HW4 Solutions. 1. Problem 2.4 (Prove Proposition 2.5) 2. Prove Proposition 2.8. Kenjiro Asami. Microeconomics I

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1 HW4 Solutions Microeconomics I May 23, 2018 Kenjiro Asami Announcement If you want to get back your answer sheets for HW4, please come to TA session on May 23, or Room 713 on the 7th floor of International Academic Research Building. 1. Problem 2.4 (Prove Proposition 2.5) Proposition 2.5: Preferences are convex if and only if, for every point x, the set NWT(x) is convex. Proof. (if part) Fix any x,y X such that x y and a [0,1]. Then, by definition of NWT( ), x NWT(y). Moreover, due to the completeness of, we have y y, which implies that y NWT(y). Since we are assuming that NWT(w) is convex for any w X, ax+(1 a)y NWT(y). Therefore, ax + (1 a)y y. Thus, is convex. (only if part) Take any y,z NWT(x) and a [0,1]. Then, by definition of NWT( ), y x and z x. Since is complete, z y or y z holds. Assume without loss of generality that y z. Due to the convexity of, ay + (1 a)z y. Since is transitive, we have ay + (1 a)z x, which implies that ay + (1 a)z NWT(x). Thus, NWT(x) is convex. 2. Prove Proposition 2.8 Proposition 2.8. a. Preferences represented by a (strictly) concave function u are (strictly) convex. b. Suppose u represents. Then, (b-i) u is quasi-concave iff are convex; (b-ii) u is strictly quasi-concave iff are strictly convex; (b-iii) u is semi-strictly quasi-concave iff are semi-strictly convex. kenjiro817@gmail.com 1

2 Proof. a. Suppose that u is concave. Fix any x, y X with y x and a [0, 1]. Then, u(y) u(x). Since u is concave, u(ay + (1 a)x) au(y) + (1 a)u(x). Moreover, u(y) u(x) implies that au(y) + (1 a)u(x) u(x). Thus, it follows that u(ay + (1 a)x) x and hence ay + (1 a)x x. Therefore is convex. Suppose next that u is strictly concave. Fix any x,y X with y x and y x and a (0,1). Then, u(y) u(x). Since u is strictly concave, if x y, u(ay + (1 a)x) > au(y) + (1 a)u(x). Moreover, u(y) u(x) implies that au(y) + (1 a)u(x) u(x). Thus, it follows that u(ay + (1 a)x) > x and hence ay + (1 a)x x. Therefore is strictly convex. b. (b-i) u is quasi-concave is convex (if part) Fix any x,y X and a [0,1]. Since is convex, x y ax + (1 a)y y y x ax + (1 a)y x Since u represents, u(x) u(y) u(ax + (1 a)y) u(y) u(y) u(x) u(ax + (1 a)y) u(x) It implies that u(ax + (1 a)y) min{u(x), u(y)}. (only if part) Fix any x,y X such that x y and a [0,1]. Since u represents, u(x) u(y). By assumption, u(ax + (1 a)y) min{u(x),u(y)} = u(y). Therefore, ax + (1 a)y y. (b-ii) u is strictly quasi-concave is strictly convex (if part) Fix any x,y X such that x y and u(x) u(y). Fix a (0,1). Then, x y. Strict convexity of implies that ax + (1 a)y y. Therefore, u(ax + (1 a)y) > u(y). (only if part) Fix any x,y X such that x y. Fix a (0,1). Suppose that x y. Since u represents, u(x) u(y). By assumption, u(ax + (1 a)y) > u(y). Therefore, ax + (1 a)y y. 2

3 (b-iii) u is semi-strictly quasi-concave is semi-strictly convex (if part) First, semi-strictly convex preference is convex. Then convexity of implies that u is quasi-concave by (b-i). Next, fix any x, y X such that u(x) > u(y) and any a (0, 1). Since u represents, x y. Then, semi-strict convexity of implies that ax + (1 a)y y. Therefore, u(ax + (1 a)y) > u(y). (only if part) First, semi-strictly quasi-concave function u is quasi-concave. Then quasi-concavity of u implies that is convex by (b-i). Next, fix any x,y X such that x y and any a (0,1). Since u represents, u(x) > u(y). By assumption, u(ax + (1 a)y) > u(y). Therefore, ax + (1 a)y y. 3. Problem *2.9 See Student s Guide. 4. (1) Example of a function f : R + R that is quasi-concave but is not concave. Consider the function f : R + R such that f (x) = x 2. f is quasi-concave: Take any x,y R + and any a [0,1]. Without loss of generality, we assume that x y. Then, ax + (1 a)y y. Since f is strictly increasing, it follows that f (x) f (y) and f (ax + (1 a)y) f (y). Therefore, f (ax + (1 a)y) min{ f (x), f (y)} holds. f is not concave: Let x = 1, y = 0, and a = 1 2. Then, f (ax + (1 a)y) = f ( 1 2 ) = 1 4 (1 a) f (y) = 1 2. Thus, a f (x) + (1 a) f (y) > f (ax + (1 a)y). and a f (x) + (2) Example of a function f : R + R that is semi-strictly quasi-concave but is not quasi-concave. Remark: The definition of semi-strict quasi-concavity that was first introduced in the lecture is different from the one in the textbook. a Let us call the former concept semi-strict* quasi-concavity. Then, (i) we can find semi-strictly* quasi-concave but not quasi-concave functions, while (ii) there does not exist semi-strictly quasi-concave but not quasi-concave function. a Definitions of semi-strict convexity/quasi-concavity in the lecture slides were corrected to be consistent with the textbook on May 16. 3

4 f : X R is: semi-strictly quasi-concave if it is quasi-concave, and for all x,y X such that f (x) > f (y), f (ax+ (1 a)y) > f (y) for all a (0,1); semi-strictly* quasi-concave if for all x,y X such that f (x) > f (y), f (ax + (1 a)y) > f (y) for all a (0,1). (i) Consider the function f : R + R such that: 1 if x 2 f (x) = 0 if x = 2 Then f is semi-strictly* quasi-concave but not quasi-concave. f is semi-strictly* quasi-concave: Take any x,y R + and any a (0,1). By construction of f, f (x) > f (y) only if x 2 and y = 2. Then, ax + (1 a)y 2 and thus f (ax + (1 a)y) = 1. Therefore, f (ax + (1 a)y) > f (y) holds. f is not quasi-concave: Let x = 1, y = 3, and a = 1 2. Then, f (x) = f (y) = 1 and f (ax + (1 a)y) = f (2) = 0. Thus, f (ax + (1 a)y) < min{ f (x), f (y)}. (ii) By definition, semi-strictly quasi-concave functions are quasi-concave. Therefore, there exists no function that is semi-strictly quasi-concave but is not quasi-concave. Remark: Semi-strict quasi-concavity does not imply quasi-concavity unless its definition requires quasiconcavity itself. 5. Problem *2.11 See Student s Guide. 6. Show that the lexicographic preference relation (defined on R 2 +) is complete, transitive, strictly monotone, strictly convex, and homothetic. Show also that it satisfies properties a and b in Proposition

5 Proof. Let the lexicographic preference relation on X R 2 + be denoted by. Completeness: x,y R 2 +: x y or y x Take any x = (x 1,x 2 ) X and y = (y 1,y 2 ) X such that x y. By definition of, x y (x 1 > y 1 (x 1 = y 1 x 2 y 2 )) y 1 x 1 (x 1 y 1 y 2 > x 2 ) y 1 > x 1 (y 1 x 1 y 2 > x 2 ) y 1 > x 1 (y 1 = x 1 y 2 > x 2 ) y 1 > x 1 (y 1 = x 1 y 2 x 2 ) y x Therefore, is complete. Note that the discussion above implies that y x if and only if y 1 > x 1 (y 1 = x 1 y 2 > x 2 ). Transitivity: x,y,z R 2 +: x y and y z x z Take any x = (x 1,x 2 ) X, y = (y 1,y 2 ) X, and z = (z 1,z 2 ) X such that x y and y z. By definition of, x y x 1 > y 1 (x 1 = y 1 x 2 y 2 ) y z y 1 > z 1 (y 1 = z 1 y 2 z 2 ) Therefore, when both x y and y z hold, there are 4 cases: 1. x 1 > y 1 y 1 > z 1, which implies that x 1 > z 1 ; 2. x 1 > y 1 (y 1 = z 1 y 2 z 2 ), which implies that x 1 > z 1 ; 3. (x 1 = y 1 x 2 y 2 ) y 1 > z 1, which implies that x 1 > z 1 ; 4. (x 1 = y 1 x 2 y 2 ) (y 1 = z 1 y 2 z 2 ), which implies that x 1 = z 1 x 2 z 2. Therefore, in either case, x z. Strict monotonicity: x,y R 2 +: x y and x y x y Take any x = (x 1,x 2 ) X and y = (y 1,y 2 ) X such that x y and x y. Then, x 1 y 1 x 2 y 2 (x 1 y 1 x 2 y 2 ) (x 1 > y 1 x 2 y 2 ) (x 1 = y 1 x 2 > y 2 ) x 1 > y 1 (x 1 = y 1 x 2 > y 2 ) x y Therefore, is strictly monotone. 5

6 Strict convexity: x,y R 2 +, a (0,1): x y and x y ax + (1 a)y y Take any x = (x 1,x 2 ) X and y = (y 1,y 2 ) X such that x y and x y and a (0,1). Then, (x 1 > y 1 (x 1 = y 1 x 2 > y 2 )) x y x 1 > y 1 (x 1 = y 1 x 2 > y 2 ) When x 1 > y 1, it follows that ax 1 + (1 a)y 1 > y 1, which implies that ax + (1 a)y y. When x 1 = y 1 and x 2 > y 2, it follows that ax 1 + (1 a)y 1 = y 1 and ax 2 + (1 a)y 2 > y 2, which implies that ax + (1 a)y y. Thus, is strictly convex. Homotheticity: x,y R 2 +, λ 0: x y λx λy Take any x = (x 1,x 2 ) X and y = (y 1,y 2 ) X such that x y. First fix any λ > 0. When x 1 > y 1, it follows that λx 1 > λy 1, which implies that λx λy. When x 1 = y 1 and x 2 y 2, it follows that λx 1 = λy 1 and λx 2 λy 2, which implies that λx λy. Next consider the case of λ = 0. Then λx = λy = 0, thus λx λy. Therefore, is homothetic. Property a in Prop. 2.16: x,m,m R + : (x,m) (x,m ) m m Take any x R + and m,m R +. It directly follows from the definition of that (x,m) (x,m ) m m Property b in Prop. 2.16: x,x,m,m,m R + : (x,m) (x,m ) (x,m + m ) (x,m + m ) Take any x,x R + and m,m,m R +. (if part) Suppose to the contrary that (x,m) (x,m ). Then, (x,m) (x,m ) x > x (x = x m > m) (For this part see the proof of completeness.) When x > x, it directly follows that (x,m + m ) (x,m+m ). When x = x and m > m, we have (x,m +m ) (x,m+m ) since m +m > m+m. (only if part) Suppose that (x,m) (x,m ). Then by definition, x > x (x = x m m ) When x > x, it follows that (x,m + m ) (x,m + m ). (x,m + m ) (x,m + m ) since m + m m + m. When x = x and m m, we have 6

7 Another Proof: Take any x,x R + and m,m,m R +. By definition of, (x,m) (x,m ) x > x (x = x m m ) ( ) x > x (x = x m + m m + m ) (x,m + m ) (x,m + m ) where (*) holds since m m if and only if m + m m + m. 7. Problem *3.1 See Student s Guide. Remark: The exposition of property b in Prop is a bit confusing because it might be interpreted in the following two ways: (A) x,x,m,m,m R + : (x,m) (x,m ) (x,m + m ) (x,m + m ) (B) x,x,m,m R + : (x,m) (x,m ) [ m R + : (x,m + m ) (x,m + m )] Actually the former interpretation is appropriate because otherwise Proposition 2.16 does not hold. See the solution for Problem *2.11 in the Student s Guide. 7