Eco 600E Advanced Microeconomics I Term: Spring (1st), 2009
Lecturer: Yosuke Yasuda
Problem Set 1: Solution Due in class on May 7
1. Question 1 (10 points)
Prove the followings (DeMorgan’s Law):
(S \ T )c = Sc[ Tc (S [ T )c = Sc\ Tc
Hint: You should use the de…nitions of union, intersection, and complement of sets. Drawing …gures (Venn diagrams) is not enough.
Answer:
First law: Take an arbitrary x in the complement of S \ T . By de…nition of the complement of a set, x is not an element of S \ T . By de…nition of the intersection of sets, x does not belong to both S and T . This is the same as saying that x belongs to either Sc or Tc, and thereby x is an element of the union of Sc and Tc. Since this is true for any x in (S \ T )c, the set of all such x’s must be contained in Sc[ Tc. Note that this reasoning is also valid in the opposite direction. Hence we verify that
(S \ T )c Sc[ Tc and
(S \ T )c Sc [ Tc. Thus, we can conclude that the two sets are equal.
Second law: Take an arbitrary x in the complement of S [ T . By de…nition of the complement of a set, x is not an element of S [ T . By de…nition of the union of sets, xbelongs to neither S nor T . This is the same as saying that x belongs to both Sc or Tc, and thereby x is an element of the intersection of Sc and Tc. Since this is true for any x in (S [ T )c, the set of all such x’s must be contained in Sc\ Tc. Note that this reasoning is also valid in the opposite direction. Hence we verify that
(S [ T )c Sc\ Tc and
(S [ T )c Sc \ Tc. Thus, we can conclude that the two sets are equal. 2. Question 2 (10 points)
Let A and B be convex sets in R2. Then, answer the following questions.
(a) Construct an example such that A [ B is not a convex set. Answer:
The following is an counter example:
A= fxj0 x1 1; x2 = 0g; B = fxjx1 = 0; 0 x2 1g:
Pick (1; 0) and (0; 1), then you can easily see that the middle point (12:12) is not an element of A [ B.
(b) Show that A \ B must be a convex set. Answer:
Pick two elements x and y from A \ B. An arbitrary convex combination of these two point, denoted by z, must be an element of A since A is convex (x and y are both elements of A). By the same argument, z must also be an element of B. Thus, z 2 A and z 2 B, which is equivalent to z 2 A \ B, which concludes that A \ B is a convex set.
3. Question 3 (10 points)
Let u : RK ! R be a concave function. Then, show that u is also a quasi-concave function.
Answer:
Suppose u is concave. Then, for all x; y 2 RK such that u(x) u(y) and 2 [0; 1], we have
u[ x + (1 )y] u(x) + (1 )u(y)
minfu(x); u(y)g + (1 ) minfu(x); u(y)g
= u(y).
This implies that the set fx 2 RKju(x) u(y)g is convex. Since y is arbitrary, u must be quasi-concave.
4. Question 4 (20 points)
Suppose % is a preference relation on X. Then, show the followings. (a) Re‡exive: For any x 2 X, x x.
Answer:
Suppose not. Then, x x, x x, which contradicts to the completeness of
%.
(b) Transitive 1: For any x; y; z 2 X, if x y and y z, then x z. Answer:
Suppose x y and y z, but x z. Then, by de…nitions, the followings must hold:
x% y and not y % x y% z and not z % y z % x or not x % z
Note that the de…nition of xz a b , a % b and not b % a (c) Transitive 2: For any x; y; z 2 X, if x y and y z, then x z.
Answer:
By de…nition of ,
x y, x % y and y % x y z , y % z and z % y. By transitivity,
x% y and y % z ) x % z z % y and y % x ) z % x Thus,
x% z and z % x , x z.
(d) Transitive 3:For any x; y; z 2 X, if x y and y % z, then x % z. Answer:
Suppose x y and y % z, then by de…nition of x y, x % y and y % x. By transitivity,
x% y and y % z ) x % z. where and are de…ned as follows:
a b , a % b and b % a a b , a % b and not b % a 5. Question 5 (20 points)
Let X = R2+. Assume that a preference relation % satis…es the following three properties:
Additive: (a1; a2) % (b1; b2) implies that (a1 + t; a2 + s) % (b1 + t; b2+ s) for all t and s.
Strictly monotone: If a1 b1 and a2 b2, then ; in addition, if either a1 > b1
or a2 > b2, then (a1; a2) (b1; b2).
Continuous
(a) Show that if % is represented by a linear utility function, i.e., u(x1; x2) = x1+ x2 with ; >0, then % satis…es the above three properties.
Answer:
(i) Additivity: Suppose u(x1; x2) u(x01; x02), that is, x1+ x2 x01 + x02. Then,
u(x1+ t; x2+ s) = (x1+ t) + (x2+ s) = x1+ x2+ ( t + s) u(x01+ t; x02+ s) = (x01+ t) + (x02+ s) = x01+ x02+ ( t + s). Therefore, u(x1; x2) u(x01; x02) if and only if u(x1+t; x2+s) u(x01+t; x02+s). (ii) Strict monotonicity: Suppose x x0 and de…ne 1; 2 0 as follows:
1 =: x1 x01 and 2 =: x2 x02 Now,
u(x1; x2) u(x01; x02) = ( x1+ x2) ( x01+ x02)
= 1 + 2,
which must be strictly positive if either 1 or 2 (or both) is positive since
; >0 by assumption.
(iii) Continuity: Since the utility function is a linear function of x1 and x2, it is clearly continuous.
(b) Find the preference relation that is 1) Additive and Strictly monotone but not Continuous, and 2) Strictly monotone and Continuous but not Additive. Answer:
There are many potential answers for this question. (1) (Any) lexicographic preferences.
(2) u(x1; x2) = minfx1; x2g. 6. Question 6 (10 points)
We say that a preference relation % is homothetic if x % y implies x % y for all 0.
Show that if a consumer has a homothetic preference relation, then her demand function is homogeneous of degree one in !. That is, x(p; !) = x(p; !) for any
>0. Answer:
By de…nition of x(p; !),
x(p; !) % y for all y 0 such that py !.
Since % is homothetic,
x(p; !) % y for all y 0 such that py !, which is equivalent to
x(p; !) % y0 for all y0 0 such that py0 !.
Thus, x(p; !) must be an optimal choice when the initial wealth is set !. 7. Question 7 (20 points)
Suppose there are three types of goods, and the consumer chooses di¤erent bundles for three di¤erent price vectors in the following way. Namely, she chooses bundle xi at prices pi, i = 1; 2; 3, where
p1 = 0
@ 1 1 2
1
A; p2 = 0
@ 1 1 1
1
A; p3 = 0
@ 1 2 1
1 A;
x1 = 0
@ 5 19
9 1
A; x2 = 0
@ 12 12 12
1
A; x3 = 0
@ 27 11 1
1 A:
In each case, the amount of expenditure is equal to her initial wealth, i.e., !i = pixi for i = 1; 2; 3.
(a) Show that the above data satisfy the Weak Axiom of revealed preference. Answer:
(i) p = p1 = (1; 1; 2) and !1 = 1 5 + 1 19 + 2 9 = 42. Then, x2 is not a¤ordable and thereby she revealed prefers x1 to x3.
(ii) p = p2 = (1; 1; 1) and !2 = 1 12 + 1 12 + 1 12 = 36. Then, x3 is not a¤ordable and thereby she revealed prefers x2 to x1.
(iii) p = p3 = (1; 2; 1) and !3 = 1 27 + 2 11 + 1 1 = 50. Then, x1 is not a¤ordable and thereby she revealed prefers x3 to x2.
It is clear the above consumer choice satis…es the Weak Axiom. (b) Show that this consumer’s behavior cannot be fully rationalized.
Hint: Assume there is some preference relation % that fully rationalizes the above data, and verify that % fails to satisfy transitivity.
Answer:
It is clear that the consumer’s revealed preference derived from her choice be- favior in (a) does not satisfy transitivity. Thus, it cannot be fully rationalized.
