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 ³ 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) = å ni=1 [d¦ (x)/dxi]·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)]·[ ¦ (x)/ xi] = [ ¦ (l x)/ (l xi)]·[ (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 = å ni=1[ ¦ (l x)/ (l xi)]·[ (l xi)/ l ]

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

¦ (x) = å ni=1[ ¦ (l x)/ (l xi)]·xi

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

¦ (x) = å ni=1[ ¦ (x)/ xi]·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 å ni=1[2¦(x)/ xixj]xi = 0 for any j.

Proof: By Euler’s Theorem, ¦ (x) = å ni=1[ ¦ (x)/ xi]·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 = å 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:

å ni=1[ 2¦ (x)/ xixj]xi = 0

which is what we sought.§

 

 

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