Violations of betweenness and choice shifts in groups

Abstract

In decision theory, the betweenness axiom postulates that a decision maker who chooses an alternative A over another alternative B must also choose any probability mixture of A and B over B itself and can never choose a probability mixture of A and B over A itself. The betweenness axiom is a weaker version of the independence axiom of expected utility theory. Numerous empirical studies documented systematic violations of the betweenness axiom in revealed individual choice under uncertainty. This paper shows that these systematic violations can be linked to another behavioral regularity—choice shifts in a group decision making. Choice shifts are observed if an individual faces the same decision problem but makes a different choice when deciding alone and in a group.

Appendix

Proof of Theorem 1

We prove first that part 1 of the theorem implies part 2.

Notice that $p_{s} = \left( {1 - a} \right)b + a > \left( {1 - a} \right)b = p_{r}$. Since $\bar{p}\left[ \varvec{s} \right] + \left( {1 - \bar{p}} \right)\left[ \varvec{r} \right] \sim \varvec{s,}$ then by setting $\alpha = \bar{p}$, $\beta = p$ and $\gamma = 1$ in part (a) of Assumption 1 we immediately obtain that mixture $p\left[ \varvec{s} \right] + \left( {1 - p} \right)\left[ \varvec{r} \right]$ is strictly preferred over $\varvec{s } \quad {\text{for all}} p \in (\bar{p},1)$. Since $\bar{p}\left[ \varvec{s} \right] + \left( {1 - \bar{p}} \right)\left[ \varvec{r} \right] \sim \varvec{r}$ then part (b) of Assumption 1 implies that $p\left[ \varvec{s} \right] + \left( {1 - p} \right)\left[ \varvec{r} \right] \prec \varvec{r }\quad {\text{for all }}p \in (0,\bar{p})$. Thus, if $p_{s} > \bar{p} > p_{r} ,$ then we must have $\varvec{s}^{\varvec{*}} \succ \varvec{s} \sim \varvec{r} \succ \varvec{r}^{\varvec{*}}$ and an individual exhibits a cautious shift. This evidence proves statement (b) iv. Therefore, an individual can exhibit a risky shift only in two cases: either when $\bar{p} \ge p_{s} > p_{r}$ or when $p_{s} > p_{r} \ge \bar{p}$.

Consider first the case when $\bar{p} \ge p_{s} > p_{r}$. Since $p_{s} = p_{r} + a,$ then this case is only possible when $a \le \bar{p}$. According to part (b) of Assumption 1, an individual has strictly quasi-convex preferences on the interval $\left[ {0,\bar{p}} \right],$ i.e. for all for $0 \le \alpha < \beta < \gamma \le \bar{p}$, the best of $\alpha \left[ \varvec{s} \right] + \left( {1 - \alpha } \right)\left[ \varvec{r} \right]$ and $\gamma \left[ \varvec{s} \right] + \left( {1 - \gamma } \right)\left[ \varvec{r} \right]$ is strictly preferred over mixture $\beta \left[ \varvec{s} \right] + \left( {1 - \beta } \right)\left[ \varvec{r} \right]$. Quasi-convexity implies that the set $L = \left\{ {p \in \left[ {0,\bar{p}} \right]:s^{*} { \succcurlyeq }p\left[ s \right] + \left( {1 - p} \right)[r]} \right\}$ is convex, i.e. an interval. Moreover, since $\varvec{s}^{\varvec{*}} { \succcurlyeq }\varvec{s}^{\varvec{*}}$ then this interval L must contain probability $p_{s}$. More specifically, interval L must be of the form either $\left[ {c, p_{s} } \right]$ or $\left[ {p_{s} , c} \right]$ for some threshold $c \in [0,\bar{p})$. An individual then exhibits a risky shift $\varvec{r}^{\varvec{*}} \succ \varvec{s}^{\varvec{*}}$ if and only if $p_{r} \in \left[ {0, { \hbox{max} }\{ c, p_{s} \} } \right]\backslash L$, which is an interval of the form $[0,{ \hbox{min} }\{ c, p_{s} \} )$, for some threshold $c \in [0,\bar{p})$. Thus, an individual exhibits a risky shift when $a \le \bar{p}$ and $p_{r} < { \hbox{min} }\{ c, p_{s} \}$. The latter inequality can be rewritten as $p_{r} = \left( {1 - a} \right)b < { \hbox{min} }\{ c, p_{s} \}$ or $b < { \hbox{min} }\{ c, p_{s} \} /\left( {1 - a} \right) \equiv b_{1} \left( a \right)$. This evidence proves part (a) i. and (b) i. of the theorem.

Now consider the second case when $p_{s} > p_{r} \ge \bar{p}$. Since $p_{s} = p_{r} + a$ then this case is only possible when $a \le 1 - \bar{p}$. According to part (a) of Assumption 1, an individual has strictly quasi-concave preferences on the interval $\left[ {\bar{p},1} \right]$, i.e. for all $\bar{p} \le \alpha < \beta < \gamma \le 1$, mixture $\beta \left[ \varvec{s} \right] + \left( {1 - \beta } \right)\left[ \varvec{r} \right]$ is strictly preferred over the worst of $\alpha \left[ \varvec{s} \right] + \left( {1 - \alpha } \right)\left[ \varvec{r} \right]$ and $\gamma \left[ \varvec{s} \right] + \left( {1 - \gamma } \right)\left[ \varvec{r} \right]$. Quasi-concavity implies that the set $U = \left\{ {p \in \left[ {\bar{p},1} \right]:p\left[ s \right] + \left( {1 - p} \right)[r]{ \succcurlyeq }r^{*} } \right\}$ is convex, i.e. an interval. Since $\varvec{r}^{\varvec{*}} { \succcurlyeq }\varvec{r}^{\varvec{*}}$ then this interval U must contain probability $p_{r}$. More specifically, interval U must be of the form either $\left[ { p_{r} ,d} \right]$ or $\left[ {d, p_{r} } \right]$ for some threshold $d \in \left[ {\bar{p},1} \right].$ An individual then exhibits a risky shift $\varvec{r}^{\varvec{*}} \succ \varvec{s}^{\varvec{*}}$ if and only if $p_{s} \in \left[ { { \hbox{min} }\{ d, p_{r} \} ,1} \right]\backslash U$, which is an interval of the form $({ \hbox{max} }\{ d, p_{r} \} ,1]$, for some threshold $d \in \left[ {\bar{p},1} \right]$. Thus, an individual exhibits a risky shift if $a \le 1 - \bar{p}$ and $p_{s} > { \hbox{max} }\{ d, p_{r} \}$. The latter inequality can be rewritten as $p_{s} = \left( {1 - a} \right)b + a > { \hbox{max} }\{ d, p_{r} \}$ or $b > ({ \hbox{max} }\{ d, p_{r} \} - a)/\left( {1 - a} \right) \equiv b_{2} \left( a \right)$. This evidence proves part (a) ii. and (b) ii. of the theorem.

To finish the proof that part 1 implies part 2 we need to show that $b_{2} \left( a \right) > b_{1} \left( a \right)$. Fix a value of $a$ s.t. $a < \bar{p}$ and $a < 1 - \bar{p}$. First note that $r^{*} \sim s^{*}$ can happen only in two cases: $\bar{p} \ge p_{s} > p_{r}$ and $p_{s} > p_{r} \ge \bar{p}$. When $\bar{p} \ge p_{s} > p_{r}$ we observe the risk shift only for all $b < b_{1} \left( a \right)$ and for $b = b_{1} \left( a \right)$ we have that $r^{*} \sim s^{*}$. Note that in this case we have $p_{r} = \left( {1 - a} \right)b_{1} \left( a \right)$ so that inequality $\bar{p} > p_{r}$ can be rewritten as:

$$ \bar{p} > \left( {1 - a} \right)b_{1} \left( a \right) $$

(1)

When $p_{s} > p_{r} > \bar{p}$ we observe the risk shift only for all $b > b_{2} \left( a \right)$ and for $b = b_{2} \left( a \right)$ we have that $r^{*} \sim s^{*}$. Note that in this case we have $p_{r} = \left( {1 - a} \right)b_{2} \left( a \right)$ so that inequality $p_{r} > \bar{p}$ can be rewritten as:

$$ \left( {1 - a} \right)b_{2} \left( a \right) > \bar{p} $$

(2)

Inequalities (1) and (2) can hold simultaneously only if $b_{2} \left( a \right) > b_{1} \left( a \right)$. Finally we have to consider the cases in which either $a \le \bar{p}$ or $a \le 1 - \bar{p}$ or both are not satisfied. Note that $a \le \bar{p}$ implies $0 < b_{1} \left( a \right) < 1$ and $a \le 1 - \bar{p}$ implies $0 < b_{2} \left( a \right) < 1$. Suppose that $a \ge \bar{p}$ in this case $b_{1} \left( a \right) = 0$. Suppose that $a \ge 1 - \bar{p}$ in this case $b_{2} \left( a \right) = 1$. Then $b_{2} \left( a \right) > b_{1} \left( a \right)$ holds for all possible values of $a$. This evidence together with the previous ones proves b) iii. of the theorem.

Now we prove that part 2 of the theorem implies part 1.

We claim that the part 2 of the theorem implies $p\left[ \varvec{s} \right] + \left( {1 - p} \right)\left[ \varvec{r} \right] \succ \varvec{r }\forall p \in \left( {\bar{p},1} \right)$ and $p\left[ \varvec{s} \right] + \left( {1 - p} \right)\left[ \varvec{r} \right] \prec \varvec{r }\forall p \in \left( {0, \bar{p}} \right)$. Consider an individual who is indifferent between $\varvec{s}$ and $\varvec{r}$. Assume $\bar{p} < a < 1$. In this case, when $b = 0$, part 2 of the theorem implies a cautious shift, i.e. $\varvec{s}^{\varvec{*}} \succ \varvec{r}^{\varvec{*}}$, $\forall a \in \left( {\bar{p}, 1} \right)$. Note that for $b = 0$, $\varvec{r}^{\varvec{*}} = \varvec{r}$. By contradiction, suppose that there exists a value $p^{*} \in (\bar{p},1)$ such that $p^{*} \left[ \varvec{s} \right] + \left( {1 - p^{*} } \right)\left[ \varvec{r} \right]{ \preccurlyeq }\varvec{r }$ and $p^{*} \left[ \varvec{s} \right] + \left( {1 - p^{*} } \right)\left[ \varvec{r} \right]{ \preccurlyeq }\varvec{s}.$ Suppose $b = 0$ and $a = p^{*} > \bar{p}$. In such a case $\varvec{r}^{\varvec{*}} = \left[ \varvec{r} \right]$ and $\varvec{s}^{\varvec{*}} = p^{*} \left[ \varvec{s} \right] + \left( {1 - p^{*} } \right)\left[ \varvec{r} \right]$. By initial assumption we have then $\varvec{r}^{\varvec{*}} { \succcurlyeq }\varvec{s}^{\varvec{*}}$, which contradicts part 2 of the theorem, that in such case states $\varvec{s}^{\varvec{*}} \succ \varvec{r}^{\varvec{*}}$.

Assume $1 - \bar{p} < a < 1$. In this case, when $b = 1$, part 2 of the theorem implies a cautious shift, i.e. $\varvec{s}^{\varvec{*}} \succ \varvec{r}^{\varvec{*}}$, $\forall a \in \left( {1 - \bar{p}, 1} \right)$. Note that for $b = 1$, $\varvec{s}^{\varvec{*}} = \varvec{s}$. By contradiction, suppose that there exists a value $p^{*} \in (0, \bar{p})$ such that $p^{*} \left[ \varvec{s} \right] + \left( {1 - p^{*} } \right)\left[ \varvec{r} \right]{ \succcurlyeq }\varvec{r }$ and $p^{*} \left[ \varvec{s} \right] + \left( {1 - p^{*} } \right)\left[ \varvec{r} \right]{ \succcurlyeq }\varvec{s}.$ Suppose $b = 1$ and $a = 1 - p^{*} > 1 - \bar{p}$. In such a case $\varvec{s}^{\varvec{*}} = \left[ \varvec{s} \right]$ and $\varvec{r}^{\varvec{*}} = p^{*} \left[ \varvec{s} \right] + \left( {1 - p^{*} } \right)\left[ \varvec{r} \right]$. By initial assumption we have then $\varvec{r}^{\varvec{*}} { \succcurlyeq }\varvec{s}^{\varvec{*}}$, which contradicts part 2 of the theorem, that in such case states $\varvec{s}^{\varvec{*}} \succ \varvec{r}^{\varvec{*}}$. The evidences above prove our claim.

Now we claim that the theorem implies the existence of $\bar{p} \in \left[ {0, 1} \right]$ such that $\varvec{s} \sim \varvec{r} \sim \bar{p}\left[ \varvec{s} \right] + \left( {1 - \bar{p}} \right)\left[ \varvec{r} \right]$. Assume an individual who is indifferent between $\varvec{s}$ and $\varvec{r}$. Consider the set $U = \left\{ {p\left[ s \right] + \left( {1 - p} \right)\left[ r \right]: p\left[ s \right] + \left( {1 - p} \right)\left[ r \right]{ \succcurlyeq }\varvec{r}} \right\}$. Using the above considerations for $\bar{p} < a < 1$ we can state that this set is closed only if lottery $\bar{p}\left[ s \right] + \left( {1 - \bar{p}} \right)\left[ r \right]{ \succcurlyeq }\varvec{r}$. Consider the set of $L = \left\{ {p\left[ s \right] + \left( {1 - p} \right)\left[ r \right]: \varvec{s}{ \succcurlyeq }p\left[ s \right] + \left( {1 - p} \right)\left[ r \right]} \right\}$. Using the above considerations for $1 - \bar{p} < a < 1$ we can state that this set is closed only if lottery $\varvec{s}{ \succcurlyeq }\bar{p}\left[ s \right] + \left( {1 - \bar{p}} \right)\left[ r \right]$. Continuity of preferences implies that sets $L$ and $U$ are closed that happen only if $s \sim r \sim \bar{p}\left[ s \right] + \left( {1 - \bar{p}} \right)\left[ r \right]$.

Now we show that statement part 2 of the theorem implies that preferences are quasi-concave for $p > \bar{p}$ and quasi-convex for $p < \bar{p}$. By definition of quasi-concavity preferences are strictly quasi-concave for $p \in \left[ {\bar{p}, 1} \right]$ if and only if are single-peaked on this interval. Suppose preferences are not quasi-concave for $p > \bar{p},$ i.e. there exist probabilities $p,q \in \left[ {\bar{p}, 1} \right]$, $p \ne q$, such that $p\left[ \varvec{s} \right] + \left( {1 - p} \right)\left[ \varvec{r} \right]\sim q\left[ \varvec{s} \right] + \left( {1 - q} \right)\left[ \varvec{r} \right]$ and for some $\alpha \in \left( {0, 1} \right)$ $\left( {\alpha \cdot p + \left( {1 - \alpha } \right) \cdot q} \right)\left[ \varvec{s} \right] + \left( {1 - \alpha \cdot p - \left( {1 - \alpha } \right) \cdot q} \right)\left[ \varvec{r} \right]{ \preccurlyeq }p\left[ \varvec{s} \right] + \left( {1 - p} \right)\left[ \varvec{r} \right]$. It follows that preferences are not single peaked in $p \in \left[ {\bar{p}, 1} \right]$ and therefore there are either (1) at least two separated intervals on $p$ where is possible to find a value of $a < 1 - \bar{p}$ such that $s^{*} \succ r^{*}$ or (2) an interval on $p$ where is possible to find a value of $a < 1 - \bar{p}$ such that $s^{*} \sim r^{*}$. In both case a violation of the part 2 of the theorem.

Using the similar arguments we can prove that part 2 of the theorem implies strictly quasi-convex preferences for $p < \bar{p}$.

${\square }$.