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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 ) Example 2 Consider a production function which shows constant return to 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 1 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). 2 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: 3 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 or in gradient notation 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) 4 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). 5

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