Homogeneity and Euler's Theorem

Blake's Newton

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Let us start with a definition:

Homogeneity: Let :Rn R be a real-valued function. Then (x1, x2 ...., xn) is homogeneous of degree k if lk(x) = (l x) where l 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(x1, x2,..., xn) = (lx1, lx2,...., lxn)

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. Euler’s 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:

Theorem: (Euler's Theorem) Given the function :Rn R, then if is positively homogeneous of degree 1 then:

(x1, x2, ...., xn) = x1 [ / x1] + x2 [ / x2] + ...... + xn [ /dxn]

or simply:

(x) = ・/font> ni=1 [d (x)/dxi]キxi

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:

[ l (x)/ (x)]キ[ (x)/ xi] = [ (l x)/ (l xi)]キ[ (l xi)/ xi]

as [ l (x)/ (x)] = l and (l xi)/ xi = l then l [ (x)/ xi] = [ (l x)/ (l xi)]l then:

(x)/ xi = (l x)/ (l xi) (E.1)

Now, differentiating both sides of the original expression l (x) = (l x) with respect to l , we get:

l (x)/ l = ・/font> ni=1[ (l x)/ (l xi)]キ[ (l xi)/ l ]

As l (xi)/ l = (xi) and (l xi)/ l = xi for all i = 1,..., n, then this expression reduces to:

(x) = ・/font> ni=1[ (l x)/ (l xi)]キxi

Now using the equality in (E.1), we can substitute (x)/ xi for (l x)/ (l xi). Thus, this becomes:

(x) = ・/font> ni=1[ (x)/ xi]キxi

which is Euler’s 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:

Corollary: if :Rn R is homogenous of degree 1, then ・/font> ni=1[2(x)/ xixj]xi = 0 for any j.

Proof: By Euler’s Theorem, (x) = ・/font> ni=1[ (x)/ xi]キxi . Differentiating with respect to xj yields:

(x)/ xj = [ 2 (x)/ x1xj]x1 + ..... + [ 2 (x)/ xjxj]xj + (x)/ xj + ..... + [ 2 (x)/ xnxj]xn

or rewriting:

(x)/ xj = ・/font> ni=1[ 2 (x)/ xixj]xi + (x)/ 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, (x)/ xj appears on both sides of the equation. Thus:

・/font> ni=1[ 2 (x)/ xixj]xi = 0

which is what we sought.

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