Default Bias in Extended Choice Rules

Introduction

Following Zhou (1997), Masatlioglu and Ok (2005, 2006) and Sagi (2006), we enrich the concept of a choice problem and include a default option along with any such choice problem. Lecture 3 of Rubinstein (2016) is an informative source for concepts discussed here. As noted there ‘experimental evidence and introspection tell us that a default option is often viewed positively by a decision maker, a phenomenon known as the status quo bias’.

In this paper we provide a reader friendly axiomatic characterization of the default bias extended choice rule, in spite of arguments against such axiomatic characterizations in Rubinstein (2016), which if we may very humbly submit, appear unconvincing to us, unless they amount to duplication of effort. It is worth emphasizing that Rubinstein (2016) contains an axiomatic characterization of the default bias extended choice rule and our framework of analysis is identical to the one developed there. Our proof makes use of the Szpilrajn’s extension theorem which for finite sets (as in our case) has a very simple proof.

The Model

Let X be a non-empty finite set of alternatives and let Ψ(X) be the set of all non-empty subsets of X. If A belongs to Ψ(X) then A is usually viewed as a choice problem or a possible list of alternatives from which a decision maker requires to choose an alternative.

A preference relation on X is a complete and reflexive binary relation ≽ on X.

When we write x ≽ y we mean ‘x is at least as good as y’. Let ≻ denote the asymmetric part of the preference relation ≽. Thus for x, y≻X, x ≻ y means ‘x is strictly preferred to y’.

A transitive preference relation on X is a preference relation on X which is also transitive.

A budget space is a non-empty subset of Ψ(X). A member of a budget space is referred to as an opportunity set/menu/budget set.

Given a budget space 𝔅 the associated extended budget space B ^ = {(A,x)|xϵAϵ𝔅}. An element of B ^ is called an extended budget set.

An extended choice rule with status quo bias is a function c: B ^ → X such that for all (A,x)ϵ B ^ , c(A,x)ϵA.

We now introduce the concept of a default bias extended choice rule, which is a special type of extended choice rule with status quo bias. The decision maker is characterized by a utility function u and a ‘bias function’ β, which assigns a positive number to each alternative. The function u is interpreted as representing the ‘true’ preferences. The number β(x) is interpreted as the bonus attached to x, when it is a default alternative.

The pair (u,β) is said to be a compatible pair on B ^ if for all (A,x)ϵ B ^ , {yϵA\{x}| u(y) > u(z) for all zϵA\{x, y} and u(y) ≥ u(x) + β(x)} is a singleton whenever {yϵA\{x}|u(y) ≥ u(x) + β(x)} ≠ ϕ.

Given a compatible pair (u,β), let c^d be an extended choice rule on B ^ defined as follows. Given (A,x)ϵ B ^ , let {c^d(A, x)} = {yϵA\{x}| u(y) > u(z) for all zϵA\{x, y} and u(y) ≥ u(x) + β(x)} if {yϵA\{x}|u(y) ≥ u(x) + β(x)} ≠ ϕ and c^d(A, x) = x, otherwise. c^d is said to be the default bias extended choice rule generated by the compatible pair (u,β).

Given an extended choice rule c on B ^ let B ^ _c = {(A, x)ϵ B ^ |c(A, x) ≠ x}. B ^ _c is the set of extended budget sets where the default alternative is not chosen.

We will axiomatically characterize the default bias extended choice rule generated by a compatible pair with the help of the following axiom.

The Main Result

In this section we present the desired axiomatic characterization which to us appears to have a simpler proof than the one in Rubinstein (2016). However, such things are always matters of personal perception and may vary from one individual to another. Further, there is always the question about the axioms appearing convincing and there again there is always scope for debate. Having said that, let us proceed to our main result.

Proposition 1: Let B ^ = {(A, x)|xϵAϵΨ(X)}. An extended choice rule c on B ^ is a default bias extended choice rule generated by a compatible pair (u,β) if and only if c satisfies WARPECR.

Proof: Suppose c on B ^ is a default bias extended choice rule generated by a compatible pair (u, β).

Let us show that c satisfies WARPCER.

Towards a contradiction suppose there exists (A, x), (B, y)ϵ B ^ _c with c(A, x) ≠ c(B, y), c(A, x)ϵB and c(B, y)ϵA. Thus, u(c(A, x)) > u(c(B, y)) > u(c(A, x)) (even if c(A, x) = y and/or c(B, y) = x, since β is a real positive valued function), which is not possible. This shows that WARPECR (i) must hold.

Thus, u(c(A, x)) ≥ u(x) + β(x) > u(x). Now, c(A, x)ϵB and u(c(A, x)) ≥ u(x) + β(x) implies{wϵB| u(w) ≥ u(x) + β(x)} ≠ ϕ. Thus, c(B, x) ≠ x leading to a contradiction. Thus WARPECR (ii) must hold.

Now suppose c is an extended choice rule on B ^ which satisfies UA and WARPECR. If B ^ _c = ϕ, then c(A, x) = x for all (A, x)ϵ B ^ _c. In this case let u(x) = 0 and β(x) = 1 for all xϵX. Then it is easy to see that c is a default bias extended choice rule generated by the compatible pair (u,β).

Hence suppose B ^ _c ≠ ϕ. Define a binary relation R on X as follows. (i) For all xϵX, we have xRx. (ii) For all x, yϵX with x ≠ y, xRy if and only if there exists (A, z)ϵ B ^ _c with c(A, z) = x and yϵA.

Clearly R is reflexive, though it may not be complete. Let us show that R is transitive. Thus, let x,y,zϵX with x ≠ y ≠ z ≠ x and both xRy and yRz. Thus, there exists (A, w¹), (B, w²)ϵ B ^ _c with x = c(A, w¹), x ≠ w¹, yϵA, y = c(B, w²), y ≠ w², zϵB. Let us show that c({x,y,z,w²}, w²) = x.

If c({x,y,z,w²}, w²) = w², then along with (B,w²)ϵ B ^ _c and c(B,w²) = yϵ{x,y,z,w²}, WARPECR (ii) stands contradicted. Thus ({x,y,z,w²},w²)ϵ B ^ _c. Suppose, c({x,y,z,w²}, w²) = z. Then along with c(B,w²) = y, WARPECR(i) stands contradicted. Thus, c({x,y,z,w²}, w²) ≠ z. If c({x,y,z,w²}, w²) = y, then along with c(A,w¹) = x, again WARPECR (i) stands contradicted. Thus, c({x,y,z,w²}, w²) = x and so we have xRz. Thus R is transitive. Let P(R) denote the asymmetric part of R.

By Szpilrajn’s extension theorem there exists a transitive preference relation (i. e., reflexive and complete binary relation) ≽ such that for all x, yϵR, xRy implies x ≽ y and xP(R)y implies x ≻ y.

Let (A, x)ϵ B ^ _c. Then clearly c(A, x)Ry for all yϵA(including x) and hence c(A, x) ≽ y for all yϵA(including x). Towards a contradiction suppose there exists yϵA\{c(A, x)} such that y ≽ c(A, x). If c(A, x)P(R)y then we have c(A, x) ≻ y contradicting y ≽ c(A, x). Thus it is not the case that c(A, x)P(R)y. Since we have c(A, x)Ry, ¬c(A, x)P(R)y implies yRc(A, x). Hence there exists (B, z)ϵ B ^ _c such that c(B, z) = y and c(A, x)ϵB. Since c(A,x) ≠ y, WARPECR(i) stands contradicted. Thus, it must be the case that c(A, x) ≻ y for all yϵA\{c(A, x)}.

Let u be a utility function representing ≽ (which exists since ≽ is a transitive preference relation on a non-empty finite set) and let β:X → ℝ + + be defined thus: β(x) = max{|u(y)||yϵX} + 100 if {AϵΨ(X)|(A, x)ϵ B ^ _c} = ϕ, and β(x) = min{u(y) − u(x)|(A, x)ϵ B ^ _c and c(A, x) = y} otherwise.

Consider xϵX.

It is easy to see that if {AϵΨ(X)|(A, x)ϵ B ^ _c} = ϕ, then β(x) > 0 and u(y) < u(x) + β(x) for all yϵX\{x}. Hence suppose {AϵΨ(X)|(A, x)ϵ B ^ _c} ≠ ϕ. Then by definition of β, there exists (B,x)ϵ B ^ _c such that β(x) = u(c(B,x)) – u(x) = min{u(c(A,x))-u(x)|(A,x)ϵc}. Let z = c(B,x) ≠ x. Since u(c(B,x)) – u(x) = β(x) and since u(c(B,x)) > u(y) for all yϵB (including x), β(x) > 0.

If (A,x)ϵ B ^ _c, then u(c(A,x)) ≥ u(x) + β(x) and u(c(A,x)) > u(y) for all yϵ{zϵA\{x,c(A,x)}| u(z) ≥ u(x) + β(x)}.

Let (A,x)ϵ B ^ \ B ^ _c. Thus, c(A,x) = x. Towards a contradiction suppose there exists yϵA\{x} such that u(y) ≥ u(x) + β(x).

Consider ({x, y, z}, x). If c({x, y, z}, x) = x, then along with c(B, x)ϵ{x, y, z} and (B, x)ϵ B ^ _c, WARPECR (ii) stands contradicted. Thus, c({x, y, z},x) ≠ x. and so ({x, y, z},x)ϵ B ^ _c. If c({x, y, z},x) = z, then u(z) > u(y) contradicting u(y) ≥ u(x) + β(x) = u(z). Thus, we must have c({x, y, z}, x) = y. But then with c({x, y, z},x) = yϵA, c(A, x) = xϵ{x, y, z} and ({x, y, z},x) ϵ B ^ _c, WARPECR(ii) stands contradicted. Thus {yϵA| u(y) ≥ u(x) + β(x)} = ϕ.

Thus c is a default bias extended choice rule generated by the compatible pair (u,β). Q.E.D.

Remark

If u is a one-to-one function and β is a bias function, then whenever {yϵA\{x}|u(y) > u(x) + β(x)} ≠ ϕ, (u,β) will be a compatible pair and the default bias extended choice rule generated by it will satisfy WARPECR. The following corollary of proposition 1 proves that the converse is also true.

Corollary of Proposition 1: Let B ^ = {(A, x)|xϵAϵΨ(X)}. If an extended choice rule c on B ^ satisfies WARPCER then it is a default bias extended choice rule generated by a compatible pair (u, β) where u is a one-to-one function.

To prove the aforementioned corollary, we use the following lemma.

Lemma 1: If c is a default bias extended choice rule then it is generated by a compatible pair (u, β) where u is a one-to-one function.

Proof: By definition of a default bias extended choice rule, we know that there is a compatible pair (v, β) such that c is generated by it.

Let S = {yϵX|v(y) ≥ v(x) + β(x) for some xϵX\{y}}. We will first show that the restriction of v to S is a one-to-one function.

Let yϵX such that v(y) ≥ v(x) + β(x) for some xϵX\{y}. Since β(x) > 0, v(y) > v(x) and so v(y) ≠ v(x). Hence suppose, zϵX\{x, y} such that v(y) = v(z). Thus, v(z) ≥ v(x) + β(x). Consider the extended budget set ({x, y, z},x). Since c is a default bias extended choice rule generated by (v,β), c({x, y, z},x) = y and c({x, y, z},x) = z contradicting y ≠ z. Thus, v(y) ≠ x(z).

Thus the restriction of v to S is a one-to-one function.

Observe that X\S = {yϵX| there does not exist any xϵX\{y} such that v(y) ≥ v(x) + β(x)}.

Thus, yϵX\S implies v(y) < v(x) + β(x) for all xϵX (including v(y) < v(y) + β(y) since β(y) > 0).

Thus, xϵS and yϵX\S implies v(x) > v(y).

For each yϵX\S consider the sets A(y) = {xϵX|v(x) ≥ v(y) + β(y)} and B(y) = {xϵX|v(x) < v(y) + β(y)}. Clearly A(y) ⊂ S and X\S ⊂ B(y). Let δ(y) = min{v (y) + β(y)- v(x) | x ∈ β(y)} 4 and δ = min{δ(y)|yϵX\S}. Clearly δ > 0. For each yϵX\S, choose ε(y)ϵ(0, δ) such that given y, zϵX\S with y ≠ z, v(y) - ε(y) ≠ v(z) − ε(z). This is clearly possible since X\S is a finite set and (0,δ) is a non-degenerate interval in the real line whose cardinality is uncountable.

Let u:X → ℝ be defined as follows: u(x) = v(x) if xϵS and u(x) = v(x) − ε(x) for all xϵX\S.

The restriction of u to S is as before and hence one-to-one. The restriction of u to X\S is by definition of u a one-to-one function. Further, xϵS and yϵX\S implies u(x) = v(x) > v(y) > v(y) − ε(y) = u(y). Thus u is a one-to-one function on X and so (u,β) is a compatible pair. It remains to verify that (u,β) generates c.

Consider (A, x)ϵ B ^ . If c(A, x) ≠ x, then v(c(A, x)) ≥ v(x) + β(x) and v(c(A, x)) > v(y) for all yϵA\{x}. Clearly c(A, x)ϵS and so u(c(A, x)) = v(c(A, x)) > v(y) = u(y) for all yϵ(A\{y})⋂S and u(c(A, x)) = v(c(A, x)) > v(y) > v(y) -ε(y) = u(y) for all yϵ(A\{y})⋂(X\S). Further u(c(A, x)) ≥ v(x) + β(x) = u(x) + β(x) if xϵS and u(c(A,x)) = v(c(A,x)) ≥ v(x) + β(x) ≥ v(x)- ε(x) + β(x) = u(x) + β(x) if X\S.

Hence suppose c(A, x) = x.

Thus, v(x) + β(x) > v(y) for all yϵA\{x}.

Case 1: xϵS.

Then u(x) + β(x) = v(x) + β(x) > v(y) = u(y) if yϵ(A\{x})⋂S and u(x) + β(x) = v(x) + β(x) > v(y) > v(y) − ε(y) = u(y) if yϵ(A\{x})⋂(X\S).

Case 2: xϵX\S.

Thus, yϵB(x) for all yϵA\{x}.

Thus, u(x) + β(x) = v(x) − ε(x) + β(x) > v(x) − δ + β(x) ≥ v(x) − δ(x) + β(x) = v(x) − min{v(x) + β(x) − v(z) | z ∈ β(x)} 4 + β ( x ) ≥ v(x) − v(x) + β(x) − v(y) 4 + β(x)= 3 ( v(x)+β(x) ) 4 + v(y) 4 = 3 ( v(x)+β(x) − v(y) ) 4 + v(y) .

Since v(x) + β(x) > v(y), we get u(x) + β(x) > 3 ( v(x)+β(x) − v(y) ) 4 + v(y) > v(y) ≥ u(y) for all yϵA\{x}.

Thus, c is generated by (u,β). Q.E.D.

The proof of the corollary now follows directly from Proposition 1 and Lemma 1.

Declaration of Conflicting Interests

The author declared no potential conflicts of interest with respect to the research, authorship and/or publication of this paper.

Funding

The author received no financial support for the research, authorship and/or publication of this paper.