Homogeneity and Euler's Theorem |
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Homogeneity and Euler's Theorem |
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Let us start with a definition:
Homogeneity: Let ｦ :Rn ｮ R be a real-valued function. Then ｦ (x1, x2 ...., xn) is homogeneous of degree k if lkｦ(x) = ｦ(l x) where l ｳ 0 (x is the vector [x1...xn]).
In other words, a function is called homogeneous of degree k if by multiplying all arguments by a constant scalar l , we increase the value of the function by lk, i.e.
lkｦ(x1, x2,..., xn) = ｦ(lx1, lx2,...., lxn)
If k = 1, we call this a linearly homogenous function. If we interpret ｦ(x) as a production function, then k = 1 implies constant returns to scale (as lk = l), k > 1 implies increasing returns to scale (as lk > l) and if 0 < k < 1, then we have decreasing returns to scale (as lk < l).
Phillip Wicksteed (1894) stated the "product exhaustion" thesis implied by the marginal productivity theory of distribution - namely, that if all agents were paid their marginal product, then total costs would exhaust the entire product. Wicksteed assumed constant returns to scale - and thus employed a linear homogeneous production function, a function which was homogeneous of degree one. It was A.W. Flux (1894) who pointed out that Wicksteed's "product exhaustion" thesis was merely a restatement of Euler's Theorem. Euler’s Theorem states that under homogeneity of degree 1, a function ｦ (x) can be reduced to the sum of its arguments multiplied by their first partial derivatives, in short:
Theorem: (Euler's Theorem) Given the function ｦ :Rn ｮ R, then if ｦ is positively homogeneous of degree 1 then:
ｦ (x1, x2, ...., xn) = x1 [ｶ ｦ /ｶ x1] + x2 [ｶ ｦ /ｶ x2] + ...... + xn [ｶ ｦ /dｶxn]
or simply:
ｦ (x) = ・/font> ni=1 [dｦ (x)/dxi]ｷxi
Proof: By definition of homogeneity of degree k, letting k = 1, then l ｦ (x) = ｦ (l x) where x is a n-dimensional vector and l is a scalar. Differentiating both sides of this expression with respect to xi and using the chain rule, we see that:
[ｶ l ｦ (x)/ｶ ｦ (x)]ｷ[ｶ ｦ (x)/ｶ xi] = [ｶ ｦ (l x)/ｶ (l xi)]ｷ[ｶ (l xi)/ｶ xi]
as [ｶ l ｦ (x)/ｶ ｦ (x)] = l and ｶ (l xi)/ｶ xi = l then l [ｶ ｦ (x)/ｶ xi] = [ｶ ｦ (l x)/ｶ (l xi)]l then:
ｶ ｦ (x)/ｶ xi = ｶ ｦ (l x)/ｶ (l xi) (E.1)
Now, differentiating both sides of the original expression l ｦ (x) = ｦ (l x) with respect to l , we get:
ｶ l ｦ (x)/ｶ l = ・/font> ni=1[ｶ ｦ (l x)/ｶ (l xi)]ｷ[ｶ (l xi)/ｶ l ]
As ｶ l ｦ (xi)/ｶ l = ｦ (xi) and ｶ (l xi)/ｶ l = xi for all i = 1,..., n, then this expression reduces to:
ｦ (x) = ・/font> ni=1[ｶ ｦ (l x)/ｶ (l xi)]ｷxi
Now using the equality in (E.1), we can substitute ｶ ｦ (x)/ｶ xi for ｶ ｦ (l x)/ｶ (l xi). Thus, this becomes:
ｦ (x) = ・/font> ni=1[ｶ ｦ (x)/ｶ xi]ｷxi
which is Euler’s Theorem.ｧ
One of the interesting results is that if ｦ(x) is a homogeneous function of degree k, then the first derivatives, ｦi(x), are themselves homogeneous functions of degree k-1. So, for the homogeneous of degree 1 case, ｦ i(x) is homogeneous of degree zero. Consequently, there is a corollary to Euler's Theorem:
Corollary: if ｦ :Rn ｮ R is homogenous of degree 1, then ・/font> ni=1[ｶ2ｦ(x)/ｶ xiｶxj]xi = 0 for any j.
Proof: By Euler’s Theorem, ｦ (x) = ・/font> ni=1[ｶ ｦ (x)/ｶ xi]ｷxi . Differentiating with respect to xj yields:
ｶ ｦ (x)/ｶ xj = [ｶ 2ｦ (x)/ｶ x1ｶxj]x1 + ..... + [ｶ 2ｦ (x)/ｶ xjｶxj]xj + ｶ ｦ (x)/ｶ xj + ..... + [ｶ 2ｦ (x)/ｶ xnｶxj]xn
or rewriting:
ｶ ｦ (x)/ｶ xj = ・/font> ni=1[ｶ 2ｦ (x)/ｶ xi ｶxj]xi + ｶ ｦ (x)/ｶ xj
where, note, the summation expression sums from all i from 1 to n (including i = j). Nonetheless, note that the expression on the extreme right, ｶ ｦ (x)/ｶ xj appears on both sides of the equation. Thus:
・/font> ni=1[ｶ 2ｦ (x)/ｶ xiｶxj]xi = 0
which is what we sought.ｧ
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