<p>No, it is not generally true that the rank of the Hessian of a CRTS production function with <span class="math-container">$K$</span> inputs is equal to <span class="math-container">$K-1$</span>.</p>
<p>Example 1. Let <span class="math-container">$x$</span> be a vector of <span class="math-container">$K$</span> inputs, and consider the quadratic production function given by:
<span class="math-container">$$ y= a+B'x + \frac{1}{2}x'Hx.$$</span>
The production function has a hessian given by the <span class="math-container">$K \times K$</span> matrix <span class="math-container">$H$</span>, and exhibits constant returns to scale iff <span class="math-container">$a=0$</span> and <span class="math-container">$H=0$</span>. So, a CRTS production function can have a hessian of rank 0.</p>
<p>Example 2. Let us consider the following CRTS function (normalized quadratic):
<span class="math-container">$$ y= B'x + \frac{1}{2}\frac{x'Cx}{\theta'x}.$$</span>
From Theorem 10 in Diewert and Wales (1987) it follows that production function is concave in <span class="math-container">$x$</span> iff <span class="math-container">$C$</span> is negative semi-definite.
Moreover, for a given level of <span class="math-container">$x=\bar{x}$</span>, the rank of the hessian is given by the rank of the matrix <span class="math-container">$C$</span>, which can be any number between 0 and <span class="math-container">$K-1$</span>.</p>
<p><strong>EDIT</strong>. A function <span class="math-container">$f,$</span> homogeneous of degree one and differentiable, satisfies for any value of <span class="math-container">$x$</span> (by one of Euler's theorem): <span class="math-container">$$f(x)=x'\frac{\partial f}{\partial x}(x),$$</span> which in turn implies that <span class="math-container">$$x'\frac{\partial^2 f}{\partial x \partial x'}(x)=0.$$</span> This means that the rank of the Hessian matrix of <span class="math-container">$f$</span> cannot be greater than <span class="math-container">$K-1$</span>. The <span class="math-container">$K$</span> column vectors of the hessian matrix are linearly dependent.</p>

CC BY-SA 4.0
2023-05-14T18:59:31.637Z
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.
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2023-05-14T20:17:13.560Z
@G.J.: I have added another example
Bertrand
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2023-05-14T21:02:02.487Z
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.
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2023-05-17T13:56:02.217Z
@G.J.: I have added few lines to show why the hessian of a linearly homogeneous function cannot be full rank.
Bertrand
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2023-05-17T23:37:42.367Z
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.
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2023-05-18T09:00:47.970Z
@G.J.: your hypothesis is not true. See the 2 couter examples above, which could easily be extended by adding a Leontief production, or a variant of a Cobb-Douglas. What do you mean by (weak/strict) concavity? If you mean that the Hessian has rank $K-1$, then of course it will have rank $K-1$, In general however, the rank of a linear homogeneous production function is at most $K-1$.
Bertrand
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2023-05-18T12:57:27.657Z
This helps me rephrase my question in the most succinct way: What assumptions do we need to make--weak/strict concavity, etc--to ensure that the rank of a linear homogenous function is exactly K - 1? The Dhrymes paper I cited a few rounds ago suggests an interior solution to a cost-minimization problem is sufficient.
G.J.
CC BY-SA 4.0
2023-05-14T18:05:14.990Z
2023-05-17T13:54:13.447Z
2023-05-17T13:54:13.447Z
Bertrand
<p>Suppose I have a constant returns production function <span class="math-container">$Q = f(X,Y,Z)$</span>, where <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, and <span class="math-container">$Z$</span> 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.</p>
<p>I'm interested in the expression <span class="math-container">$f_{xx}f_{yy} - f_{xy}^2$</span>, 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 <span class="math-container">$Z$</span>.</p>
<p>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?</p>

CC BY-SA 4.0
2023-05-13T10:42:29.920Z
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.
Bertrand
CC BY-SA 4.0
2023-05-13T11:05:42.380Z
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.
<microeconomics><production-function><economies-of-scale>
Constant returns and (weak/strict) concavity
G.J.
# Constant returns and (weak/strict) concavity

(Asked by: username_0 on 2023-05-11)

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?

**Tags:** microeconomics, production-function, economies-of-scale
> **Comments:**
> **username_1**: 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.  
> **username_0**: 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?  

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## 1 answers

### Answer by username_1 on 2023-05-14. Score: 2

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.

> **Comments:**
> **username_0**: 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.  
> **username_1**: @G.J.: I have added another example  
> **username_0**: 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?  
> **username_1**: @G.J.: I have added few lines to show why the hessian of a linearly homogeneous function cannot be full rank.  
> **username_0**: 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.  
> **username_1**: @G.J.: your hypothesis is not true. See the 2 couter examples above, which could easily be extended by adding a Leontief production, or a variant of a Cobb-Douglas. What do you mean by (weak/strict) concavity? If you mean that the Hessian has rank $K-1$, then of course it will have rank $K-1$, In general however, the rank of a linear homogeneous production function is at most $K-1$.  
> **username_0**: This helps me rephrase my question in the most succinct way: What assumptions do we need to make--weak/strict concavity, etc--to ensure that the rank of a linear homogenous function is exactly K - 1? The Dhrymes paper I cited a few rounds ago suggests an interior solution to a cost-minimization problem is sufficient.  

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