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ECON6003/6703 Mathematical Methods for Economics Problem Set 3
发布时间:2022-05-20
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Problem Set 3
ECON6003/6703 Mathematical Methods for Economics
Exercise 1. Verify that is represented by a quasi-concave utility function if and only if for every x, the weakly preferred set
fy 2Xjyxg
is convex.
Solution. Suppose can be represented by a quasi-concave utility function. In
fact let U : X ¡! R be such a utility representation. Next, pick any x 2 X and let C = fy 2 Xj y xg denote the above set. We need to show that C is a convex set. For this, for any given y; y0 2 C, we need to show that z =
y + (1 ¡
) y0 2 C for all
2 (0; 1), which is to show that z
x.
By definition of C ,
Since U is quasi-concave, we have
U(z) = U(y+(1¡
)y0)
minfU(y); U(y0)g
U(x)
where the last two inequalities follows from (1) .
To complete the answer, we must also show that the converse also holds. That is suppose C is a convex set for any x and show that U must be quasi-concave. That is, for y; y0 , where without loss of generality y y0 , we then need to show that
U(z) U(y0):
But this is immediate upon setting x = y0 in the definition of the above set C: Since C is convex, z 2 C and hence the above inequality holds.
Exercise 2. Suppose is represented by a quasi-concave utility function U . Construct a new utility function V (x) = f(U(x)) wheref: R ¡! R is strictly increasing. Verify that a) V also represents
and b). V is quasi-concave.
Solution. (Please note the correction in the definition of f , the domain is the set of real numbers, not X)
Part (a). For any x; y 2 X
x y , U(x)
U(y) (since U is a utility repre-
sentation of )
, f(U(x)) f(U(y)) (since f is increasing)
or V (x)V (y)
Hence V is also a utility representation of .
Part (b). Use the previous exercise. is represnted by the quasi-concave implies C is convex, which in turn implies that any other representation of
, in partic- ular the representation V must also be quasi-concave.
Exercise 3. Here is an equivalent definition of a continuous preference relation to the one given in lecture slides.
A preference relation on Rn is said to be continous if for any y 2 Rn and every infinite sequence xn
xn
y 8n and n(l)im!1xn =x
) x
y:
Use the above definition (or the original definition given in class) to answer the following:
Take to be the so called lexicographic order on R
: That is, given x = (x1 ; x2) and y=(y1 ; y2)
x⃞y ,
Verify that is complete and transitive but not continuous .
Solution. Note that with these preferences, (x1 ; x2) (y1 ; y2) if and only if x = y. In other words the indifference
curve
passing through (x1 ; x2) is simply that point! No two elements are indifferent.
I will leave it to you to verify that is complete and transitive.
The more interesting thing here is this preference is not continuous! To see this, take for instance y = (0; 0) and the sequence xn =¡ ; 1
. Since the first coordi- nate of xn is greater than that of y, i.e.
> 0, we have xn ⃞ y for all n. However, limn!1 xn =x = (0; 1). At x, both x and y has an equal first coordinate, by y has a strictly greater value for its second coordinate. Hence y ⃞ x. In other words, at every point in the sequence, xn is strictly preferred to y but the preference switches at the limit.
Note. There is a somewhat deep theorem in math which shows that Lexicographic preferences do not admit a real represenation. As such, Lexicographic preferences constitute an important example to demonstrate that the assumption of conti- nuity of preferences cannot be dispensed with in Debreu's Utility Representation Theorem.
Exercise 4. Let B R be a convex feasible set and f : B ¡! R be strictly quasi- concave. If f admits a maximum, then it must be unique.
Solution. Assume, by way of contradiction, that y and y0 are two distinct maxi- mizers of f on the domain B . Then, by definition,
f(y)= f(y0) f(x) 8x 2 B :
However, since B is a convex set, z = y +
y0 2 B and since f is strictly quasi- concave,
f(z) > minff(y); f(y0)g= f(y)
Moreover, z y (since y
y0). Therefore the above inequality contradicts that y maximizes f on B .
Exercise 5. Show by counter-examples that Brouwer's Fixed Point Theorem does not hold by dropping exactly one of its assumption in each of them.
Solution. The theorem requires as its hypotheses continuous function f that maps a convex and compact subset S of Rm to itself.
Drop continuity alone: Let f : [0; 1] ¡! [0; 1] where f(x) = 0 if x > and f(x) = 0 if x <
.
Drop f mapping S to itself: Easy. S =[0; 0.5] and f(x) = 1 + x.
Drop compactness alone: S =(¡1; 1) and f(x) = .
Drop convexity alone: Let S be the unit circle in R2 centered at the origin, and for each x=(x1 ; x2) 2 S, define f(x) = ¡x.
Exercise 6. Let X Rm and Y
Rn be a pair of convex sets. Show that their Cartesian product, namely X
Y = f(x ; y) 2 Rm+nj x 2 X ; y 2 Y g is a convex set.
Solution. Choose any two elements (x; y) and (x0 ; y0) from X Y . We need to show that (
x + (1 ¡
) x0 ;
y + (1 ¡
) y0) 2 X
Y for all
2 [0; 1]. But this is immediate since
x + (1 ¡
) x0 2 X , since X is convex and
y + (1 ¡
) y0 2 Y , since Y is convex.
Exercise 7. In class, I explained how we can use the Brouwer's Fixed point The- orem to offer sufficient conditions for the existence of a Nash Equilibrium. Present (write down) all of that material formally as Theorem followed by a Proof. Be as formal as you can, mimic the style of proofs written in the textbook.