Selective attribute rules

Abstract

We develop an extension of Luce’s (Individual Choice Behavior: A Theoretical Analysis. Wiley, New York, 1959) model to apply multiattributes to stochastic choice. We consider an agent who focuses on selective (salient) attributes standing out in a choice set. These attributes are endogenously determined according to the criterion of Just Noticeable Difference. The criterion selects attributes whose variation impacts the agent enough so that she cares about each of them. When such selective attributes vary with the choice set, we adjust the choice probabilities for the elements of the choice set. We find that all the violations of Luce’s axioms can be attributed to a change in selective attributes. When the global choice set is of the product form, selective attribute rules are characterized. In particular, the distinguished case of (non-selective) attribute rules can be characterized by two axioms, Independence from Irrelevant Alternatives and Separability.

Appendices

Appendices

Appendix A: Generalized attribute rules of Gul, Natenzon and Pesendorfer

Gul et al. (2014) consider another generalization of Luce’s random choice model which they call “attribute rule”. Since we already use this term with a different meaning, we shall call their concept a Generalized Attribute Rule (GAR). Their analysis is based on a “rich” and consequently infinite set of alternatives whereas ours assumes a finite set. But even in the case of finitely many alternatives, not all selective attribute rules are generalized attribute rules and vice versa. We are going to use the notation ${\widehat{\rho }}$ for generalized attribute rules and $\rho$ for selective attribute rules.

Let $X\subseteq {\mathbb {R}}^n_{++}$ with n-dimensional attributes $A=\{1, \ldots , n\}$.¹⁰ For an attribute $i\in A$ and a choice set C, define $w_i(C)=\sum _{y\in C}w_i(y_i)$. For $i, \, C$ and $x\in C$, define $\sigma _i(x, C)=w_i(x_i)/w_i(C)$. $\sigma _i(x, C)$ is the probability that x is chosen from C if only attribute i is used in a Luce-type formula. A generalized attribute rule ${\widehat{\rho }}(x, C)$ is obtained as the weighted sum of the $\sigma _i(x, C)$ as follows: There exists a vector $q=(q_1, \ldots , q_n)\in {\mathbb {R}}^n_{++}$ such that $\sum _{i\in A}q_i =1$ and ${\widehat{\rho }}(x, C)= \sum _{i\in A} q_i \sigma _i(x, C)$ for x and C with $x\in C$. In short, we can write ${\widehat{\rho }}(x, C)=q\cdot \sigma (x, C)$ where $\sigma (x, C)=(\sigma _1(x, C), \ldots , \sigma _n(x,C))\in {\mathbb {R}}^n_{++}$.

Both generalized attribute rules and selective attribute rules include traditional Luce rules. But neither set contains the other.

First, we show that not every selective attribute rule can be obtained as generalized attribute rule. To show this, let us consider $n=2$ and the SAR $\rho$ given by $X=\{x,y,z\}$ with $x=(10,30), y=(10,10), z=(30,10)$, $\, k=(15,15)$, $w(u)=(u_1, u_2)$ for $u=(u_1,u_2)\in X$. Then $x=(x_1, x_2),\, y=(x_1,z_2), z=(z_1, z_2)$ and $z_1>x_1>0, x_2>z_2>0$. For an arbitrary attribute value function ${\widehat{w}}$ and weight vector $q=(q_1,q_1)$, we obtain

$$\begin{aligned}\begin{array}{ll} \sigma _1(x, \{x,y\})=1/2; &{} \sigma _2(x, \{x,y\})={\widehat{w}}_2(x_2)/({\widehat{w}}_2(x_2)+{\widehat{w}}_2(y_2));\\ \sigma _1(z, \{y,z\})={\widehat{w}}_1(z_1)/({\widehat{w}}_1(y_1)+{\widehat{w}}_1(z_1)); &{} \sigma _2(z, \{y,z\})=1/2. \end{array}\end{aligned}$$

${\widehat{\rho }}=\rho$ requires ${\widehat{\rho }}(x, \{x,y\})=\rho (x, \{x,y\})$ and ${\widehat{\rho }}(z, \{y,z\})=\rho (z, \{y,z\})$, that is

$$\begin{aligned} q_1\cdot \frac{1}{2}+ q_2\cdot \frac{{\widehat{w}}_2(x_2)}{{\widehat{w}}_2(x_2)+{\widehat{w}}_2(y_2)}=\frac{3}{4};~ ~q_1\cdot \frac{{\widehat{w}}_1(z_1)}{{\widehat{w}}_1(y_1)+{\widehat{w}}_1(z_1)}+ q_2\cdot \frac{1}{2}=\frac{3}{4}. \end{aligned}$$

Adding the two equations yields

$$\begin{aligned} q_1\cdot \frac{{\widehat{w}}_1(z_1)}{{\widehat{w}}_1(y_1)+{\widehat{w}}_1(z_1)} +q_2\cdot \frac{{\widehat{w}}_2(x_2)}{{\widehat{w}}_2(x_2)+{\widehat{w}}_2(y_2)}=1, \end{aligned}$$

which is a contradiction because of $\dfrac{{\widehat{w}}_1(z_1)}{{\widehat{w}}_1(y_1)+{\widehat{w}}_1(z_1)}<1$ and $\dfrac{{\widehat{w}}_2(x_2)}{{\widehat{w}}_2(x_2)+{\widehat{w}}_2(y_2)}<1$. Hence ${\widehat{\rho }}\ne \rho$ has to hold. This shows that $\rho$ cannot be obtained as generalized attribute rule.

Second, we show that not every generalized attribute rule can be obtained as selective attribute rule. To see this, let $n=2$, $q=(1/2,1/2)$, $X=\{x,y,z\}$ with $x=(10,30), y=(10,10), z=(30,10)$, and attribute values ${\widehat{w}}(u)=(u_1, u_2)$ for $u=(u_1,u_2)\in X$. Then

C	$\sigma _1(x,C)$	${\widehat{\rho }}(x, C)$	$\sigma _1(y,C)$	${\widehat{\rho }}(y,C)$	$\sigma _1(z, C)$	${\widehat{\rho }}(z,C)$
	$\sigma _2(x,C)$		$\sigma _2(y,C)$		$\sigma _2(z, C)$
X	1/5	2/5	1/5	1/5	3/5	2/5
	3/5		1/5		1/5
$\{x,y\}$	1/2	5/8	1/2	3/8
	3/4		1/4
$\{x,z\}$	1/4	1/2			3/4	1/2
	3/4				1/4
$\{y,z\}$			1/4	3/8	3/4	5/8
			1/2		1/2

Now consider any attribute value function w and some $k=(k_1,k_2)\ge 0$. Then $x=(x_1, x_2),\, y=(x_1,z_2), z=(z_1, z_2)$ and $z_1>x_1>0, x_2>z_2>0$. Suppose that the resulting selective attribute rule $\rho$ satisfies $\rho ={\widehat{\rho }}$. Then $A_k(X)\ne \emptyset$.

$\bullet$ If $k\gg (0,0)$, then three potential cases result:

• CASE 1: $A_k(X)=\{1\}$. In that case, $A_k(\{x,y\})=\{1\}$ has to hold as well whereas $x_1=y_1$ implies $A_k(\{x,y\})\ne \{1\}$.

• CASE 2: $A_k(X)=\{2\}$. In that case, $A_k(\{y,z\})=\{2\}$ has to hold as well whereas $y_2=z_2$ implies $A_k(\{y,z\})\ne \{2\}$.

• CASE 3: $A_k(X)=\{1,2\}$. Then further $A_k(\{x,y\})=\{2\}$ and $A_k(\{y,z\})=\{1\}$. From $\rho (x,\{x,y\})={\widehat{\rho }}(x,\{x,y\})$ and $\rho (y,\{y,z\})={\widehat{\rho }}(y,\{y,z\})$, we have

  $w_2(x_2)=(5/3)w_2(y_2)=(5/3)w_2(z_2)~~\text {and}~~w_1(z_1)=(5/3)w_1(y_1)=(5/3)w_1(x_1),$ which is equivalent to

  $w_2(y_2)=w_2(z_2)= (3/5)w_2(x_2)$ and $w_1(x_1)=w_1(y_1)=(3/5)w_1(z_1)$. It follows that

  $$\begin{aligned} \rho (y, X)=\dfrac{(3/5)w_1(z_1)+(3/5)w_2(x_2)}{(11/5)w_1(z_1)+(11/5)w_2(x_2)}=\dfrac{3}{11},\end{aligned}$$

  whereas ${\widehat{\rho }}(y, X)=1/5$.

  $\bullet$ If $k_1=0, k_2>0$, then two more potential cases result:

• CASE 4: $A_k(X)=\{1\}$. Then $A_k(C)=\{1\}$ for all C and $\rho$ is a Luce rule and satisfies IIA whereas ${\widehat{\rho }}$ violates IIA.

• CASE 5: $A_k(X)=\{1,2\}$. Then $A_k(\{x,y\})=\{1,2\}$ holds as well so that IIA prevails for $\rho$ when alternative z is added to the choice set $\{x,y\}$, which is not the case for ${\widehat{\rho }}$.

  $\bullet$ If $k_1>0, k_2=0$, the resulting potential cases are similar to CASE 4 and CASE 5.

  $\bullet$ If $k=(0,0)$, then $\rho$ is an attribute rule and satisfies IIA whereas ${\widehat{\rho }}$ violates IIA.

Thus in all cases, $\rho ={\widehat{\rho }}$ cannot hold. Therefore, ${\widehat{\rho }}$ cannot be obtained as selective attribute rule.

Appendix B: PALMs versus SARs

PALMs and SARs simply differ because an SAR satisfies $\sum _{x\in C} \rho (x,C)=1$ for all $C\in {\mathcal {X}}$ whereas a PALM can yield $\sum _{x\in C}\rho (x,C)<1$ for some $C\in {\mathcal {X}}$, because of the outside option.¹¹

Regardless of the normalization of probabilities, one can find specific differences in specific examples.

Let us first consider an extended Luce model in the sense of Echenique et al. (2018). Let $X=\{x^1,x^2,x^3,x^4\}$ and $x^0\notin X$ be the “outside option”. Let $v(x)=1$ for $x\in X$ and $v(x^0)=0$. We obtain “extended Luce values” given as

$$\begin{aligned} \mu (C,x)=\dfrac{v(x)}{\sum _{z\in C}v(z)+v(x^0)}=\dfrac{1}{|C|}\end{aligned}$$

for $C\in {\mathcal {X}}, \, x\in C$. Moreover, let $\succ$ be s strict order on X such that $x^1\succ x^2\succ x^3\succ x^4$. Then formula (2) in Echenique et al. (2018) yields

$$\begin{aligned} \rho (x, C)=\mu (x,C)\cdot \prod _{z\in C, z \succ x}(1-\mu (z,C)).\end{aligned}$$

In particular, we get

$$\begin{aligned} \rho (x^1,\{x^1,x^4\})= & {} \mu (x^1, \{x^1,x^4\})=\frac{1}{2},\\ \rho (x^4,\{x^1,x^4\}= & {} \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4},\\ \rho (x^1,\{x^1,x^3,x^4\})= & {} \mu (x^1,\{x^1,x^3,x^4\})=\frac{1}{3},\\ \rho (x^4,\{x^1,x^3,x^4\})= & {} \frac{1}{3}\cdot \frac{2}{3}\cdot \frac{2}{3}=\frac{4}{27},\\ \rho (x^1,X)= & {} \mu (x^1,X)=\frac{1}{4},\\ \rho (x^4,X)= & {} \frac{1}{4}\cdot \frac{3}{4}\cdot \frac{3}{4}\cdot \frac{3}{4}=\frac{27}{256}. \end{aligned}$$

Hence when the choice set is expanded first from $\{x^1, x^4\}$ to $\{x^1, x^3, x^4\}$ and then to X, two violations of IIA occur whereas $n=1$ allows only one increase in the number of salient attributes.

Next reconsider the SAR $\rho$ of Example 1. Then there do not exist a finite set of alternatives $X'$ and a PALM specified by $(v,\succeq )$ with the set of alternatives $X'$ and outside option $x_0\notin X'$ such that

• (a) $X\subset X'$,

• (b) $\succeq$ is “rich”,

• (c) the stochastic choice rule $\rho _{(v,\succeq )}$ defined by $(v,\succeq )$ coincides with $\rho$ on ${\mathcal {X}}$.

For suppose there exist a finite set of alternatives $X'$ and a PALM $(v,\succeq )$ with set of alternatives $X'$ and outside option $x_0\notin X'$ such that (a)–(c) hold. By Theorem 2 of Echenique et al. (2018), $\succeq ^*=\succeq$ holds for the revealed preference relation $\succeq ^*$ and IIA holds for the hazard rate function q given by $\rho _{(v,\succeq )}$ and $\succeq ^*=\succeq$. In particular,

$$\begin{aligned} \frac{q(x,\{x,y\})}{q(y,\{x,y\})}=\frac{q(x,X)}{q(y,X)}\end{aligned}$$

would hold. Because of (b) and (c), the pertinent values of q can be calculated using $\rho$ and the weak order induced by $\succeq$ on X. There are thirteen potential weak orders on X. In each case, the above IIA identity is violated; see table below. Hence a finite set of alternatives $X'$ and a PALM $(v,\succeq )$ with set of alternatives $X'$ and outside option $x_0\notin X'$ such that (a)–(c) hold does not exist. On the other hand, the SAR can be extended to an arbitrary finite set $X'\supset X$.

$\succsim$	$q(x,\{x,y\})$	$q(y,\{x,y\})$	$\frac{{{\varvec{q}}}({\varvec{x}},\{{\varvec{x}},{\varvec{y}}\})}{{\varvec{q}}({\varvec{y}},\{{\varvec{x}},{\varvec{y}}\})}$	q(x, X)	q(y, X)	$\frac{{\varvec{q}}({\varvec{x}},{\varvec{X}})}{{\varvec{q}}({\varvec{y}},{\varvec{X}})}$
$z\succ x\succ y$	1/4	1	1/4	4/9	1	4/9
$z\sim x\succ y$	1/4	1	1/4	4/10.1	1	4/10.1
$x\succ y\succ z$	1/4	1	1/4	4/10.1	5/6.1	24.4/50.5
$x\succ y\sim z$	1/4	1	1/4	4/10.1	5/6.1	24.4/50.5
$x\succ z\succ y$	1/4	1	1/4	4/10.1	1	4/10.1
$z\succ x\sim y$	1/4	3/4	1/3	4/9	5/9	4/5
$z\sim x\sim y$	1/4	3/4	1/3	4/10.1	5/10.1	4/5
$x\sim y\succ z$	1/4	3/4	1/3	4/10.1	5/10.1	4/5
$z\succ y\succ x$	1	3/4	4/3	1	5/9	9/5
$z\sim y\succ x$	1	3/4	4/3	1	5/10.1	10.1/5
$y\succ x\succ z$	1	3/4	4/3	4/5.1	5/10.1	40.4/25.5
$y\succ x\sim z$	1	3/4	4/3	4/5.1	5/10.1	40.4/25.5
$y\succ z\succ x$	1	3/4	4/3	1	5/10.1	10.1/5