<p>In general the statement is <strong>wrong</strong>. Here is a counterexample:</p>

<p>Suppose you have <span class="math-container">$f(k,l) = -k l^\beta$</span> with <span class="math-container">$\beta &gt;0$</span> and <span class="math-container">$(k,l)\in\mathbb{R}^2_{++}$</span> (you can interpret <span class="math-container">$f$</span> as a production function for a "<em>bad</em>" commodity). Then you have:
<span class="math-container">$$f(tk,tl) = - t^{1+\beta} k l^\beta = t^{1+\beta} f(k,l) &lt; t f(k,l)$$</span>
so that <span class="math-container">$f(k,l)$</span> is decreasing returns to scale. </p>

<p>Now let's evaluate <span class="math-container">$ f_{ll}f_{kk} - f^2_{kl} $</span>. Since <span class="math-container">$f$</span> is linear in <span class="math-container">$k$</span> we have that <span class="math-container">$f_{kk}$</span> is zero, so it is <span class="math-container">$ f_{ll}f_{kk}$</span>.  On the other hand, we have <span class="math-container">$f_{kl} = -\beta l^{\beta-1}$</span> so that overall we have:
<span class="math-container">\begin{equation}
f_{ll}f_{kk} - f^2_{kl} = 0 - \left[ -\beta l^{\beta-1} \right]^2 = - \beta^2 l^{2(\beta -1)} &lt;0
\end{equation}</span></p>

<p>So then, <strong>where is the trick</strong>?</p>

<p>The fact is that <span class="math-container">$ f_{ll}f_{kk} - f^2_{kl} &gt; 0$</span> is <strong>not a sufficient condition</strong> to conclude that the hessian of <span class="math-container">$f(k,l)$</span> is negative semi-definite and so <span class="math-container">$f$</span> is concave because you also have to impose restrictions on the sign of the principal minor of order 1 of the hessian matrix.</p>

<p>However, provided the hessian matrix is indeed negative semi-defintie than you conclude <span class="math-container">$f$</span> is concave and decreasing returns to scale will follow, the statement then holds.</p>

CC BY-SA 4.0
2019-05-10T19:09:55.917Z
2019-05-10T19:09:55.917Z
GabMac
<p>We examine the function <span class="math-container">$F(K,L)$</span> that is homogeneous of degree <span class="math-container">$\lambda &lt; 1$</span>. Then we have that its partial derivatives are homogeneous of degree <span class="math-container">$\lambda -1 $</span>.</p>

<p>For a homogeneous function <span class="math-container">$F(K,L)$</span> of degree <span class="math-container">$\lambda$</span> it holds that</p>

<p><span class="math-container">$$K\cdot F_K + L\cdot F_L = \lambda \cdot F(K,L)  \tag{1}$$</span></p>

<p>Analogously for the partial derivatives we have</p>

<p><span class="math-container">$$F_L:\;K\cdot F_{LK} + L\cdot F_{LL} = (\lambda -1)F_L &lt;0 \implies K\cdot F_{LK} &lt; -L\cdot F_{LL} \tag{2} $$</span></p>

<p>and</p>

<p><span class="math-container">$$F_K: L\cdot F_{KL} + K\cdot F_{KK} = (\lambda -1)F_K &lt;0 \implies L\cdot F_{KL} &lt; -K\cdot F_{KK} \tag{3} $$</span></p>

<p><em>If</em> the cross-partial is non-negative then <em>necessarily</em> the second partials are negative, and also both sides in the inequalities above are positive. Then multiplying per side we have</p>

<p><span class="math-container">$$K\cdot F_{LK} \cdot L\cdot F_{KL} &lt; L\cdot (-F_{LL}) \cdot K\cdot (-F_{KK})$$</span></p>

<p><span class="math-container">$$\implies F_{KK}\cdot F_{LL} &gt; F^2_{KL}$$</span></p>

<p>So a <em>sufficient</em> condition for (joint) concavity of a production function in two variables that exhibits decreasing returns to scale, is that the cross-partial derivative is non-negative. </p>

<p>Note that this sufficient condition related to the cross-partial guarantees also the "first step" for joint concavity, namely, that the second-order partials are negative (they alone provide for partial concavity of the production function).</p>

CC BY-SA 4.0
2019-05-10T19:10:27.010Z
2019-05-10T19:31:11.087Z
2019-05-10T19:31:11.087Z
Alecos Papadopoulos
<p>For a single output production function, decreasing returns to scale does imply concavity of <span class="math-container">$f$</span>. I believe you can find this result in Ch5 of MWG, either as a proposition or an exercise. The condition you have comes from considering the Hessian of <span class="math-container">$f$</span>, and showing that the determinant is positive. For more on the Hessian Conditions, this is a handy reference: <a href="https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t" rel="nofollow noreferrer">https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t</a></p>

CC BY-SA 4.0
2018-10-11T15:03:16.473Z
2018-10-11T15:03:16.473Z
Mathew Knudson
<p>For the productions <span class="math-container">$f(k,l) $</span> that are continuously differentiable, is the proposition  that</p>

<p>"<span class="math-container">$f(k,l)$</span> is decreasing return to scale <span class="math-container">$\Leftrightarrow f_{ll}f_{kk}-f_{kl}^2&gt;0$</span>"</p>

<p>always true, I have checked for Cobb-Douglas functions and believe it will also hold for all Constant-Elasticity fuction, can someone give me some hints or show some counter-examples?</p>

<p>Here we define <span class="math-container">$f(k,l)$</span> to be decreasing return to scale if <span class="math-container">$ f(tk,tl)&lt;tf(k,l)$</span> for <span class="math-container">$t&gt;1$</span> for all <span class="math-container">$(k,l)$</span> in the domain.</p>

<production-function><profit-maximization><cost-functions>
Is "$f(k,l)$ is decreasing return to scale $\Leftrightarrow f_{ll}f_{kk}-f_{kl}^2>0$" always true?
Jake ZHANG Shiyu
# Is "$f(k,l)$ is decreasing return to scale $\Leftrightarrow f_{ll}f_{kk}-f_{kl}^2>0$" always true?

(Asked by: username_0 on 2018-10-11)

For the productions $f(k,l) $ that are continuously differentiable, is the proposition that

"$f(k,l)$ is decreasing return to scale $\Leftrightarrow f\_{ll}f\_{kk}-f\_{kl}^2>0$"

always true, I have checked for Cobb-Douglas functions and believe it will also hold for all Constant-Elasticity fuction, can someone give me some hints or show some counter-examples?

Here we define $f(k,l)$ to be decreasing return to scale if $ f(tk,tl)<tf(k,l)$ for $t>1$ for all $(k,l)$ in the domain.

**Tags:** production-function, profit-maximization, cost-functions


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## 3 answers

### Answer by username_1 on 2019-05-10. Score: 2 | Accepted answer

In general the statement is **wrong**. Here is a counterexample:

Suppose you have $f(k,l) = -k l^\beta$ with $\beta >0$ and $(k,l)\in\mathbb{R}^2\_{++}$ (you can interpret $f$ as a production function for a "*bad*" commodity). Then you have:
$$f(tk,tl) = - t^{1+\beta} k l^\beta = t^{1+\beta} f(k,l) < t f(k,l)$$
so that $f(k,l)$ is decreasing returns to scale.

Now let's evaluate $ f\_{ll}f\_{kk} - f^2\_{kl} $. Since $f$ is linear in $k$ we have that $f\_{kk}$ is zero, so it is $ f\_{ll}f\_{kk}$. On the other hand, we have $f\_{kl} = -\beta l^{\beta-1}$ so that overall we have:
\begin{equation}
f\_{ll}f\_{kk} - f^2\_{kl} = 0 - \left[ -\beta l^{\beta-1} \right]^2 = - \beta^2 l^{2(\beta -1)} <0
\end{equation}

So then, **where is the trick**?

The fact is that $ f\_{ll}f\_{kk} - f^2\_{kl} > 0$ is **not a sufficient condition** to conclude that the hessian of $f(k,l)$ is negative semi-definite and so $f$ is concave because you also have to impose restrictions on the sign of the principal minor of order 1 of the hessian matrix.

However, provided the hessian matrix is indeed negative semi-defintie than you conclude $f$ is concave and decreasing returns to scale will follow, the statement then holds.


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### Answer by username_2 on 2019-05-10. Score: 1

We examine the function $F(K,L)$ that is homogeneous of degree $\lambda < 1$. Then we have that its partial derivatives are homogeneous of degree $\lambda -1 $.

For a homogeneous function $F(K,L)$ of degree $\lambda$ it holds that

$$K\cdot F\_K + L\cdot F\_L = \lambda \cdot F(K,L) \tag{1}$$

Analogously for the partial derivatives we have

$$F\_L:\;K\cdot F\_{LK} + L\cdot F\_{LL} = (\lambda -1)F\_L <0 \implies K\cdot F\_{LK} < -L\cdot F\_{LL} \tag{2} $$

and

$$F\_K: L\cdot F\_{KL} + K\cdot F\_{KK} = (\lambda -1)F\_K <0 \implies L\cdot F\_{KL} < -K\cdot F\_{KK} \tag{3} $$

*If* the cross-partial is non-negative then *necessarily* the second partials are negative, and also both sides in the inequalities above are positive. Then multiplying per side we have

$$K\cdot F\_{LK} \cdot L\cdot F\_{KL} < L\cdot (-F\_{LL}) \cdot K\cdot (-F\_{KK})$$

$$\implies F\_{KK}\cdot F\_{LL} > F^2\_{KL}$$

So a *sufficient* condition for (joint) concavity of a production function in two variables that exhibits decreasing returns to scale, is that the cross-partial derivative is non-negative.

Note that this sufficient condition related to the cross-partial guarantees also the "first step" for joint concavity, namely, that the second-order partials are negative (they alone provide for partial concavity of the production function).


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### Answer by username_3 on 2018-10-11. Score: -1

For a single output production function, decreasing returns to scale does imply concavity of $f$. I believe you can find this result in Ch5 of MWG, either as a proposition or an exercise. The condition you have comes from considering the Hessian of $f$, and showing that the determinant is positive. For more on the Hessian Conditions, this is a handy reference: <https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t>


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