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Suppose we have a production function with constant returns to scale. Let us denote it by $F(A,K,L)$ where $A$ is the technology, $K$ the capital and $L$ Labor. Further assume the first partial derivative of $L,K$ are both positive and the second partial negative. How to show the production function is concave in $K$ and $L$ but not strictly so?

This problem is from Acemoglu's Intro to Modern Economic Growth. I don't quite understand what it means to say the function is concave in $K$ and $L$. Do I need to show that $F(K,L)$ (treating A as constant) is concave? or do I show $F(K)$ and $F(L)$ are concave (treating L,A or K,A as constants)?

If it is the latter, it seems the problem seems trivial by the second derivative test.

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  • $\begingroup$ With multivariate functions you need to use the Hessian - the matrix of second orders.The univariate 2nd derivative test doesn't work. $\endgroup$
    – VCG
    Sep 10, 2016 at 2:10
  • $\begingroup$ @VCG so what do you think the question is asking for? Can you please give a mathematical definition please? $\endgroup$
    – Kun
    Sep 10, 2016 at 4:15
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    $\begingroup$ The question is straightforward. It is asking you to find the Hessian and show that the Hessian has the properties of a weak concave function (NSD). The links that Alvaro provided give you the definitions. I can write up an answer if you are still struggling with the definitions. $\endgroup$
    – VCG
    Sep 10, 2016 at 13:08
  • $\begingroup$ @VCG yes, I looked at the links. But since I am not given partial F over partial L partial K, how would I even derive the hessian matrix? Oh, I am also given constant return to scale.Thank you. $\endgroup$
    – Kun
    Sep 10, 2016 at 13:10

1 Answer 1

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So we want the Hessian to be NSD, so we need the PMs to alternate weakly.

$H=\begin{bmatrix} F_{kk} & F_{kl} \\ F_{kl} & F_{ll} \end{bmatrix}~~NSD \iff~~~F_{kk},F_{ll}\leq0~~~\&~~F_{kk}F_{ll}-F_{kl}^2\geq0$

We are given that $F_{kk},F_{ll}<0$ so we need to figure out the cross partials.

Constant returns to scale implies that we have a homogeneous of degree 1 function:

$F(K,L)=KF_k+LF_l \implies F_l=KF_{kl}+LF_{ll}+F_l~~ \&~~ F_k=LF_{kl}+KF_{kk}+F_k$

Now we can substitute these equations into the expression for the 2nd PM:

$\big(-\frac{L}{K}F_{kl}\big)\big(-\frac{K}{L}F_{kl}\big)-F_{kl}^2=0~~~~~$ and we're done.

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  • $\begingroup$ I would like to point a mistake... $F^2_{kl}$ is wrong, unless we are talking when $F_{kl}=F_{lk}$. However, this is not always true $\endgroup$
    – Yorgos
    Jan 15, 2017 at 19:30
  • $\begingroup$ @Yorgos $F_{KL}=F_{LK}$ by Young's theorem? $\endgroup$
    – VCG
    Jan 18, 2017 at 5:00
  • $\begingroup$ Consider a quadaratic productin function $f(k,l)=3kl^{2}+4lk^{2}$, then $f_{kl}\neq f_{lk}$. My point is that when you have a Cobb-Douglas production function then $f_{kl}=f_{lk}$, but it might be cases where young's theorem is not valid $\endgroup$
    – Yorgos
    Jan 18, 2017 at 8:13
  • $\begingroup$ @Yorgos I believe you are mistaken. $f_{kl}=6l+8k$ and $f_{lk}=6l+8k$. Cross derivatives are always identical (given regularity conditions). That is a general mathematical result: en.wikipedia.org/wiki/Symmetry_of_second_derivatives $\endgroup$
    – VCG
    Jan 18, 2017 at 16:28
  • $\begingroup$ from the link you provided there is a statement "In most "real-life" circumstances the Hessian matrix is symmetric, although there are a great number of functions that do not have this property.". My point was not to say that you were wrong, but rather it would be better to keep is as $f_{kl}f_{lk}$ $\endgroup$
    – Yorgos
    Jan 18, 2017 at 21:52

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