On Machina’s paradoxes and limited attention

Abstract

Limited attention and similarity of some of the states of the world together may nudge an agent to perceive the “grand world” as a collection of “small worlds”. We use this idea as an explanation for some of the ambiguity paradoxes posed by Machina (Am Econ Rev 99(1):385–392, 2009; Am Econ Rev 104(12):3814–3840, 2014) as a challenge to the prominent ambiguity theories. In addition, we propose a measure of rationality based on a number of potential violations of preference for the first-order stochastic dominance. We use this measure to explore how departures from perfect attention can be improved through changes to the structure of the “small worlds”.

Appendix

Consider two states $s_i,s_j\in \varOmega$ and lotteries $l_i(x)$ and $l_j(x)$ with positive payoff x, and let $P(s_i)\ge P(s_j)$. Also normalize $u(0)=0$ and $\phi (u(0))=0$ for simplicity. Denote $s_i\in A_i$ and $s_j\in A_j$.

Lemma 1

Suppose that $s_i$ and $s_j$ are from two events with at least one of them being non-trivial. If, in addition to $P(s_i)\ge P(s_j)$, we have one of the following:

• 1. $P(s_i|A_i)=P(s_j|A_j)$

• 2. $P(s_i|A_i)\ge P(s_j|A_j)$ and $P(A_i)\ge P(A_j),$

then $s_i$ and $s_j$ never generate possible FOSD violation.

Proof

For a potential violation of FOSD, we need

$$\begin{aligned}&V_\pi (l_i(x))<V_\pi (l_j(x))\Leftrightarrow P(A_i)\phi (P(s_i|A_i)u(x))<P(A_j) \phi (P(s_j|A_j)u(x))\\&\quad \phi \left( \frac{P(s_i|A_i)}{P(s_j|A_j)}P(s_j|A_j)u(x)\right) <\frac{P(A_j)}{ P(A_i)}\phi (P(s_j|A_j)u(x)). \end{aligned}$$

Denote $y=P(s_j|A_j)u(x)>0$; then:

$$\begin{aligned} \phi \left( \frac{P(s_i)}{P(s_j)}\frac{P(A_j)}{P(A_i)}y\right) <\frac{P(A_j)}{ P(A_i)}\phi (y). \end{aligned}$$

(2)

First, suppose that $P(s_i|A_i)=P(s_j|A_j)$. In this case, inequality 2 simplifies to $P(A_j)>P(A_i)$, which leads to a contradiction, because $1<\frac{P(s_i) }{P(s_j)}=\frac{P(A_i)}{P(A_j)}$.

Second, consider the case $P(s_i|A_i)\ge P(s_j|A_j)$ and $P(A_i)\ge P(A_j)$. Notice $\frac{P(s_i)}{P(s_j)}\frac{P(A_j)}{P(A_i)}\ge 1$, so the left side in inequality 2 is greater than $\phi (y)$, because $\phi$ is strictly increasing:

$$\begin{aligned} \phi \left( \frac{P(s_i)}{P(s_j)}\frac{P(A_j)}{P(A_i)}y\right) \ge \phi (y) >\frac{P(A_j)}{P(A_i)}\phi (y). \end{aligned}$$

Hence, we get a contradiction. $\square$

Proof of Proposition 1

Three cases are possible:

• 1. At least one of the states belongs to a non-trivial event and both states belong to different events. In this case, it is always possible to find a probability distribution P and a strictly increasing function $\phi$, such that the inequality 2 holds. Keeping in mind Lemma 1, choose the probability distribution P, such that $P(s_i|A_i)<P(s_j|A_j)$ and $P(s_i)\ge P(s_j)$. In addition, for example, let $\phi (y)=y^t$ with $t\in (0,+\infty )$. Then, inequality 2 becomes

  $$\begin{aligned} \left( \frac{P(s_i)}{P(s_j)}\frac{P(A_j)}{P(A_i)}y\right) ^t< & {} \frac{P(A_j)}{ P(A_i)}y^t\\ \left( \frac{P(s_i)}{P(s_j)}\frac{P(A_j)}{P(A_i)}\right) ^t< & {} \frac{P(A_j)}{ P(A_i)}. \end{aligned}$$

  We have $\frac{P(s_i)}{P(s_j)}\frac{P(A_j)}{P(A_i)}<1$; thus, there exists t large enough for the inequality to hold. Hence, two states with at least one of them from a non-trivial event may generate the possible FOSD violation.

• 2. Both states are singletons. Consider inequality 2 again. Note that for two singletons, we have $P(A_i)=P(s_i)$ and $P(A_j)=P(s_j)$; then, the inequality simplifies to

  $$\begin{aligned} \phi (y)<\frac{P(s_j)}{P(s_i)}\phi (y)\Rightarrow P(s_i)<P(s_j), \end{aligned}$$

  leading to contradiction. Hence, two singletons never generate FOSD violation.

• 3. Both states are from the same event A. Consider the inequality 2 again. In the case of the same event, we have $P(A_i)=P(A_j)=P(A)$ and

  $$\begin{aligned} \phi \left( \frac{P(s_i)}{P(s_j)}y\right) <\phi (y). \end{aligned}$$

  However, $\frac{P(s_i)}{P(s_j)}y>y$ and $\phi$ is a strictly increasing function; thus, we get a contradiction. Hence, two states from the same event never generate FOSD violation.

To summarize, the FOSD violations may come only from states belonging to different events with at least one of the events being non-trivial. Now, suppose that $\pi$ consists of $n_s$ singletons and also non-trivial events $A_1,\dots ,A_k$ with the state numbers $n_1,\dots ,n_k$. Note that each state from each non-trivial event $A_i$ generates FOSD violations with all states that do not belong to $A_i$, hence delivering $n_i(N-n_i)$ potential FOSD violations. Also, singletons generate FOSD violations with all other states except singletons, implying $n_s(N-n_s)$ FOSD violations. Finally, we simply need to add all FOSD violations and take into account that we are double counting them:

$$\begin{aligned} \begin{aligned} \mu (\pi )&=0.5(n_1(N-n_1)+\dots +n_k(N-n_k)+n_s(N-n_s))\\&=0.5\left( N^2-\sum _{i=1}^kn_i^2-n_s^2\right) . \end{aligned} \end{aligned}$$

$\square$