Euler's Theorem: f(x) is homogeneous of degree m, if and ... have Euler's
Theorem. ... ˆ . Z sx sx. = +. (. 1 1. 1. a x s. Z. = = share of 1). If ai's also change,
then.
Kwan Choi, International Trade, Spring 2008 0. Preliminaries Euler’s Theorem and Constant Returns to Scale Definition: f(x) is said to be homogenous of degree m [HD(m)], if and only if, for λ > 0,
f (λ x) = λ m f ( x),
for all x = ( x1 , x2 ,..., xn ).
(1)
Euler’s Theorem: f(x) is homogeneous of degree m, if and only if mf ( x) =
∂f ∂f ∂f x1 + x2 + ... + xn . ∂x1 ∂x2 ∂xn
Proof: Fix x and let g (λ ) = LHS of (1) with respect to λ,
g '(λ ) =
f (λ x ). Differentiate the
∂f (λ x) ∂f (λ x) x1 + x2 + ... ∂ (λ x1 ) ∂ (λ x2 )
When evaluated at λ =1,
1
(2)
n ∂f ( x) ∂f ( x) g '(1) = x1 + x2 + ... = ∑ fi xi . ∂x1 ∂x2 i =1
(3)
Differentiating the RHS of (1), we get
g '(λ ) = mλ m −1 f ( x). When evaluated at λ = 1, have Euler’s Theorem.
(4)
g '(1) = mf ( x ). Thus, we
Application to trade theory Constant Returns to Scale Production Function = homogeneous of degree 1. (1) Production function,
Y = F ( K , L ) , is HD(1). ⇒
FK K + FL L = mY = Y . ⇒ p ( FK K + FL L) = pY , or in competitive markets,
wL + rK = pY . ⇒ Zero Profits always (even in the short run).
2
(5)
⇒ This is different from zero profits of competitive firms in Long Run Equilibrium.
(2) If F(K,L) is HD(1), then Marginal Products are HD(0). Differentiate (5) with respect to K,
FKK K + FLK L + FK = FK . Or
FKK K + FLK L = 0i FK = 0. The same is true with MPL. Similarly, if F(K,L,T) is HD(1), then
FK K + FL L + FT T = Y . Differentiating (6) with respect to K gives
FKK K + FK + FLK L + FTK T = FK . Again, we get
FKK K + FLK L + FTK T = 0. 3
(6)
● In general, even with many factors (n > 2), if the production function exhibits CRS, then marginal product of every input is HD(0). Hat Calculus Z = XY ⇔
Zˆ = Xˆ + Yˆ.
Z = X/Y ⇔
Zˆ = Xˆ − Yˆ.
Z = 1/Y ⇔
Zˆ = −Yˆ.
Z = f ( X ,Y )
ˆ + ε Yˆ. ε ε X Y ⇔ X (
Z = a1 x1 + a2 x2 . ⇔ Zˆ = s1 xˆ1 + s2 xˆ2 . ax
( s1 = Z1 1 = share of 1). If ai’s also change, then
Zˆ = s1 ( aˆ1 + xˆ1 ) + s2 ( aˆ2 + xˆ2 ) .
4
X
=
∂Z X ∂X Z )
Z = f ( g ( X ), Y )
ˆ + ε Yˆ. ε ε X Y ⇔ g X
∂Z g ∂g X , εX = ∂g Z ∂X g ∂Z g ∂g X ∂Z X ε ε × = × = = ε ZX . gX ⇒ Zg ∂g Z ∂X g ∂X g
εg =
5
