applet-magic.com
Thayer Watkins
Silicon Valley
& Tornado Alley
USA

Euler's Theorem
and Its Generalization

A function F(L,K) is homogeneous of degree n if for any values of the parameter λ


F(λL, λK) = λnF(L,K)
 

The analysis is given only for a two-variable function because the extension to more variables is an easy and uninteresting generalization.

When F(L,K) is a production function then Euler's Theorem says that if factors of production are paid according to their marginal productivities the total factor payment is equal to the degree of homogeneity of the production function times output. The case of n=1 is an important special case. For that case if factors of production are paid according to their marginal productivities then output will exactly cover the factor payments.

A corollary to Euler's Theorem for production functions is that the sum of the elasticities of output with respect to factor inputs is equal to the degree of homogeneity of the production function; i.e.,


L(∂F/∂L)/F + K(∂F/∂K)/F = n.
 

This result is obtained simply dividing through the equation for Euler's Theorem by the level of output.

Generalizations

The equation that was obtained by differentiating the defining condition for homogeneity of degree n with respect to the parameter λ can be differentiated a second time with respect to λ and the value of λ set equal to unity. The result is:


(∂2F/∂L2)L2 + (∂F2/∂K∂L)KL + (∂2F/∂L∂K)LK + (∂F2/∂L2)L2
= n(n-1)F(L,K)
 

Since the cross derivatives are equal the above can be expressed as:


(∂2F/∂L2)L2 + 2(∂F2/∂K∂L)KL + (∂F2/∂L2)K2 = n(n-1)F(L,K)
 

For the special case of n=1 the above equation reduces to:


(∂2F/∂L2)L2 + 2(∂F2/∂K∂L)KL + (∂F2/∂L2)L2 = 0
 


 

The process of differentiating with respect to λ and setting λ equal to unity can be continued. The result is:

for m any positive integer less than or equal to n+1 and where C(m i) are the binomial coefficients m!/(m-i)!i!. In the above formula a partial derivative of the form (∂mF/∂K0∂Lm) is just (∂mF/∂Lm).


Illustration

Let F(L,K)=L2K3. This is a homogeneous function of degree 5. Then


∂F/∂L = 2LK3 and ∂F/∂K = 3L2K3
so
(∂F/∂L)L + (∂F/∂K)K = 2L2K3 + 3L2K3 = 5L2K3
 

The second derivatives are


(∂F2/∂L2) = 2K3
(∂F2/∂K∂L) = (∂F2/∂L∂K) = 6LK2
(∂F2/∂K2) = 6L2K
 

Thus


(∂2F/∂L2)L2 + 2(∂F2/∂K∂L)KL + (∂F2/∂L2)K2
[1(2) + 2(6) + 1(6)]L2K3
= 20L2K3 = 5(4)L2K3
 

The third order derivatives are, without distinguishing between the equal cross derivatives,


(∂F3/∂L3) = 0
(∂F3/∂K∂L2) = 6K2
(∂F3/∂K2∂L) = 12LK
(∂F3/∂K3) = 6L2
 

Thus


(∂F3/∂L3)L3 + 3(∂F3/∂K∂L2)L2K + 3(∂F3/∂K2∂L)LK2 + (∂F3/∂K3)K3
= [1(0) + 3(6) + 3(12) + 1(6)]L2K3 = 60L2K3
= 5(4)(3)L2K3
 


HOME PAGE OF applet-magic