ECON711-Microeconomic Theory. Solution to Homework #4. By Enrique Martínez García †.
∗
Due Date: 10/01/2004. (Fall 2004).
Enrique Martínez Martínez García, García, Ph.D Student Student,, Departme Department nt of Economic Economics, s, Univers University ity † Enrique of WisconsinWisconsin-Madis Madison. on. 743 7439 9 Social Social Science Building, Building, 118 1180 0 Observa Observator tory y Drive, Drive, MadiMadison, son, WI 53 5370 706. 6. Tel.: el.: (608 (608)) 262 262-43 -4389 89.. E-mai E-mail: l: ema emart rtin inezg ezgar ar@wi @wisc. sc.edu edu.. Webpag ebpage: e: http://mywebspace.wisc.edu/emartinezgar/web/. ∗
1
Problem 1 (1) (JR, 3.32) The Translog Cost Function . It has been shown that the translog cost function is a (local) second-order approximation to an arbitrary cost function such that c (w1 ,...,wn , y) = c (w1,...,wn ) y. It is given implicitly by the linear-in-logs form: n
1 ln (c) = α0 + αi ln (wi ) + 2 i=1
X
n
n
XX
γ ij ln (wi ) l n (wj ) + ln (y) .
i=1 j =1
If γ ij = γ ji , for i = 1 ,...,n, the substitution matrix is symmetric, as required. Solution 2 (1) (a) Show that the translog cost function is in fact a (local) second-order approximation to a cost function c (w1 ,...,wn ) y . Take the logarithms of this expression,
ln c (w1,...,wn , y) = ln c (w1 ,...,wn ) + ln y = ln c eln w ,...,eln wn + ln y,
¡
1
¢
and take a second-order Taylor approximation of ln c eln w ,...,eln wn with respect to (ln w1 , ..., ln wn ) around the point (ln w1∗ , ..., ln wn∗ ) = (0,..., 0) (alternatively, (w1∗ ,...,wn∗ ) = (1, ..., 1)),
³
∗
∗
ln c (w1 ,...,wn , y) = ln c eln w ,...,eln wn 1 + 2
n
n
XX
1
d2 ln c eln w ,...,eln wn
i=1 j =1
= ln c (1, ..., 1) +
¡ X
∗
∗
1
d ln wi d ln wj n
i=1
¢
¡
1
n
´¯¯¯
wi =1, ∗
∗
∀i
+
i=1
∗
∗
eln w1 ,...,eln wn
X ¡ d ln c
¢
d ln wi
∗
¯¯ ¯¯
¢
∗
eln wi (ln wi − ln wi∗ )
eln wi +ln wj (ln wi − ln wi∗ )(ln wi − ln wi∗ )
d ln c (1, ..., 1) 1 ln wi + d ln wi 2
n
n
XX i=1 j =1
d2
2
n
1 αi ln wi + ln c (w1 ,...,wn , y) = α0 + 2 i=1
n
n
XX
γ ij ln wi ln wj + ln y.
i=1 j =1
(b) What restrictions on the parameters are required to ensure homogeneity of degree one in input prices?. Take the logarithm of the cost function evaluated at the new vector of inputprices (tw1 ,...,twn ), n
1 αi ln (twi ) + ln c (tw1 ,...,twn , y) = α0 + 2 i=1
X
2
n
n
XX i=1 j =1
wi =1,
∀i
ln c (1,..., 1) ln wi ln wi + ln y. d ln wi d ln wj
Denote the constant terms in the above expression as: α0 = ln c (1,..., 1), d ln c(1,...,1) d ln c(1,...,1) ∀i, and γ ij = d ln w d ln w . If we substitute this terms back into αi = d ln wi i j the equation, we get the translog cost function:
X
+ ln y = ∗
γ ij ln (twi ) l n (twj ) + ln y,
¯¯ ¯
and apply the properties of the logarithm to rearrange this expression as follows: n
1 αi ln wi + ln c (tw1 ,...,twn , y) = α0 + 2 i=1 n
+
ÃX ! αi
i=1
ln t +
1 2
X X X n
n
n
XX
γ ij ln wi ln wj + ln y+
i=1 j =1
n
γ ij
n
(ln t)2 = ln c (w1 ,...,wn , y) +
i=1 j =1
ÃX ! αi
i=1
1 ln t + 2
Let’s suppose that the parameter restrictions we need to impose to ensure homogeneity in prices are: n
n
X
X
αi = 1 ,
i=1
γ ij = 0 .
j =1
It is easy to check that, indeed this conditions will su ffi ce to ensure homogeneity. Substituting back into the equation for ln c (tw1 ,...,twn , y), we will obtain that:
ln c (tw1 ,...,twn , y) = ln c (w1 ,...,wn , y) + ln t = ln [tc (w1 ,...,wn , y)] Using the exponential function on both sides will not change the equality, c (tw1 ,...,twn , y) = tc (w1 ,...,wn , y) ,
but it is enough to prove homogeneity, as required. (c) For what values of the parameters does the translog reduce to the CobbDouglas form? Recall that the Cobb-Douglas cost function is: n
c (w1 ,...,wn , y) = K
Y
wiai y ,
i=1
n
n
X
Y
ai = 1 , K =
i=1
ai a− . i
i=1
Take the logarithm of this expression, n
ln c (w1 ,...,wn , y) = ln K +
X
ai ln wi + ln y.
i=1
By the method of matching coe ffi cients we can argue easily that the Translog reduces to a simple Cobb-Douglas cost function whenever: α0 αi γ ij
= ln K, = ai , ∀i = 1 ,...,n, = 0, ∀i, j = 1 ,...,n.
i (w ,...,wn ,y ) (d) Show that input shares si (w1 ,...,wn , y) = wicx(w in the translog ,...,wn ,y) cost function are linear in the logs of input prices. 1
1
3
n
n
XX i=1 j =1
γ ij
(ln t)
First, notice that an easy way of computing si (w1 ,...,wn , y) uses the inputprice elasticity of the cost function: d ln c (w1 ,...,wn , y) dc (w1 ,...,wn , y) wi , = d ln wi dwi c (w1 ,...,wn , y)
so by Shephard’s lemma (i.e.
dc(w1 ,...,wn ,y) dwi
= xi (w1 ,...,wn , y)) we can argue:
d ln c (w1 ,...,wn , y) wi = xi (w1 ,...,wn , y) = si (w1 ,...,wn , y) . d ln wi c (w1 ,...,wn , y)
Then, once we have shown this result, it is easy to prove that under the assumption of symmetry of the substitution matrix (i.e., γ ij = γ ji , ∀i, j = 1,...,n), si (w1 ,...,wn , y )
=
1 d ln c (w1 ,...,wn , y) = αi + 2 d ln wi n
= αi +
X
n
n
X
X
1 γ ij ln wj + 2 i=1
γ ji ln wj =
i=1
γ ij ln wj .
i=1
Indeed, input shares in the translog cost function are linear in the logs of input prices. Remark: If instead of approximating a cost function like c (w1 ,...,wn ) y , we took a second order linear approximation over any well-behaver sot function c (w1 ,...,wn , y), we would still fi nd similar results. In that case, it is easy to show that input shares would be linear in the logs of input prices and output. Problem 3 (2) (JR, 3.32). Let u (x) represent some consumer’s monotonic preferences over x ∈
P
P
4
cannot be expressed in terms of u (x). When could we fi nd a transformation such that f = v ◦ u? If and only if for all bundles x , x ∈
00
n 0
00
0
00
0
u (x ) ≥ u (x ) ⇔ f (x ) ≥ f (x ) ⇔ u (x ) +
n
X
0
00
xi ≥ u (x ) +
i=1
X
0
xi .
i=1
If u (x) = h ( ni=1 xi ) where h : <+ → < is a strictly increasing function, then the transformation v exists. And v = y + h−1 (y ) for all y ∈ R (u) = <+ would be in fact a strictly monotonic transformation such that f = v ◦ u. In this case, f would represent the same preferences as u. Otherwise, most likely it will not represent the same preferences.
P
Problem 5 (3) (JR, 3.32). A consumer of two goods faces positive prices and has a positive income. His utility function is: u (x1 , x2 ) = max [ax1 , ax2 ] + min [x1 , x2 ] , where 0 < a < 1.
Derive the Marshallian demand functions. Solution 6 (3) First, consider the case where x1 ≥ x2 . Then, it follows that u (x) = ax1 + x2 . Second, consider the case where x1 ≤ x2 . Then, it follows that u (x) = ax2 + x1 . (Obviously, for x1 = x2 it must be true that u (x) = ax1 + x2 = ax2 + x1 ) It is evident, then, that we are dealing with linear preferences whose slope changes after crossing the 45o -line. In other (more technical) words, the indi ff erence curves are piecewise linear with slopes −a (if x1 > x2 ) and − a1 (if x1 < x2 ). Furthermore, the utility is not d ff erentiable at points where x1 = x2 . Note that − a1 < −a or, equivalently, a1 > a, given the fact that we assume a ∈ (0, 1). For the Marshallian demands, this implies fi ve di ff erent cases must be considered. (Use a grapical representation to gain further intuition on this results). Case 1. If pp < a, it must follow that: 1 2
x ( p,y) =
µ ¶
y ,0 . p1
Case 2. If pp = a, it must follow that: 1 2
x ( p, y )
= =
½µ ½ α
y
y
, p1 + p2 p1 + p2
¶
µ ¶
y + (1 − α) ,0 p1
(x1 , x2 ) : x1 ≥ x2 and x2 = 1 2
x ( p,y) =
µ
y
y
, p1 + p2 p1 + p2
5
: α ∈ [0, 1] =
y p1 y − − ax1 x1 = p2 p2 p2
Case 3. If a1 > pp > a , it must follow that:
¶
.
¾ ¾
.
Case 4. If pp = a1 , it must follow that: 1 2
x ( p, y )
½µ ½
=
α
y
y
, p1 + p2 p1 + p2
¶
µ ¶
y + (1 − α) 0, p2
(x1 , x2 ) : x1 ≤ x2 and x1 =
= 1 2
x ( p,y) =
µ ¶
6
0,
y p2
: α ∈ [0, 1] =
y p2 y − − ax2 x2 = p1 p1 p1
Case 5. If pp > a1 , it must follow that: .
¾ ¾
.
