Categorical consideration and perception complementarity

Abstract

We introduce the notion of perception complementarity in a representation for random choice rules, which we call Complemented Perception Categorical Consideration Representation. The model is similar in nature to the perception-adjusted Luce model of Echenique et al. (Math Soc Sci 93:67–76, 2018), but allows perception to be menu-dependent. We provide a representation theorem and an identification result. Furthermore, we consider special cases of the model, including the case when it reduces down to the Random Consideration Set Rule of Manzini and Mariotti (Econometrica 82(3):1153–1176, 2014).

Appendix

1.1 Proof of Observation 1

Proof

Suppose Axioms 1–3 are satisfied. Then, we can derive a revealed visibility order $\overset{{r^{*}}}{\geqslant }$ on X. Now, for any $A\in K$, we can use $\overset{{r^{*}}}{\geqslant }$ to define $\overset{A}{>^*}$. Since we restricted attention to distinct revealed categories $R_i\ne R_j$ when defining $\overset{A}{>^*}$, it is irreflexive. Now, we prove it is semiconnex. Take any $A\in K$ and take any $R_i\in \mathcal{{C}}^r$ and $R_j\in \mathcal{{C}}^r$ such that $R_i\ne R_j$ and $R_i \cap A \ne \emptyset$ and $R_j\cap A \ne \emptyset$. Now, take the set $Y=\lbrace x\in X:\, x\in \left( R_i \cap A \right) \cup \left( R_j \cap A \right) \rbrace$. Since $(X,\overset{r}{\geqslant ^*})$ is a totally ordered set and Y is a finite, nonempty subset of X, it has a unique greatest element, call it, $x^*$. Now, we must have one of two cases. If $x^*\in R_i \cap A$, we have $x^* \overset{r}{\geqslant ^*} x'$ for all $x' \in R_j \cap A$. If $x^*\in R_j \cap A$, then we have $x^* \overset{r}{\geqslant ^*} x'$ for all $x' \in R_i \cap A$. In the former case, by definition of $\overset{A}{>^*}$, we have $R_i \overset{A}{>^*} R_j$. In the latter, we have $R_j \overset{A}{>^*} R_i$. Therefore, $\overset{A}{>^*}$ is semiconnex. Finally, we prove transitivity. Suppose $R_i \overset{A}{>^*} R_j$ and $R_j \overset{A}{>^*} R_k$. Then, we know that there is a $x_j^* \in R_j \cap A$ such that $x_j^* \overset{r}{\geqslant ^*} x_k$ for all $x_k\in R_k \cap A$ and $R_j\ne R_k$. Similarly, there is a $x_i^* \in R_i \cap A$ such that $x_i^* \overset{r}{\geqslant ^*} x_j$ for all $x_j\in R_j \cap A$ and $R_i\ne R_j$. In particular, $x_i^*\overset{r}{\geqslant ^*} x_j^*$. Now, if $R_i=R_k$, then we would also have that $x_j^* \overset{r}{\geqslant ^*} x_i^*$, and since $x_i^*\ne x_j^*$, this would violate antisymmetry of $\overset{r}{\geqslant ^*}$, which is a total order. Therefore, $R_i\ne R_k$. Furthermore, by transitivity of $\overset{r}{\geqslant ^*}$, we have that $x_i^* \overset{r}{\geqslant ^*} x_k$ for all $x_k \in R_k \cap A$. By definition, this gives us $R_i \overset{A}{>^*} R_k$. Therefore, $\overset{A}{>^*}$ is transitive. $\square$

1.2 Proof of Observation 2

Proof

Let Axiom 1 be satisfied so that we can derive $\mathcal{{C}}^r$ and $\overset{r}{\geqslant }$. Suppose the binary relation $\overset{r}{\geqslant }$ satisfies Axioms 2–3. Take two extensions of $\overset{r}{\geqslant }$ that are total orders on X. Call these two extensions $\overset{r}{\geqslant '}$ and $\overset{r}{\geqslant ^*}$. Now, for any menu $A\in K$, we denote the menu-dependent perception derived from $\overset{r}{\geqslant '}$ as $\overset{A}{>'}$, and the one derived from $\overset{r}{\geqslant ^*}$ as $\overset{A}{>^*}$. We want to show that $\overset{A}{>'} = \overset{A}{>^*}$. So, take any $R_i,R_j\in \mathcal{{C}}^r$ such that $R_i\ne R_j$, and we need to show that $R_i \overset{A}{>^*} R_j$ iff $R_i \overset{A}{>'} R_j$ (if such $R_i$ and $R_j$ does not exist, there is nothing to show, so we consider the case that they exist). We need to consider only $R_i \ne R_j$ by Observation 1 (in particular, irreflexivity). By definition,

$$\begin{aligned} R_i\overset{A}{>'} R_j \iff \, \exists \, x_i\in R_i \cap A {\text { such \,that }}\, x_i \overset{r}{\geqslant '} x_j \, \text{ }\forall x_j\in R_j \cap A. \end{aligned}$$

Now, by quasi-completeness, for any $x\in R_i \cap A$ and any $x'\in R_j \cap A$ such that $R_i\ne R_j$, we have either $x_i \overset{r}{\geqslant } x_j$ or $x_j \overset{r}{\geqslant } x_i$. Since $\overset{r}{\geqslant '}$ and $\overset{r}{\geqslant ^*}$ are extensions of $\overset{r}{\geqslant }$, if $x \overset{r}{\geqslant } x'$, then we have $x \overset{r}{\geqslant '} x'$ and $x \overset{r}{\geqslant ^*} x'$. Also, if $x \overset{r}{\geqslant '} x'$, then we cannot have $x' \overset{r}{\geqslant } x$ because this would mean that we also have $x' \overset{r}{\geqslant '} x$, and this would violate antisymmetry of $\overset{r}{\geqslant '}$, but since $\overset{r}{\geqslant }$ is quasi-complete, we must have $x \overset{r}{\geqslant } x'$. The last sentence works for $\overset{r}{\geqslant '}$ replaced with $\overset{r}{\geqslant ^*}$ as well. Therefore, for any $x_i\in R_i \cap A$ and $x_j\in R_j \cap A$ such that $R_i \ne R_j$, we have $x_i \overset{r}{\geqslant '} x_j \iff x_i \overset{r}{\geqslant } x_j \iff x_i \overset{r}{\geqslant ^*} x_j .$ Therefore, the above becomes

$$\begin{aligned} R_i\overset{A}{>'} R_j&\iff \, \exists \, x_i\in R_i \cap A {\text { such \, that }}\, x_i \overset{r}{\geqslant '} x_j \, \text{ }\forall x_j\in R_j \cap A \\&\iff \, \exists \, x_i\in R_i \cap A {\text { such \, that }}\, x_i \overset{r}{\geqslant } x_j \,\text{ } \forall x_j\in R_j \cap A \\&\iff \,\exists \, x_i\in R_i \cap A {\text { such\, that }} \, x_i \overset{r}{\geqslant ^*} x_j \,\text{ } \forall x_j\in R_j \cap A \\&\iff R_i\overset{A}{>^*} R_j. \end{aligned}$$

Then, it is immediate that $q^*=q'$. $\square$

1.3 Proof of Theorem 1

Proof

Sufficiency:

We will first prove the sufficiency of Axioms 1–4 for the CPCC Representation. Take a nondegenerate random choice rule $\rho$ such that the induced binary relation $\overset{c}{\sim }$ satisfies Axiom 1 and $\overset{r}{\geqslant }$ satisfies Axioms 2–3, and such that the hazard rate function defined by using any total order $\overset{r}{\geqslant ^*}$, which is an extension of $\overset{r}{\geqslant }$, satisfies Axiom 4.

Step 1: Constructing the total order and partition for a CPCC

This step is easy as we let any $\overset{r}{\geqslant ^*}$, which is an extension of $\overset{r}{\geqslant }$ to a total order on X, be the total order on X for the definition of a CPCC. We also let the set of revealed categories $\mathcal{{C}}^r$ be the partition of X for the definition of a CPCC.

Step 2: Constructing the function $\gamma$

By the proof of Observation 1, for any menu $A\in K$, the menu-dependent perception $\overset{A}{>^*}$ induced by $\overset{r}{\geqslant ^*}$ is a strict total order on the set $\lbrace R\in \mathcal{{C}}^r : R\cap A\ne \emptyset \rbrace .$ Therefore, the hazard rate function is well defined.

By Axiom 4, for any $R_i \in \mathcal{{C}}^r$ and any $A,B\in K$ such that $A\cap R_i \ne \emptyset$ and $B\cap R_i\ne \emptyset$, we have

$$\begin{aligned} \sum _{x'\in R_i} q(x',A)=\sum _{x'\in R _i}q(x',B) . \end{aligned}$$

For each $A\in K$, let $x^i_A$ be any alternative in $A\cap R_i$. By definition of the hazard rate, this means that

$$\begin{aligned} \frac{\rho (R_i,A)}{1-\rho (A_{x^i_A},A)}=\frac{\rho (R_i,B)}{1-\rho (B_{x^i_B},B)}=c_i \end{aligned}$$

for some $c_i\in (0,1)$. (We know it is in this interval by definition of a nondegenerate random choice rule.) We let $\gamma (R_i)=c_i$, and will do the same for any $R_j\in \mathcal{{C}}^r$ by using some $D\in K$ such that $D\cap R_j\ne \emptyset$. That is, let $\gamma (R_j)=c_j ,$ where

$$\begin{aligned} c_j=\frac{\rho (R_j,D)}{1-\rho (D_{x^j_D},D)} \end{aligned}$$

for some $D\in K$ such that $D\cap R_j\ne \emptyset$. Therefore, we have the function $\gamma :\mathcal{{C}}^r \rightarrow (0,1)$, and for any $R_i \in \mathcal{{C}}^r$ and $A\in K$ such that $R_i \cap A \ne \emptyset$, we have

$$\begin{aligned} \rho (R_i,A)=\gamma (R_i)[1-\rho (A_{x^i_A},A)]. \end{aligned}$$

(1)

Step 3: Constructing the function u

For any $R_i\in \mathcal{{C}}^r$, we define the set $\mathcal{{M}}_i=\lbrace A\in K| \,A\cap R_i\ne \emptyset \rbrace$ and the function $\rho _{R_i}:R_i \times \mathcal{{M}}_i \rightarrow [0,1]$ such that for any $M\in \mathcal{{M}}_i$ and $x\in R_i$,

$$\begin{aligned} \rho _{R_i}(x, M)=\frac{\rho (x,M)}{\rho (R_i,M)}. \end{aligned}$$

Since $\sum _{x\in R_i \cap M}\rho _{R_i}(x,M)=1$ for all $M\in \mathcal{{M}}_i$, and $\rho _{R_i}(x,M)>0$ for all $M\in \mathcal{{M}}_i$ and $x\in M\cap R_i$, and

$$\begin{aligned} \frac{\rho _{R_i}(x,M_1)}{\rho _{R_i}(y,M_1)}=\frac{\frac{\rho (x,M_1)}{\rho (R_i,M_1)}}{\frac{\rho (y,M_1)}{\rho (R_i,M_1)}}=\frac{\rho (x,M_1)}{\rho (y,M_1)}=\frac{\rho (x,M_2)}{\rho (y,M_2)}=\frac{\frac{\rho (x,M_2)}{\rho (R_i,M_2)}}{\frac{\rho (y,M_2)}{\rho (R_i,M_2)}}=\frac{\rho _{R_i}(x,M_2)}{\rho _{R_i}(y,M_2)} \end{aligned}$$

for all $M_1,M_2\in \mathcal{{M}}_i$ and x and y such that $x,y\in M_1\cap R_i$ and $x,y\in M_2\cap R_i$ by construction of $\mathcal{{C}}^r$, there is a function $u_i:R_i\rightarrow {\mathbb {R}}_{++}$ such that for any $M\in \mathcal{{M}}_i$ and any $x\in R_i\cap M$,

$$\begin{aligned} \rho _{R_i}(x,M)=\frac{u_i(x)}{\sum _{\lbrace y:\,y\in R_i \cap M \rbrace } u_i(y)}. \end{aligned}$$

Since each $u_i$ was defined on $R_i$ for all $R_i$ in $\mathcal{{C}}^r$, which partitions X, we can redefine $u:X\rightarrow {\mathbb {R}}_{++}$ where for each $x\in X$, if $x\in R_i$, we let $u(x)=u_i(x)$.

Step 4: Showing Steps 1–3 gives CPCC Representation of $\rho$

Now, for any $A\in K$, let $\lbrace R_{i_1},\ldots ,R_{i_a}\rbrace$ be a collection of revealed categories such that it contains every revealed category with nonempty intersection with A and $R_{i_s} \overset{A}{>^*} R_{i_{s+1}}$ for all $1\le s <a$. (We can do this because $\overset{A}{>^*}$ is a strict total order on a finite set.) Now, take any $R_{i_j}$ from the collection and any $x_j\in R_{i_j}\cap A$. If $j=1$, then $A_{x_j}$ is empty and we have

$$\begin{aligned} \rho (R_{i_j},A)=\gamma (R_{i_j}). \end{aligned}$$

Otherwise, take $j-1$, and let $x_{j-1}$ be any element in $R_{i_{j-1}}\cap A$. Then, we have

$$\begin{aligned} \rho (R_{i_j},A)&=[1-\rho (A_{x_{j-1}},A)-\rho (R_{i_{j-1}},A)]\gamma (R_{i_j}) \\&=\left( 1-\rho (A_{x_{j-1}},A)-[1-\rho (A_{x_{j-1}},A)]\gamma (R_{i_{j-1}})\right) \gamma (R_{i_j}) \,\,\,{\text { by (1)}}\\&=[1-\rho (A_{x_{j-1}},A)]\left( 1-\gamma (R_{i_{j-1}})\right) \gamma (R_{i_{j}})\,\,{\text { by factoring.}} \end{aligned}$$

Again, if $j-1=1$, this expression becomes just

$$\begin{aligned} \left( 1-\gamma (R_{i_{j-1}})\right) \gamma (R_{i_{j}}). \end{aligned}$$

Otherwise, take $j-2$, and if we take any $x_{j-2}\in R_{i_{j-2}}\cap A,$ we will have

$$\begin{aligned}\rho (R_{i_{j}},A)&=\left( 1-\rho (A_{x_j},A)\right) \gamma (R_{i_{j}}) \\&=\left[ 1-\rho (A_{x_{j-1}},A)-\rho (R_{i_{j-1}},A)\right] \gamma (R_{i_{j}}) \\&=\left[ 1-\rho (A_{x_{j-1}},A)-\left( 1-\rho (A_{x_{j-1}},A)\right) \gamma (R_{i_{j-1}})\right] \gamma (R_{i_{j}}) \\&=\left[ (1-\rho (A_{x_{j-1}},A))\left( 1-\gamma (R_{i_{j-1}})\right) \right] \gamma (R_{i_{j}})\\&= \left[ (1-\rho (A_{x_{j-2}},A)-\rho (R_{i_{j-2}},A) )\left( 1-\gamma (R_{i_{j-1}})\right) \right] \gamma (R_{i_{j}})\\&=\left[ \left( 1-\rho (A_{x_{j-2}},A)-\left( 1-\rho (A_{x_{j-2}},A)\right) \gamma (R_{i_{j-2}})\right) \left( 1-\gamma (R_{i_{j-1}}) \right) \right] \gamma (R_{i_{j}})\\&=\left[ \left( 1-\rho (A_{x_{j-2}},A)\right) \left( 1-\gamma (R_{i_{j-2}})\right) \left( 1-\gamma (R_{i_{j-1}})\right) \right] \gamma (R_{i_{j}})\\&=\left[ \left( 1-\rho (A_{x_{j-2}},A)\right) \right] \left( 1-\gamma (R_{i_{j-2}})\right) \left( 1-\gamma (R_{i_{j-1}}) \right) \gamma (R_{i_{j}}). \end{aligned}$$

Again, if $j-2=1$ is the greatest element, this expression becomes just

$$\begin{aligned} \rho (R_{i_{j}},A)= \left( \prod _{ \lbrace R'\in \mathcal{{C}}^r : \, R'\overset{A}{>^*} R_{i_{j}} \rbrace } 1-\gamma (R')\right) \gamma (R_{i_{j}}). \end{aligned}$$

(2)

If $j-2\ne 1$, we can repeat this process (taking the element with the index one lower and factoring out) T times until we get to $j-T=1$, and the expression thus becomes this equation (2).

Now, take any $A\in K$ and any $x\in A$. Let $R^x$ be the revealed category that contains x. Then, we have

$$\begin{aligned}\rho _{R^x}(x,A)&=\frac{\rho (x,A)}{\rho (R^x,A)}=\frac{u(x)}{\sum _{x'\in R^x\cap A}u(x')} \\&\implies \rho (x,A)=\rho (R^x,A)\frac{u(x)}{\sum _{x'\in R^x \cap A}u(x')}, \end{aligned}$$

and by equation (2), this is equal to

$$\begin{aligned} \left( \prod _{ \lbrace R\in {\mathcal{{C}}}^r : \, R\overset{A}{>^*} R^x \rbrace } 1-\gamma (R)\right) \gamma (R^x) \frac{u(x)}{\sum _{x'\in R^x \cap A}u(x')}. \end{aligned}$$

Therefore, $\gamma ,u,\overset{r}{\geqslant ^*}, {\text {and }} \mathcal{{C}}^r$ that we just constructed gives a CPCC Representation.

Necessity:

For the necessity direction, suppose a nondegenerate random choice rule $\rho$ has a CPCC Representation with $\overset{v}{\geqslant },\mathcal{{C}},u$, and $\gamma$.

Step 1: Necessity of Axiom 1

Step 1.1: For any $C_i\in \mathcal {C}$ and $x_i\in C_i$ and $x_j\in {X\setminus C_i}$ , we cannot have $x_i \overset{c}{\sim } x_j$ or $x_j \overset{c}{\sim } x_i$ .

Take any $C_i\in \mathcal {C}$ and $x_i\in C_i$ and $x_j\in {X \setminus C_i}$. Denote the category of $x_j$ as $C_j$ for some $C_j\in \mathcal {C}$. Consider the case that $x_i\overset{v}{\geqslant } x_j$. Take the menu $A=\lbrace x_i,x_j \rbrace$. Then, we have $C_i \overset{A}{\succ } C_j$, so $\rho (x_i,\lbrace x_i \rbrace )=\gamma (C_i)$ and $\rho (x_i, \lbrace x_i,x_j \rbrace )=\gamma (C_i).$ Therefore, we do not have $\rho (x_i, \lbrace x_i,x_j \rbrace )<\rho (x_i,\lbrace x_i \rbrace )$, so we do not have $x_i \overset{c}{\sim } x_j$ or $x_j \overset{c}{\sim } x_i$. We can reverse the role of $x_i$ and $x_j$ in this argument for the case that $x_j\overset{v}{\geqslant } x_i$ and show that we do not have $x_i \overset{c}{\sim } x_j$ or $x_j \overset{c}{\sim } x_i$. Therefore, we can conclude that if $x_i$ and $x_j$ in X are not from the same $C_i\in \mathcal {C}$, then we cannot have $x_i \overset{c}{\sim } x_j$ or $x_j \overset{c}{\sim } x_i$.

Step 1.2: Use 1.1 to show necessity of Axiom 1

Take any $x_i,x'_i,x^*_i$ such that $x_i \overset{c}{\sim } x'_i$ and $x'_i \overset{c}{\sim } x^*_i$. If $x_i=x^*_i$, then it is immediate that $x_i \overset{c}{\sim } x_i^*$ by the way we defined $\overset{c}{\sim }$. So, take the case that $x_i\ne x^*_i$. By Step 1.1, it must be the case that $x_i,x'_i,x^*_i\in C_i$ for some $C_i\in \mathcal {C}$, meaning that they have to be from the same category. Take any menu $A,B\in K$ such that $x_i \in A\cap B$ and $x_i^* \in A\cap B.$ Then, we have

$$\begin{aligned} \frac{\rho (x_i,A)}{\rho (x^*_i,A)}&= \frac{\left( \prod _{ \lbrace C'\in \mathcal{{C}} : \, C'\overset{A}{\succ } C^{x_i} \rbrace } 1-\gamma (C')\right) \gamma (C^{x_i})\frac{u(x_i)}{\sum _{x'\in C^{x_i} \cap A}u(x')}}{\left( \prod _{ \lbrace C'\in \mathcal{{C}} : \, C'\overset{A}{\succ } C^{x^*_i} \rbrace } 1-\gamma (C')\right) \gamma (C^{x^*_i})\frac{u(x^*_i)}{\sum _{x'\in C^{x_i^*} \cap A}u(x')} } \\&=\frac{u(x_i)}{u(x^*_i)} \,\text { since }C^{x_i^*}=C^{x_i}\\&= \frac{\left( \prod _{ \lbrace C'\in \mathcal{{C}} : \, C'\overset{B}{\succ } C^{x_i} \rbrace } 1-\gamma (C')\right) \gamma (C^{x_i})\frac{u(x_i)}{\sum _{x'\in C^{x_i} \cap B}u(x')}}{\left( \prod _{ \lbrace C'\in \mathcal{{C}} : \, C'\overset{B}{\succ } C^{x^*_i} \rbrace } 1-\gamma (C')\right) \gamma (C^{x^*_i})\frac{u(x^*_i)}{\sum _{x'\in C^{x_i^*} \cap B}u(x')} } \\&=\frac{\rho (x_i,B)}{\rho (x^*_i,B)} \,{\text { by definition of CPCC Representation.}} \end{aligned}$$

Also,

$$\begin{aligned} \rho (x_i,\lbrace x_i \rbrace )=\gamma (C_i), \end{aligned}$$

while

$$\begin{aligned} \rho (x_i,\lbrace x_i,x^*_i \rbrace )=\gamma (C_i) \left( \frac{u(x_i)}{u(x_i)+u(x^*_i)}\right) , \end{aligned}$$

so $\rho (x_i,\lbrace x_i,x^*_i \rbrace )<\rho (x_i,\lbrace x_i \rbrace )$, since $\frac{u(x_i)}{u(x_i)+u(x^*_i)}<1$. An identical argument gives $\rho (x^*_i,\lbrace x_i,x^*_i \rbrace )<\rho (x^*_i,\lbrace x^*_i \rbrace )$. Thus, by definition of $\overset{c}{\sim }$, we have $x_i \overset{c}{\sim } x_i^*$. Therefore, the binary relation $\overset{c}{\sim }$ derived from $\rho$ is transitive, and Axiom 1 is satisfied.

Step 2: $\mathcal{{C}}=\mathcal{{C}}^r$

The equation and inequality that we just derived to show the transitivity also shows that if $x_i$ and $x^*_i$ are in the same category $C_i\in \mathcal {C}$, then we must have $x_i \overset{c}{\sim } x^*_i$, and by symmetry of $\overset{c}{\sim }$, we also have $x_i^* \overset{c}{\sim } x_i$. (Again, these are immediate if $x_i=x_i^*$.) We also have from Step 1.1 that if $x^*_i\overset{c}{\sim } x_i$, then $x_i,x_i^*$ are in the same $C_i\in \mathcal {C}$. This gives us the if and only if statement:

$$\begin{aligned} x_i {\text { and }}x_i^* {\text { are in the same }}C_i\in \mathcal{{C}} {\text { iff }} x^*_i\overset{c}{\sim } x_i {\text { and }} x_i\overset{c}{\sim } x^*_i. \end{aligned}$$

Therefore, $x_i$ and $x^*_i$ are in the same $C_i\in \mathcal {C}$ iff $x_i$ and $x^*_i$ are in the same $R_i\in \mathcal{{C}}^r$. So, we have $\mathcal{{C}}=\mathcal{{C}}^r$. Having established this, showing necessity of Axioms 2–4 is trivial in the case that $\mathcal {C}$ is a singleton. Therefore, for the rest of the proof, we focus on the case where there is more than 1 element in $\mathcal {C}$.

Step 3: Necessity of Axiom 2

We prove this by contradiction. So suppose we induce the binary relation $\overset{r}{\geqslant }$ using $\rho$ and the revealed categories. Assume $\exists$ a finite sequence $x_1,\ldots ,x_n$ such that $x_1 \overset{r}{\geqslant } x_2, \ldots ,x_{n-1} \overset{r}{\geqslant } x_n$ and $x_n \overset{r}{\geqslant } x_1$. We first show that for any $x,x'\in X$ such that $x\ne x'$ and $x\overset{v}{\geqslant } x'$, we cannot have $x' \overset{r}{\geqslant } x$. So, take any $x \overset{v}{\geqslant } x'$ such that $x\ne x'$ and suppose that $x' \overset{r}{\geqslant } x$. By definition of $\overset{r}{\geqslant }$, x and $x'$ are in distinct revealed categories. That is, if $x\in R$ and $x'\in R'$ for some $R,R'\in \mathcal{{C}}^r$, then $R\ne R'$. By Step 2, this means that x and $x'$ are in distinct categories. That is, if $x\in C$ and $x'\in C'$ for some $C,C'\in \mathcal {C}$, then $C\ne C'$. Then, since $x \overset{v}{\geqslant } x'$, we have $\rho (x',\lbrace x'\rbrace )=\gamma (C'),$ while if we take $\lbrace x,x' \rbrace$, we have $\rho (x',\lbrace x,x'\rbrace )=\left( 1-\gamma (C) \right) \gamma (C'),$ so that

$$\begin{aligned} \frac{\rho (x',\lbrace x'\rbrace )}{\rho (x',\lbrace x,x'\rbrace )} =\frac{1}{1-\gamma (C)}=1 {\text { (since }} x' \overset{r}{\geqslant } x \text {)}, \end{aligned}$$

which is a contradiction since $\gamma (C)\in (0,1)$. Therefore, we conclude that if $x\ne x'$ and $x \overset{v}{\geqslant } x'$, then we do not have $x' \overset{r}{\geqslant } x$. Taking the contraposition gives us that if $x' \overset{r}{\geqslant } x$, then we do not have $x \overset{v}{\geqslant } x'$, and since $\overset{v}{\geqslant }$ is a total order on X, we must have $x' \overset{v}{\geqslant } x$. Repeating this argument gives $x_1 \overset{v}{\geqslant } x_2,\ldots ,x_{n-1} \overset{v}{\geqslant } x_n.$ By transitivity of $\overset{v}{\geqslant }$, we have $x_1 \overset{v}{\geqslant } x_n$. Since we assumed $x_n \overset{r}{\geqslant } x_1$, we also have $x_n \overset{v}{\geqslant } x_1$. Since $x_n \overset{r}{\geqslant } x_1$, they must be in different revealed categories by definition of $\overset{r}{\geqslant }$, and step 2 gives us $\mathcal{{C}}=\mathcal{{C}}^r$, so we must have that $x_1$ and $x_n$ are in distinct categories, and so $x_1\ne x_n$. This contradicts antisymmetry of $\overset{v}{\geqslant }$. Therefore, we conclude that we cannot have such a sequence $x_1,\ldots ,x_n$. Hence, $\overset{r}{\geqslant }$ satisfies acyclicity.

Step 4: Necessity of Axiom 3

Take any $R_i\in \mathcal{{C}}^r$ and any $x_i\in R_i$ and any $x_j\in {X \setminus R_i}$. By Step 2, we know that if $x_i\in C_i$, where $C_i\in \mathcal {C}$, then $x_j\notin {C_i}$. Therefore, $x_i \ne x_j$, and since $\overset{v}{\geqslant }$ is complete, we must have $x_i \overset{v}{\geqslant } x_j$ or $x_j \overset{v}{\geqslant } x_i$, but not both since $\overset{v}{\geqslant }$ is antisymmetric. Take the former case. We already showed that if $x_i \overset{v}{\geqslant } x_j$, then we do not have $x_j \overset{r}{\geqslant } x_i$, and we have

$$\begin{aligned} \rho (x_i,\lbrace x_i \rbrace ) =\gamma (C_i)=\rho (x_i,\lbrace x_i,x_j \rbrace ){\text { by definition of CPCC Representation}},\\ \end{aligned}$$

so we have just $x_i \overset{r}{\geqslant } x_j$. If we take the latter case, the same type of argument gives us that we have just $x_j \overset{r}{\geqslant } x_i$. So, we have $x_i \overset{r}{\geqslant } x_j$ or $x_j \overset{r}{\geqslant } x_i$ but not both. Hence, $\overset{r}{\geqslant }$ satisfies quasi-completeness.

Step 5: Necessity of Axiom 4

We now take any total order $\overset{r}{\geqslant ^*}$ on X that is an extension of $\overset{r}{\geqslant }$.

Lemma 1

For any $x\in R_i$ and $z\in R_j$ such that $R_i\ne R_j$, $z \overset{r}{\geqslant } x \iff z \overset{r}{\geqslant ^*} x$.

Proof of Lemma 1

For any $x\in R_i$ and $z\in R_j$ such that $R_i\ne R_j$, since $\overset{r}{\geqslant ^*}$ is an extensions of $\overset{r}{\geqslant }$, if $z \overset{r}{\geqslant } x$, then we have $z \overset{r}{\geqslant ^*} x$. For the other direction, if $z \overset{r}{\geqslant ^*} x$, then we cannot have $x \overset{r}{\geqslant } z$ because this would mean that we also have $x \overset{r}{\geqslant ^*} z$, and this would violate antisymmetry of $\overset{r}{\geqslant ^*}$, but since $\overset{r}{\geqslant }$ is quasi-complete by Step 4 we must have $z \overset{r}{\geqslant } x$. $\square$

Lemma 2

For any $x\in C_i$ and $z\in C_j$ such that $C_i \ne C_j$, $z \overset{v}{\geqslant } x$ iff $z \overset{r}{\geqslant } x$.

Proof of Lemma 2

We showed in Step 3 that for any $x\in X$ and $z\in X$, if $z \overset{r}{\geqslant } x$, then we must have $z \overset{v}{\geqslant } x$. For the other direction, if $z\in C_j$ and $x\in C_i$ and $C_i\ne C_j$, and $z \overset{v}{\geqslant } x$, then we will have $\rho (z,\lbrace z \rbrace )=\gamma (C_j)$ and $\rho (z, \lbrace x,z \rbrace )=\gamma (C_j),$ so that $\rho (z,\lbrace z \rbrace )=\rho (z,\lbrace x,z\rbrace ).$ Since $z\in C_j$ and $x\notin C_j$, if $z\in R_j$ for some $R_j\in \mathcal{{C}}^r$, then $x\notin {R_j}$ because $\mathcal{{C}}=\mathcal{{C}}^r$. Therefore, we must have $z \overset{r}{\geqslant } x$ by definition of $\overset{r}{\geqslant }$. $\square$

We use $\overset{r}{\geqslant ^*}$ to define menu-dependent perception $\overset{A}{>^*}$ for all $A\in K$, and define the hazard rates according to these strict total orders. Now, take any $R_i\in \mathcal{{C}}^r$ and any $A,B\in K$ such that $R_i\cap A\ne \emptyset$ and $R_i\cap B\ne \emptyset$. Then,

$$\begin{aligned} \sum _{x'\in R_i}q(x',A)&=\sum _{x'\in R_i}\frac{\rho (x',A)}{1-\rho (A_{x'},A)} \,{\text { by definition of }}q.\\ \end{aligned}$$

Now, for any $x'\in R_i\cap A$ and $y\in A$, we will have $y\in A_{x'}$ iff $y \in R_j$ for some $R_j\in \mathcal{{C}}^r$ such that $R_j \overset{A}{>^*} R_i$. By definition, this means that $y\in A_{x'}$ iff $y \in R_j$ for some $R_j\ne R_i$ such that there exists $z\in R_j\cap A$ such that $z \overset{r}{\geqslant ^*} x$ for all $x\in R_i\cap A$. So, for any $x',x^*\in R_i \cap A$, the set $A_{x'}=A_{x^*}$ because this set does not depend on which element in $R_i\cap A$ we have. Therefore, the expression becomes

$$\begin{aligned} \frac{1}{1-\rho (A_{{\tilde{x}}},A)}\sum _{x'\in R_i}\rho (x',A) \end{aligned}$$

for some ${\tilde{x}}\in R_i\cap A$. Now, by definition, the set $A_{{\tilde{x}}}$ is equal to

$$\begin{aligned}&\lbrace y\in A|\,y \overset{A}{\succ ^*} {\tilde{x}}\rbrace {\text {,\, and\, by \,definition\, this \,is }}\,\\&\lbrace y\in A|\, y\in R_j {\text { and }} R_j\overset{A}{>^*} R_i \rbrace {\text {,\, and\, by\, definition\, this \,is }}\,\\&\lbrace y\in A|\, y\in R_j\ne R_i {\text { and }}\exists z\in R_j \cap A{\text { s.t. }} z \overset{r}{\geqslant ^*} x\,\text{ } \forall \,x\in R_i \cap A\rbrace . \end{aligned}$$

Therefore, by Lemma 1, the set is equal to

$$\begin{aligned} \lbrace y\in A|\,y\in R_j\ne R_i {\text { and }}\exists z\in R_j \cap A {\text { s.t. }} z \overset{r}{\geqslant } x\,\text{ }\forall \,x\in R_i \cap A \rbrace . \end{aligned}$$

Using Step 2, this set is

$$\begin{aligned} \lbrace y\in A|\, y\in C_j\ne C_i {\text { and }}\exists z\in C_j\cap A {\text { s.t. }} z \overset{r}{\geqslant } x\, \text{ }\forall \,x\in C_i \cap A \rbrace , \end{aligned}$$

where $R_i=C_i$.

Using Lemma 2, the set becomes

$$\begin{aligned} \lbrace y\in A|\, y\in C_j\ne C_i {\text { and }}\exists z\in C_j \cap A {\text { s.t. }} z \overset{v}{\geqslant } x\,\text{ }\forall \,x\in C_i\cap A \rbrace . \end{aligned}$$

By definition, this is $\lbrace y\in A|\,y\in C_j {\text { such that }} C_j \overset{A}{\succ } C_i \rbrace$. So, $A_{{\tilde{x}}}=\lbrace y\in A|\,y\in C_j {\text { s.t. }} C_j \overset{A}{\succ } C_i \rbrace$.

Now, since X is finite and $\overset{A}{\succ }$ is a strict total order on the categories that have nonempty intersection with A, we can enumerate the set $\lbrace C_j\in \mathcal{{C}}| \,C_j \overset{A}{\succ } C_i \rbrace$ as $\lbrace C_1^*,\ldots ,C_M^* \rbrace$, where $C_1^* \overset{A}{\succ } C_2^* \overset{A}{\succ } \cdots \overset{A}{\succ } C_M^*$. (Of course, it can be empty.) Then, $A_{{\tilde{x}}}$ can be partitioned into $\lbrace \lbrace C_1^*\cap A\rbrace ,\ldots , \lbrace C_M^*\cap A \rbrace \rbrace$. (Again, this collection can be empty.) So, we have

$$\begin{aligned} \frac{1}{1-\rho (A_{{\tilde{x}}},A)}\sum _{x'\in R_i}\rho (x',A)&=\frac{1}{1-\rho (C^*_1\cap A,A)-\cdots -\rho (C^*_M\cap A,A)}\sum _{x'\in C_i}\rho (x',A)\\&=\frac{\sum _{x'\in C_i\cap A}\rho (x',A)}{1-\rho (C^*_1\cap A,A)-\cdots -\rho (C^*_M\cap A,A)}, \end{aligned}$$

since $\rho (x',A)=0$ when $x'\notin {A}$. From definition of CPCC, the numerator of the fraction is

$$\begin{aligned}\sum _{x'\in C_i\cap A}&\left[ \left( \prod _{C' \in \mathcal{{C}}: C'\overset{A}{\succ }C_i} \left( 1-\gamma (C') \right) \right) \gamma (C_i)\frac{u(x')}{\sum _{x^*\in C_i\cap A}u(x^*)}\right] \\&=\gamma (C_i) \prod _{C' \in \mathcal{{C}}: C'\overset{A}{\succ }C_i} \left( 1-\gamma (C')\right). \end{aligned}$$

(This is just equal to $\gamma (C_i)$ when $A_{{\tilde{x}}}$ is empty.) Now, for each $C^*_k$ for $1\le k \le M$ (if $A_{{\tilde{x}}}$ is empty we can ignore this), we have

$$\begin{aligned} \rho (C^*_k\cap A,A)&=\sum _{x'\in C^*_k\cap A} \frac{u(x')}{\sum _{x^*\in C_k^*\cap A}u(x^*)}\gamma (C^*_k) \prod _{n=1}^{k-1}\left( 1-\gamma (C^*_n)\right) \\&= \gamma (C^*_k)\prod _{n=1}^{k-1}(1-\gamma (C^*_n)). \\ \end{aligned}$$

So, we can expand the denominator one by one and factor in a way similar to the proof of the sufficiency and the denominator becomes (when $A_{{\tilde{x}}}$ is empty the denominator is just equal to 1)

$$\begin{aligned} \prod _{n=1}^M 1-\gamma (C^*_n)=\prod _{C' \in \mathcal{{C}}: C'\overset{A}{\succ }C_i} (1-\gamma (C')) \end{aligned}$$

by how we defined the sets $C^*_j$ for $1\le j\le M$, so that the fraction becomes

$$\begin{aligned} \sum _{x'\in R_i}q(x',A)=\gamma (C_i). \end{aligned}$$

We can do the same type of derivation for the menu B to get $\sum _{x'\in R_i}q(x',B)=\gamma (C_i)$, which implies $\sum _{x'\in R_i}q(x',A)=\sum _{x'\in R_i}q(x',B).$ Since $R_i$, A, and B were arbitrary, we conclude that Axiom 4 is satisfied. $\square$

1.4 Proof of Theorem 2

Proof

Take $(\mathcal{{C}},\overset{v}{\geqslant },u,\gamma )$ and $(\mathcal{{C}'},\overset{v}{\geqslant '},u',\gamma ')$ as in the statement of the theorem.

Step 1: $\mathcal{{C}}=\mathcal{{C}}'$

Suppose that $\mathcal{{C}}\ne \mathcal{{C}}'$. Then there exists a $C\in \mathcal {C}$ such that $C\notin \mathcal{{C}'}$. Take any such C. Take any $x\in C$. Since $\mathcal{{C}'}$ is also a partition of X, x must belong to some $C'\in \mathcal {C}'$. Since C is assumed to not be in $\mathcal {C}'$, there exists a $y\in X$ such that $y\ne x$, and either $y\in C$ but $y\notin {C'}$, or $y\in C'$ but $y \notin {C}$. Otherwise, C and $C'$ would be equal and this would violate our assumption. Now, take the menus $\lbrace x \rbrace ,\lbrace y \rbrace , \lbrace x,y \rbrace$, and consider the latter case and denote the category in $\mathcal {C}$ that contains y as $C^*$. Now, if $x \overset{v}{\geqslant } y$, then we have $\rho (x, \lbrace x \rbrace )=\gamma (C),$ while $\rho (x, \lbrace x,y \rbrace )=\gamma (C)$ as well, since $C \overset{\lbrace x,y \rbrace }{\succ } C^*$. Now, consider the case that $y \overset{v}{\geqslant } x$. Then, we have $\rho (y, \lbrace y \rbrace )=\gamma (C^*)$ and $\rho (y, \lbrace x,y \rbrace )=\gamma (C^*).$ In either case, there will be a $z\in \lbrace x,y \rbrace$ such that $\rho (z, \lbrace x,y \rbrace )=\rho (z, \lbrace z \rbrace )$. For $\rho '$, in either of the cases of $x \overset{v}{\geqslant } y$ or $y \overset{v}{\geqslant } x$, since they are in the same $C'\in \mathcal {C}'$, we will have $\rho '(x, \lbrace x \rbrace )=\gamma '(C')$ and $\rho '(x, \lbrace x,y \rbrace )=\gamma '(C') \frac{u'(x)}{u'(x)+u'(y)},$ while $\rho '(y, \lbrace y \rbrace )=\gamma '(C')$ and $\rho '(y, \lbrace x,y \rbrace )=\gamma '(C') \frac{u'(y)}{u'(x)+u'(y)} .$ So in either case, there does not exist a $z\in \lbrace x,y \rbrace$ such that $\rho '(z, \lbrace x,y \rbrace )=\rho '(z, \lbrace z \rbrace )$, so $\rho \ne \rho '$. This is a contradiction. We can derive a contradiction with the same type of construction for the former case with $y\in C$. Therefore, we conclude that $\mathcal{{C}}=\mathcal{{C}}'$.

Step 2: $\overset{v}{\geqslant } \overset{TS}{=} \overset{v}{\geqslant '}$

Now, assume that $\overset{v}{\geqslant } \overset{TS}{\ne } \overset{v}{\geqslant '}$. Since we already showed equivalence of $\mathcal{{C}}=\mathcal{{C}}'$, we will only refer to $\mathcal {C}$ for the remainder of the proof. By definition of two orders on X being $TS\text {-}equivalent$, we must have that there exists a $x,y \in X$ such that $x\in C$ for some $C\in \mathcal {C}$ and $y\notin {C}$ and $(x,y)\in \overset{v}{\geqslant }$, but $(x,y)\notin \overset{v}{\geqslant '}$. Then, $x\overset{v}{\geqslant } y$, but $y \overset{v}{\geqslant '} x$. So, we again take the menus $\lbrace x \rbrace$ and $\lbrace x,y \rbrace$. Denote the category that contains y as $C'$. Then, $\rho (x, \lbrace x \rbrace )=\gamma (C)$, while $\rho (x, \lbrace x,y \rbrace )=\gamma (C)$ as well, so that we have $\rho (x, \lbrace x \rbrace )=\rho (x, \lbrace x,y \rbrace ).$ Now, $\rho '(x, \lbrace x \rbrace )=\gamma '(C)$, while $\rho '(x, \lbrace x,y \rbrace )=\left( 1-\gamma '(C')\right) \gamma '(C),$ and since $\left( 1-\gamma '(C')\right) \in (0,1)$, we have $\rho '(x, \lbrace x,y \rbrace )< \rho '(x, \lbrace x \rbrace )$. This contradicts $\rho =\rho '$. Therefore, we must have $\overset{v}{\geqslant } \overset{TS}{=} \overset{v}{\geqslant '}$.

Step 3: $\gamma =\gamma '$

For any $C\in \mathcal {C}$, let x be an element in C. We must have

$$\begin{aligned} \gamma (C)=\rho (x, \lbrace x \rbrace )=\rho '(x, \lbrace x \rbrace )=\gamma '(C) . \end{aligned}$$

Step 4: $u\overset{{\mathcal {C}}}{=} u'$

Now, since X is finite, $\mathcal {C}$ must be finite. So we can enumerate $\mathcal {C}$ as $\mathcal{{C}}=\lbrace C_1,\ldots ,C_k \rbrace$. For any $C_j\in \mathcal {C}$ such that $|C_j|=1$, it is obvious that $u(x)=\alpha _j u'(x)$ for all $x\in C_j$. Now, for any $C_k$ such that $|C_k|\ge 2$, fix $x\in C_k$ and take any $y\in C_k$ such that $y\ne x$, and take the menu $\lbrace x,y \rbrace$. We must have $\rho (x, \lbrace x,y \rbrace )=\rho '(x, \lbrace x,y \rbrace )$, which means that

$$\begin{aligned} \gamma (C_k) \left( \frac{u(x)}{u(x)+u(y)}\right) =\gamma '(C_k)\left( \frac{u'(x)}{u'(x)+u'(y)}\right) , \end{aligned}$$

and by Step 3, $\gamma (C_k)=\gamma '(C_k)$, so call this value f, and let the value of $\rho (x, \lbrace x,y \rbrace )=\rho '(x, \lbrace x,y \rbrace )$ be $p_{xy}$. Then, we must have

$$\begin{aligned} \left( {\frac{f}{p_{xy}}}-1 \right) u(x)=u(y) \end{aligned}$$

(3)

and

$$\begin{aligned} \left( {\frac{f}{p_{xy}}}-1 \right) u'(x)=u'(y). \end{aligned}$$

(4)

Now, let $\frac{u(x)}{u'(x)}=\alpha _k$. Since $u(x)\in {\mathbb {R}}_{++}$ and $u'(x)\in {\mathbb {R}}_{++}$, we know that $\alpha _k \in {\mathbb {R}}_{++}$. Dividing (3) by (4) gives us $\frac{u(y)}{u'(y)}=\alpha _k$. Since y was arbitrary, for any $x_k\in C_k$, we must have $\frac{u(x_k)}{u'(x_k)}=\alpha _k$, or $u(x_k)=\alpha _k u'(x_k)$. By definition, this means that $u|_{C_k}=\alpha _k u'|_{C_k}$ . Since $C_k$ was arbitrary, we conclude that $u\overset{{\mathcal {C}}}{=} u'$. $\square$