu '_1 > 0$, we must have $\hat{\lambda } \cdot s < 0$ and thus $\mu _1 = 0$. Accordingly, $\pi /2 = \theta (E_{p-q}[u(x)], s) > \theta (E_{p'-q'}[u(x)], s)$.

Case 2: $\hat{\mu }_1 > 0$

First, we consider $\mu _1 > 0$. By construction, $\hat{\lambda } \cdot r = \hat{\mu }_1 \hat{\lambda } \cdot s + \hat{\mu }_2 \hat{\lambda } \cdot \lambda ^\infty = 0$, or

$$\begin{aligned} \hat{\lambda } \cdot s = -\frac{\hat{\mu }_2}{\hat{\mu }_1} \hat{\lambda } \cdot \lambda ^\infty . \end{aligned}$$

(A.2)

Now, $\hat{\lambda } \cdot \lambda ^\infty \ge 0$ because $\hat{\lambda }$, $\lambda ^\infty \in \varLambda ^{\tau ^*(p, q)} \subseteq \varDelta ^{n-1}$. Suppose $\hat{\lambda } \cdot \lambda ^\infty = 0$, which follows from (A.2) that $\hat{\lambda } \cdot s = 0$. This implies that $\hat{\lambda } \cdot E_{p-q}[u(x)] = \hat{\lambda } \cdot E_{p'-q'}[u(x)] = 0$, contradicting (A.1). Thus, we conclude that $\hat{\lambda } \cdot \lambda ^\infty > 0$. Substituting (A.2) into $\hat{\lambda } \cdot E_{p-q}[u(x)] = \mu _1 \hat{\lambda } \cdot s + \mu _2 \hat{\lambda } \cdot \lambda ^\infty \ge 0$ gives $\mu _2/\mu _1 \ge \hat{\mu }_2/\hat{\mu }_1$, whereas substituting (A.2) into $\hat{\lambda } \cdot E_{p'-q'}[u(x)] = \mu '_1 \hat{\lambda } \cdot s + \mu '_2 \hat{\lambda } \cdot \lambda ^\infty < 0$ obtains $\mu '_2/\mu '_1 < \hat{\mu }_2/\hat{\mu }_1$. (Note that we already have $\mu '_1 > 0$.) Accordingly, $\mu _2/\mu _1 \ge \hat{\mu }_2/\hat{\mu }_1 > \mu '_2/\mu '_1$. Thus, the slope of the vector difference $E_{p-q}[u(x)]$ is greater than that of $E_{p'-q'}[u(x)]$, that is, $\theta (E_{p-q}[u(x)], s) > \theta (E_{p'-q'}[u(x)], s)$. Second, consider $\mu _1 = 0$. Then, we obtain $\pi /2 = \theta (E_{p-q}[u(x)], s) > \theta (E_{p'-q'}[u(x)], s)$, as in Case 1.

In both cases, we conclude that $\theta (E_{p-q}[u(x)], t) > \theta (E_{p'-q'}[u(x)], t)$ for the projection t of $E_{p-q}[u(x)]$ onto $h^\infty$ because the vector t has the same direction as s by construction. Statement (b) is straightforward because it is the contraposition of (a). $\square$

1.4 Proof of Theorem 4

We first note that, if the alternative pairs (p, q) and $(p', q')$ are strongly congruent with respect to some $h^\infty$ that has normal vector $\lambda ^\infty \in \cap ^{\infty }_{\tau = 0} \varLambda ^\tau$, then (p, q) and $(p', q')$ are congruent with respect to $h^\infty$; that is, there exist $s \in \mathfrak {R}^n$, $\psi \in \{-1, 1\}$, $\mu _1$, $\mu _2$, $\mu '_1$, $\mu '_2 \ge 0$ such that $\lambda ^\infty \cdot s = 0$, $E_{p-q}[u(x)] = \mu _1 s + \mu _2 \psi \lambda ^\infty$, and $E_{p'-q'}[u(x)] = \mu '_1 s + \mu '_2 \psi \lambda ^\infty$.

Now, we prove statement (a). Assume that $\tau ^*(p, q) < \tau ^*(p', q')$. Because (p, q) and $(p', q')$ are congruent with respect to $h^\infty$, Theorem 3 implies that $\theta (E_{p-q}[u(x)], t) > \theta (E_{p'-q'}[u(x)], t)$ for the projection t of $E_{p-q}[u(x)]$ onto $h^\infty$, which in turn implies that $\theta (E_{p-q}[u(x)], s) > \theta (E_{p'-q'}[u(x)], s)$ for the vector s defined in the previous paragraph because s has the same direction as t. From the last condition, it follows that $\mu _2/\mu _1 > \mu '_2/\mu '_1$. The strong congruence of (p, q) and $(p', q')$ then implies that $\mu _2 > \mu '_2$, and thus $|\lambda ^\infty \cdot E_{p-q}[u(x)]| = |\mu _2 \psi | \Vert \lambda ^\infty \Vert ^2 > |\mu '_2 \psi | \Vert \lambda ^\infty \Vert ^2 = |\lambda ^\infty \cdot E_{p'-q'}[u(x)]|$ because $\lambda ^\infty \cdot s = 0$. Statement (b) is straightforward because it is the contraposition of (a). $\square$