<p>C.E.S production functions are indeed tiresome to differentiate, and it is very easy to make a mistake. But hey, this (algebraic manipulation) is part of the job, and one needs to practice moving around symbols , creating common factors, etc. </p>

<p><strong>But there is another way to show that $F_{KK} &lt;0$ in totally abstract terms.</strong></p>

<p><strong>First,</strong> prove 4. (linear homogeneity), it is easy.  </p>

<p><strong>Second</strong>, exploit Euler's Homogeneous Function Theorem, and decompose $Y$ using the first derivatives and the variables (this is a very well known result).</p>

<p><strong>Third</strong>, re-arrange the previous to obtain an alternative abstract expression of the first derivative of say, capital.</p>

<p><strong>Fourth</strong>, differentiate what you got in step Three, to obtain an expression for $F_{KK}$. </p>

<p>All the above are done using abstract notation $F_K, F_{KK}$,etc and nowhere the actual C.E.S functional form. After step Four, you will need to calculate the cross-partial, which is easy, even for the C.E.S. function.</p>

CC BY-SA 3.0
2018-04-02T19:17:05.363Z
2018-04-02T19:38:02.307Z
2018-04-02T19:38:02.307Z
Alecos Papadopoulos
<p>Say I'm given the production function of the form</p>

<p>\begin{equation*}
Y= F(K,L,A) = (\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1}{\epsilon}}
\end{equation*}</p>

<p>Where $\epsilon, \alpha \in (0,1)$. </p>

<p>Now I need to check three things:</p>

<ol>
<li>That marginal products are positive and they are decreasing</li>
<li>Inada conditions are satisfied</li>
<li>Essentiality</li>
<li>Linear homogeneity</li>
</ol>

<p>The conditions 2 and 4 imply the condition 3 so no need to check for that. When taking derivatives with respect to $K$ I see that</p>

<p>\begin{equation*}
F_K = \alpha  K^{\epsilon -1}(\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1 - \epsilon}{\epsilon}} &gt;0
\end{equation*}</p>

<p>Which is positive. Now I need to show that $F_{KK}&lt; 0$. See:</p>

<p>\begin{equation*}
F_{KK} = \alpha (\epsilon -1) K^{\epsilon -2}(\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1 - \epsilon}{\epsilon}} + \alpha^2 K^{2\epsilon -2} \frac{1- \epsilon}{\epsilon} (\alpha K^{\epsilon} + (1- \alpha) L^{\epsilon})^{\frac{1}{\epsilon}-2}
\end{equation*}</p>

<p>However I couldn't find a way to decide on the sign of this expression. Any help would be appreciated. Thanks.</p>

CC BY-SA 3.0
2018-04-02T16:02:42.090Z
Have you tried simplifying it? There are a lot of common factors in this sum, all of which have positive signs. E.g.: $\alpha$ appears in both.
Giskard
CC BY-SA 3.0
2018-04-02T16:16:01.380Z
@denesp I couldn't simplify it that's why it is on here
Kuantew
CC BY-SA 3.0
2018-04-02T19:04:03.380Z
It appears there is a typo, you write $Y$ in the right-hand-side instead of $L$ ? (on three different occasions)
Alecos Papadopoulos
<macroeconomics><microeconomics><economic-growth>
How to check whether the production function below is neoclassical or not
Kuantew
# How to check whether the production function below is neoclassical or not

(Asked by: username_0 on 2018-04-02)

Say I'm given the production function of the form

\begin{equation\*}
Y= F(K,L,A) = (\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1}{\epsilon}}
\end{equation\*}

Where $\epsilon, \alpha \in (0,1)$.

Now I need to check three things:

1. That marginal products are positive and they are decreasing
2. Inada conditions are satisfied
3. Essentiality
4. Linear homogeneity

The conditions 2 and 4 imply the condition 3 so no need to check for that. When taking derivatives with respect to $K$ I see that

\begin{equation\*}
F\_K = \alpha K^{\epsilon -1}(\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1 - \epsilon}{\epsilon}} >0
\end{equation\*}

Which is positive. Now I need to show that $F\_{KK}< 0$. See:

\begin{equation\*}
F\_{KK} = \alpha (\epsilon -1) K^{\epsilon -2}(\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1 - \epsilon}{\epsilon}} + \alpha^2 K^{2\epsilon -2} \frac{1- \epsilon}{\epsilon} (\alpha K^{\epsilon} + (1- \alpha) L^{\epsilon})^{\frac{1}{\epsilon}-2}
\end{equation\*}

However I couldn't find a way to decide on the sign of this expression. Any help would be appreciated. Thanks.

**Tags:** macroeconomics, microeconomics, economic-growth

> **Comments:**
> **username_1**: Have you tried simplifying it? There are a lot of common factors in this sum, all of which have positive signs. E.g.: $\alpha$ appears in both.  
> **username_0**: @denesp I couldn't simplify it that's why it is on here  
> **username_2**: It appears there is a typo, you write $Y$ in the right-hand-side instead of $L$ ? (on three different occasions)  

---

## 1 answers

### Answer by username_2 on 2018-04-02. Score: 3 | Accepted answer

C.E.S production functions are indeed tiresome to differentiate, and it is very easy to make a mistake. But hey, this (algebraic manipulation) is part of the job, and one needs to practice moving around symbols , creating common factors, etc.

**But there is another way to show that $F\_{KK} <0$ in totally abstract terms.**

**First,** prove 4. (linear homogeneity), it is easy.

**Second**, exploit Euler's Homogeneous Function Theorem, and decompose $Y$ using the first derivatives and the variables (this is a very well known result).

**Third**, re-arrange the previous to obtain an alternative abstract expression of the first derivative of say, capital.

**Fourth**, differentiate what you got in step Three, to obtain an expression for $F\_{KK}$.

All the above are done using abstract notation $F\_K, F\_{KK}$,etc and nowhere the actual C.E.S functional form. After step Four, you will need to calculate the cross-partial, which is easy, even for the C.E.S. function.


---

