<p>I think your intuition is correct and can be shown formally under your constraint.</p>
<p>When you say that:</p>
<p><span class="math-container">$$\frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} &gt; 0, \phantom{xxx}i\not = j$$</span></p>
<p>You are forcing all the off-diagonal elements of the Hessian (<span class="math-container">$H$</span>) to be positive. Now we know that <span class="math-container">$H$</span> is negative semi-definite. If we assume that it is negative definite (which will also force <span class="math-container">$f(.)$</span> to be <em>strictly</em> concave) then <span class="math-container">$H$</span> is invertible and its inverse is also symmetric negative definite. So we can write the Jacobian <span class="math-container">$(\nabla_qx^*)$</span> using SOC as:</p>
<p><span class="math-container">$$\nabla_qx^* = (1/p)H^{-1}$$</span></p>
<p>Now use the <a href="https://math.stackexchange.com/questions/3524906/non-diagonal-elements-of-positive-definite-matrix">result</a> (it took me a while to find this but couldn't prove myself) that if off-diagonal elements of <span class="math-container">$H$</span> are non-negative then for <span class="math-container">$H^{-1}$</span> 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.</p>

CC BY-SA 4.0
2022-02-28T13:04:47.173Z
Thx. That makes it very clear.
Jesper Hybel
CC BY-SA 4.0
2022-02-28T11:13:20.540Z
2022-03-02T01:14:37.390Z
2022-03-02T01:14:37.390Z
Dayne
<p>Considering a firm taking prices for granted and maximizing profits</p>
<p><span class="math-container">$$pf(x_1,...,x_K) - \sum_{i=1}^K q_i x_i,$$</span></p>
<p>where <span class="math-container">$f$</span> is strictly concave. Furthermore, let the factor demand curves be the solutions <span class="math-container">$x^*_i(p,q)$</span> and the slopes the derivative with respect to factor prices <span class="math-container">$\frac{\partial x^*_i(p,q)}{\partial q_j}$</span>.</p>
<p>Is it then possible to say anything about the slope of factor demand curves if it is assumed that factors are <em>complements</em> in the sense of all</p>
<p><span class="math-container">$$(A) \ \ \frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} &gt; 0, \phantom{xxx}i\not = j$$</span></p>
<p><strong>Motivation of the question</strong></p>
<p>Intuitively, if some factor becomes more expensive less of the factor is used <span class="math-container">$\frac{\partial x^*_j(p,q)}{\partial q_j}&lt;0$</span>. Using less of the factor <span class="math-container">$j$</span> would decrease the marginal productivity of the factor <span class="math-container">$i$</span> if <span class="math-container">$\frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} &gt; 0$</span> which in itself would lead one to assume that <span class="math-container">$\partial x_i^*(p,q)/\partial q_j &lt;0$</span>. However, if I remember correctly this result does not hold in general for <span class="math-container">$K&gt;2$</span>.</p>
<p>However, it seems to me that condition (A) would rule out the existence of <strong>indirect effects</strong> resulting from a price increase in <span class="math-container">$q_j$</span> leading to lower demand of <span class="math-container">$x_j$</span> and with less use of <span class="math-container">$x_j$</span> less use of <span class="math-container">$x_i$</span> being counteracted by the fact that less use of some third factor <span class="math-container">$x_s$</span> having a positive effect on the marginal productivity of <span class="math-container">$x_i$</span>. Hence, contrary to condition (A) requiring that</p>
<p><span class="math-container">$$ \frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_s} &lt; 0.$$</span></p>

<macroeconomics><demand><profit-maximization>
Complement in production and the slope of factor demand curves
Jesper Hybel
# Complement in production and the slope of factor demand curves

(Asked by: username_0 on 2022-02-27)

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

$$ \frac{\partial^2 f(x\_1,...,x\_K)}{\partial x\_i \partial x\_s} < 0.$$

**Tags:** macroeconomics, demand, profit-maximization


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## 1 answers

### Answer by username_1 on 2022-02-28. Score: 3 | Accepted answer

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](https://math.stackexchange.com/questions/3524906/non-diagonal-elements-of-positive-definite-matrix) (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.

> **Comments:**
> **username_0**: Thx. That makes it very clear.  

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