Class 4
1.
Completeness
2.
Reflexivity
3.
Transitivity
4.
Continuity
5.
Insatiability
6.
Convexity
How to check the axioms for a function?
Axioms 1-3 hold as long as f : R

R
. Then consumers behave rationally. Axiom 4 is satisfied as long as a functions consists of multiplication, division, summation and subtraction. Axiom 5 is satisfied if 1 st partial derivatives are positive. In axiom 6 we assume that the average of any two baskets is better than both of these baskets.
Mathematically, it means that indifference curve has to be convex which means that the absolute value of the slope (
MRS xy

MU
MU y x
  dy dx
) has to diminish: d dy
MRS xy

0 y
A
B x
Problem 1. Test whether the following functions could be utility functions:. a.) x

2 xy
 y , x , y

0

1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU x
MU y


U
 x


U
 y

1


1
 y x x y

0
 x , y

R

Function has positive 1 st partial derivatives. Axiom

0
 x , y

R

5 is satisfied.
MRS xy
  dy dx
MRS xy

MU
MU y x
 x
 x
 x y y
 y y x dMRS xy dx

2
1 y dy dx x x

2
1 x
MRS falls. Axiom 6 is satisfied. b.) u
 
 x
0 .
5 y
2
, x , y

0 y

2 x y
 y x x

2 y x
 
1
2 x
1
 y x

0 ,
 x , y

R

1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).

MU x
MU y



U
 x

U
 y

0 .
5 x

0 .
5 y
2 
0 ,
 x , y

R

Function has positive 1 st partial derivatives.

2 x
0 .
5 y

0 ,
 x , y

R

Axiom 5 is satisfied.
MRS xy
  dy dx dMRS xy dx
MRS xy

MU
MU y x 
0 .
5 x

0 .
5 y
2
2 x
0 .
5 y
 y
4 x
 dy
4 x

4 y dx
16 x
2

 y
4 x
4 x

4 y x
 
5 y
16 x
2

0 ,
 x , y

R

MRS falls. Axiom 6 is satisfied. c.) u
 xy
4
, x , y

0
1°: A function of real numbers, hence axioms 1-3 are satisfied.

A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU x
MU y



U
 x

U
 y
 y
4 
0 ,
 x , y

R


4 xy
3 
0 ,
 x , y

R

Function has positive 1 st partial derivatives. Axiom
5 is satisfied.
MRS xy
  dy dx dMRS xy dx
MRS xy

MU
MU y x y
4

4 xy
3
 y
4 x
 dy
4 x

4 y dx
16 x
2

 y
4 x
4 x

4 y x
 
5 y
16 x
2

0 ,
 x , y

R

MRS falls. Axiom 6 is satisfied. d.) u
 
 x (
 y )
0 .
5
, x , y

0
This function has complex values f : R


C
(square root of a negative value).
Axiom 1 does not hold so there is no need to continue the check. e.) u
 

2 x

4 y , x , y

0
1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU x
MU y



U
 x

U
 y

2

0 ,
 x , y

R


4

0 ,
 x , y

R

Function has positive 1 st partial derivatives. Axiom 5 is satisfied.

MRS xy
  dy dx
MRS xy

MU
MU y x

2
4

1
2 dMRS xy dx

0 ,
 x , y

R

MRS is constant. Axiom 6 does not hold. f.) u
 
 x

2 y , x , y

0
1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU x
MU y



U
 x

U
 y

1

0 ,
 x , y

R


2

0 ,
 x , y

R

Function has positive 1 st partial derivatives. Axiom 5 is satisfied.
MRS xy
  dy dx dMRS xy dx
MRS xy

MU
MU y x

1
2

0 ,
 x , y

R

MRS is constant. Axiom 6 does not hold.
Problem 2. Which functions from problem 1 have the same ordinal properties (the same MRS)?

MRS is identical for functions in b) and c), and for functions in e) and f) (but these functions do not satisfy axiom 6).
Problem 3. Test whether the following functions could be utility functions:. a.) u
 
 x
 y

, x , y

0 ,

,
 
0
1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU x
MU y



U
 x

U
 y


 x
 
1 y

 x
 y
 
1

0 ,
 x , y ,

,


0 ,
 x , y ,

,


R


R

Function has positive 1 st partial derivatives. Axiom 5 is satisfied.
MRS xy dMRS xy dx
  dy dx


MRS xy dy dx


2 x x

2
 y


MU
MU y
  x

 x
 
1
 x
 y
 x

2
 x x
2

 y y

 
1
 y

 y
 x
 
 y

2 x
2



 x
 



0 ,
 x , y ,

,
 
R

MRS is constant. Axiom 6 does not hold. b.) u
 
  x
  y , x , y

0 ,

,
 
0
1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU x
MU y



U
 x

U
 y
 
 

0 ,
 x , y

R


0 ,
 x , y

R

Function has positive 1 st partial derivatives. Axiom 5 is satisfied.
MRS xy
  dy dx dMRS xy dx
MRS xy

MU
MU y x 



0 ,
 x , y

R

MRS is constant. Axiom 6 does not hold.

Problem 4. Answer and explain: a.) The shape of a utility function is ordinal property. Is it correct?
yes. b.) Basket A is twice as good as the basket B?
It is a cardinal property. Ordinal property would be if we could say that A is better than be, but not how mush better. c.) Basket A is better than basket B. Is it ordinal or cardinal property?
Ordinal. d.) Which axioms are not satisfied for the following indifference curves: y y y
A B C y
D x y
E x y
F x x x
A: Not convex (Axiom 6)
B: Concave in one part (Axiom 6)
C: Concave (Axiom 6)
D: For the same amount of x there are 2 values of y which give the consumer the same amount of utility (Axiom 5)
E: Not continuous (Axiom 4)
F: More x and y give consumer the same satisfaction as less x and y (Axiom 5).
Problem 5. Check whether the function 𝑼 = 𝒙 𝟒 preferences. 𝒚 𝟒 satisfies the axioms of x

1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU
MU x y



U
 x

U
 y


4
4 x x
3
4 y y
4
3


0
 x
0
 x ,
, y y


R

R

Function has positive 1 st partial derivatives. Axiom
5 is satisfied.
MRS

MU
MU y x

4 x
3 y
4
4 x
4 y
3
 y x dy x
 y dMRS dx
 dx x 2
MRS xy dMRS dx
  dy dx dy dx
 
MRS xy

 y x x x
2
 y
 
2 y x
2

0
 x , y

R

MRS decreases. Axiom 6 is satisfied.
.
Problem 6. Check whether the function 𝑼 = 𝒙
𝟏 preferences.
𝟒 𝒚
𝟒
𝟏
satisfies the axioms of
1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU x
MU y



U
 x

U
 y


1
4
1
4 x

3
4 y 4
1 x 4
1 y

3
4


0
 x ,
0
 x , y

R
 y

R

Function has positive 1 st partial derivatives.
Axiom 5 is satisfied.
1 x

3
4 y 4
1
MRS xy

MU x
MU y
 4
1 x 4
1 y

3
4
 y x
MRS decreases. Axiom 6 is satisfied.
4
Problem 7. Check whether the function
U preferences.
 xy
2
, x , y

0
satisfies the axioms of

1°: A function of real numbers, hence axioms 1-3 are satisfied.
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU
MU x y



U
 x

U
 y

 y
2
2 xy


0
 x ,
0
 x , y y


R

R

Function has positive 1 st partial derivatives. Axiom 5 is satisfied.
MRS

MU x
MU y
 y
2
2 xy
 y
2 x dMRS
 dx satisfied. dy dx
2 x

4 x
2
2 y

 y
2 x
2 x
4 x
2

2 y


3 y
4 x
2
Problem 8. Check whether the function
U

0
 x ,

5 x y

R


3 y ,
MRS decreases. Axiom 6 is x , y

of preferences.
1°: A function of real numbers, hence axioms 1-3 are satisfied.
0
satisfies the axioms
A function is obtained by elementary operations. Hence it is continuous (axiom 4).
MU
MU x y



U
 x

U
 y


5
3


0
 x ,
0
 x , y y


R

R

Function has positive 1 st partial derivatives. Axiom 5 is satisfied.
.
MRS

MU
MU y x

5
3
Axiom 6 not satisfied. dMRS

0 dx
Problem 9. Check whether the function
U
 x xy
 y
, x , y

0 of preferences.
1°: A function of real numbers, hence axioms 1-3 are satisfied.
satisfies the axioms
Function is not continuous (axiom 4 not satisfied).

MU x


U
 x
MU y


U
 y

 y ( x
 x

 y ) y


2 xy

2 y
 x
 y

2 x ( x
 x

 y ) y


2 xy

2 x
 x
 y

2

0
 x , y

R

Function has positive 1 st partial

0
 x , y

R
 derivatives. Axiom 5 is satisfied.
.
MRS

MU
MU y x
 y
2 x
2 dMRS
2 y dy x
2 x
4

2 xy
2 2 y


 y x
2
2 
 x
2 
 dx
 x
4 dx dMRS


2 y
3 
4
2 xy
2


2 y (
4 y
 x ) dx x x
MRS decreases. Axiom 6 is satisfied.

0
 x ,
2 xy
2 y

R

Problem 10. Check whether the Cobb-Douglasov function 𝒖(𝒒
𝟏 where 0<α<1 satisfies the axioms of preferences.
, 𝒒
𝟐
) = 𝒒 𝜶
𝟏 𝒒 𝟏−𝜶
𝟏
1°: A function of real numbers, hence axioms 1-3 are satisfied.
Function is not continuous (axiom 4 not satisfied).
𝑀𝑈
𝑀𝑈
2
1
= 𝑑𝑢
= 𝑑𝑞
1 𝑑𝑢
= 𝛼𝑞 𝛼−1
1 𝑞 1−𝛼
2
> 0
= (1 − 𝛼)𝑞 𝛼
1 𝑞 −𝛼
2
> 0 𝑑𝑞
2
Function has positive 1 st partial derivatives. Axiom 5 is satisfied. 𝑑𝑀𝑈
1 𝑑𝑞 𝑑𝑀𝑈
2 𝑑𝑞
1
2
=
= 𝑑
2 𝑢 𝑑𝑞
2
1 𝑑
2 𝑢 𝑑𝑞
2
2
= 𝛼(𝛼 − 1)𝑞 𝛼−2
1
= −𝛼(1 − 𝛼)𝑞
× 𝑞 𝛼−2
1 𝑞
1−𝛼
2
1−𝛼
2
MRS decreases. Axiom 6 is satisfied.
< 0 jer je 𝛼 − 1 < 0
< 0 jer je −𝛼 < 0
𝑀𝑅𝑆
12
= −
𝑀𝑈
1
=
𝑀𝑈
2
MRS decreases. Axiom 6 is satisfied.
−𝛼𝑞 𝛼−1
1 𝑞 1−𝛼
2
(1 − 𝛼)𝑞 𝛼
1 𝑞 −𝛼
2
= − 𝛼𝑞
2
(1 − 𝛼)𝑞
1
< 0