Homework 4 Solution - Econ 503 Professor John H Nachbar...

Econ 503
Professor John H. Nachbar
Fall 2007
Homework 4
Answers
1. By Slutsky, D
p
φ ( p
?
, m
?
) = D
p
h ( p
?
, m
?
) - D
m
φ ( p
?
, m
?
) x
*0
. Pre and post mul-
tiplying by p
?
, since D
p
h ( p
?
, m
?
) p
?
= 0, x
? 0
p
?
= p
?
· x
?
= m
?
(Walras’s Law),
and p
?
· D
m
φ ( p
?
, m
?
) = 1 (differentiate Walras’s Law with respect to m ), the
result follows.
Alternatively, you can differentiate Walras’s Law with respect to p to get x
? 0
+
p
? 0
D
p
φ ( p
?
, m
?
) = 0, which implies p
? 0
D
p
φ ( p
?
, m
?
) = - x
? 0
; post multiplying
by p
?
and applying Walras’s Law again gives the result.
2. By the symmetry of D
p
h , β = 3, hence
A =



- 10 α 3
γ - 4 δ
3 ζ



From the fact that D
p
h p
*
= 0, we have - 10 * 1 + α * 2 + 3 * 6 = 0, hence
α = - 4, hence
A =



- 10 - 4 3
γ - 4 δ
3 ζ



By symmetry, γ = - 4, hence
A =



- 10 - 4 3
- 4 - 4 δ
3 ζ



Again from the fact that D
p
h p
*
= 0, - 4 * 1 - 4 * 2 + δ * 6 = 0 implies δ = 2,
hence
A =



- 10 - 4 3
- 4 - 4 2
3 ζ



Once again applying symmetry, = 2. Once again from the fact that D
p
h p
*
=
0, 3 * 1 + 2 * 2 + ζ * 6 = 0 implies, ζ = - 7 / 6. In summary,
A =



- 10 - 4 3
- 4 - 4 2
3 2 -
7
6



1

3. (a) Recall that for each n ,
φ
n
( p, m ) =
α
n
p
n
m.
Therefore,
D
p
φ ( p, m ) =


-
α
1
p
2
1
m 0
0 -
α
2
p
2
2
m


.
(b)
v ( p, m ) = α
1
ln
α
1
p
1
m + α
2
ln
α
2
p
2
m
= α
1
[ln( α
1
) - ln( p
1
)] + α
2
[ln( α
2
) - ln( p
2
)] + ln( m ) ,
where I have used the fact that α
1
+ α
2
= 1.