Introduction to Economic Analysis Fall 2009 Problems on Chapter 1: Rationality

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1 Introduction to Economic Analysis Fall 2009 Problems on Chapter 1: Rationality Alberto Bisin October 29, Decision Theory After a brief review of decision theory, on preference relations, we list a series of problems. Some of the problems have solutions, the others are left as exercises for the reader. 1.1 Review: Basic de nitions and results Let X be a choice set. De nition. The relation is said to be complete if for all x; y 2 X, either x y or y x (or both). De nition. The relation is said to be transitive if for all x; y; z 2 X such that x y and y z, we have x z. De nition. The strict preference relation,, is de ned by x y, x y and y x De nition. The indi erence relation,, is de ned by x y, x y and y x Claim. Assume the weak preference relation,, is complete and transitive. Then the strict preference relation,, de ned as above is transitive. Proof of claim. Let x; y; z 2 X such that x y and y z. Firstly we have x y and y z ) x z (by transitivity of ) Secondly we have y x and z y 1

2 From here we would like to conclude that z x, but we not to shown the transitivity of the negation of the weak preference relation. Suppose it was the case that z x. We also have from above that x y. By transitivity of it must be the case that z y. However y z implies z y, and z y and z y cannot hold at the same time. Thus z x.summing up we have x z and z x =) x z (by de nition of ) and so we have the transitivity of. Claim. Assume the weak preference relation,, is complete and transitive. Then the indi erence preference relation,, de ned as above is transitive. Proof. Let x; y; z 2 X such that x y and y z. Firstly we have Secondly we have Summing up we have x y and y z (by de nition of ) ) x z (by transitivity of ) z y and y x (by de nition of ) ) z x (by transitivity of ) x z and z x =) x z (by de nition of ) and thus we have shown the transitivity of. 2 Problems A few of the problems follow straightforwardly from the analysis of Chapter 1. Some others require a signi cant amount of ingenuity. 2.1 Problem 1 (i) Consider a group of individuals A; B and C and the relation at least as tall as, as in A is at least as tall as B. Does this relation satisfy the completeness and transitivity properties? (ii) Consider instead a group of individuals A; B and C and the relation taller than, as in A is taller than B. Does this relation satisfy the completeness and transitivity properties? (iii) Consider the same group of individuals A; B and C and the relation at least as smart as, as in A is at least as smart as B. Does this relation satisfy the completeness and transitivity properties? 2

3 2.1.1 Solution. Part (i) The relation at least as tall as is complete and transitive. To verify completeness, pick any two individuals A and B. Clearly, either individual A is at least as tall as individual B or individual B is at least as tall as individual A or both. For transitivity, pick three individuals A, B and C and suppose that individual A is at least as tall as individual B and individual B is at least as tall as individual C. Obviously, individual A must be at least as tall as individual C. Thus, the relation at least as tall as satis es the transitivity property. Part (ii) is left to the reader. Part (iii) is as (i) if smartness is de ned, e.g., by the I.Q. number. But we can certainly fudge the de nition so as to generate an incomplete or a non-transitive relationship. Try this. 2.2 Problem 2 Determine if completeness and transitivity are satis ed for the following preferences de ned on x = (x 1 ; x 2 ) and y = (y 1 ; y 2 ) i) x % y iff (if and only if) x 1 y 1 and x 2 y 2 ii) x % y iff (if and only if) x 1 y 1 and x 2 y 2 iii) x % y iff x 1 > y 1 or x 1 = y 1 and x 2 > y 2 iv) x % y iff (if and only if) x 1 < y 1 and x 2 y 2 v) x % y iff maxfx 1 ; x 2 g maxfy 1 ; y 2 g vi) x % y iff (if and only if) x 1 y 1 and x 2 y Solution. Part (i) The preference relation x % y iff (if and only if) x 1 y 1 and x 2 y 2 is not complete. Consider the following counter example: x = (0; 1) and y = (1; 0). Clearly, neither x i y i for all i nor y i x i for all i. So, neither x % y nor y % x. Hence, the bundles x = (0; 1) and y = (1; 0) can not be compared. The preference relation x % y iff (if and only if) x 1 y 1 and x 2 y 2 is however transitive. Pick x = (x 1 ; x 2 ), y = (y 1 ; y 2 ) and z = (z 1 ; z 2 ) and suppose that x % y and y % z towards showing that x % z. By assumption, x % y then x i y i for all i and since y % z, y i z i for all i. That is, and Hence, x 1 y 1 and x 2 y 2 y 1 z 1 and y 2 z 2 x 1 z 1 and x 2 z 2 Therefore, x % y and y % z imply that x % z. 3

4 Part (ii) Preferences de ned by x % y iff (if and only if) x 1 y 1 and x 2 y 2 are not complete. Consider the following counter example: x = (0; 1) and y = (1; 0). Clearly, neither x i y i for all i nor y i x i for all i. So, neither x % y nor y % x. Hence, the bundles x = (0; 1) and y = (1; 0) can not be compared. On the other hand, preferences de ned by x % y iff (if and only if) x 1 y 1 and x 2 y 2 are transitive. Pick x = (x 1 ; x 2 ), y = (y 1 ; y 2 ) and z = (z 1 ; z 2 ) and suppose that x % y and y % z. Need to show that this implies x % z. By assumption, Since x % y then x i y i for all i and since y % z then y i z i for all i. That is, x 1 y 1 and x 2 y 2 and Hence, y 1 z 1 and y 2 z 2 x 1 z 1 and x 2 z 2 Therefore, x % y and y % z imply that x % z. Note that this is called Pareto preference relation. Part iii). Preferences de ned by x % y iff x 1 > y 1 or x 1 = y 1 and x 2 > y 2 are not complete. For a counter example pick two bundles x and y such that x = y. For example, x = (1; 1) and y = (1; 1). Clearly, since x 1 = y 1 and x 2 = y 2 neither x % y nor y % x. Hence, the two bundles x = (1; 1) and y = (1; 1) can not be compared by this preference relation. On the other hand, preferences de ned by x % y iff x 1 > y 1 or x 1 = y 1 and x 2 > y 2 are transitive. Pick x = (x 1 ; x 2 ), y = (y 1 ; y 2 ) and z = (z 1 ; z 2 ) and suppose that x % y and y % z. Need to show this implies x % z. By assumption, Since x % y then and since y % z then Hence, it must hold that either x 1 > y 1 or if x 1 = y 1 then x 2 > y 2 either y 1 > z 1 or if y 1 = z 1 then y 2 > z 2 either x 1 > z 1 or if x 1 = z 1 then x 2 > z 2 which implies that x % z. Note this is called Lexicographic preference relation. Part (iv) The preference relation x % y iff (if and only if) x 1 < y 1 and x 2 y 2 is not complete. Consider the following counter example: x = (0; 1) and y = (1; 0). Although x 1 < y 1, x 2 > y 2. Therefore x = (0; 1) and y = (1; 0) can t be compared. However, it is transitive. Pick x = (x 1 ; x 2 ), y = (y 1 ; y 2 ) and z = (z 1 ; z 2 ) and suppose that x % y and y % z towards showing that x % z. By assumption, x % y then x 1 < y 1 and x 2 y 2 and since y % z, y 1 < z 1 and y 2 z 2. That is, x 1 < y 1 and x 2 y 2 and y 1 < z 1 and y 2 z 2 4

5 Hence, x 1 < z 1 and x 2 z 2 Therefore, x % y and y % z imply that x % z. Part (v) The preference relation x % y iff maxfx 1 ; x 2 g maxfy 1 ; y 2 g is complete. Pick any x = (x 1 ; x 2 ) and y = (y 1 ; y 2 ). Clearly, either holds or, maxfx 1 ; x 2 g maxfy 1 ; y 2 g maxfx 1 ; x 2 g maxfy 1 ; y 2 g holds or both. Hence, either x % y or y % x or both. It is also transitive. Pick any x = (x 1 ; x 2 ), y = (y 1 ; y 2 ) and z = (z 1 ; z 2 ) and suppose that x % y and y % z. To show that transitivity we need that x % z. Since x % y and since y % z So, we conclude that maxfx 1 ; x 2 g maxfy 1 ; y 2 g maxfy 1 ; y 2 g maxfz 1 ; z 2 g maxfx 1 ; x 2 g maxfz 1 ; z 2 g which implies that x % z. Therefore, x % y and y % z imply that x % z. Part (vi) is left to the reader. 2.3 Problem 3 For the following choice environments and preference relations, decide if the relation is complete and transitive 1. For X R 2, x y if and only if max(x 1 ; x 2 ) max(y 1 ; y 2 ) 2. For X R 2 ; x y if and only if x 1 y 1 3. For X R, x y if a/ x is a rational number and and y is not or b/ if neither or both x and y are rational but x y 4. For X R 2, x y if and only if x 1 > y 2 5. For X R, x y if and only if x y 2 6. For X R 3, x y if and only if x i > y i for 2 of the 3 i 7. For X R 2, x y if and only if x 1 + x 2 = y 1 y 2 8. For X R 2, x y if and only if p (x 1 3) 2 + (x 2 7) 2 p (y 1 3) 2 + (y 2 7) 2 9. For X R 2, x y if and only if x 1 + x 2 y 1 + y 2 and x 1 x 2 y 1 y For X R 2, x y if and only if x 1 + x 2 y 1 + y 2 or x 1 x 2 y 1 y For X R 2 ++, x y if and only if x 1 + x 2 y 1 + y 2 and x 1 x 2 y 1 y 2 5

6 2.4 Problem 4 For the following preferences, determine whether or not there is a utility representation, and if there is, write down two utility functions that represent the preferences 1. When choosing between bundles of containing both apples and oranges, I always prefer the bundle with the most apples. 2. When choosing between bundles containing both apples and oranges, I prefer the bundle with the most items of fruit. 3. When choosing between bundles containing both apples and oranges, I prefer the bundle that is closest to having the same number of apples and oranges. 4. When choosing between bundles containing both apples and oranges, I ip a coin. If it comes up heads, I prefer the bundle with the most apples. If it comes up tails I prefer the bundle with the most oranges. 2.5 Problem 5 Consider the following examples of choice sets X and associated preference relations : (a) X = f1; 2; 3g and is de ned by 1 1; 1 2; (b) X = fall the people in the worldg and is de ned by for any x; y 2 X (c) X = R and is de ned by (d) X = R and is de ned by x y if x "shares at least one given name with" y for any x; y 2 X for any x; y 2 X x y if x y x y if jx yj > 1 6

7 X = R and is de ned by for any x; y 2 X x y if jx yj is an integer multiple of 2 (i) Show the following: Example (a) is NOT complete and NOT transitive; Example (b) is NOT complete and NOT transitive; Example (c) is complete and transitive; Example (d) is NOT complete and NOT transitive; Example (e) is NOT complete but is transitive. (ii) [Note that this exercise goes beyond the level of this course and is only for those who are interested] Consider the following potential properties of the above preference relations: Weakly complete: on X is weakly complete if for all x; y 2 X, either x = y, x y or y x. Negatively transitive: on X is negatively transitive if for all x; y; z 2 X such that x y and y z, it is the case that x z. Re exive: on X is re exive if for all x 2 X, x x. Irre exive: on X is irre exive if for all x 2 X, x x. Symmetric: on X is symmetric if for all x; y 2 X such that x y, it is the case that y x. Asymmetric: on X is asymmetric if for all x; y 2 X such that x y, it is the case that y x. Antisymmetric: on X is antisymmetric if for all x; y 2 X such that x y and y x, it is the case that x = y. Acyclic: on X is acyclic if for all x 1 ; :::; x n 2 X, for n 2 N, such that x 1 x 2 ::: x n 1 x n, then it is the case that x 1 6= x n. For each of the examples (a)-(e), show whether or not the preference relation has any of the properties listed above. That is, if the preference relation has the property, provide a proof of this; if not, provide a counter-example. 2.6 Problem 6 The DM cannot detect small di erences. She consumes two goods: x and y and she strictly prefers bundle (x 0 ; y 0 ) to bundle (x; y), (x 0 ; y 0 ) (x; y), if and only if x 0 y 0 xy > 1 (if the di erence is less than one in absolute value she is indi erent). (i) Show that the strict part of her preferences is transitive, i.e, that if (x 1 ; y 1 ) (x 2 ; y 2 ) 7

8 and (x 2 ; y 2 ) (x 3 ; y 3 ) then (x 1 ; y 1 ) (x 3 ; y 3 ) : (ii) Show that the indi erence part of the preferences is not transitive, i.e., that if (x 1 ; y 1 ) (x 2 ; y 2 ) and (x 2 ; y 2 ) (x 3 ; y 3 ) then it is not necessary that (x 1 ; y 1 ) (x 3 ; y 3 ) Solution. Part (i) Consider three bundles (x 1 ; y 1 ); (x 2 ; y 2 ) and (x 3 ; y 3 ):If (x 1 ; y 1 ) (x 2 ; y 2 ); then it follows that x 1 y 1 x 2 y 2 > 1: If (x 2 ; y 2 ) (x 3 ; y 3 ); then x 2 y 2 x 3 y 3 > 1:Summing up these two inequalities we obtain that x 1 y 1 x 3 y 3 > 2; which implies that x 1 y 1 x 3 y 3 > 1: Notice now that this last inequality holds only if (x 1 ; y 1 ) (x 3 ; y 3 ): Part (ii) Consider the following counterexample: (x 1 ; y 1 ); (x 2 y 2 ) and (x 3 ; y 3 ) : x 1 y 1 = 1; x 2 y 2 = 1:8 and x 3 y 3 = 2:2: Observe that (x 1 ; y 1 ) (x 2 ; y 2 ) since x 2 y 2 x 1 y 1 = 0:8 < 1: At the same time, (x 2 ; y 2 ) (x 3 ; y 3 ) since x 3 y 3 x 2 y 2 = 0:4 < 1: However, (x 3 ; y 3 ) (x 1 ; y 1 ) since x 3 y 3 x 1 y 1 = 1:2 > Problem 7 Consider the preference ordering over the composition of Congress induced by the following procedure: I ip a coin; if heads I strictly prefer a Congress with a Democratic majority; if tails I strictly prefer a Congress with a Republican majority. (i) Is this preference relationship complete? (ii) Is it transitive? 8