Problem 2: ii) Completeness of implies that for any x X we have x x and thus x x. Thus x I(x).

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1 Bocconi University PhD in Economics - Microeconomics I Prof. M. Messner Problem Set 1 - Solution Problem 1: Suppose that x y and y z but not x z. Then, z x. Together with y z this implies (by transitivity) that y x which contradicts our starting assumption. Problem 2: i) We have to show that I(x) I(y) implies I(x) = I(y). So assume that z I(x) I(y). Then z x and z y. Moreover since x x for all x I(x) transitivity implies z x for all x I(x) and thus (once more by transitivity) y x for all x I(x). But this means that I(x) I(y). By a perfectly analogous argument it can be shown that transitivity also implies I(y) I(x). Hence, I(x) = I(y). ii) Completeness of implies that for any x X we have x x and thus x x. Thus x I(x). Problem 3: Plausibility of properties i) and ii) given our interpretation of (Our interpretation: the ranking over the sets is generated by an underlying ranking over X as follows: A is at least as good as B if and only if the best element in A is at least as good as the best element in B according to the ranking defined over X): i) The first part fits with our interpretation (adding C to both sets A and B cannot invert the ranking). The second part instead is not consistent with our interpretation: If the best thing in C is better than the best thing in A B then the decision maker should be indifferent between A C and B C. ii) Again the first part of the condition is compatible with our interpretation (adding something to a set which strictly dominates all elements of the set should make the set strictly better). The second part though is again at odds with our interpretation. Adding things to a set A which would never be chosen should not make A strictly worse. It is easily seen that the universal indifference preference satisfies both conditions (under universal indifference part two of i) and ii) are void conditions; part one of i) is satisfied since we have that Y Y for all Y, Y X). Example where i) is satisfied and ii) is violated: Let Z = {x, y} so that X = {{x}, {y}, {x, y})}. Notice, that in this case the only way to have sets A, B and C such that C is disjoint to A and B is that A, B, C {x, y} and A = B. This means that i) is a vacuous condition and is thus satisfied by any preference. On the other hand, any preference which 1) ranks the three elements of X strictly and 2) puts {x, y} either at the top or at the bottom of the ranking (as for example in the case {x, y} {x} {y}) must violate ii). In order to see this just observe that if we have a preference such that {x} {y} then the first part of ii) implies {y} {x} = {x, y} {y}. Moreover, if {x} {y} then the second part of ii) implies {x} {x} {y} = {x, y}. Thus, whenever {x} {y}, then property ii) can be satisfied only if {x} {x, y} {y}. 1

2 Example where ii) is satisfied and i) is violated: Let Z = {x, y, z} and consider the preference on X which is defined as follows: {x, y, z} {x, y} {x, z} {y, z} {x} {y} {z}. Given that a decision maker with such preferences is indifferent between all singleton sets condition ii) is vacuously satisfied. In order to see that condition i) is not satisfied by this preference, simply observe that we have {x, y} {x} but not {x, y} {z} {x} {z} (which means that the second part of i) is violated). Finally, we have to show that if Z contains at least three elements, then no preference which ranks three (or more) singleton sets strictly can satisfy both i) and ii). So let Z = {x, y, z} and take a preference on X such that {x} {y} {z}. Moreover, assume - by contraposition - that the preference satisfies both i) and ii). Then, since {y} {z} the first part of ii) implies {y} {z} = {y, z} {z}. If we add to both these sets the element x then by (second part of ) i) the strict preference should continue to hold, i.e. {y, z} {x} = {x, y, z} {z} {x} = {x, z}. Similarly, (the second part of) ii) also implies that {x} {x, y} and thus by i) we must have {x, z} {x, y, z}. So whenever we consider a set of alternatives with at least three alternatives then no preference profile which ranks at least three singleton sets strictly can be compatible with both conditions (since assuming that the conditions are satisfied implies both {x, y, z} {x, z} and {x, z} {x, y, z}). Problem 4: i) The statement is not fully correct as it is stated in the problem set. In order to see this, consider the following example: X = R, u(x) = x for all x 1 and u(x) = x + 1 for all x > 1 and v(x) = x for all x. Then any function f for which the condition f(u(x)) = v(x) for all x holds, must satisfy f(z) = z for z 1 and f(z) = z 1 for all z > 2. But that means that the co-domain of (, 1] (2, ) under f coincides with R and thus there is no way to associate to any element z in the interval (1, 2] a number f(z) such that f would be strictly increasing over the entire real line. The statement of the problem is true if we slightly weaken the strict monotonicity condition: If f(u(x)) = v(x) holds for all x, then f must be strictly increasing over the co-domain of u. In order to see this, take any two elements, z and z, from the co-domain of u, such that z < z. Then there must exist (distinct) x, x X such that u(x) = z and u(x ) = z, which implies u(x) < u(x ). Since both u and v represent it follows that we must also have v(x) < v(x ) or equivalently f(z) < f(z ) which is what we had to show. ii) Yes: Let X = R and take the preference be defined as follows: x y x y. This preference can be represented by the discontinuous utility function u, { x if x 1 u (x) = x + 1 if x > 1. iii) Consider the two alternatives x = 1.5 and y = 2. Since u(x) = 1 < 2 = u(y) we must have x y. Now notice that any neighborhood of y, B y, contains a point y < 2 whose utility is 1, and thus y x. But if any neighborhood of y contains points which are just as good as x then the pair of alternatives x and y must violate condition C1 and so the preference described by the utility function u cannot be continuous. 2

3 iv) A proof for the equivalence between C1 and C2 can be found in the textbook of Rubinstein. Before showing that C3 and C4 are equivalent to equivalent to C1 and C2 it is convenient to recall the definitions of open and closed sets. Definition 1. The set A is open if for every a A there exists a ɛ > 0 such that the neighborhood B(a, ε) is a subset of A. Definition 2. The set A is closed if for any convergent sequence {a n } in A the limit point is contained in A (i.e. if {a n } A, a n a implies a A). It is easy to show that if A is closed then A c is open and vice versa. This immediately implies that the definitions C3 and C4 are equivalent. It remains to be shown that C1/C2 is equivalent to C3/C4. In what follows we use the following notation: UCS(x) = {y : y x}, UCS(x) = {y : y x},lcs(x) = {y : x y} and LCS(x) = {y : x y}. C1 C3: We have to show that for every x, the sets UCS(x) and LCS(x) are open. So assume that y LCS(x). By C1 there exists a neighborhood of y, B(y), such that x y for all y B(y) and thus B(y) LCS(x), which is what we had to show. C3/C4 C1: Suppose - by contraposition - that C3/C4 holds but not C1. If C1 does not hold, then there must exist x and y such that x y and such that for any sequence {ε n } which converges to zero there are x n B(x, ε n ) and y n B(y, ε n ) such that y n x n. By openness of UCS(y) and LCS(x) there must be some N such that B(y, ε n ) LCS(x) and B(x, ε n ) UCS(y) for all n > N. Hence, in particular for all n > N we must have x y n and x n y. Now observe that for the sequence {y n } n>n y n y M there either exists a M > N such that for all n > M or the sequence contains a subsequence {y nk } which satisfies y nk y nk+1 for all k. In the second case we have (by transitivity) LCS(y nk+1 ) LCS(y nk ) for all k and thus LCS(y nk ) LCS(y n1 ) for all k > 1. Since x nk LCS(y nk ) for all k we have {x nk } LCS(y n1 ). Given that LCS(y n1 ) is closed, x nk x implies that x LCS(y n1 ). But now observe that y n1 x is in contradiction with our assumption that x y n for all n > N. In the first case we have y n y M x M for all n > M and thus {y n } n>m UCS(x M ). Closedness of UCS(x M ) implies that the limit of the sequence {y n } n>m, y, is an element of UCS(x M ). This of course means that y x M which again contradicts the assumption that x n y for all n > N. Since we reach a contradiction in either of the two possible cases we can conclude that C1 must hold whenever C3/C4 do. v) A proof of this statement can be found in the textbook of Rubinstein. Problem 5: We have to show that max{x 1, x 2 } > max{y 1, y 2 } implies that there must be some N such that x n 1 + x n 2 > y n 1 + y n 2 for all n N. Since X = R 2 + we know that x i, y i 0 and max{x 1, x 2 } > 0. Thus the condition x n 1 + x n 2 > y n 1 + y n 2 can equivalently be stated as x n 1 (max{x 1, x 2 }) n + x n 2 (max{x 1, x 2 }) n > y n 1 (max{x 1, x 2 }) n + y n 2 (max{x 1, x 2 }) n. 3

4 Clearly the lhs of this expression is for all n larger or equal to 1. On the other hand, since max{x 1, x 2 } is strictly larger than both components of y, the rhs of this inequality converges to 0 as n grows. Hence, we can indeed conclude that whenever x y then there must exist some N such that x n y for all n N. Problem 6: i) Note that u (x 1, x 2 ) = αx 1 + βx 2 with α, β > 0 satisfies the three properties: 1. The following inequalities are all equivalent given our assumptions: (x 1, x 2 ) (y 1, y 2 ) αx 1 + βx 2 αy 1 + βy 2 αx 1 + αt + βx 2 + βs αy 1 + αt + βy 2 + βs (x 1 + t, x 2 + s) (y 1 + t, y 2 + s) 2. If x 1 y 1 and x 2 y 2 then αx 1 +βx 2 αy 1 +βy 2 or equivalently (x 1, x 2 ) (y 1, y 2 ). If at least one of the inequalities x i y i is strict then the strict inequality holds also for the weighted sum of the two components which means (x 1, x 2 ) (y 1, y 2 ). 3. Since the preference is represented by a continuous utility function (linear functions on R 2 are continuous!) it follows that the preference must be continuous (see part v) of Problem 4). ii) 1. The preference relation defined by x y x 1 x 2 y 1 y 2, satisfies properties C ( is represented by the continuous function u(x) = x 1 x 2 ) and M (if x i y i, then x 1 x 2 y 1 y 2 and the inequality becomes strict if either x 1 > y 1 or x 2 > y 2 ) but not A ((2, 1) (1, 2) but (2 + 2, 1 + 1) (1 + 2, 2 + 1)). 2. Consider the preference which satisfies x y for all x, y X (universal indifference). It is trivial to see that this preference is continuous and additive but not monotonic. 3. Lexicographic preferences satisfy A (adding the same constant to both components of two vectors does not change the ranking of the two components) and M but - as we have shown in class - not C. iii) The statement can be proved by showing that the indifference sets for a preference which satisfies the above discussed properties coincide with the contour sets of a linear utility function. Of course the level sets of a linear function are straight lines which are parallel to each other. Thus, what we have to prove is that the indifference sets for a preference which satisfies the three properties are straight parallel lines. Take any x and pick some y I(x) such that x y (assume for the moment that there is such a y). We have to show that any point z X which lies on the straight line which runs through these two points satisfies z x. We show this in three steps. First, we argue that if z is the midpoint of the section connecting x] and y (i.e. z = (x + y)/2) then z x y. In a second step we will argue that z x holds for all points z which can be written in the form z = αx + (1 α)y for some α [0, 1]. Finally, we will show that z x holds also for all z X such that z = αx + (1 α)y for some α R. 4

5 Step 1: Let z = (x + y)/2. If z y then either z y or y z. In the first case condition A implies that z + (x y)/2 y + (x y)/2. Since z + (x y)/2 = x and y + (x y)/2 = z we thus have x z which together with z y contradicts our assumption that x y. The case y z is treated analogously and is thus omitted. Step 2: Suppose z = αx + (1 α)y for some α [0, 1]. Construct two sequences, {x n } and {y n }, as follows: Set x 0 = x and y 0 = y. Given x n and y n let m n = (x n + y n )/2 and define x n+1 = m n, y n+1 = y n if z belongs to the segment which connects m n with y n ; otherwise (i.e. if z lies closer to x n than to y n ) set x n+1 = x n and y n+1 = m n. By Step 1 we know that x n x for all n. Moreover, by construction x n z. But then continuity implies that z x. Step 3: Let z be on the line which runs both through x and y (but not on the segment in between x and y) and suppose wlog that it is closer to x. Then there must be some natural number n such that w = z +n(y x) is on the segment which connects x and y. By Step 2 we know that w y. But then condition A implies that w (y x) y (y x) = x y. Applying the same argument once more then delivers that w 2(y x) y and so on. In particular, itarating this procedure n times, yields z = w n(y x) y x and so we are done. In order to conclude our proof we have to argue a) that our assumption that I(x) contains a point different from x is warranted (whenever x is different from the origin) and b) that the indifference curves are parallel. In order to show that a) holds it is convenient to use the following fact: If is continuous and a x and x b then there exists a α (0, 1) such that x αa + (1 α)b. We will prove this in Problem 6 of PS 2. If x 0 then either x 1 0 or x 2 0 (or both). In the second case (the first one is treated in a perfectly analogous way) there is a ε > 0 such that x = (x 1 + ε, x 2 ) and x = (x 1, x 2 ε) both belong to X. By condition M we have that x x and x x. Since X is convex it must contain all points on the segment which connects x and x (this segment does not contain x). Thus by the above mentioned fact, we know that there is a point z on this segment such that x z which is what we had to show. Finally, suppose that the indifference curves for the two points x and x are not parallel to each other. Wlog let x x. Define t = x x and consider the set I(x) + t = {z X : z = z + t for some z I(x)}. Condition A implies that according to the preference all elements in I(x) + t are equivalent. The fact that I(x) is not parallel to I(x ) means that I(x) + t does not coincide with I(x ). The two sets share though the point x. Assuming that the indifference sets are not parallel thus leads to the conclusion that there are points y I(x ) and y I(x) + t such that y x, y x but not y y. Since this contradicts transitivity it follows that the indifference curves must be parallel. 5