Rational choices: an ecological approach

Abstract

We address the oft-repeated criticism that the demands which the rational choice approach makes on the knowledge and cognition of a decision-maker (DM) are way beyond the capabilities of typical human intelligence. Our key finding is that it may be possible to arrive at this ideal of rationality by means of cognitively less demanding, heuristic-based ecological reasoning that draws on information about others’ choices in the DM’s environment. Formally, we propose a choice procedure under which, in any choice problem, the DM, first, uses this information to shortlist a set of alternatives. The DM does this shortlisting by a mental process of categorization, whereby she draws similarities with certain societal members—the ingroup—and distinctions from others—the outgroup—and considers those alternatives that are similar (dissimilar) to ingroup (outgroup) members’ choices. Then, she chooses from this shortlisted set by applying her preferences, which may be incomplete owing to limitations of knowledge. We show that, if a certain homophily condition connecting the DM’s preferences with her ingroup–outgroup categorization holds, then the procedure never leads the DM to making bad choices. If, in addition, a certain shortlisting consistency condition holds vis-a-vis non-comparable alternatives under the DM’s preferences, then the procedure results in rational choices.

A Appendix

1.1 Preliminaries

We first prove two lemmas that we use to prove our results.

Lemma A.1

If$(\succ ,\succ ')$is an (xy)-TOP pair on$S \in {\mathcal {P}}(X)$with$x \succ z$, for all$z \in S{\setminus }\{x\}$, and$\succ ^*$is an asymmetric, transitive binary relation onSwith$x \succ ^* y$, then

$$\begin{aligned} |\succ -\succ ^*|_S<|\succ '-\succ ^*|_. \end{aligned}$$

Proof

Since $(\succ ,\succ ')$ is an (xy)-TOP pair, $w\succ z$ if and only if $w\succ ' z$ for all $w,z\in S{\setminus }\{x,y\}$. Hence:

$$\begin{aligned} |\succ -\succ ^*|_{S{\setminus }\{x,y\}}-|\succ '-\succ ^*|_{S{\setminus }\{x,y\}} =0, \end{aligned}$$

and it follows that:

$$\begin{aligned} |\succ -\succ ^*|_{S}-|\succ '-\succ ^*|_{S} =&|\succ -\succ ^*|_{\{x,y\}}-|\succ '-\succ ^*|_{\{x,y\}}\\&+\sum _{z\in S{\setminus }\{x,y\}}\left[ |\succ -\succ ^*|_{\{x,z\}}-|\succ '-\succ ^*|_{\{x,z\}}\right] \\&+\sum _{z\in S{\setminus }\{x,y\}}\left[ |\succ -\succ ^*|_{\{y,z\}}-|\succ '-\succ ^*|_{\{y,z\}}\right] . \end{aligned}$$

Define the sets $S_1=\{z\in S: x\succ z \succ y\}$ and $S_2=\{z\in S: y \succ z\}$. Clearly, $S_1 \cup \ S_2 \cup \{x,y\} = S$. Furthermore, since $x \succ z$, $x \succ ' z$, $y \succ z$ and $y \succ ' z$, for all $z \in S_2$, it follows that:

$$\begin{aligned} \sum _{z\in S_2} |\succ -\succ ^*|_{\{x,z\}}&=\sum _{z\in S_2} |\succ '-\succ ^*|_{\{x,z\}}, \text{ and } \\ \sum _{z\in S_2} |\succ -\succ ^*|_{\{y,z\}}&=\sum _{z\in S_2} |\succ '-\succ ^*|_{\{y,z\}}. \end{aligned}$$

Therefore:

$$\begin{aligned} \sum _{z\in S{\setminus }\{x,y\}}\left[ |\succ -\succ ^*|_{\{x,z\}}-|\succ '-\succ ^*|_{\{x,z\}}\right] =&\sum _{z\in S_1}\left[ |\succ -\succ ^*|_{\{x,z\}}-|\succ '-\succ ^*|_{\{x,z\}}\right] \\ \sum _{z\in S{\setminus }\{x,y\}}\left[ |\succ -\succ ^*|_{\{y,z\}}-|\succ '-\succ ^*|_{\{y,z\}}\right] =&\sum _{z\in S_1}\left[ |\succ -\succ ^*|_{\{y,z\}}-|\succ '-\succ ^*|_{\{y,z\}}\right] . \end{aligned}$$

Next, define the sets

$$\begin{aligned} V_1 = \{z\in S_1: z \succ ^* x\},&V_2 = \{z\in S_1: x \succ ^* z\} \\ {\hat{V}}_1 = \{z\in S_1: z \succ ^* y\},&{\hat{V}}_2 = \{z\in S_1: y \succ ^* z\} \end{aligned}$$

and verify that

• $|\succ -\succ ^*|_{\{x,y\}}-|\succ '-\succ ^*|_{\{x,y\}}=-1$

• $\sum _{z\in S_1}\left[ |\succ -\succ ^*|_{\{x,z\}}-|\succ '-\succ ^*|_{\{x,z\}}\right] = \# V_1 - \# V_2$

• $\sum _{z\in S_1}\left[ |\succ -\succ ^*|_{\{y,z\}}-|\succ '-\succ ^*|_{\{y,z\}}\right] = \# {\hat{V}}_2 - \# {\hat{V}}_1$.

Now, since $\succ ^*$ is transitive and $x \succ ^* y$, we have that $z \succ ^* x \Rightarrow z \succ ^* y$. Hence, $V_1 \subseteq {\hat{V}}_1$ and $\# V_1 \le \#{\hat{V}}_1$. Furthermore, $y \succ ^* z \Rightarrow x \succ ^* z$. Hence, ${\hat{V}}_2 \subseteq V_2$ and $\#{\hat{V}}_2 \le \#V_2$. Accordingly, $[\# V_1 - \#{\hat{V}}_1] + [\#{\hat{V}}_2 - \#V_2] \le 0$ and, therefore:

$$\begin{aligned} 0\ge & {} [\# V_1 - \# V_2] + [\#{\hat{V}}_2 - \# {\hat{V}}_1]\\= & {} \sum _{z\in S_1}\left[ |\succ -\succ ^*|_{\{x,z\}}-|\succ '-\succ ^*|_{\{x,z\}}\right] \\&+ \sum _{z\in S_1}\left[ |\succ -\succ ^*|_{\{y,z\}}-|\succ '-\succ ^*|_{\{y,z\}}\right] \end{aligned}$$

Hence, $|\succ -\succ ^*|_{S}-|\succ '-\succ ^*|_{S} < 0$.$\square$

Before stating the next lemma, we introduce some notation. For any choice problem $S \in {\mathcal {P}}(X)$ and any alternative $x\in S$ define the ecological support for x in this choice problem by the following:

$$\begin{aligned} \gamma _S(x)=\alpha \#\{i\in I_n: c_i(S)=x\}-(1-\alpha )\#\{j\in I_n^c: c_j(S)=x\}. \end{aligned}$$

That is:

$$\begin{aligned} \Gamma (S)=\mathop {\hbox {argmax}}\limits _{x \in S} \gamma _S(x). \end{aligned}$$

Lemma A.2

Suppose that$c_i$, $i=1,\dots , n-1$, satisfies WARP, $c_n$is an ESH, and Homophily holds. If$x \succ ^* y$, $x, y \in X$, then for any$S \in {\mathcal {P}}(X)$with$x, y \in S$, $\gamma _S(x) \ge \gamma _S(y)$.

Proof

Let $x, y \in S$ and $x \succ ^* y$. First, note that if there exists no $i \in {\mathcal {I}}$, such that $c_i(S) = x$ or $c_i(S) = y$, then $\gamma _S(x)= \gamma _S(y)=0$ and our desired conclusion follows immediately. Therefore, consider the case where there exists $i \in {\mathcal {I}}$ with $c_i(S) = x$ or $c_i(S) = y$. This means that there exists at least one (xy)-TOP pair, $\succ$ and $\succ '$, on S, such that $\succ _{iS} = \succ$ or $\succ _{iS}$$=~\succ '$. Furthermore, since $x \succ ^* y$, from the last Lemma, we know that for any (xy)-TOP pair, $\succ$ and $\succ '$, on S with, say, x the top element of $\succ$ and y of $\succ '$, we have $|\succ -\succ ^*_S|_S<|\succ '-\succ ^*_{S}|_S$. Therefore, under this case, the collection of relevant (xy)-TOP pairs on S is non-empty. Let $(\succ ^k, \succ '^k)_{k=1}^K$ be the set of all relevant (xy)-TOP pairs on S, where $x\succ ^k z$ for all $z \in S{\setminus } \{x\}$ and $y \succ '^k z$, for all $z \in S{\setminus } \{y\}$, for all $k=1, \dots , K$. As such, the Homophily condition implies that $\sigma ^S_{xy}$, the homophily index defined with respect to this collection of relevant (xy)-TOP pairs, is such that

$$\begin{aligned} \sigma ^S_{xy} = \sum _{k=1}^K w^S_{xy}(\succ ^k, \succ '^k)\cdot \sigma ^S_{xy}(\succ ^k, \succ '^k) \ge \frac{1}{2}. \end{aligned}$$

That is:

$$\begin{aligned}&\sum _{k=1}^K \left[ \frac{\alpha \#[I_n\cap (N(\succ ^k)\cup N(\succ '^{k}))]+ (1-\alpha )\#[I_n^c\cap (N(\succ ^k)\cup N(\succ '^{k}))]}{\sum _{{\hat{k}}=1}^K \{\alpha \#[I_n\cap (N(\succ ^{{\hat{k}}})\cup N(\succ '^{{\hat{k}}}))]+ (1-\alpha )\#[I_n^c\cap (N(\succ ^{{\hat{k}}})\cup N(\succ '^{{\hat{k}}}))]\}}\right] \\&\quad \times \left[ \frac{\alpha \#[I_n\cap N(\succ ^k)]+ (1-\alpha )\#[I_n^c\cap N(\succ '^{k})]}{\alpha \#[I_n\cap (N(\succ ^k)\cup N(\succ '^{k}))]+ (1-\alpha )\#[I_n^c\cap (N(\succ ^k)\cup N(\succ '^{k}))]}\right] \ge \frac{1}{2}. \end{aligned}$$

That is:

$$\begin{aligned}&\sum _{k=1}^K \left[ \frac{\alpha \#[I_n\cap (N(\succ ^k)\cup N(\succ '^{k}))]+ (1-\alpha )\#[I_n^c\cap (N(\succ ^k)\cup N(\succ '^{k}))]}{\sum _{{\hat{k}}=1}^K \{\alpha \#[I_n\cap (N(\succ ^{{\hat{k}}})\cup N(\succ '^{{\hat{k}}}))]+ (1-\alpha )\#[I_n^c\cap (N(\succ ^{{\hat{k}}})\cup N(\succ '^{{\hat{k}}}))]\}}\right] \\&\quad \times \left[ \frac{\alpha \#[I_n\cap N(\succ ^k)]+ (1-\alpha )\#[I_n^c\cap N(\succ '^{k})]}{\alpha \#[I_n\cap (N(\succ ^k)\cup N(\succ '^{k}))]+ (1-\alpha )\#[I_n^c\cap (N(\succ ^k)\cup N(\succ '^{k}))]}- \frac{1}{2}\right] \ge 0. \end{aligned}$$

Denote:

$$\begin{aligned} \theta =\sum _{{\hat{k}}=1}^K \{\alpha \#[I_n\cap (N(\succ ^{{\hat{k}}})\cup N(\succ '^{{\hat{k}}}))]+ (1-\alpha )\#[I_n^c\cap (N(\succ ^{{\hat{k}}})\cup N(\succ '^{{\hat{k}}}))]\}, \end{aligned}$$

and we have that

$$\begin{aligned}&\frac{1}{2\theta }\sum _{k=1}^K\{2(\alpha \#[I_n\cap N(\succ ^k)]+ (1-\alpha )\#[I_n^c\cap N(\succ '^{k})]) -\alpha \#[I_n\cap (N(\succ ^k)\cup N(\succ '^{k}))]\\&\qquad - (1-\alpha )\#[I_n^c\cap (N(\succ ^k)\cup N(\succ '^{k}))]\} \ge 0\\&\quad \Rightarrow \sum _{k=1}^K\{2\alpha \#[I_n\cap N(\succ ^k)]+ 2(1-\alpha )\#[I_n^c\cap N(\succ '^{k})] -\alpha \#[I_n\cap N(\succ ^k)]\\&\qquad -\alpha \#[I_n\cap N(\succ '^{k})] - (1-\alpha )\#[I_n^c\cap N(\succ ^k)] - (1-\alpha )\#[I_n^c\cap N(\succ '^{k})]\} \ge 0. \end{aligned}$$

That is:

$$\begin{aligned}&\sum _{k=1}^K\{\alpha \#[I_n\cap N(\succ ^k)]+ (1-\alpha )\#[I_n^c\cap N(\succ '^{k})]\\&\qquad -\alpha \#[I_n\cap N(\succ '^{k})] - (1-\alpha )\#[I_n^c\cap N(\succ ^k)]\} \ge 0\\&\quad \Rightarrow \alpha \sum _{k=1}^K\#[I_n\cap N(\succ ^k)] - (1-\alpha )\sum _{k=1}^K\#[I_n^c\cap N(\succ ^k)]\\&\quad \ge \alpha \sum _{k=1}^K\#[I_n\cap N(\succ '^{k})]-(1-\alpha )\sum _{k=1}^K\#[I_n^c\cap N(\succ '^{k})] \\&\quad \Rightarrow \alpha \sum _{k=1}^K\#\{i\in I_n:\;\succ _{iS}=\succ ^k\} - (1-\alpha )\sum _{k=1}^K\#\{j\in I_n^c:\;\succ _{jS}=\succ ^k\}\\&\quad \ge \alpha \sum _{k=1}^K\#\{i\in I_n:\;\succ _{iS}=\succ '^{k}\}-(1-\alpha )\sum _{k=1}^K\#\{j\in I_n^c:\;\succ _{jS}=\succ '^{k}\}\\&\quad \Rightarrow \alpha \#\{i\in I_n: x\succ _i z,\forall z\in S\backslash \{x\}\} - (1-\alpha )\#\{j\in I_n^c: x\succ _j z,\forall z\in S\backslash \{x\}\}\\&\quad \ge \alpha \#\{i\in I_n: y\succ _i z,\forall z\in S\backslash \{y\}\} - (1-\alpha )\#\{j\in I_n^c: y\succ _j z,\forall z\in S\backslash \{y\}\}. \end{aligned}$$

That is:

$$\begin{aligned}&\alpha \#\{i\in I_n: c_i(S)=x\} - (1-\alpha )\#\{j\in I_n^c: c_j(S)=x\}\\&\quad \ge \alpha \#\{i\in I_n: c_i(S)=y\} - (1-\alpha )\#\{j\in I_n^c: c_j(S)=y\} \end{aligned}$$

Or, $\gamma _S(x) \ge \gamma _S(y)$. $\square$

1.2 Proof of Propositions

1.2.1 Proof of Proposition 4.1

Consider $x\in \Gamma (S)$ and $y\in S{\setminus }\Gamma (S)$. That is, $\gamma _S(x) > \gamma _S(y)$. Clearly, it cannot be the case that $y \succ ^* x$, for then, by Lemma A.2, $\gamma _S(y) \ge \gamma _S(x)$.

Let $x^* \in S$ be such that $x^* \succ ^* y$, for all $y \in S{\setminus }\{x^*\}$. By Lemma A.2, $\gamma _S(x^*) \ge \gamma _S(y)$, for all $y \in S{\setminus }\{x^*\}$. Hence, $x^* \in \Gamma (S)$ and $c_n(S) = x^*$.

Let $x \succ ^* y$. By Lemma A.2, $\gamma _{\{x,y\}}(x) \ge \gamma _{\{x,y\}}(y)$. Therefore, either $\Gamma (\{x,y\})=\{x,y\}$ or $\Gamma (\{x,y\})=\{x\}$. In either case, $c_n(\{x,y\})=x$, which implies $x \succ _n y$. Hence, $\succ ^*~\subseteq ~\succ _n$.

1.2.2 Proof of Proposition 4.2

Consider any sets $S,T\in {\mathcal {P}}(X)$, such that $c_n(S)=x$, $y\in S$, $x\in T$. To establish that $c_n$ satisfies WARP, we need to show that $c_n(T)\ne y$. If $y \notin T$, the conclusion is immediate. Therefore, assume that $y \in T$.

Our desired conclusion follows if we can show that $c_n(S\cap T)=x$. We first show that $x\in \Gamma (S\cap T)$. Clearly, $\Gamma (S\cap T)\ne \emptyset$. Pick $w \in \Gamma (S\cap T)$. If $w=x$, we have reached the desired conclusion. Therefore, consider $w\ne x$. First, note that it cannot be the case that $w\succ ^* x$. Clearly, this cannot be the case if $w \in \Gamma (S)$ since $c_n(S)=x \succ ^* y$, for all $y \in \Gamma (S){\setminus }\{x\}$. On the other hand, if $w\in S{\setminus }\Gamma (S)$, then $\gamma _S(x) > \gamma _S(w)$, by the definition of $\Gamma (S)$. However, if $w\succ ^* x$, then by Lemma A.2, $\gamma _S(w)\ge \gamma _S(x)$! Second, if $x \succ ^* w$ then, by Lemma A.2, we have $\gamma _{S\cap T}(x) \ge \gamma _{S\cap T}(w)$. Hence, $x \in \Gamma (S\cap T)$. Finally, consider the case that neither $x \succ ^* w$ nor $w\succ ^* x$. In this case, first, it has to be that $w\in S{\setminus }\Gamma (S)$ for if $w \in \Gamma (S)$, then $x\succ ^* w$. Second, $\Gamma (\{x,w\})\ne \{x,w\}$ for if this were so, the ESH cannot determine what $c_n(\{x,w\})$ is. Therefore, $\Gamma (\{x,w\}) = \{x\}$ or $\Gamma (\{x,w\}) = \{w\}$. However, $\Gamma (\{x,w\}) \ne \{w\}$, since $\Gamma (\{x,w\}) = \{w\}$, $x\in \Gamma (S)$ and $w\in S{\setminus }\Gamma (S)$ violates Ecological Shortlisting Consistency. Hence, $\Gamma (\{x,w\}) = \{x\}$. Therefore, by the same condition, $w \in \Gamma (S\cap T)$ implies that $x \in \Gamma (S\cap T)$. That is, we have shown that $x \in \Gamma (S\cap T)$ under all possible cases.

Finally, note that since $c_n$ is an ESH, the set $\Gamma (S\cap T)$ has a $\succ ^*$-maximal element. Arguments made above establish that this element has to be x and not some $w\ne x$. To reiterate, if it was such a w, then clearly $w\in S{\setminus }\Gamma (S)$, i.e., $\gamma _S(x) > \gamma _S(w)$. However, $w\succ ^* x$ would imply by Lemma A.2 that $\gamma _S(w)\ge \gamma _S(x)$! Hence, $x \succ ^* y$, for all $y \in \Gamma (S\cap T){\setminus }\{x\}$ and $c_n(S\cap T)=x$.

We have already shown above in the proof of Proposition 4.1 that if Homophily holds, then $\succ ^*$$\subseteq ~\succ _n$. Furthermore, it can be shown that if $c_n$ satisfies WARP, then the base relation $\succ _n$ is complete and transitive and, indeed, rationalizes $c_n$. We omit those details here as they are quite standard.