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Consumer Theory – Cobb-Douglas Preferences, Lecture notes of Economics

Cobb-Douglass preferences are one of the simplest algebraic representations of well-behaved preferences. Dividing (4) by (5): 1 See the Appendix to Chapter 4 ...

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Economics 3030: Intermediate Microeconomic Theory Topic 2: Consumer Theory Cobb-Douglas Preferences
1
Economics 3030: Intermediate Microeconomic Theory
Topic 2: Consumer Theory
Cobb-Douglas Preferences
Reference: Varian, p.93
Outline: 1. Introduction
2. Cobb-Douglass Preferences
3. Positive Monotonic Transformations
4. Marginal Rate of Substitution
1. Introduction
Cobb-Douglass preferences are one of the simplest algebraic representations of well-behaved
preferences.
2. Cobb-Douglas Preferences
Assume the consumer’s utility function is given by:
u x1,x2
( )
=x1
cx1
d
(1)
We will maximise this utility function subject to the following budget constraint:
p1x1+p2x2=m
(2)
Thus, setting up the Lagrange multiplier:1
max
x1,x2,
λ
L=x1
cx2
d+
λ
mp1x1p2x2
( )
(3)
L
x1
=cx1
c1x2
d
λ
p1=0
(4)
L
x2
=dx1
cx2
d1
λ
p2=0
(5)
L
λ
=mp1x1p2x2=0
(6)
Dividing (4) by (5):
1 See the Appendix to Chapter 4 of Varian for details.
pf3
pf4
pf5

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Economics 30 30: Intermediate Microeconomic Theory

Topic 2: Consumer Theory

Cobb-Douglas Preferences

Reference: Varian, p.

Outline: 1. Introduction

_2. Cobb-Douglass Preferences

  1. Positive Monotonic Transformations
  2. Marginal Rate of Substitution_

1. Introduction

Cobb-Douglass preferences are one of the simplest algebraic representations of well-behaved

preferences.

2. Cobb-Douglas Preferences

Assume the consumer’s utility function is given by:

u x 1

, x 2 ( ) =^ x 1

c x 1

d (1)

We will maximise this utility function subject to the following budget constraint:

p 1

x 1

  • p 2

x 2

= m (2)

Thus, setting up the Lagrange multiplier:

1

max

x 1 , x 2 , λ

L = x 1

c x 2

d

  • λ mp 1

x 1

p 2

x 2 ( ) (3)

∂ L

x 1

= cx 1

c − 1 x 2

d − λ p 1

∂ L

x 2

= dx 1

c x 2

d − 1 − λ p 2

∂ L

∂ λ

= mp 1

x 1

p 2

x 2

Dividing (4) by (5):

1 See the Appendix to Chapter 4 of Varian for details.

cx 1

c − 1 x 2

d

dx 1

c x 2

d − 1

p 1

p 2

c

d

x 1

c − 1 − c x 2

dd + 1

p 1

p 2

c

d

x 2

x 1

p 1

p 2

x 1

c

d

p 2

p 1

x 2

Also note from (6) that:

x 2

m

p 2

p 1

p 2

x 1

We now have two equations - (7) and (8) – and two unknowns x 1

, x 2 ( ). Substituting (8) into (7)

implies:

x 1

c

d

p 2

p 1

x 2

x 1

c

d

p 2

p 1

m

p 2

p 1

p 2

x 1

x 1

c

d

m

p 1

c

d

x 1

x 1

c

d

c

d

m

p 1

x 1

d

c + d

c

d

m

p 1

x 1

c

c + d

m

p 1

p 1

x 1

m

p 2

x 2

m

c

c + d

d

c + d

3. Positive Monotonic Transformation

Recall equation (1):

u x 1

, x 2

( ) =^ x

1

c x 1

d (14)

If we make a positive monotonic transformation of the Cobb-Douglas utility function by raising

it to the 1 ( c + d ) power, we derive:

v x 1

, x 2

( ) =^ u^ x

1

, x 2

1

c + d = x 1

c x 1

d

1 c + d = x 1

c c + d x 1

d c + d (15)

Define a new number a = c ( c + d ) enables us to rewrite (15) as:

v x 1

, x 2

( ) = x

1

c c + d x 1

d c + d = x 1

a x 1

1 − a (16)

Thus, we can always take a positive monotonic transformation of Cobb-Douglass preferences

that will make the exponents sum to 1. This is particularly useful, because it implies:

x 1

c

c + d

m

p 1

= a

m

p 1

x 2

d

c + d

m

p 2

= ( 1 − a )

m

p 2

Such that the consumer’s expenditure shares are given by:

p 1

x 1

m

= a

p 2

x 2

m

= 1 − a

Another useful positive monotonic transformation of the Cobb- Douglas utility is to take logs vis :

w x 1

, x 2

( ) =^ ln^ u^ x

1

, x 2

( ) =^ ln^ x

1

c x 1

d

( ) =^ ln^ x

1

c

( ) +^ ln^ x

1

d

( ) =^ c^ ln^ x

1

  • d ln x 2

4. Marginal Rate of Substitution

Generally, if u = u x 1

, x 2

( ) , then:

du x 1

, x 2 ( ) =^ u x 1

dx 1

  • u x 2

dx 2

dx 2

dx 1 MRS

u x 1

u x 2

d

2 x 2

dx 1

2

MRS

u x 1 x 1

u x 2

From (19):

dx 2

dx 1 MRS

u x 1

u x 2

c x 1

d x 2

cx 2

dx 1

Equivalently, from (1):

dx 2

dx 1 MRS

u x 1

u x 2

cx 1

c − 1 x 2

d

dx 1

c x 2

d − 1

cx 1

c − 1 − c

dx 2

d − 1 − d

cx 2

dx 1