# Constant returns and (weak/strict) concavity

Suppose I have a constant returns production function $$Q = f(X,Y,Z)$$, where $$X$$, $$Y$$, and $$Z$$ are the inputs. Because of constant returns, the Hessian matrix of second-order partial derivatives (f_ij) that can be derived from the production function has determinant equal to zero.

I'm interested in the expression $$f_{xx}f_{yy} - f_{xy}^2$$, which can be interpreted as the determinant of the two-by-two minor matrix that can be derived from the Hessian by deleting the row and column having to do with input $$Z$$.

I've worked out numerical examples (e.g., the Cobb-Douglas) and the expression I'm interested in is positive. Is this is a general property of constant returns production functions? If so, is there a simple proof of this result?

• If you have a production function with $K$ inputs, the rank of the hessian can be anything between 1 and $K$. If you initially have a production function with full rank hessian and and impose CRTS then the rank of your Hessian is reduced by one. May 13, 2023 at 10:42
• Thank you so much Bertrand. I believe I understand the point. Is there a reference you can point me to that would elaborate and perhaps provide a proof?
– G.J.
May 13, 2023 at 11:05

No, it is not generally true that the rank of the Hessian of a CRTS production function with $$K$$ inputs is equal to $$K-1$$.

Example 1. Let $$x$$ be a vector of $$K$$ inputs, and consider the quadratic production function given by: $$y= a+B'x + \frac{1}{2}x'Hx.$$ The production function has a hessian given by the $$K \times K$$ matrix $$H$$, and exhibits constant returns to scale iff $$a=0$$ and $$H=0$$. So, a CRTS production function can have a hessian of rank 0.

Example 2. Let us consider the following CRTS function (normalized quadratic): $$y= B'x + \frac{1}{2}\frac{x'Cx}{\theta'x}.$$ From Theorem 10 in Diewert and Wales (1987) it follows that production function is concave in $$x$$ iff $$C$$ is negative semi-definite. Moreover, for a given level of $$x=\bar{x}$$, the rank of the hessian is given by the rank of the matrix $$C$$, which can be any number between 0 and $$K-1$$.

EDIT. A function $$f,$$ homogeneous of degree one and differentiable, satisfies for any value of $$x$$ (by one of Euler's theorem): $$f(x)=x'\frac{\partial f}{\partial x}(x),$$ which in turn implies that $$x'\frac{\partial^2 f}{\partial x \partial x'}(x)=0.$$ This means that the rank of the Hessian matrix of $$f$$ cannot be greater than $$K-1$$. The $$K$$ column vectors of the hessian matrix are linearly dependent.

• Once you impose the restriction that a = 0 and that the Hessian H = 0, the function is CRTS. However, the function is no longer concave as well. So maybe the result in your original answer applies only to concave production functions. That's why I was trying to find a generalized proof of the original result. I think this paper by Dhrymes (On a Class of Utility and Production Functions Yielding Everywhere Differentiable Demand Functions, Propositions 4 and 5) in the Oct. 67 ReStud may come close to it, but it is certainly not transparent.
– G.J.
May 14, 2023 at 18:59
• @G.J.: I have added another example May 14, 2023 at 20:17
• Thanks so much for taking the time. I actually found another reference (Theil-Tilanus, Int. Economic Review, 1964) that seems to conclude that for "neoclassical" production functions, the rank of the Hessian for a linear homogeneous function is N-1. Maybe it all has to do with the assumptions made about the technology. I've never worked with the functions in your examples. It is the case that with functions like the Cobb-Douglas or CES that second-order minor is positive, so the rank is K-1. So what is the restriction in these functions that lead to that rank?
– G.J.
May 14, 2023 at 21:02
• @G.J.: I have added few lines to show why the hessian of a linearly homogeneous function cannot be full rank. May 17, 2023 at 13:56
• Thanks for following up. If you look at my original post, the question is not about the hessian of the linear homogeneous function. As I said in that post, I know it has zero determinant. My question was whether the rank was N-1 (in my case with 3 inputs, a rank of 2). it seems to be true for all commonly used production functions I've tried out.
– G.J.
May 17, 2023 at 23:37