Considering a firm taking prices for granted and maximizing profits

$$pf(x_1,...,x_K) - \sum_{i=1}^K q_i x_i,$$

where $f$ is strictly concave. Furthermore, let the factor demand curves be the solutions $x^*_i(p,q)$ and the slopes the derivative with respect to factor prices $\frac{\partial x^*_i(p,q)}{\partial q_j}$.

Is it then possible to say anything about the slope of factor demand curves if it is assumed that factors are complements in the sense of all

$$(A) \ \ \frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} > 0, \phantom{xxx}i\not = j$$

Motivation of the question

Intuitively, if some factor becomes more expensive less of the factor is used $\frac{\partial x^*_j(p,q)}{\partial q_j}<0$. Using less of the factor $j$ would decrease the marginal productivity of the factor $i$ if $\frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} > 0$ which in itself would lead one to assume that $\partial x_i^*(p,q)/\partial q_j <0$. However, if I remember correctly this result does not hold in general for $K>2$.

However, it seems to me that condition (A) would rule out the existence of indirect effects resulting from a price increase in $q_j$ leading to lower demand of $x_j$ and with less use of $x_j$ less use of $x_i$ being counteracted by the fact that less use of some third factor $x_s$ having a positive effect on the marginal productivity of $x_i$. Hence, contrary to condition (A) requiring that

$$ \frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_s} < 0.$$


1 Answer 1


I think your intuition is correct and can be shown formally under your constraint.

When you say that:

$$\frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} > 0, \phantom{xxx}i\not = j$$

You are forcing all the off-diagonal elements of the Hessian ($H$) to be positive. Now we know that $H$ is negative semi-definite. If we assume that it is negative definite (which will also force $f(.)$ to be strictly concave) then $H$ is invertible and its inverse is also symmetric negative definite. So we can write the Jacobian $(\nabla_qx^*)$ using SOC as:

$$\nabla_qx^* = (1/p)H^{-1}$$

Now use the result (it took me a while to find this but couldn't prove myself) that if off-diagonal elements of $H$ are non-negative then for $H^{-1}$ they will be non-positive (in the link it is for positive definite but should be directly extendible). This proves your intuition that the off-diagonal elements of the Jacobian will also be negative (or at least non-positive) which is what you are looking for.

  • 1
    $\begingroup$ Thx. That makes it very clear. $\endgroup$ Commented Feb 28, 2022 at 13:04

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