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ECON6001/6701 Microeconomic Analysis 1 Problem Set 3
发布时间:2022-09-17
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Problem Set 3 (Solutions) (Solutions)
ECON6001/6701 Microeconomic Analysis 1
Exercise 1. Consider the following utility functions:
U (x1 ; x2) = min{x1 ; x2g (2)
U (x1 ; x2) = x x
1 ¡
) where 0 < a < 1. (3)
Derive the a) Marshallian Demand, b) Indirect Utility function, and where appropriate, verify Roy's identity.
Solution. Linear Preferences in Eq. (1) Here goods are perfect substitutes. If p1 <p2 , income
is spent entirely on good 1, and entirely on good 2 if p1 > p2 . The entire budget line is optimal when p1 = p2 .
Hence the Marshallian Demands are the functions
x1 (p; y) = ( if p1 < p2
0 if p1 > p2
x2 (p; y) = (y(0) if p1 < p2
provided
1. If this were so, then anything on the budget line is optimal, and the Marshallian Demand then ceases to be a function when p1 = p2 .
Indirect Utility
V (p; y) = x1 (p; y)+x2 (p; y)
=
:
y
= min{p1 ; p2g
Verifying Roy 's Identity: Suppose (p; y) is such that p1 < p2 . Here, V (p; y) = and this is a differentiable function. So, Roy's identity must hold for these parameter values.
V (p; y) = :
@V(p; y) y @V(p; y)
@p1 p1(2) @p2
@V(p; y)
¡/
=
= x1 (p; y)
¡/
= 0 = x2 (p; y)
Similarly, when p1 > p2 . (Those of you more mathematically inclined will notice that V (p; y) is not differentiable when p1 = p2 . For such parameter values Roy's identity cannot be verified.)
Complements in Eq. (2) This is the case of perfect complements. Optimal choice must lie on
the 45 degree line, i.e. consumption must be equal, and it must satisfy the the budget constraint. This gives,
x1 (p; m) = ; x2 (p; m) =
Indirect Utility
V (p; y) = min{x1 (p; y); x2 (p; y)g
y
This is clearly differentiable for all prices and income. Hence Roy's Identity may be verified.
@V(p; y) 1
@y = p1 + p2
@V(p; y) y @V(p; y) y
¡/
=
= x1 (p; y) = x2 (p; y)
Cobb-Douglas Preferences in (3) Standard case, set MRS equal price ratio etc. and obtain
the Marshallian Demands:
x1 (p; y) = a x2 (p; y) = (1¡a)
:
Indirect Utility
V (p; y) = x1 (p; y)a x2 (p; y)a
= a
)a
(1¡a)
)(1 ¡ a)
= K
where K = aa (1¡a)(1 ¡ a), a constant.
Roy 's Identity The Indirect Utility function given above is clearly differentiable for all p>0 and y > 0. Hence Roy's Identity must hold for all parameter values.
= ¡a K
= ¡(1¡a) K
@V(p; y) = K 1
@y p1(a)p2(1) ¡ a
You can verify directly that ¡ /
= xi (p; y).
Exercise. The following two functions were estimated by observing a consumers purchases for her consumption of three goods at various income levels y (sufficiently high) and prices p = (p1 ; p2 ; p3).
x1 = 100 ¡ 5 + p
+ 6
x2 = 100 + p + V
+ 6
:
The Greek letters are (as yet undetermined) constants.
1) If this is a utility maximizing locally non-satiated consumer, what must be the demand for good 3?
I By Walras Law, (i.e. the budget constraint binds) p1x1 + p2x2 + p3x3 = y, which gives
(y ¡ p1x1 + p2x2)
3
2) Are these functions homogeneous of degree zero in income and prices?
I Yes, if we scale income y to 入 y and each price pi to 入 pi , the RHS of each of the
above expressions remains unchanged. Since both x1 and x2 are homogeneous, then by construction, x3 is also homogeneous.
3) What further restrictions do we need on p ; V and 6 to conclude that the above functions are generated by a utility maximizing consumer?
Hint: Look up Theorem 1 on (See Lec 3 Slides!) You have already verified two of the required three in the previous parts. Also remember, even though there are 3 goods here, the demand for good 3 is not independently given, so, for all purposes it is a 2 good economy.
I We need to calculate the Slutsky Matrix and verify that it is symmetric and negative semi-
definite. The entry in the ith row and the jth column of the Slutsky matrix is
sij = +xj
:
That is, we need to calculate the above terms and verify that
s11 < 0 |
det |
|
|
s12 s22 s32 |
s13 1
|
and also sij = sji for all i j .
Since x3 is not independently chosen but defined to satisfy Walras law, it is not necessary to verify the third inequality it will automatically hold! So, below, we shall check only the first two inequalities and that s12 = s21 .
We begin with s11 . Direct differentiation gives us
s11 = + x1
= ¡ + x1
:
Now observe that x1 can be made to be a very large number by increasing m if 6
0. So, the only way we can ensure that s11 < 0, for all prices and incomes is if, from the above expression, 6 = 0. Hence
s11 = ¡ < 0
This simplifies the expression for x1 ; x2
x1 = 100 ¡ 5 + p
x2 = 100 + p + V
:
That is, goods 1 and 2 do not directly depend on income, only on
s12 = + x2
=
s21 = + x2
=
s22 = =
We now check the sign of det ).
det
) = det
5¡p3bp3
ACC(1)
= (¡5V ¡ p2)
We therefore need
¡5 V p2
One solution is to set V = ¡5 and p = 5, in which case x1 = x2 for all prices.