Solution Homework 1 - EconS 501

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1 Solution Homework 1 - EconS [Checking properties of preference relations-i]. For each of the following preference relations in the consumption of two goods (1 and 2): describe the upper contour set, the lower contour set, the indifference set of bundle (2,1), and interpret them. Then check whether these preference relations are rational (by separately examining whether they are complete and transitive), monotone, and convex. (a) Bundle (x 1, x 2 ) is weakly preferred to (y 1, y 2 ), i.e., (x 1, x 2 ) (y 1, y 2 ), if and only if max {x 1, x 2 } max {y 1, y 2 }. Intuitively, this relation states that a bundle (x 1, x 2 ) is preferred to an alternative bundle (y 1, y 2 ) if and only if the most abundant component of the first bundle exceeds the most abundant component of the second bundle. In this preference relation, for a given bundle (2, 1), the upper contour set is defined as all those bundles that contain more than two units in its most abundant component, i.e. (x 1, x 2 ) where x i 2 for at least one good i. Figure 1.1 depicts this UCS, which contains all the bundles to the right-hand side of x 1 = 2 (since they all have more than two units in the first component), but also those points that, despite being to the left of x 1 = 2, contain more than two units of good 2, i.e., all bundles (x 1, x 2 ) for which x 2 2 in the left-hand corner of the figure. In contrast, the LCS embodies all bundles for which its most abundant component is weakly lower than 2, i.e., (x 1, x 2 ) where x i 2 for every good i, and are graphically depicted by the points below x 2 = 2 and to the left of x 1 = 2 on figure 1.1. Figure 1.1. UCS, LCS and IND of bundle (2,1). 1

2 As a consequence, the IND set of bundle (2, 1) contains all those bundles for which its most abundant component has exactly two units, i.e., (2, x 2 ) and (x 1, 2) for all x 1 [0, 2] and x 2 [0, 2], as depicted in the square area of figure 1.1, thus including the origin (0, 0) as part of the IND set of bundle (2, 1). Completeness. For completeness to hold, we need to show that, for any two bundles (x 1, x 2 ), (y 1, y 2 ) R 2, either max {x 1, x 2 } max {y 1, y 2 }, or max {y 1, y 2 } max {x 1, x 2 }, or both, i.e., max {x 1, x 2 } = max {y 1, y 2 }. 1 Let max {x 1, x 2 } = a, and max {y 1, y 2 } = b. Since a, b R, and real numbers are totally ordered, then either a b, or b a, or both (a = b). It then follows that either (x 1, x 2 ) (y 1, y 2 ), or (y 1, y 2 ) (x 1, x 2 ), or both, i.e., (x 1, x 2 ) (y 1, y 2 ). Hence, this preference relation is complete. Note that this property could be anticipated by our description of the UCS, LCS and IND sets in figure 1.1 whereby, for a given bundle (2, 1), we can rank any other bundle in the positive quadrant (x 1, x 2 ) as being preferable to (2, 1), i.e., (x 1, x 2 ) (2, 1), or (2, 1) being preferred to this bundle, i.e., (2, 1) (x 1, x 2 ), or both, i.e. (x 1, x 2 ) (2, 1). Transitivity. Take three bundles (x 1, x 2 ), (y 1, y 2 ) and (z 1, z 2 ) R 2 satisfying (x 1, x 2 ) (y 1, y 2 ) and (y 1, y 2 ) (z 1, z 2 ). Then, they must satisfy that, on one hand, max {x 1, x 2 } max {y 1, y 2 } and, on the other hand, max {y 1, y 2 } max {z 1, z 2 }. Therefore, by transitivity of the greater than or equal relation ( ), max {x 1, x 2 } max {z 1, z 2 }, which implies (x 1, x 2 ) (z 1, z 2 ). Hence, the preference relation is transitive. Since, it is also complete as shown above, this preference relation is rational. Monotonicity. Recall that, in order to check if a preference relation is monotonic we need to show that, if we increase all components of a given bundle (x 1, x 2 ), the newly created bundle is strictly preferred to (x 1, x 2 ). Take a bundle (y 1, y 2 ) such that y 1 > x 1 and y 2 > x 2, e.g., y 1 = x 1 + ε and y 2 = x 2 + δ where ε, δ > 0. Then, max {y 1, y 2 } > max {x 1, x 2 } with strict inequality, and it follows that (y 1, y 2 ) (x 1, x 2 ). Hence, the preference relation satisfies monotonicity, i.e., if y i > x i for all components i = {1, 2}, then y x. Convexity. Recall that, for a preference relation to satisfy convexity, we need that, for any three bundles (x 1, x 2 ), (y 1, y 2 ) and (z 1, z 2 ); where the last two bundles are weakly preferred to (x 1, x 2 ), that is (y 1, y 2 ) (x 1, x 2 ) and (z 1, z 2 ) (x 1, x 2 ) 1 Intuitively, the last case occurs when the amount of the most abundant item in bundle x coincides with the amount of the most abundant item in bundle y. 2

3 it must hold that the convex combination of bundles (y 1, y 2 ) and (z 1, z 2 ) is also weakly preferred to (x 1, x 2 ), λ(x 1, x 2 ) + (1 λ)(y 1, y 2 ) (x 1, x 2 ) In order to check whether this preference relation satisfies convexity, take three bundles (x 1, x 2 ), (y 1, y 2 ) and (z 1, z 2 ) R 2 such that (y 1, y 2 ) (x 1, x 2 ) and (z 1, z 2 ) (x 1, x 2 ). Therefore, it must be that max {y 1, y 2 } max {x 1, x 2 }, and similarly that max {z 1, z 2 } max {x 1, x 2 }. The convex combination of (y 1, y 2 ) and (z 1, z 2 ) yields a first component of λy 1 + (1 λ)z 1 and a second component of λy 2 + (1 λ)z 2. However, the largest of these two components, max {λy 1 + (1 λ)z 1, λy 2 + (1 λ)z 2 }, is not necessarily higher than max {x 1, x 2 }. In order to see that, consider an example in which the two premises of convexity hold, that is, max {y 1, y 2 } max {x 1, x 2 } and max {z 1, z 2 } max {x 1, x 2 }, such as (y 1, y 2 ) = (0, 4) and (x 1, x 2 ) = (3, 3) and (z 1, z 2 ) = (4, 0). Indeed, note that max {y 1, y 2 } = 4, max {x 1, x 2 } = 3 and max {z 1, z 2 } = 4, implying that max {y 1, y 2 } = 4 3 = max {x 1, x 2 } and max {z 1, z 2 } = 4 3 = max {x 1, x 2 }. Now, note that the convex combination of (y 1, y 2 ) and (z 1, z 2 ) with λ, will give us values between 0 and 4. For intermediate values of λ (such as λ = 1 ) the convex combination of bundles y and x yields 2 { 1 max , } 2 0 = max {2, 2} = 2, which is lower than max {x 1, x 2 } = 3. Hence, the preference relation is not convex. Alternative proof of non-convex preferences without using a counterexample. We could anticipate the non-convexity of preferences, since figure 1.1 depicted a non-convex upper contour set (which implies non-convex preferences). As we see in figure 1.2, we can find bundles, like x, for which its upper contour set is not convex. That is, y x but αx + (1 α)y x for all α [0, 1] 3

4 Figure 1.2. Non-convex upper contour sets. In order to formally demonstrate that the upper contour set is not convex, we need to show that, for a convex combination of two bundles, i.e., αx + (1 α)y, we have that max{αx 1 + (1 α)y 1, αx 2 + (1 α)y 2 } < max{x 1, x 2 } = x 2 (1) which would imply that the upper contour set is non-convex and, therefore, preferences are not convex. In order to prove that, we focus on the case in which y x, i.e., bundles x and y lie on the same indifference curve, which means that x 1 = y 2 and that x 2 = y 1 (as depicted in figure 1.3). To show that inequality (1) is indeed satisfied, we will first focus on the case in which the convex combination of bundles x and y lies below the 45 0 line (i.e., low values of α), and then on the case in which this convex combination lies above the 45 0 line (i.e., high values of α): Case 1: The convex combination lies below the 45 0 line, as depicted in figure

5 Figure 1.3. The linear combination lies below the line. In particular, since the convex combination lies below the 45 0 line, we have that αx 1 + (1 α)y 1 αx 2 + (1 α)y 2 which implies that max{αx 1 + (1 α)y 1, αx 2 + (1 α)y 2 } = αx 1 + (1 α)y 1 In addition, we have that αx 1 + (1 α)y 1 αx 2 + (1 α)y 1 since x 1 < x 2, given that bundle x lies above 45 0 line. In addition, αx 2 + (1 α)y 1 = x 2 since x 2 = y 1 given that both bundles lie on the same indifference curve (see figure 1.3). Therefore, max{αx 1 + (1 α)y 1, αx 2 + (1 α)y 2 } < max{x 1, x 2 } implying that the consumer s utility from the convex combination of bundles x and y is strictly lower than from consuming bundle x alone, i.e., a violation of convexity in preferences. Case 2: The convex combination now lies above the 45 0 line, as depicted in figure 1.4. Figure 1.4. The linear combination lies above the line. 5

6 In particular, since the convex combination lies above the 45 0 line, we have that αx 1 + (1 α)y 1 αx 2 + (1 α)y 2, which implies that max{αx 1 + (1 α)y 1, αx 2 + (1 α)y 2 } = αx 2 + (1 α)y 2 We furthermore have that αx 2 + (1 α)y 2 αx 2 + (1 α)y 1 since y 1 > y 2, given that bundle y lies below the 45 0 line. Finally, αx 2 + (1 α)y 1 = x 2 since bundles x and y lie on the same indifference curve, thus entailing that x 2 = y 1. As a consequence, max{αx 1 + (1 α)y 1, αx 2 + (1 α)y 2 } < max{x 1, x 2 } which also suggests that the consumer s utility from the convex combination of bundles x and y is strictly lower than from consuming bundle x alone, which constitutes a violation of convexity in preferences for Case 2 as well. Therefore, preferences are not convex. Remark: Note that this preference relation satisfies local nonsatiation. In order to show this property, take any bundle (x 1, x 2 ) R 2 and ε > 0. Let us now generate a new bundle (y 1, y 2 ) in which both components of (x 1, x 2 ) have been increased by ε 2, i.e., (y 1, y 2 ) ( x 1 + ε 2, x 2 + ε 2). Hence, max {y1, y 2 } > max {x 1, x 2 }. By the preference relation in this example, it follows that y x, but not x y, so that y x. We just need to confirm that bundles x and y are ε close, by finding their Cartesian distance x y = [ x 1 ( x 1 + ε )] 2 [ ( + x 2 x 2 + ε )] 2 ε = which is smaller than ε, implying that the preference relation satisfies local nonsatiation. 2. [Checking properties of preference relations-ii]. Consider the following prefer- 6

7 ence relation defined in X = R 2 +. A bundle (x 1, x 2 ) is weakly preferred to another bundle (y 1, y 2 ), i.e., (x 1, x 2 ) (y 1, y 2 ), if and only if min {3x 1 + 2x 2, 2x 1 + 3x 2 } min {3y 1 + 2y 2, 2y 1 + 3y 2 } (a) For any given bundle (y 1, y 2 ), draw the upper contour set, the lower contour set, and the indifference set of this preference relation. Interpret. Take a bundle (2, 1). Then, min { , } = min {8, 7} = 7. The upper contour set of this bundle is given by UCS(2, 1) = {(x 1, x 2 ) (2, 1)} = {min {3x 1 + 2x 2, 2x 1 + 3x 2 } 7 min {8, 7}} which is graphically represented by all those bundles in R 2 + which are strictly above both lines 3x 1 + 2x 2 = 7 and 2x 1 + 3x 2 = 7. That is, for all (x 1, x 2 ) strictly above both lines x 2 = x 1 and x 2 = x 1. (See figure 1.9, which depicts these two lines and shades the set of bundles lying weakly above both lines.) On the other hand, the lower contour set is defined as LCS(2, 1) = {(2, 1) (x 1, x 2 )} = {7 min {3x 1 + 2x 2, 2x 1 + 3x 2 }}, which is graphically represented by all bundles (x 1, x 2 ) weakly below the maximum of the lines described above. For instance, bundle (y 1, y 2 ) = (2.5, 0), which lies on the horizontal axis and between both lines horizontal intercept, implies min{ , } = min{7.5, 5} = 5 thus implying that this consumer prefers bundle (x 1, x 2 ) = (2, 1) than (y 1, y 2 ) = (2.5, 0). A similar argument applies to all other bundles lying above x 2 = 7 3x and below x 2 = 7 2x 3 3 1, where bundle (2.5, 0) also belongs; see the triangle that both lines form at the right-hand side of the figure. Similarly, 7

8 bundles such as (0, 2.5) yield min{ , } = min{5, 7.5} = 5, which implies that the consumer also prefers bundle (2, 1) to (0, 2.5). An analogous argument applies to all bundles above line x 2 = 7 3x but below x 2 = 7 2x in the triangle at the left-hand side of figure 2.1. Figure 2.1. UCS and LCS of bundle (2,1). Finally, those bundles for which the UCS and LCS overlap are those in IND of bundle (2,1). (b) Check if this preference relation satisfies: (i) completeness, (ii) transitivity, and (iii) weak convexity. Completeness. First, note that both of the elements in the min{ } operator are real numbers, i.e., (3x 1 + 2x 2 ) R + and (2x 1 + 3x 2 ) R +, thus implying that the minimum min {3x 1 + 2x 2, 2x 1 + 3x 2 } = a exists and it is also a real number, a R +. Similarly, the minimum min {3y 1 + 2y 2, 2y 1 + 3y 2 } = b exists and is a real number, b R +. Therefore, we can easily compare a and b, obtaining that either a b, which implies (x 1, x 2 ) (y 1, y 2 ); or a b, which implies (y 1, y 2 ) (x 1, x 2 ), or both, a = b, which entails (x 1, x 2 ) (y 1, y 2 ). 8

9 Hence, the preference relation is complete. Transitivity. We need to show that, for any three bundles (x 1, x 2 ), (y 1, y 2 ) and (z 1, z 2 ) such that (x 1, x 2 ) (y 1, y 2 ) and (y 1, y 2 ) (z 1, z 2 ), then (x 1, x 2 ) (z 1, z 2 ) First, note that (x 1, x 2 ) (y 1, y 2 ) implies a min {3x 1 + 2x 2, 2x 1 + 3x 2 } min {3y 1 + 2y 2, 2y 1 + 3y 2 } b and (y 1, y 2 ) (z 1, z 2 ) implies that b min {3y 1 + 2y 2, 2y 1 + 3y 2 } min {3z 1 + 2z 2, 2z 1 + 3z 2 } c Combining both conditions we have that a b c, which implies that a c. Hence, we have that min {3x 1 + 2x 2, 2x 1 + 3x 2 } min {3z 1 + 2z 2, 2z 1 + 3z 2 } and thus (x 1, x 2 ) (z 1, z 2 ), implying that this preference relation is transitive. Weak Convexity. This property implies that the upper contour set must be convex. That is, if bundle (x 1, x 2 ) is weakly preferred to (y 1, y 2 ), (x 1, x 2 ) (y 1, y 2 ), then the convex combination of these two bundles is also weakly preferred to (y 1, y 2 ), λ(x 1, x 2 ) + (1 λ) (y 1, y 2 ) (y 1, y 2 ) for any λ [0, 1] For compactness, let a 3x 1 + 2x 2, b 2x 1 + 3x 2, c 3y 1 + 2y 2 and d 2y 1 +3y 2. Hence, the property that (x 1, x 2 ) (y 1, y 2 ) implies min {a, b} min {c, d}. We therefore need to show that min {λa + (1 λ) c, λb + (1 λ) d} min {c, d} 1. First case: min {a, b} = a, min {c, d} = c and a c. Therefore, min {λa + (1 λ) c, λb + (1 λ) d} = λa + (1 λ) c and λa + (1 λ) c > min {c, d} = c. For this case, convexity is satisfied. 2. Second case: min {a, b} = a, min {c, d} = d and a d. Hence, a > b and 9

10 c > d, implying that min {λa + (1 λ) c, λb + (1 λ) d} = λa + (1 λ) d and λa + (1 λ) d min {c, d} = d given that a d. For this case, convexity is satisfied as well. An analogous argument applies in the other two cases, in which min {a, b} = b and min {c, d} = c, and in which min {a, b} = b but min {c, d} = d. 3. [Monotonicity and Strong monotonicity.] Explain monotonicity and strong monotonicity in preference relations, and compare them. Provide an example where a bundle x is (strictly) preferred to bundle y when preferences satisfy strong monotonicity, but x is not necessarily preferred to y under monotonicity. Monotonicity states that increasing the amount of some commodities cannot hurt, and increasing the amount of all commodities is strictly preferred. Formally, if we take bundle y R L and weakly increase all k components, so that we generate a new bundle x R L satisfying x k y k for all k, then an individual with monotonic preferences would prefer the newly created bundle to the original bundle, i.e., x y. (Note that this implies that at least one component of the bundle has been strictly increased while the remaining components can be left unaffected.) In addition, if we strictly increase the amount of all components in bundle y, this individual would strictly prefer the new bundle, i.e., if x k > y k for all k, then x y. Strong monotonicity. On the other hand, strong monotonicity states that the consumer is strictly better off with additional amounts of any commodity. That is, if we strictly increase the amount of at least one commodity, the consumer strictly prefers the newly created bundle x to his original bundle y. That is, if x k y k for all good k and x y, then x y. (Note that this implies that x j > y j for at least one commodity j, since otherwise both bundles would coincide.) Comparison. Then, a consumer s preference relation can satisfy monotonicity (if additional amounts of one of his commodity do not harm his utility), but does not need to satisfy strong monotonicity (since for that to occur, he would need to become strictly better off as a consequence of the additional amounts in one of his commodities). However, if a consumer s preferences satisfy strong monotonicity, they must also satisfy monotonicity. That is why strong monotonicity is a more restrictive ( stronger ) assumption on preferences than monotonicity. 10

11 Example: Consider bundles x = (1, 2) and y = (1, 1). If preferences satisfy strong monotonicity, x y since the second component in bundle x is higher than the corresponding component in y, i.e., x j y j for some good j. However, if preferences only satisfy monotonicity, we cannot state that x y (strictly), since x k > y k does not hold for all k commodities. 4. [Lexicographic preferences and WARP]. Does the lexicographic preference relation induce a choice structure that satisfies the weak axiom of revealed preference (WARP)? [Hint: Prove it in two steps. First, show whether the lexicographic preference relation is a rational preference relation, and then show that every rational preference relation implies a choice structure which satisfies WARP]. We want to prove this statement in two steps: 1) Every lexicographic preference relation is rational, and 2) Every rational preference relation satisfies WARP The first step was already shown in exercise 9 of this chapter. Let us prove here the second step (that every rational preference relation satisfies WARP). Let us first recall the definition of a choice rule satisfying WARP: if for some B B with x, y B, we have x C(B), then for any B B with x, y B, and where y C(B ), we must have x C(B ) First, take some budget set B B. Let us assume that x, y B, and that bundle x is revealed preferred to all other bundles in B, i.e., x C (B, ). Hence, x y. To check whether WARP is satisfied, suppose that for other budget set B B that also contains bundles x and y, i.e., x, y B, we have that bundle y is revealed preferred, i.e., y C (B, ). This implies that y z for any z B. Since we also had that x y. Hence, by transitivity (note that here we can use transitivity because the preference relation is rational) we obtain x y and y z, then x z and x z for all z B implies that x C (B, ). Therefore, WARP is satisfied. 11

12 5. [MWG 1.C.1] Consider the choice structure (β, C ( )) with β = ({x, y}, {x, y, z}) and C ({x, y}) = {x}. Show that if (β, C ( )) satisfies the weak axiom, then we must have C ({x, y, z}) = {x}, = {z}, or {x, z}. Let us define B = {x, y} and B = {x, y, z}, so that B, B β. Given that the choice structure (β, C ( )) satisfies the weak axiom of revealed preference, we have the following property holds. For B β with x, y B, we have x C (B). Then for B β with x, y B, with x, y B, we must also have x C (B ). This implies that x y, but not possible to have y x. Assuming the preference relation is rational, then 1. if y z, then x y z such that C (B ) = {x}. 2. if z x, then z x y such that C (B ) = {z}. 3. if z x and also x z, then x z y such that C (B ) = {x, z}. 6. [MWG 1.C.3] Suppose that choice structure (β, C ( )) satisfies the weak axiom. Consider the following two possible revealed preference relations, and : x y there is some B β such that x, y B, x C (B), and y / C (B) x y x y but not y x where is the revealed at-least-as-good-as relation defined in Definition 1.C.2. [(a)] 1. Show that and give the same relation over X; that is, for any x, y X, x y x y. Is this still true if (β, C ( )) does not satisfy the weak axiom? [" "] Suppose that clause x y holds. Assume that there exists B β, where x, y B, and y C (B ), then the weak axiom implies that x C (B ), which violates y x in clause x y. [" "] Suppose that clause x y holds. Then B B, where x, y B, x C (B ) implies that y / C (B ), and this is suffi cient for some B β in clause x y. Therefore, the weak axiom guarantees that x y x y. However, if the weak axiom does not hold, then it is possible for the existence of some B B, where B β and x, y B, such that y C (B ) but x / C (B ) that does 12

13 not violate clause x y but contradicts clause x y. Therefore, and are not equivalent without the weak axiom. 2. Must be transitive? Let X = {x, y, z}, and β = {{x, y}, {y, z}}. Then x C ({x, y}) x y, and y C ({y, z}) y z. However, since {x, z}, {x, y, z} / β, we do not have x z; and therefore, need not be transitive. 3. Show that if β includes all three-element subsets of X, then is transitive. Let x, y, z X, x y and y z. Then, {x, y, z} β, and by (a), x y and y z. Hence, we have neither y x nor z y. Since rationalizes (β, C ( )), this implies that y / C ({x, y, z}) and z / C ({x, y, z}). Furthermore, since C ({x, y, z}), C ({x, y, z}) = {x}; and as a result, x z. 13