Econ 310 Microeconomic Theory II Solutions to Assignment 1

Econ 310 Microeconomic Theory II
Solutions to Assignment 1
Solution 1 (a) There are two ways to solve this problem. The student gets full credit by
pursuing EITHER one of the following two methods:
Method 1: Directly solving the maximization problem using the Lagrangean method.
The consumer’s problem is:
max 2 ln x1 + ln x2
x1 ;x2
s:t:
p1 x1 + p2 x2
y:
By non-satiation, the constraint holds with equality. First, set up the Lagrangian:
L = 2 ln x1 + ln x2 + (y
p 1 x1
p2 x 2 ) :
The …rst-order conditions are:
2
@L
=
@x1
x1
1
@L
=
@x2
x2
@L
= y
@
Equations (1) and (2) yield
using (3):
2
=1
x1 x2
=
p1
.
p2
p1 = 0
(1)
p2 = 0
(2)
p 1 x1
(3)
p2 x 2 = 0
Therefore, x2 =
p1
x.
2p2 1
Then we can solve for x1
p1
x1 = y
2p2
1
p 1 x1 + p 1 x1 = y
2
p 1 x1 + p 2
)
Denote p = (p1 ; p2 ; p3 ). The Marshallian demand functions are:
x1 (p; y) =
y
2y
; x2 (p; y) =
:
3p1
3p2
Substituting the Marshallians into the utility function, we get the indirect utility function:
v (p; y)
= 2 ln
2y
3p1
+ ln
= 2 ln 2 + 2 ln y
= 2 ln 2 + 3 ln y
1
y
3p2
2 ln 3p1 + ln y ln 3p2
2 ln 3p1 ln 3p2
(4)
Method 2: Notice that the utility function is just a log transformation of the CobbDouglas utility function:
2=3 1=3
2 ln x1 + ln x2 = ln x21 x2 = ln x1 x2
3
2=3 1=3
= 3 ln x1 x2
:
Since the log function is a strictly increasing function, maximizing the value of 2 ln x1 +ln x2
2=3 1=3
is equivalent to maximizing x1 x2 . Thus one can directly write down the Marshallian
demand functions for the Cobb-Douglas utility function:
x1 (p; y) =
y
2y
; x2 (p; y) =
:
3p1
3p2
Similarly, the indirect utility function is given by (4).
(b) Given v (p; y) from (a), derive the expenditure function from duality:
v (p; e (p; u)) = u
)
2 ln 2 + 3 ln e (p; u) 2 ln 3p1
u + 2 ln 3p1 + ln 3p2
)
3
ln 3p2 = u
2 ln 2
= ln e (p; u)
Therefore, the expenditure function is
e (p; u) = exp
u + 2 ln 3p1 + ln 3p2
3
2 ln 2
:
From Shephard’s lemma, the Hicksian demand functions are:
xh1 (p; u) =
@e (p; u)
2
=
exp
@p1
3p1
u + 2 ln 3p1 + ln 3p2
3
2 ln 2
:
(c) Consider any t > 0. Then,
v (tp; ty) = 2 ln 2 + 3 ln ty
2 ln 3tp1
= 2 ln 2 + 3 (ln t + ln y)
= 2 ln 2 + 3 ln y
2 ln 3p1
2
ln 3tp2
2 (ln t + ln 3p1 )
(ln t + ln 3p2 )
ln 3p2 = v (p; y) ;
and
xh1 (tp; u) =
=
=
=
=
u + 2 ln 3tp1 + ln 3tp2 2 ln 2
2
exp
3tp1
3
2
u + 2 (ln 3p1 + ln t) + (ln 3p2 + ln t)
exp
3tp1
3
2
u + 2 ln 3p1 + ln 3p2 2 ln 2
exp
+ ln t
3tp1
3
2
u + 2 ln 3p1 + ln 3p2 2 ln 2
exp
3p1
3
xh1 (p; u) :
2 ln 2
Solution 2 (a) The consumer’s cost-minimization problem is:
min
p 1 x1 + p 2 x2
p
p
s:t:
x1 + 2 x 2
x1 ;x2
u:
Since the utility function is strictly increasing, the constraint must hold with equality at the
optimum. The Lagrangian is:
p
p
( x1 + 2 x2
L = p 1 x1 + p 2 x2
u) :
The …rst-order conditions are:
1
@L
1=2
= p1
x1
=0
@x1
2
@L
1
1=2
= p2
2 x2
=0
@x2
2
p
p
@L
x1 + 2 x2 u = 0
=
@
By (5) and (6), we get x2 =
functions:
)
2p1
p2
(5)
(6)
(7)
2
x1 . Substitute it into (7) and solve for Hicksian demand
p
2p1 p
x1 + 2
x1 u = 0
p2
2
p2 u
xh1 (p1 ; p2 ; u) =
;
xh2 (p1 ; p2 ; u) =
4p1 + p2
3
2p1 u
4p1 + p2
2
:
Hence, the expenditure function:
e (p1 ; p2 ; u) = p1
p2 u
4p1 + p2
2
+ p2
2p1 u
4p1 + p2
2
=
p1 p 2
u2 :
4p1 + p2
(b) Derive the indirect utility function using duality:
e (p1 ; p2 ; v (p1 ; p2 ; y)) = y
p1 p2
[v (p1 ; p2 ; y)]2 = y
4p1 + p2
)
v (p1 ; p2 ; y) =
s
(4p1 + p2 ) y
:
p1 p2
Solution 3 (a) The utility-maximization problem is given by
max fmin [
x1 ;x2
s:t:
p
p
3
x1 ; a 3 x2 ]g
p1 x1 + p2 x2 = y;
where the constraint holds with equality because of non-satiation. Consider the following
two cases:
p
p
(i) Suppose 3 x1
a 3 x2 at the optimum. Then the maximization problem becomes
p
max a 3 x2
x1 ;x2
s:t:
p1 x1 + p2 x2 = y:
p
Obviously it comes down to maximizing the value of x2 . Given the supposition 3 x1
p
p
p
a 3 x2 , it must be the case that 3 x1 = a 3 x2 .
p
p
a 3 x2 at the optimum. Then the maximization problem
(ii) Now suppose 3 x1
becomes
p
max 3 x1
s:t: p1 x1 + p2 x2 = y:
x1 ;x2
p
Obviously it comes down to maximizing the value of x1 . Given the supposition 3 x1
p
p
p
a 3 x2 , again it must be the case that 3 x1 = a 3 x2 .
p
p
Therefore, it must be true that 3 x1 = a 3 x2 at the optimum. Then the budget constraint
becomes
p1 a3 x2 + p2 x2 = y:
Thus,
x1 (p1 ; p2 ; y) =
a3 y
;
a3 p 1 + p 2
x2 (p1 ; p2 ; y) =
a3 p
y
:
1 + p2
(8)
Given a > 0, obviously the Marshallian demand strictly decreases with price. Moreover,
4
both demand functions increase with income y. They are indeed normal goods.
(b) There are three ways to derive the Hicksian demand. [A student gets full credit for
presenting ANY one of the following methods.]
Method 1: Given the utility function, set up the cost-minimization problem directly:
min p1 x1 + p2 x2
s:t:
x1 ;x2
min [
p
p
3
x1 ; a 3 x2 ] = u;
where the constraint holds with equality because the objective function is strictly increasing.
p
p
p
p
Again, we can consider the two cases where 3 x1
a 3 x2 and 3 x1 a 3 x2 at the optimum.
p
p
By the same logic as in part (a), it must be true that 3 x1 = a 3 x2 . Then the Hicksian
p
p
demands can be solved from 3 x1 = a 3 x2 = u:
xh1 (p1 ; p2 ; u) = u3 ;
xh2 (p1 ; p2 ; u) =
u
a
3
:
Obviously, here the Hicksian demand is not a¤ected by any of the prices.
Method 2: Given the Marshallians in (8), the indirect utility function is
v (p1 ; p2 ; y) =
p
3
r
p
3
x1 (p1 ; p2 ; y) = a x2 (p1 ; p2 ; y) = a 3
a3 p
y
:
1 + p2
Recall one of the dualities v (p; e (p; u)) = u; which implies
s
a3
e (p1 ; p2 ; u)
= u:
a3 p 1 + p 2
We use this to solve for the expenditure function:
e (p1 ; p2 ; u) =
u
a
3
a3 p 1 + p 2 :
(9)
Then by Shephard’s lemma,
xh1 (p1 ; p2 ; u) =
@e (p1 ; p2 ; u)
= u3
@p1
xh2 (p1 ; p2 ; u) =
and
@e (p1 ; p2 ; u)
u
=
@p2
a
3
:
Method 3: Similarly to the …rst part of Method 2, we can derive the expenditure function
(9). Recall one of the dualities, we have xhi (p; u) = xi (p; e (p; u)). Given that we have the
Marshallian demand functions from part (a) and also the expenditure function, this duality
5
implies
xh1 (p1 ; p2 ; u) =
a3
a3 e (p1 ; p2 ; u)
=
a3 p 1 + p 2
xh2 (p1 ; p2 ; u) =
a2
a2 e (p1 ; p2 ; u)
=
p 1 + p 2 a2
6
u 3
(a3 p1 + p2 )
a
a3 p 1 + p 2
u 2
(p1 + p2 a2 )
a
p 1 + p 2 a2
= u3
=
u
a
3
:

Download Report