Let us start with a definition:
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.
If k = 1, we call this a linearly homogenous function. If we interpret ¦(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. Eulers Theorem states that under homogeneity of degree 1, a function ¦ (x) can be reduced to the sum of its arguments multiplied by their first partial derivatives, in short:
Proof: By definition of homogeneity of degree k, letting k = 1, then l ¦ (x) = ¦ (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:
as [¶ l ¦ (x)/¶ ¦ (x)] = l and ¶ (l xi)/¶ xi = l then l [¶ ¦ (x)/¶ xi] = [¶ ¦ (l x)/¶ (l xi)]l then:
Now, differentiating both sides of the original expression l ¦ (x) = ¦ (l x) with respect to l , we get:
As ¶ l ¦ (xi)/¶ l = ¦ (xi) and ¶ (l xi)/¶ l = xi for all i = 1,..., n, then this expression reduces to:
Now using the equality in (E.1), we can substitute ¶ ¦ (x)/¶ xi for ¶ ¦ (l x)/¶ (l xi). Thus, this becomes:
which is Eulers Theorem.§ One of the interesting results is that if ¦(x) is a homogeneous function of degree k, then the first derivatives, ¦i(x), are themselves homogeneous functions of degree k-1. So, for the homogeneous of degree 1 case, ¦ i(x) is homogeneous of degree zero. Consequently, there is a corollary to Euler's Theorem:
Proof: By Eulers Theorem, ¦ (x) = å ni=1[¶ ¦ (x)/¶ xi]·xi . Differentiating with respect to xj yields:
or rewriting:
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, ¶ ¦ (x)/¶ xj appears on both sides of the equation. Thus:
which is what we sought.§
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