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```					    A Brief Introduction to Homogenous Functions
Reference: Simon & Blum, chapter 20

1    Deﬁnition and Examples
Homogenous functions arise naturally in many economic models.

Deﬁnition 1 For any scalar k, a real-valued function f(x1 , ..., xn ) : Rn → R is
homogenous of degree k, denoted by HD(k), if for all x1 , .., .xn and all t > 0

f(tx1 , ..., txn ) = tk f(x1 , ..., xn )

scale, i.e. F (tK, tL) = tF (K, L) .This is an example of a HD(1) function. It
means, for example if you double the inputs of production, your output will be-
come doubled. If the production function is HD(k) for k > 1, then F (2K, 2L) =
2k F (K, L) > 2F (K, L). Therefore if you double your inputs your output in-
creases more than double. We say this production function shows increasing
returns to scale. Similarly if k < 1 , the production function exhibits decreasing
returns to scale.

Example 3 Let’s v(p1 , ..., pn , w) denote the value function corresponding to a
utility maximization problem

maxx1 ,...,xn u(x1 , ..., xn )
s.t.     p1 x1 + ... + pn xn ≤ w

then if you change the prices to tp1 , ..., tpn and the endowment to tw then the
budget constraint will become

t p1 x1 + ... + tpn xn ≤ tw

which is the same as
p1 x1 + ... + pn xn ≤ w

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therefore the result of the utility maximization does not change, i.e.

v(tp1 , ..., tpn , tw) = v(p1 , ..., pn , w)

therefore v(p1 , ..., pn , w) is a HD(0) function.

Example 4 Let’s Π(p, w) be the proﬁt function of a ﬁrm which is the optimized
value of the proﬁt solved from

max pf (z) − wz
z≥0

then if both the price of input and output change to tp and tw then the maxi-
mization problem becomes
max tpf (z) − twz
z≥0

which implies Π(tp, tw) = tΠ(p, w), i.e. the proﬁt function is a HD(1) func-
tion. Notice that in this example, the input demand z(p, w) remains the same,
Z(tp, tw) = Z(p, w), i.e. the input demand function of the ﬁrm is HD(0).

Example 5 It is easy to determine the degree of homogeneity of polynomials
and multinomials. For example f(x) = xn is HD(n). The function

f(x, y) = xn y m

is HD(n + m), check this!

Proposition 6 If f, g : Rn → R are homogenous of degree m then so is f + g.

Proof. simply

(f + g)(tx1 , ..., txn ) = f (tx1 , ..., txn ) + g(tx1 , ..., txn )

= tm f (x1 , ..., xn ) + tm g(x1 , ..., xn )

= tm (f + g)(x1 , ..., xn )

Proposition 7 If f is HD(m) and g is HD(n) then fg is HD(m + n).

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Proof. simply

(f g)(tx1 , ..., txn ) = f (tx1 , ..., txn ) × g(tx1 , ..., txn )

= tm f (x1 , ..., xn ) × tn g(x1 , ..., xn )

= tm+n (f g)(x1 , ..., xn )

Notice that if f, g do not have the same degree of homogeneity then f + g is
not a homogenous function. For example, f (x) = x is HD(1) and g(x) = x2 is
HD(2) but x2 + x is not a homogenous function.
The above two propositions give simple criteria to ﬁnd homogeneity degrees
of function. For products you need to sum up the degrees of homogeneity and
for the sums of homogenous functions, the degree remains the same if both
functions have the same degree.

Example 8

• f (x, y, z) = x2 y + xyx is HD(3).

x3 y+y 2 zx
• f (x, y, z) =    xy−zx        has degree 4 − 2 = 2.

• f (x1 , ..., xn ) = Axα1 ...xαn (Cobb-Douglas function) has degree α1 + ... +
1      n

αn .

The homogenous functions have some important properties that we mention
two of them

Theorem 9 Suppose f (x1 , ..., xn ) is a continuously diﬀerentiable function, and
∂f
HD(k), then    ∂xi   is HD(k − 1).

Proof. Without loss of generality we assume i = 1. We need to show

∂f                           ∂f
(tx1 , ..., txn ) = tk−1     (x1 , ..., xn )
∂x1                          ∂x1

. Find the partial derivative of the two sides of the following identity with respect
to x1 , using chain rule for diﬀerential of the left side:

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f(tx1 , ..., txn ) = tk f(x1 , ..., xn )

you get
∂f                         ∂f
t       (tx1 , ..., txn ) = tk     (x1 , ..., xn )
∂x1                        ∂x1
which is the desired result.
The next theorem has wide applications in economics and is called the Euler
Formula for homogenous functions.

Theorem 10 Let f(x1 , ...xn ) be a C 1 function and HD(k), then for all x1 , ..., xn
∂f              ∂f
x1       + .... + xn     = kf(x1 , ..., xn )
∂x1             ∂xn
x. f = kf (x)

Proof. Let’s denote x = (x1 , ..., xn ). We have

f(tx1 , ..., txn ) = tk f(x1 , ..., xn )

taking the derivative of both sides with resect to t yields
∂f                  ∂f
x1       (tx) + ... + xn     (tx) = ktk−1 f(x1 , ..., xn )             (1)
∂x1                 ∂xn
∂f
Now notice that each function            ∂xi   is HD(k − 1) by previous theorem, therefore
∂f                ∂f
we can set   ∂xi (tx)   = tk−1 ∂xi (x). Therefore (1) becomes
∂f                      ∂f
tk−1 x1       (x) + ... + tk−1 xn     (x) = ktk−1 f(x1 , ..., xn )
∂x1                     ∂xn
or equivalently
∂f              ∂f
x1        + .... + xn     = kf (x1 , ..., xn ) .
∂x1             ∂xn

This formula has a lot of applications in economics.

Example 11 Consider the production function F (K, L) = K α L1−α . It is easy
to see F is HD(1). Applying the Euler formula for F implies

KFk (K, L) + LFL (K, L) = F (K, L)                           (2)

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notice that Fk (K, L) and FL (K, L) are the marginal products of capital and
labor. If we are modelling a competitive economy, then capital and labor are
paid their marginal productivity, therefore KFk (K, L) and LFL (K, L) are paid
to capital and labor respectively. By (2) this exactly equals the whole output,
therefore the ﬁrm will gain no proﬁt.        This no proﬁt condition for ﬁrms in a
competitive economy therefore is implied by having a HD(1) production function.

The converse of the Euler Formula also holds but the proof is diﬀerent and
involves the use of diﬀerential equations.

Theorem 12 Suppose f (x1 , ..., xn ) is a C 1 function on Rn and there is an
+

scalar k such that for all x1 , ..., xn ∈ R+ we have

∂f              ∂f
x1       + .... + xn     = kf(x1 , ..., xn )
∂x1             ∂xn

then f is HD(k).

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