From lemma 2, we have a unique (up to a positive affine transformation) linear function U : M (≽) → R. Note that by Axiom 4, for any two-element set we have either {a} ∼ {a, b}, {a, b} ∼ {b}, or {a} ≻ {a, b} ≻ {b}. Hence, if there is a unique U defined over M (≽), then we have a unique U defined over all two-element sets. Using Lemma 1, then we obtain a unique U defined over A .
Now we will move to define u, v over all points in ∆. First, we define the linear function u: ∆ → R as
u(x) := U ({x})
For any a, b, δ with a, b ∈ ∆, δ ∈ (0, 1), if (a, b, δ) such that {a, b} ∈ M (≽) and {a, (1 −
δ)b + δx} ∈ M (≽) for all x ∈ ∆, we can define the function v : ∆ → R.
v(x; a, b, δ) := U({a, b}) − U({a, (1 − δ) b + δx) δ
For properties on v we adopt Lemma 4 of Gul and Pesendorfer (2001) but change the domain of U from A to M (≽). Note that the proof in GP(2001) is still valid here since all two-element sets using in their proof of Lemma 4 is in M (≽).
Lemma 4. Let U be a linear function that represents some ≽ over M (≽). Suppose that {a, (1 − δ)b + δx} ∈ M (≽) for all z ∈ ∆. Then:
(i) ∀z such that {a, z} ∈ M (≽) , v (z; a, b, δ) = U ({a, b}) − U ({a, z}) (ii) v(a; a, b, δ) = U ({a, b}) − U ({a})
(iii) v(αz + (1 − α) z′; a, b, δ) = αv (z; a, b, δ) + (1 − α) v (z′; a, b, δ) . (iv) v(z; a, b, δ) = v (z; a, b, δ′) , ∀δ′ ∈ (0, δ)
(v) Suppose that {a, (1 − δ)b + δx} ∈ M (≽) for all z ∈ ∆.
Since L (≽) is not a mixture space, we don’t have U is linear over L (≽). However, using an argument similar to the proof of Lemma 5.6 in Kreps (1988), we obtain a weaker version of linearity of U.
Lemma 5. Let U be a function that represent some ≽ satisfying Axiom 1,2a-c, 3 and 4. If {x, y} ∈ L (≽), then
U(α{x, y} + (1 − α) A) = αU({x, y}) + (1 − α) U(A) for all A ∈ M (≽) .
Proof. If {x, y} satisfies {x} ≻ {x, y} ≻ {y}, the linearity is already proven in lemma 4. We have to deal with {x, y} with {x} ∼ {x, y} ≻ {y} or {x} ≻ {x, y} ∼ {y}. If {x, y} satisfies {x} ≻ {x, y} ∼ {y}, we claim that hα({x, y}, A) ∼ hα({y}, A) for all A∈ M (≽). By Axiom 3, {x} ≻ {y} implies hα({x}, A) ≻ hα({y}, A) , and Axiom 4 further implies hα({x}, A) ≽ hα({x, y}, A) ≽ hα({y}, A). In this case, we only have to show that hα({x, y}, A) ≻ hα({y}, A) will lead to a contradiction. Let us take A = {a, b}
have {x} ≻ {x, y} ∼ {y}. Since {x} ≻ {y}, letting α, β ∈ (0, 1) and applying Axiom 3, we obtain
β{x} + (1 − β){y} ≻ β{y} + (1 − β){y} = {y} ∼ {x, y},
and
α{x′} + (1 − α)A ≻ α{x, y} + (1 − α)A, where {x′} = β{x} + (1 − β){y}.
Since hα({x′}, A) ≻ hα({x, y}, A) ≻ hα({y}, A), von Neumann-Morgenstern continu-ity implies there exists some γ ∈ (0, 1) such that
hα({x, y}, A)
≻ hγ(hα({x′}, A), hα({y}, A)).
We only deal with the case when A= {a, b} with {a} ≻ {a, b} ≻ {b} because when A is a singleton set, the argument is similar but easier. Since hα({x′}, {a, b}) and hα({y}, {a, b}) are both in M (≽), we uses Lemma 2 to obtain
hα({x, y}, A) ≻ hγ(hα({x′}, {a, b}), hα({y}, {a, b}))
∼ ˜hγ(hα({x′}, {a, b}), hα({y}, {a, b}))
∼ hα(hγ({x′}, {y}), {a, b})
≻ hα({x, y}, {a, b}),
which yields a contradiction. The last “≻” uses the fact that {x′} ≻ {y}, hγ({x′}, {y}) ≻ {y} ∼ {x, y} and Axiom 4. Since hα({x, y}, A) ∼ hα({y}, A), we have U(hα({x, y}, A)) = U(hα({y}, A)). Using Lemma 2, we have U (hα({y}, A)) = αU ({y})+(1 − α) U (A) = αU({x, y}) + (1 − α) U (A) for all A ∈ M (≽)
For {x, y} with {x} ∼ {x, y} ≻ {y}, Axioms 3 and 4 implies hα({x}, {a, b}) ≽ hα({x, y}, {a, b}) in the previous discussion. By using the fact that {x} ≻ {y}, and
apply-ing a similar argument, we can rule out the possibility that hα({x}, {a, b}) ≻ hα({x, y}, {a, b}) and further obtain U(hα({x, y}, A)) = U (hα({x}, A)) = αU ({x}) + (1 − α) U (A) = αU({x, y}) + (1 − α) U (A) for all A ∈ M (≽)
The following lemma is our version of lemma 5 of Gul and Pesendorfer (2001).
Lemma 6. Let U be a function that represents some ≽ satisfying Axioms 1, 2a-c, 3 and 4.
Suppose L(a) = φ, b ∈ S(a) and δ ∈ (0, 1) satisfy U ({a}) > U ({a, (1 − δ)b + δz}) >
U({(1 − δ)b + δz}) for all z ∈ ∆. Then
U({a, y}) = max
x∈{a,y}{u (x) + v (x; a, b, δ)} − max
z∈{a,y}v(z; a, b, δ) ,
Proof. For the case where U({a}) > U({a, y}) > U ({y}), by GP(2001), we know v(y; a, b, δ) ≥ v (a; a, b, δ) and u (a)+v (a; a, b, δ)−v (y; a, b, δ) > u (y)+v (y; a, b, δ)−
v(y; a, b, δ).
Let A = (1 − δ){a, b} + δ{a, y}. Since {a, b} ∈ M (≽), by Lemma 5 we have U(A) = (1 − δ) U ({a, b}) + δU ({a, y}) for {a, y} ∈ L (≽). For the case where U({a}) = U({a, y}) > U ({y}) , the first part of Lemma 2 establishes that U (A) = minz∈AU({a, z}) . Hence, we have v (a; a, b, δ) ≥ v (y; a, b, δ) for the by the same argu-ment in GP(2001). For the case where U({a}) > U({a, y}) = U ({y}) and y ∈ Lc(a), we will show v(y; a, b, δ) ≥ v (a; a, b, δ) + u (a) − u (y). From Gul and Pesendorfer (2001) this is equivalent to show that
U({a, (1 − δ) b + δy}) ≤ (1 − δ) U ({a, b}) + δU ({a, y}) = U(A)
The above inequality holds because of the second part of Lemma 2 U(A) = maxw∈A({w, (1−
δ)b + δy}).
Chapter 3
Main theorem
The previous chapter focuses on sets without the constraint of willpower and obtains results similar to GP(2001). Since the main difference between our model and GP(2001) is on menus with no self-control due to lack of willpower, to get our representation function U over A , we add one more axiom to regulate the willpower constraints across different menus. Let A be the closure of A.
Axiom 5. For any a, x ∈ ∆, if b ∈ L(a) ∩ S(a) and y ∈ S (x), then 12{a} + 12{x, y} ≽
12{a, b} + 12{x}.
Axiom 5 requires that given the same chance of having to perform costly self-control, DM prefers facing a temptation that he does not have to exhaust his entire willpower.
From Lemma 3, we know that S(a) is convex and δa + (1 − δ) b ∈ S (a) if b ∈ S (a).
Since we are interested in the preferences that could reveal limited willpower and costly self-control, we add the following axiom.
Axiom 6 (Richness). There exists an a in the interior of∆, such that L (a) ̸= φ and S(a) contains a non-empty open ball.
First let us introduce a lemma before we show our main theorem.
Lemma 7. Let U be a function that represents some ≽ satisfying Axioms 1, 2a-c, and 3-6. Take a in the interior of∆, such that L (a) ̸= φ and b ∈ S(a), δ ∈ (0, 1) satisfy
U({a}) > U ({a, (1 − δ)b + δz}) > U ({(1 − δ)b + δz}) for all z ∈ ∆. Then
Hence, by Lemma 6, we have the desired result for all {a, y} ∈ L (≽). The remaining part of the proof shows that if y ∈ L (a), then we must have v(y; a, b, δ) − v(a; a, b, δ) ≥
Now we are ready to prove our main theorem.
Theorem 3. A preference ≽ satisfies Axiom 1, 2a-c, 3-6 if and only if there are continuous linear function u, v and a constant w such that the function U defined as
U(A) = max
x∈A{u(x) + v(x)} − maxy
∈A v(y) s.t. max
y∈A v(y) − v(x) ≤ w
Proof. From Axiom 6, there exists x in the interior of∆, such that L (x) ̸= Φ and y ∈ S(x), δ ∈ (0, 1) satisfy U ({x}) > U ({x, (1 − δ)y + δz}) > U ({(1 − δ)y + δz}) for all z ∈ ∆. By Lemma 7, we can let u (z) := U ({z}), v (z) := v (z; x, y, δ) for all z∈ ∆ and w = maxz∈S(x)v(z; x, y, δ) − v (x; x, y, δ). Now consider the set A = {a, b}, where a and b are in the relative interior of∆. Assume without loss of generality, that u(a) ≥ u (b). Since a is in the interior of ∆, we can select α and a′ such that {a, c} = α{a′} + (1 − α){x, y}. Hence, u ({a}) > U({a, c}) > u(c). Then for δ′ sufficiently small we have
U({a}) > U ({a, (1 − δ′) c + δ′z}) > U ({(1 − δ′) c + δ′z}) for all z ∈ ∆.
If L(a) = Φ, then {a, b} ∈ L (≽). Hence, we can apply Lemma 6, U({a, b}) = max
z∈{a,b}{u (a) + v (a; a, c, δ′)} − max
z∈{a,b}v(z; a, c, δ′) ,
Since the willpower constraint is relevant only when b ∈ S (a), we need to show that v(b; a, c, δ′) − v (a; a, c, δ′) ≤ w for b ∈ S (a). Since b ∈ S (a), by Axiom 5, we have
12{x} + 12{a, b} ≽ 12{a} + 12{x, z}, for z ∈ L (x) ∩ S(x). This implies that 12u(x) +
12U({a, b}) ≥ 12u(a) + 12U({x, z}), U ({a, b}) = u (a) + v (a; a, b, δ′) − v (b′; a, b, δ′) and U({x, z}) = u (x) − w. Hence, we have v (b; a, c, δ′) − v (a; a, c, δ′) ≤ w. Let δ∗ = min{δ, δ′}. By Lemma 4 (iv), v (·; x, y, δ∗) = v (·; x, y, δ) and v (·; a, c, δ∗) = v(·; a, c, δ′) . By Lemma 4 (v), for an appropriate constant k, v (·; x, y, δ∗) = v (·; a, c, δ∗)+
k and hence it follows that
U({a, b}) = max
z∈{a,b}{u (z) + v (z)} − max
z′∈{a,b}v(z′) s.t. max
z′∈{a,b}v(z′) − v (z) ≤ w
If L(a) ̸= Φ, we can apply Lemma 7, Now consider an arbitrary finite set A. We know that
U(A) = max con-straint maxminmax problem because for the pair {a∗, x} we would choose x instead of a∗. Hence, a∗will not survive after the second requirement, i.e.,minb∈Awhen we take b= x.
If v(x) − v (a∗) ≤ w, then for any pair {a∗, z} where z ∈ A, a∗ satisfies the constraint.
Chapter 4 Conclusion
This dissertation is the first one to characterize both costly self-control and limited willpower with representation theorem. We provide a method to distinguish the two pos-sible reasons. It separates the set T(a) into two parts: L(a) and Lc(a), the reason of willpower and the reason of resisting cost respectively. The method we provide here is to construct a menu with a lottery and a less tempting prize. There is a possible alter-native definition for L(a) which can separate T (a) into two identical partitions. Sup-pose b ∈ T (a). If there exists a sequence of lotteries {x}ni=0 (x0 = a, xn = b), and {xi, xi+1} ∈ M (≽) for all i, then b ∈ T (a). In this version of definition, no self-control caused by lack of willpower implies that we can find a series of lotteries like a bridge so that decision maker can perform self-control in every step of this bridge. This definition may be more nature for some readers. We are going to verify the equivalence of these two definitions.
Besides, we are going to compare preferences with different characteristic in our fur-ther study. For example, decision makers with same measurement of benefit and temp-tation, u and v, and with different level of willpower, the constant w, may have some relationship in the preference domain.
Furthermore, if possible, we want to find some axioms which are more likely to be realized in place of the current version. Our main purpose is finding a series of verifiable axioms corresponding to the model with costly self-control and limited willpower. Hence,
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