Mathematical methods for economic theory

Martin J. Osborne
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2.5 Homogeneous functions

Definition

Multivariate functions that are “homogeneous” of some degree are often used in economic theory. For a given number k, a function is homogeneous of degree k if, when each of its arguments is multiplied by any number t > 0, the value of the function is multiplied by tk. For example, a function is homogeneous of degree 1 if, when all its arguments are multiplied by any number t > 0, the value of the function is multiplied by the same number t.

Here is a precise definition. Because the definition involves the relation between the value of the function at (x1, ..., xn) and its values at points of the form (tx1, ..., txn) where t is any positive number, it is restricted to functions for which (tx1, ..., txn) is in the domain whenever t > 0 and (x1, ..., xn) is in the domain. (Some domains that have this property are the set of all real numbers, the set of nonnegative real numbers, the set of positive real numbers, the set of all n-tuples (x1, ..., xn) of real numbers, the set of n-tuples of nonnegative real numbers, and the set of n-tuples of positive real numbers.)

Definition
Let f be a function of n variables defined on a set S for which (tx1, ..., txn) ∈ S whenever t > 0 and (x1, ..., xn) ∈ S, and let k be a number. Then f is homogeneous of degree k if
f(tx1, ..., txn) = tkf(x1, ..., xn) for all (x1, ..., xn) ∈ S and all t > 0.

Example 2.5.1
For the function f(x1, x2) = Axa
1xb
2 with domain {(x1, x2): x1 ≥ 0 and x2 ≥ 0} we have
f(tx1, tx2) = A(tx1)a(tx2)b = Ata+bxa
1xb
2 = ta+bf(x1, x2),
so that f is homogeneous of degree a + b.

Example 2.5.2
Let f(x1, x2) = x1 + x2
2, with domain {(x1, x2): x1 ≥ 0 and x2 ≥ 0}. Then
f(tx1, tx2) = tx1 + t2x2
2.
It doesn't seem to be possible to write this expression in the form tk(x1 + x2
2) for any value of k. But how do we prove that there is no such value of k? Suppose that there were such a value. That is, suppose that for some k we have
tx1 + t2x2
2 = tk(x1 + x2
2) for all (x1, x2) ≥ (0, 0) and all t > 0.
Then in particular, taking t = 2, we have
2x1 + 4x2  =  2k(x1 + x2
2) for all (x1, x2).
Taking (x1, x2) = (1, 0) and (x1, x2) = (0, 1) we thus have
2 = 2k and 4 = 2k,
which is not possible. Thus f is not homogeneous of any degree.
In economic theory we often assume that a firm's production function is homogeneous of degree 1 (if all inputs are multiplied by t then output is multiplied by t). A production function with this property is said to have “constant returns to scale”.

Suppose that a consumer's demand for goods, as a function of prices and her income, arises from her choosing, among all the bundles she can afford, the one that is best according to her preferences. Then we can show that this demand function is homogeneous of degree zero: if all prices and the consumer's income are multiplied by any number t > 0 then her demands for goods stay the same.

Partial derivatives of homogeneous functions

The following result is sometimes useful.
Proposition 2.5.1
Let f be a differentiable function of n variables that is homogeneous of degree k. Then each of its partial derivatives f'i (for i = 1, ..., n) is homogeneous of degree k − 1.
Proof
The homogeneity of f means that
f(tx1, ..., txn) = tkf(x1, ..., xn) for all (x1, ..., xn) and all t > 0.
Now differentiate both sides of this equation with respect to xi, to get
tf'i(tx1, ..., txn) = tkf'i(x1, ..., xn) for all (x1, ..., xn) and all t > 0,
and then divide both sides by t to get
f'i(tx1, ..., txn) = tk−1f'i(x1, ..., xn) for all (x1, ..., xn) and all t > 0,
so that f'i is homogeneous of degree k − 1.

Application: level curves of homogeneous functions

This result can be used to demonstrate a nice result about the slopes of the level curves of a homogeneous function. As we have seen, the slope of the level curve of the function F through the point (x0, y0) at this point is
−
F'1(x0, y0)
F'2(x0, y0)
.
Now suppose that F is homogeneous of degree k, and consider the level curve through (cx0, cy0) for some number c > 0. At (cx0, cy0), the slope of this curve is
−
F'1(cx0, cy0)
F'2(cx0, cy0)
.
By the previous result, F'1 and F'2 are homogeneous of degree k−1, so this slope is equal to
−
ck−1F'1(x0, y0)
ck−1F'2(x0, y0)
 = −
F'1(x0, y0)
F'2(x0, y0)
.
That is, the slope of the level curve through (cx0, cy0) at the point (cx0, cy0) is exactly the same as the slope of the level curve through (x0, y0) at the point (x0, y0), as illustrated in the following figure.

x → y ↑ 0 x0 cx0 y0 cy0

In this figure, the red lines are two level curves, and the two green lines, the tangents to the curves at (x0, y0) and at (cx0, cy0), are parallel.

We may summarize this result as follows.

Let F be a differentiable function of two variables that is homogeneous of some degree. Then along any given ray from the origin, the slopes of the level curves of F are the same.

Euler's theorem

A function homogeneous of some degree has a property sometimes used in economic theory that was first discovered by Leonhard Euler (1707–1783).
Proposition 2.5.2 (Euler's theorem)
Let f be a differentiable function of n variables defined on an open set S for which (tx1, ..., txn) ∈ S whenever t > 0 and (x1, ..., xn) ∈ S. Then f is homogeneous of degree k if and only if
∑n
i=1xif'i(x1, ..., xn) = kf(x1, ..., xn) for all (x1, ..., xn) ∈ S.       (*)
Condition (*) may be written more compactly, using the notation ∇f for the gradient vector of f and letting x = (x1, ..., xn), as
x·∇f(x) = kf(x) for all x ∈ S.
Proof
I first show that if f is homogeneous of degree k then (*) holds. If f is homogeneous of degree k then
f(tx1, ..., txn) = tkf(x1, ..., xn) for all (x1, ..., xn) and all t > 0.
Differentiate each side of this equation with respect to t, to give
x1f'1(tx1, ..., txn) + x2f'2(tx1, ..., txn) + ... + xnf'n(tx1, ..., txn) = ktk−1f(x1, ..., xn).
Now set t = 1, to obtain (*).

I now show that if (*) holds then f is homogeneous of degree k. Suppose that (*) holds. Fix (x1, ..., xn) and define the function g of a single variable by

g(t)  =  t−kf(tx1, ..., txn) − f(x1, ..., xn).
We have
g'(t)  =  −kt−k−1f(tx1, ..., txn) + t−k∑n
i=1 xif'i(tx1, ..., txn).
By (*), we have
∑n
i=1txifi'(tx1, ..., txn) = kf(tx1, ..., txn),
so that g'(t) = 0 for all t. Thus g(t) is a constant. But g(1) = 0, so g(t) = 0 for all t, and hence f(tx1, ..., txn) = tkf(x1, ..., xn) for all t > 0, so that f is homogeneous of degree k.
Example 2.5.3
Let f be the production function of a firm that uses n inputs to produce a single output. Suppose that f is homogeneous of degree 1 (i.e. has “constant returns to scale”). Euler's theorem shows that if the price (in terms of units of output) of each input i is its “marginal product” f'i(x1, ..., xn), then the total cost, namely
∑n
i=1xif'i(x1, ..., xn)
is equal to the total output, namely f(x1, ..., xn).