Self-Control Preferences and Status-Quo Bias

Proof of Theorem 1

The “if part” is obvious and explained in the main text. For the “only if part,” let ≿ ́ be a binary relation on X defined by x ≿ ́ y if x = c({y}, x). We will show that ≿ ́ is complete and transitive. Assume y ≠ c({x}, y), then, since x = c({x}, x), we have by E-WARP that y ≠ c({y}, x), and we find that x = c({y}, x), and ≿ ́ is complete. For transitivity, take x, y, and z such that x = c({y}, x), y = c({z}, y) but x ≠ c({z}, x). Then, y = c({z}, y) but x ≠ c({z}, x) imply by E-WARP that x ≠ c({y}, x) – a contradiction. We now show that ≿ ́ rationalizes c is the desired sense. First, assume that s = c(A, s). Then, s ≿ ́ x for all x ∈ A, for otherwise s ≠ c({x}, s) for some x ∈ A, and by x = c({x}, x), we have a contradiction to E-WARP. We have left to show that y = c(A, x) implies (i) y ≻ ́ x , and (ii) y ≿ ́ z for all z ∈ A. y = c(A, x) implies x ≠ c(A, x), and by E-WARP, x ≠ c({y}, x) and y ≻ ́ x . z ≻ ́ y implies z = c({z}, y) and if z ∈ A, we have by E-WARP that y ≠ c(A, x) – a contradiction. This completes the proof. □

Lemma 1

Let χ be the set of all nonempty, finite subsets of X and let c : χ × X → X be a choice function that satisfies E-WARPS. Then, there exists a preference relation ≿ on χ satisfying DFC such that x = c(A, x) if and only if {x} ≿ A.

Proof

Define ≿ by A ≿ B if [x = c(A, x) ⇒ x = c(B, x)], and note that ≿ is complete, as x = c(A, x) and x ≠ c(B, x) imply, by E-WARPS, that if y = c(B, y), then y = c(A, y). Clearly, ≿ is also transitive, as if A ≿ B ≿ D, then x = c(A, x) ⇒ x = c(B, x) and x = c(B, x) ⇒ x = c(D, x), hence, we have, x = c(A, x) ⇒ x = c(D, x).

We now show that for the constructed ≿, x = c(A, x) if and only if {x} ≿ A. The fact that {x} ≿ A implies x = c(A, x) follows directly from x = c({x}, x). For the other direction, assume for the sake of contradiction that x = c(A, x), but there exists y ∈ X such that y ≠ c(A, y) and y = c({x}, y), then we have a contradiction to E-WARPS.

We have left to show that ≿ satisfies DFC. Assume not, then x ≠ c(A, x) for all x ∈ A. That is, for any x ∈ A, there is y ∈ A such that y = c(A, x). By E-WARPS, we then have that y = c({y}, x), and by completeness of ≿, {y} ≻ {x}. This means that for any x ∈ A, there is y ∈ A such that {y} ≻ {x}. But this is impossible, as A ∈ χ is finite and ≿ is transitive. □

Proof of Theorem 2

The “if part:” to see that RSP implies E-WARPS, note that x = c(A, s) for some s ∈ X implies by RSP either {x} ≿ {y} for all y ∈ A, which by DFC implies {x} ≿ A or that {s} = {x} ≿ A, hence, {x} ≿ A holds in any case. In addition, y ≠ c(A, y) imply that A ≻ {y}, thus, we find that {x} ≻ {y}. Now, for and s′ ∈ X such that B ∪ {s′} ∋ x and y = c(B, s′), we have s′ ≠ y, then {y} ≿ {x} – a contradiction. If, on the other hand, s′ = y, then we have {y} ≿ B, and each of B = {x} and x ≠ c(B, x) implies again {y} ≿ {x}. This completes the “if part”.

For the “only if part”, let ≿ be as defined in Lemma 1. Then, we only need to show that if x = c(A, s) and x ≠ s, then {x} ≿ {y} for all y ∈ A, or equivalently that for any y = c(B, s) ≠ s, we have {y} ≿ {x} for all x ∈ B. This follows by the contrapositive of condition (i) in E-WARPS, which implies that y = c(B, s′) ≠ s′ and x ∈ B ∪ s′ imply [x = c(A, x) ⇒ y = c(A, y)], and by taking A = {x}, we find y = c({x}, y) (i.e. {y} ≿ {x}). This completes the proof. □

Proof of Corollary 2

The facts that x = c({y}, x) if and only if {x} ≿ {y} and that c(A, x) ≠ x = c(B, x) for some x ∈ X implies A ≻ B follow directly from the Proof of Theorem 2. We have, thus, left to show that if x = c(A, x) ⇔ x = c(B, x), then there exists a preference relation ≿ ̃ such that B ≻ ̃ A . Assume that x = c(A, x) ⇔ x = c(B, x), and let B ∼ 1 ≔ { D ∈ χ \ σ ∣ D ∼ B } \ B and B ∼ 2 ≔ { { x } ∣ { x } ∼ B } ∪ B where ≿ is as defined in Lemma 1. Now, define ≿ ̃ ≔ ≿ \ ( D , E ) ∣ D ∈ B ∼ 1 and E ∈ B ∼ 2 . We show that the constructed ≿ ̃ rationalizes c in the desired sense. First note that because ≿ is complete and transitive, ≿ ̃ is complete and transitive by construction. Second, note that {x} ≿ A implies { x } ≿ ̃ A for all x ∈ X and A ∈ χ. It thus follows from the fact that ≿ satisfies DFC, that ≿ ̃ satisfies DFC. Third, because ≿ rationalizes c in the desired sense, ≿ ̃ rationalizes c for any menu D ∈ χ such ¬(D ∼ B). Now, assume that x = c(D, x) and D ∼ B or { x } ∼ B , then {x} ≿ D and { x } ≿ ̃ D . Thus, assume that x ≠ c(D, x) for some D ∼ B, then D ≻ {x} and, hence, D ≻ ̃ { x } . Similarly, x = c(D, s) and {x} ∼ B implies x ∈ max_≿ D and, hence, x ∈ max ≿ ̃ D . Finally, x ≠ c(D, s), x ∈ D, and {x} ∼ B implies {y} ≻ {x} for some y ∈ D, and, hence { y } ≻ ̃ { x } . This completes the proof. □

Lemma 2

Let χ be the set of all nonempty, compact subsets of X and let c : χ × X → X be a choice function that satisfies E-WARPS(a) and CA. Then, there exists a preference relation ≿ on χ satisfying DFC such that x = c(A, x) if and only if {x} ≿ A.

Proof

Given Lemma 1, it remains to show that ≿ defined by A ≿ B if [x = c(A, x) ⇒ x = c(B, x)] satisfies DFC (note that the only part in the proof of Lemma 1, where we use the fitness of χ is when proving that DFC holds). First note that CA implies that ≿ is continuous on σ (i.e. the sets {x∣{x} ≿ {y}} and {x∣{y} ≿ {x}} are closed for all y ∈ X): take two convergent sequences x _n and y _n such that {x _n } ≿ {y _n } for all n, that is, x _n = c({y _n }, x _n ) for all n. Then, by CA, we have lim x _n = c(lim{y _n }, lim x _n ); hence, lim{x _n } ≿ lim{y _n }. Thus, the restriction of ≿ to σ (denoted ≿ ́ ) is complete, transitive and continuous and because A ∈ χ is compact, ≿ ́ attains a maximum on A. The rest of the proof follows as in that of Lemma 1. □

Proof of Theorem 3

The “if part:” assume that c satisfies RSP, with ≿ that is continuous on σ × χ. Then, E-WARPS holds as in Theorem 2. To see that CA holds, note that x _n = c(A _n , x _n ) for each n implies by RSP that {x _n } ≿ A _n for all n, and by the continuity of ≿ on σ × χ, lim{x _n } ≿ lim A _n , hence lim x _n = c(lim A _n , lim x _n ).

For the “only if part”, first note that by Lemma 2 a binary relation ≿ ́ on X defined by x ≿ ́ y if and only if x = c({y}, x) is continuous on X. And since X is a compact set, ≿ ́ has a continuous utility representation u ́ : X → R .

Now, define ≿ as in Lemma 1, recall that (by Lemma 2) ≿ satisfies DFC, that {x} ≿ A if and only if x = c(A, x), and note that ≿ is continuous on σ × χ. Take two convergent sequences x _n and A _n such that {x _n } ≿ A _n for all n, that is, x _n = c(A _n , x _n ) for all n. Then, by CA, lim x _n = c(lim(A _n , x _n )); hence, lim{x _n } ≿ lim A _n . Note that this also means that {x∣{x} ≿ A} is closed set and since X is bounded, {x∣{x} ≿ A} is compact and u ́ , defined above, attains a minimum on this set. We can therefore assume without loss of generality that u ́ ( { x α } ) = 1 and u ́ ( { x α } ) = 0 .

Next, define u : χ → R such that for all A ∈ χ:

We show that u satisfies:

By the W-Axiom, for any A ∈ χ, u(A) ≔ β ≥ 0, and by DFC, β ≤ 1 also holds. By the continuity of u ́ on X, for any such 0 ≤ β ≤ 1, there exists x ∈ X such that u ́ ( { x } ) = β , and because u ( { x } ) = u ́ ( x ) for all x ∈ X, (5) holds. However, using (5), we can show that u represents ≿. Indeed, u(A) ≥ u(B) if and only if {x} ≿ {y} for {x} ∼ A and {y} ∼ B. Thus, by Theorem 2 the first part of the theorem is complete.^[21]

For the second part, we only left to show that the constructed u is unique. Assume that c admits RSP both for the constructed u and for some v. Then, u(A) > u(B) implies c(B, x) ≠ x = c(A, x) for some x ∈ X, and v(A) > v(B) must hold. For the other direction, first note that since v rationalizes c, v represents a preference relation that admits DFC, and in addition, it satisfies that v({x}) ≥ v(A) if and only if x = c(A, x). Thus, by the argument above v also satisfies property (5). Now, assume that v(A) > v(B), then x ≠ c(A, x) for v({x}) = v(B) (i.e. x = c(B, x)), hence, u(A) > u(B). Thus, v = u and the proof is complete. □

Proof of Corollary 3

To see that (3) implies E-WARPS(a), assume the former, then x = c(A, x) and y ≠ c(A, y) implies {x} ≿ A ≻ {y} and y = c(B, y) implies {y} ≿ B, hence, {x} ≻ {y} ≿ B, and we have B ≠ {x} and x = c(B, x). Conversely, let ≿ be as defined in Lemma 1, and ≿ ́ defined by x ∼ ́ y for all x, y ∈ X. Now, the fact that s = c(A, s) if and only if {s} ≿ A follows by Lemma 1 (note that Lemma 1 uses only the “(a) part” of E-WARPS). Thus, A ≻ {s} implies c(A, s) ∈ A, and the result follows by m a x ≿ ́ A = A . □

Proof of Theorem 4

We first show that (iii) implies (i): assume that (4) holds, then xPy implies u ́ ( x ) ≥ θ ( A ) > u ́ ( y ) and hence u ́ ( x ) > u ́ ( y ) . Thus, xP _T y, with P _T being the transitive closure of P also implies u ́ ( x ) > u ́ ( y ) and hence yPx is impossible. We now show that (i) implies (ii): define R ⊆ σ × χ by { x } R A if x ∈ C(A), and P ⊆ χ × σ by A P { x } if x ∈ A\C(A). Let ▹ ⊆ χ × χ ≔ R ∪ P and note that by definition, ⊳ is antisymmetric. Moreover, x ⊳ A ⊳ y implies xPy. Thus, if P contains no cycles, ⊳ cannot have cycles either. By a standard result ⊳ can then be extended to a complete, antisymmetric, and transitive preference relation ≿ on χ. Thus, ≿ has a utility representation u : χ → R . It is easy now to verify that ≿ rationalizes C in the desired sense: x ∈ C(A) implies x R A , and thus, u({x}) ≥ u(A); and x ∈ A\C(A) implies A P x ; therefore, u(A) > u({x}) follows. The fact that (ii) implies (iii) is obvious and, hence, the proof of the theorem is complete. □

Definition 5

A choice function c admits unconstrained overwhelming temptation if there exist a preference relation ≿ on χ and a preference relation ≿ ́ on X, such that for all (A, s) ∈ χ × X:

Proposition 1

A choice function c admits unconstrained overwhelming temptation if and only if R* is acyclic.

Proof

As we showed, given Definition 5, xR*y implies {x} ≻ {y}. Thus, the acyclicty of R* follows by the transitivity of ≿ and the “only if part” is complete. For the other direction, let R be a binary relation on X × χ defined by x R A if x = c(A, x) ⊄ A, let P be a binary relation on χ × X defined by A P x if x ≠ c ( A , x ) or A = { x } , and let R * be a binary relation on χ defined by A R * B if there exists x ∈ X such that A P x R B. Note that R * is consistent in the sense that for all A _i , i = 0, …, n such that A i − 1 R * A i for all i > 0 and ¬ ( A i R * A i − 1 ) for some i > 0, we have ¬ ( A n R * A 0 ) , because A n R * A 0 creates, in this situation, an R*-cycle.^[23] By Theorem 3 in Suzumura (1976), being consistent, R * can be extended to a preference relation ≿ on χ such that A R * B and ¬ ( B R * A ) imply A ≻ B.

Now, let ≿ ́ defined by x ∼ ́ y for all x, y ∈ X, and note that to show that the representation holds for the constructed ≿ and for ≿ ́ , we only need to show that (i) {y} ≿ A implies y = c(A, y), and (ii) A ≻ {y} ⊄ A implies y ≠ c(A, y). Assume y ≠ c(A, y), then by y = c({y}, y) ⊄ {y}, we have A R * { y } and ¬ ( { y } R * A ) , hence, A ≻ {y}. This shows that (i) holds. Similarly, y = c(A, y) ⊄ A implies { y } R * A , and {y} ≿ A. This completes the proof. □

Definition 6

A choice function admits status-quo bias (SQB) if there exists a complete and transitive preference relation ≿ ́ on X and a strict partial order ▹ ⊆ ≿ ́ such that:^[25]

Proposition 2

The (G-)RSP and SQB models are not nested.

Proof

Let X = {x, y, z} and note that x = c({x}, y) = c({x, y}, y) = c({x, y, z}, y) and s = c(A, s) for any (A, s) ≠ ({x}, y), ({x, y}, y), ({x, y, z}, y) is consistent with SQB for ▹ = {(x, y)} and ≿ ́ such that x ≿ ́ y ≿ ́ z but it is not G-RSP; as z = c({x, y, z}, z), y ≠ c({x, y, z}, y), and y = c({z}, y) violate E-WARPS(a). Now let X = {x, y, z} and c be an RSP choice function on X such that {x} ≻ {x, y} ≻ {y} ≻ {x, y, z} ≻ {x, z} ≻ {z}. Then, x = c({x}, y), and if c admits SQB, then x ▹ y must hold. However, we also have y = c({x, y, z}, y), which is a contradiction. Thus, we have shown that there are SQB choice functions that are not G-RSP, and there are RSP choice functions that are not SQB, and the proof follows by the fact that any RSP choice function is also G-RSP. □