A Note on Concavity Conditions of Cobb–Douglas and CES Production Function with at Least Two Inputs

Introduction

In microeconomic theory, the production function represents a very important concept, particularly in solving profit maximization problem. One of the properties that the production function must satisfy is a strict quasiconcavity (see Jehle and Reny, 2011). In Avvakumov, Kiselev, Orlov & Taras’ev (2010), the authors considered two most common production functions in microeconomic literature, a Cobb–Douglas and a constant elasticity of substitution (CES). Since every strict concave function defined on a convex set is strictly quasiconcave, a necessary and sufficient condition for the strict concavity of Cobb–Douglas and CES production function is given in Avvakumov et al. (2010). To prove negative definiteness of the Hessian matrices for both production functions certain recurrences are used in Avvakumov et al. (2010). In this note, we want to show that the negative definiteness of the Hessians can be proved directly, without the use of recurrences and mathematical induction.

We use the notation of Avvakumov et al. (2010) to the extent convenient to us, while the rest of the used notation is explained in what follows.

Let n ≥ 2 be a positive integer. We start with definitions of the Cobb–Douglas and CES production functions. A Cobb–Douglas production function is a real function defined as

F ¯ ( x ) = A ⋅ ∏ j = 1 n x j ε j

(1)

where x = ( x 1 , x 2 , ... , x n ) , x j > 0 , j = 1 , 2 , ... , n , is an input vector of n factors of production, A = F ¯ ( 1 , 1 , ... , 1 ) > 0 is a known calibration factor that determines the level of production for unit values of all factors, and the exponents ε j , 0 < ε j < 1 , j = 1 , 2 , ... , n , are the elasticity parameters.

A constant elasticity of substitution (CES) production function is a real function defined as

F ˜ ( x ) = A ˜ ⋅ ( ∑ j = 1 n α j x j − ρ ) − σ ρ

(2)

where A ˜ = F ˜ ( 1 , 1 , ... , 1 ) > 0 is a known calibration factor, similarly as in the case of Cobb–Douglas production function. The α j ’s are positive real numbers (the allocation coefficients) that satisfy

∑ j = 1 n α j = 1

(3)

The σ (the degree of homogeneity) is assumed positive

σ > 0

(4)

while the substitution coefficient ρ is subject to the constraints

ρ ≠ 0 , σ + ρ > 0 , 1 + ρ > 0.

(5)

In the next section the results of Avvakumov et al. (2010) are analysed, while in the last section the complementary results are given.

The Results of Avvakumov et al. (2010)

Both theorems in this section are proved in Avvakumov et al. (2010). Here we give their statements and a brief overview of the proofs.

Theorem 1 (Avvakumov et al. [2010], Theorem 2.1.). A necessary and sufficient condition for the strict concavity in ℝ + n of the Cobb–Douglas production function is

σ n < 1

(6)

where σ n = ε 1 + ε 2 + ⋯ + ε n is the sum of elasticity parameters.

Proof.

To check the property of strict concavity of function (1), Avvakumov et al. (2010) showed that its Hessian can be written as

H F ¯ ( x ) = − F ¯ ( x ) H ( x )

(7)

where H(x) is a symmetrical n × n matrix of the form

H ( x ) = [ ε 1 ( 1 − ε 1 ) x 1 2 − ε 1 ε 2 x 1 x 2 ⋯ − ε 1 ε n x 1 x n − ε 1 ε 2 x 1 x 2 ε 2 ( 1 − ε 2 ) x 2 2 ⋯ − ε 2 ε n x 2 x n ⋮ ⋮ ⋱ ⋮ − ε 1 ε n x 1 x n − ε 2 ε n x 2 x n ⋯ ε n ( 1 − ε n ) x n 2 ]

(8)

By formula (7) the condition of strict negative definiteness of the Hessian H F ¯ ( x ) is equivalent to the condition of positive definiteness of the matrix H(x). Formula (8) shows that this matrix is representable as a product of three n × n matrices

H F ¯ ( x ) = X H ⌢ X

(9)

where

H ⌢ = [ ε 1 ( 1 − ε 1 ) − ε 1 ε 2 ⋯ − ε 1 ε n − ε 1 ε 2 ε 2 ( 1 − ε 2 ) ⋯ − ε 2 ε n ⋮ ⋮ ⋱ ⋮ − ε 1 ε n − ε 2 ε n ⋯ ε n ( 1 − ε n ) ]

(10)

and

X = diag { 1 x 1 , 1 x 2 , ... , 1 x n }

(11)

The matrix H(x) from (9) is positive definite simultaneously with the matrix H ⌢ from (10), since the quadratic forms with the matrices H(x) and H ⌢ are related by

z T H F ¯ ( x ) z = z ⌢ T H ⌢ z ⌢

where the n dimensional vectors z and z ⌢ = X z with the coordinates z i and z ⌢ i = z i x i , i=1,2,…,n, respectively, are simultaneously nonzero. It thus remains to check the conditions for the positive definiteness of the matrix (10) that depends on the positive elasticity parameters ε 1 , ε 2 , ... , ε n .

To check positive definiteness of the matrix H ⌢ , Avvakumov et al. (2010) used Sylvester’s criterion, which is expressed in terms of positiveness of all the principal corner minors of the matrix H ⌢ :

Δ 1 > 0 , Δ 2 > 0 , ... , Δ n > 0

(12)

The minors Δ 1 , Δ 2 , ... , Δ n in (12) can be expressed as follows

{ Δ 1 = ε 1 ( 1 − ε 1 ) = ε 1 ( 1 − σ 1 ) , Δ 2 = ε 1 ε 2 ( 1 − ε 1 − ε 2 ) = ε 1 ε 2 ( 1 − σ 2 ) , ⋮ Δ n = ε 1 ε 2 ⋯ ε n ( 1 − ε 1 − ε 2 ⋯ − ε n ) = ε 1 ε 2 ⋯ ε n ( 1 − σ n ) .

(13)

Avvakumov et al. (2010) state that the formulas in (13) are easily proved by induction using the recurrence

Δ k = ε k ⋅ Δ k − 1 − ε k ⋅ ( ε 1 ε 2 ⋯ ε k ) , k = 2 , ... , n

(14)

Furthermore, they state that from (13), using assumption 0 < ε j < 1 , j = 1 , 2 , ... , n , it follows that the system of inequalities (12) is equivalent to the condition (6) for the sum of elasticities σ n , whereupon the proof of the Theorem 1 is completed. Q.E.D.

Here we want to emphasize that the recurrence (14) and induction are not necessary to find formulas (13), which will be shown in the next section.

Theorem 2 (Avvakumov et al. [2010], Theorem 3.1.). A necessary and sufficient condition for the strict concavity in ℝ + n of the CES production function is that

0 < σ < 1 ,

(15)

where σ is the degree of homogeneity of the CES production function.

Proof.

Similarly to the proof of the previous theorem, Avvakumov et al. (2010) have shown that the Hessian of the CES function (2) can be written as

H F ˜ ( x ) = − M ( x ) σ + ρ q ( x ) G ( x )

(16)

where

q ( x ) = ∑ j = 1 n α j x j − ρ

(17)

M ( x ) = A ˜ σ ⋅ q ( x ) − σ ρ − 1

(18)

and

G ( x ) = [ ( α 1 x 1 − ρ − 1 ) 2 d 1 ( x ) − α 1 x 1 − ρ − 1 α 2 x 2 − ρ − 1 ⋯ − α 1 x 1 − ρ − 1 α n x n − ρ − 1 − α 1 x 1 − ρ − 1 α 2 x 2 − ρ − 1 ( α 2 x 2 − ρ − 1 ) 2 d 2 ( x ) ⋯ − α 2 x 2 − ρ − 1 α n x n − ρ − 1 ⋮ ⋮ ⋱ ⋮ − α 1 x 1 − ρ − 1 α n x n − ρ − 1 − α 2 x 2 − ρ − 1 α n x n − ρ − 1 ⋯ ( α n x n − ρ − 1 ) 2 d n ( x ) ]

(19)

here

d i ( x ) = 1 + ρ σ + ρ ⋅ q ( x ) α i x i − ρ − 1 , i = 1 , 2 , ... , n

Furthermore, by (5) we have

M ( x ) ⋅ σ + ρ q ( x ) > 0

The conditions of strict negative definiteness of the Hessian H F ˜ ( x ) is equivalent, by (16), to the condition of positive definiteness of the matrix G(x). From (19), we have

G ( x ) = X G ⌢ ( x ) X

(20)

where

G ⌢ ( x ) = [ d 1 ( x ) − 1 ⋯ − 1 − 1 d 1 ( x ) ⋯ − 1 ⋮ ⋮ ⋱ ⋮ − 1 − 1 ⋯ d 1 ( x ) ]

(21)

X = diag { α 1 x 1 − ρ − 1 , α 2 x 2 − ρ − 1 , ... , α n x n − ρ − 1 }

(22)

Since the matrix G(x) is positive definite simultaneously with the matrix G ⌢ ( x ) , it remains to check the conditions of positive definiteness for the matrix (21). To investigate positive definiteness of the matrix G ⌢ ( x ) , Avvakumov et al . (2010) examined conditions under which all the principal corner minors of the matrix G ⌢ ( x ) , Δ i , i = 1 , 2 , ... , n , are positive. They have shown that the first minor is

Δ 1 = d 1 ( x ) = 1 − σ σ + ρ + 1 + ρ σ + ρ ⋅ α 2 x 2 − ρ + ⋯ + α n x n − ρ α 1 x 1 − ρ

and that the rest of the minors can be found by the recurrence

Δ k = 1 + ρ σ + ρ ⋅ q ( x ) α k x k − ρ Δ k − 1 − ( ( 1 + ρ ) q ( x ) σ + ρ ) k − 1 ⋅ 1 α 1 x 1 − ρ ⋯ α k − 1 x k − 1 − ρ , k = 2 , ... , n ,

(23)

from where they got

{ Δ 1 = 1 − σ σ + ρ + 1 + ρ σ + ρ ⋅ α 2 x 2 − ρ + ⋯ + α n x n − ρ α 1 x 1 − ρ , Δ 2 = ( 1 + ρ σ + ρ q ( x ) ) 1 α 1 x 1 − ρ α 2 x 2 − ρ ( 1 − σ σ + ρ ∑ j = 1 2 α j x j − ρ + 1 + ρ σ + ρ ∑ j = 3 n α j x j − ρ ) , Δ 3 = ( 1 + ρ σ + ρ q ( x ) ) 2 1 ∏ j = 1 3 α j x j − ρ ( 1 − σ σ + ρ ∑ j = 1 3 α j x j − ρ + 1 + ρ σ + ρ ∑ j = 4 n α j x j − ρ ) , ⋮ Δ n = ( 1 + ρ σ + ρ q ( x ) ) n − 1 ⋅ 1 ∏ j = 1 n α j x j − ρ ⋅ 1 − σ σ + ρ ⋅ q ( x ) .

(24)

From formulas (24), Avvakumov et al. (2010) concluded that the condition (15) is equivalent to positivity of the principal corner minors of the matrix G ⌢ ( x ) . That is the key argument which completes the proof of Theorem 2. Q.E.D.

Complementary Approach

We now show that to find formulas (13) and (24), it is not necessary to use recurrences (14) and (23) to calculate all the principal corner minors of (10) and (21), respectively.

Lemma 1. Let p be a positive integer, and a 1 , a 2 , ... , a p be real numbers. Let

A = [ a 1 − 1 − 1 ⋯ − 1 − 1 a 2 − 1 ⋯ − 1 − 1 − 1 a 3 ⋯ − 1 ⋮ − 1 ⋮ − 1 ⋮ − 1 ⋱ ⋯ ⋮ a p ]

(25)

If there exist i , j ∈ { 1 , 2 , ... , p } such that i ≠ j and a i = a j = − 1 , then

det A = 0

Suppose that there exists unique q ∈ { 1 , 2 , ... , p } such that a q = − 1 . Then

det A = − ∏ i = 1 i ≠ q p ( 1 + a i )

(26)

If a i ≠ − 1 for all i = 1 , 2 , ... , p , then for any q ∈ { 1 , 2 , ... , p } we have

det A = ∏ i = 1 p ( 1 + a i ) ⋅ [ a q 1 + a q − ∑ i = 1 i ≠ q p 1 1 + a i ]

(27)

Proof.

In this case, the matrix A obviously has at least two equal rows (or columns), which implies detA = 0.

If q ≠ 1 then, by switching the first and the qth row, and by subsequently switching the first and the qth column, we can replace element a_q at the position (1,1), and detA does not change. Let us now find detA. By multiplying the first column of A by -1 and adding it to its jth column, j = 2 , 3 , ... , p , we get

det A = | a q − 1 − 1 ⋯ − 1 − 1 a 2 − 1 ⋯ − 1 − 1 − 1 a 3 ⋯ − 1 ⋮ − 1 ⋮ − 1 ⋮ − 1 ⋱ ⋯ ⋮ a p | = | a q − a q − 1 − a q − 1 ⋯ − a q − 1 − 1 1 + a 2 0 ⋯ 0 − 1 0 1 + a 3 ⋯ 0 ⋮ − 1 ⋮ 0 ⋮ 0 ⋱ ⋯ ⋮ 1 + a p |

(28)

Since a q = − 1 , from (28) we get a determinant of a lower triangular matrix:

det A = | − 1 0 0 ⋯ 0 − 1 1 + a 2 0 ⋯ 0 − 1 0 1 + a 3 ⋯ 0 ⋮ − 1 ⋮ 0 ⋮ 0 ⋱ ⋯ ⋮ 1 + a p | = − ∏ i = 1 i ≠ q p ( 1 + a i )

(29)

First of all, let us notice that for any s , t ∈ { 1 , 2 , ... , p } , s ≠ t , we have

a s + a s a t − 1 − a s ( 1 + a s ) ( 1 + a t ) = a t + a t a s − 1 − a t ( 1 + a t ) ( 1 + a s ) ⇔ a s 1 + a s − 1 1 + a t − ∑ i = 1 i ≠ s , t p 1 1 + a i = a t 1 + a t − 1 1 + a s − ∑ i = 1 i ≠ s , t p 1 1 + a i ⇔ a s 1 + a s − ∑ i = 1 i ≠ s p 1 1 + a i = a t 1 + a t − ∑ i = 1 i ≠ t p 1 1 + a i .

Now, without loss of generality, we prove formula (27) for q = 1. Thus, in case a i ≠ − 1 , i = 1 , 2 , ... , p , from (28) by factoring out the term ( 1 + a j ) from the jth row, j = 1 , 2 , ... , p , we get

det A = ∏ i = 1 p ( 1 + a i ) ⋅ | a 1 1 + a 1 − 1 − 1 ⋯ − 1 − 1 1 + a 2 1 0 ⋯ 0 − 1 1 + a 3 0 1 ⋯ 0 ⋮ − 1 1 + a p ⋮ 0 ⋮ 0 ⋱ ⋯ ⋮ 1 |

(30)

Now, by multiplying the jth column in (30) by 1 1 + a j , j = 2 , 3 , ... , p , and adding it to the first column, we get a determinant of an upper triangular matrix

det A = ∏ i = 1 p ( 1 + a i ) ⋅ | a 1 1 + a 1 − ∑ i = 2 p 1 1 + a i − 1 − 1 ⋯ − 1 0 1 0 ⋯ 0 0 0 1 ⋯ 0 ⋮ 0 ⋮ 0 ⋮ 0 ⋱ ⋯ ⋮ 1 |

(31)

det A = ∏ i = 1 p ( 1 + a i ) ⋅ [ a 1 1 + a 1 − ∑ i = 2 p 1 1 + a i ] ,

(32)

that is equal to (27) for q = 1 . Q.E.D.

Let us now calculate the pth, p = 1 , 2 , ... , n , principal corner minor of the matrix H ⌢ from (10):

Δ p = | ε 1 ( 1 − ε 1 ) − ε 1 ε 2 ⋯ − ε 1 ε p − ε 1 ε 2 ε 2 ( 1 − ε 2 ) ⋯ − ε 2 ε p ⋮ ⋮ ⋱ ⋮ − ε 1 ε p − ε 2 ε p ⋯ ε p ( 1 − ε p ) | = { f a c t o r o u t ε j f r o m t h e j th c o l u m n a n d t h e n f r o m t h e j th r o w f o r j = 1 , 2 , ... , p } = ∏ j = 1 p ε j 2 ⋅ | 1 − ε 1 ε 1 − 1 ⋯ − 1 − 1 1 − ε 2 ε 2 ⋯ − 1 ⋮ ⋮ ⋱ ⋮ − 1 − 1 ⋯ 1 − ε p ε p | .

Since 1 − ε j ε j ≠ − 1 for all j = 1 , 2 , ... , p , ... , n , the last determinant can be found by formula (32) with a i = 1 − ε i ε i , i = 1 , 2 , ... , p . Thus,

Δ p = ∏ j = 1 p ε j 2 ⋅ ∏ i = 1 p ( 1 + 1 − ε i ε i ) ⋅ [ 1 − ε 1 ε 1 1 + 1 − ε 1 ε 1 + ∑ k = 2 p 1 1 + 1 − ε k ε k ] = ∏ j = 1 p ε j 2 ⋅ ( 1 − ∑ i = 1 p ε i ) = ε 1 ε 2 ⋯ ε p ( 1 − σ p ) .

Therefore, we can see that the last formula is equivalent to (13).

Finally, let us calculate the pth ( p = 1 , 2 , ... , n ) principal corner minor of the matrix G ⌢ ( x ) from (21). Since for all x ∈ ℝ + n we have

d i ( x ) = 1 + ρ σ + ρ ⋅ q ( x ) α i x i − ρ − 1 ≠ − 1 , i = 1 , 2 , ... , p , ... , n ,

by letting a i = d i ( x ) , and using the formulas q ( x ) = ∑ i = 1 n α i x i − ρ and (32) we have

Δ p = ∏ i = 1 p ( 1 + 1 + ρ σ + ρ ⋅ q ( x ) α i x i − ρ − 1 ) ⋅ [ 1 − σ + ρ 1 + ρ ⋅ α 1 x 1 − ρ q ( x ) − ∑ k = 2 p σ + ρ 1 + ρ ⋅ α k x k − ρ q ( x ) ] = ∏ i = 1 p ( 1 + ρ σ + ρ ⋅ q ( x ) α i x i − ρ ) ⋅ [ 1 − ∑ k = 1 p σ + ρ 1 + ρ ⋅ α k x k − ρ q ( x ) ] = ∏ i = 1 p ( 1 + ρ σ + ρ ⋅ q ( x ) α i x i − ρ ) ⋅ σ + ρ 1 + ρ ⋅ 1 q ( x ) ⋅ [ 1 + ρ σ + ρ ⋅ q ( x ) − ∑ k = 1 p α k x k − ρ ] = ( 1 + ρ σ + ρ ⋅ q ( x ) ) p − 1 ⋅ ∏ i = 1 p 1 α i x i − ρ ⋅ [ 1 − σ + σ + ρ σ + ρ ⋅ q ( x ) − ∑ k = 1 p α k x k − ρ ] = ( 1 + ρ σ + ρ ⋅ q ( x ) ) p − 1 ⋅ ∏ i = 1 p 1 α i x i − ρ ⋅ [ 1 − σ σ + ρ ⋅ q ( x ) + ( q ( x ) − ∑ k = 1 p α k x k − ρ ) ] = ( 1 + ρ σ + ρ ⋅ q ( x ) ) p − 1 ⋅ ∏ i = 1 p 1 α i x i − ρ ⋅ [ 1 − σ σ + ρ ⋅ ( ∑ j = 1 p α j x j − ρ + ∑ j = p + 1 n α j x j − ρ ) + ∑ k = p + 1 n α k x k − ρ ] = ( 1 + ρ σ + ρ ⋅ q ( x ) ) p − 1 ⋅ ∏ i = 1 p 1 α i x i − ρ ⋅ [ 1 − σ σ + ρ ⋅ ∑ j = 1 p α j x j − ρ + 1 + ρ σ + ρ ⋅ ∑ j = p + 1 n α k x k − ρ ] ,

which is equivalent to (24).