<p>The problem is that, with an elasticity of substitution <span class="math-container">$\sigma&lt;1$</span>, in the CES production function we have negative exponents.<span class="math-container">$^1$</span></p>
<p>I rewrite your <strong>CES production function:</strong></p>
<p><span class="math-container">$$Y = f(K, L) = (a \cdot K^\rho + (1 - a) \cdot L^\rho )^{1/\rho}.\;\;\;\;\;(1)$$</span></p>
<p><strong>Elasticity of substitution <span class="math-container">$\sigma = 1/(1-\rho) &lt;1$</span> implies <span class="math-container">$\rho&lt;0$</span>,</strong> so that the exponents <span class="math-container">$\rho$</span> and <span class="math-container">$1/\rho$</span> in equation <span class="math-container">$(1)$</span> are negative.</p>
<p>In particular, this implies that the factors <span class="math-container">$K$</span> and <span class="math-container">$L$</span> in <span class="math-container">$(1)$</span> are the denominator of a fraction. Therefore, <strong>we cannot set <span class="math-container">$K=0$</span> or <span class="math-container">$L=0$</span>.</strong></p>
<p>To see it clearly, let’s rewrite <span class="math-container">$(1)$</span> as</p>
<p><span class="math-container">$$Y = f(K, L) = (a \cdot \frac {1} {K^{-\rho}} + (1 - a) \cdot \frac {1} {L^{-\rho}}  )^{1/\rho}\;\;\;(2)$$</span></p>
<p>where, obviously, <span class="math-container">$-\rho &gt;0$</span> if <span class="math-container">$\rho &lt;0$</span>.</p>
<p>Even though we cannot set <span class="math-container">$K$</span> or <span class="math-container">$L$</span> equal to <span class="math-container">$0$</span> to show that the factors are all essential, we can, instead, <strong>take the limit of the production function as <span class="math-container">$K$</span> or <span class="math-container">$L$</span> go to 0.</strong></p>
<p>For example, let's take the limit  of <span class="math-container">$(1)$</span>  as <span class="math-container">$L \rightarrow0$</span> (for a fixed level of <span class="math-container">$K$</span>).</p>
<p>Rewrite <span class="math-container">$(2)$</span> as</p>
<p><span class="math-container">$$Y = f(K, L) = \frac {1}{(a \cdot \frac {1} {K^{-\rho}} + (1 - a) \cdot \frac {1} {L^{-\rho}}  )^{-1/\rho}}\;\;\;\;(3)$$</span></p>
<p>(remember that <span class="math-container">$-\rho$</span> and <span class="math-container">$-1/\rho$</span> are positive numbers).</p>
<p>As <span class="math-container">$L\rightarrow0$</span> then  <span class="math-container">$1/L^{-\rho}\rightarrow \infty$</span>, and the overall denominator of <span class="math-container">$(3)$</span> goes to <span class="math-container">$\infty$</span>. Therefore,  the fraction <span class="math-container">$(3)$</span> representing the production <span class="math-container">$Y$</span>  goes to <span class="math-container">$0$</span>.</p>
<p>That is,  <strong>as <span class="math-container">$L \rightarrow 0$</span> production <span class="math-container">$Y \rightarrow 0$</span>, irrespective of the level of the other factor <span class="math-container">$K$</span>.</strong></p>
<p>A similar, symmetric, argument applies to <span class="math-container">$K$</span>.</p>
<p>Hence, <strong>both factors are essential</strong>, because <strong>any of them going to zero implies that production goes to zero</strong>.</p>
<hr />
<p><span class="math-container">$^1$</span> <sup>And we can clash with the awful mathematical prohibition: “You cannot divide by zero”, the mortal sin in mathematics . </sup></p>

CC BY-SA 4.0
2023-02-05T21:08:20.470Z
That's it. Thank you for the clear and comprehensive answer.
MoreQuestionsThanAnswers
CC BY-SA 4.0
2023-02-05T21:24:54.193Z
You're welcome! 
I'm happy to have been useful.
BakerStreet
CC BY-SA 4.0
2024-03-28T07:43:31.537Z
Hello @BakerStreet, if I understand correctly your argument, it means that if $\rho \in (0,1)$, then we cannot speak about essentiality as $1 / K^{-\rho}$ would tend to $K^{\rho} \xrightarrow{} 0$? So, it would still depend upon the choice of $L$? BUT THEN... Wouldn't this be contradictory with intuition behind redundancy of non-negativity constraints in using CES? From those we know that for $\rho \neq 1$, we would always have both variables non-zero, right? Such as the answers here: https://economics.stackexchange.com/questions/57714/ces-function-lagrangian-or-kuhn-tucker
Athaeneus
CC BY-SA 4.0
2024-03-28T11:46:17.043Z
Hello @Athaeneus I mean a very simple thing: if the exponent is negative we can't set the factors to $0$, as in mathematics it is not possible divide by $0$. But we can take the limit of $Y$ for $K$ and $L$  $\to 0$: we analyse what happens to production when a factor approaches $0$, and we see that also production approaches $0$. In this sense the factors can be said 'essentials '.
BakerStreet
CC BY-SA 4.0
2024-03-28T12:05:01.620Z
But you have shown this for $\rho <0$ (i.e., complements). But with substitutes, this intuition changes as $\rho >0$, then $\lim_{L \to 0} 1/L^{-\rho} \to 0$, because $-\rho$ is negative number. Then, the whole denominator of (3) does not tend to infinity but rather: $a \frac{1}{K^{-\rho}}$... But then... $Y(K,0) \neq 0$, but is some positive number instead, right? So, in case of substitutes, you will never have essentiality? Or do you need to use another trick?
Athaeneus
CC BY-SA 4.0
2024-03-28T13:12:33.550Z
I hope to understand correctly your observation. Yes, I answered the question, that concerned negative exponents only, the question  assumed $\sigma<1$, my answer concerned this case only. Of course, in $\sigma \geq 1$ it could be different, it is another matter. I think you are right, in the latter  case, there is no essentiality.
BakerStreet
CC BY-SA 4.0
2023-02-02T14:15:09.927Z
2023-07-27T16:39:44.707Z
2023-07-27T16:39:44.707Z
BakerStreet
<p>Consider the CES production function:
<span class="math-container">$$Y = f(K, L) = (a \cdot K^\rho + (1 - a) \cdot L^\rho )^{1/\rho}$$</span>
The elasticity of substitution is <span class="math-container">$\sigma = 1/(1 - \rho)$</span>. I remember that, if the elasticity of substitution is less than one, all factors are <em>essential</em>. A factor is essential if zero input gives zero output, i.e. <span class="math-container">$f(K = 0, L &gt; 0) = 0$</span> or <span class="math-container">$f(K &gt; 0, L = 0) = 0$</span>.</p>
<p>How to show that <span class="math-container">$\sigma &lt; 1$</span> implies essentialness?</p>
<p>Setting <span class="math-container">$K = 0$</span> gives:
<span class="math-container">$$Y = ( (1 - a) \cdot L^\rho )^{1/\rho} = (1 - a)^{1/\rho} \cdot L &gt; 0$$</span>
<span class="math-container">$K$</span> does not seem to be essential. Where's my mistake?</p>

<microeconomics><production-function><ces-function>
CES production function: How to show that $\sigma < 1$ implies essentialness?
MoreQuestionsThanAnswers
# CES production function: How to show that $\sigma < 1$ implies essentialness?

(Asked by: username_0 on 2023-01-31)

Consider the CES production function:
$$Y = f(K, L) = (a \cdot K^\rho + (1 - a) \cdot L^\rho )^{1/\rho}$$
The elasticity of substitution is $\sigma = 1/(1 - \rho)$. I remember that, if the elasticity of substitution is less than one, all factors are *essential*. A factor is essential if zero input gives zero output, i.e. $f(K = 0, L > 0) = 0$ or $f(K > 0, L = 0) = 0$.

How to show that $\sigma < 1$ implies essentialness?

Setting $K = 0$ gives:
$$Y = ( (1 - a) \cdot L^\rho )^{1/\rho} = (1 - a)^{1/\rho} \cdot L > 0$$
$K$ does not seem to be essential. Where's my mistake?

**Tags:** microeconomics, production-function, ces-function

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## 1 answers

### Answer by username_1 on 2023-02-02. Score: 4

The problem is that, with an elasticity of substitution $\sigma<1$, in the CES production function we have negative exponents.$^1$

I rewrite your **CES production function:**

$$Y = f(K, L) = (a \cdot K^\rho + (1 - a) \cdot L^\rho )^{1/\rho}.\;\;\;\;\;(1)$$

**Elasticity of substitution $\sigma = 1/(1-\rho) <1$ implies $\rho<0$,** so that the exponents $\rho$ and $1/\rho$ in equation $(1)$ are negative.

In particular, this implies that the factors $K$ and $L$ in $(1)$ are the denominator of a fraction. Therefore, **we cannot set $K=0$ or $L=0$.**

To see it clearly, let’s rewrite $(1)$ as

$$Y = f(K, L) = (a \cdot \frac {1} {K^{-\rho}} + (1 - a) \cdot \frac {1} {L^{-\rho}} )^{1/\rho}\;\;\;(2)$$

where, obviously, $-\rho >0$ if $\rho <0$.

Even though we cannot set $K$ or $L$ equal to $0$ to show that the factors are all essential, we can, instead, **take the limit of the production function as $K$ or $L$ go to 0.**

For example, let's take the limit of $(1)$ as $L \rightarrow0$ (for a fixed level of $K$).

Rewrite $(2)$ as

$$Y = f(K, L) = \frac {1}{(a \cdot \frac {1} {K^{-\rho}} + (1 - a) \cdot \frac {1} {L^{-\rho}} )^{-1/\rho}}\;\;\;\;(3)$$

(remember that $-\rho$ and $-1/\rho$ are positive numbers).

As $L\rightarrow0$ then $1/L^{-\rho}\rightarrow \infty$, and the overall denominator of $(3)$ goes to $\infty$. Therefore, the fraction $(3)$ representing the production $Y$ goes to $0$.

That is, **as $L \rightarrow 0$ production $Y \rightarrow 0$, irrespective of the level of the other factor $K$.**

A similar, symmetric, argument applies to $K$.

Hence, **both factors are essential**, because **any of them going to zero implies that production goes to zero**.

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$^1$ And we can clash with the awful mathematical prohibition: “You cannot divide by zero”, the mortal sin in mathematics .

> **Comments:**
> **username_0**: That's it. Thank you for the clear and comprehensive answer.  
> **username_1**: You're welcome!  I'm happy to have been useful.  
> **username_2**: Hello @BakerStreet, if I understand correctly your argument, it means that if $\rho \in (0,1)$, then we cannot speak about essentiality as $1 / K^{-\rho}$ would tend to $K^{\rho} \xrightarrow{} 0$? So, it would still depend upon the choice of $L$? BUT THEN... Wouldn't this be contradictory with intuition behind redundancy of non-negativity constraints in using CES? From those we know that for $\rho \neq 1$, we would always have both variables non-zero, right? Such as the answers here: https://economics.stackexchange.com/questions/57714/ces-function-lagrangian-or-kuhn-tucker  
> **username_1**: Hello @Athaeneus I mean a very simple thing: if the exponent is negative we can't set the factors to $0$, as in mathematics it is not possible divide by $0$. But we can take the limit of $Y$ for $K$ and $L$  $\to 0$: we analyse what happens to production when a factor approaches $0$, and we see that also production approaches $0$. In this sense the factors can be said 'essentials '.  
> **username_2**: But you have shown this for $\rho <0$ (i.e., complements). But with substitutes, this intuition changes as $\rho >0$, then $\lim_{L \to 0} 1/L^{-\rho} \to 0$, because $-\rho$ is negative number. Then, the whole denominator of (3) does not tend to infinity but rather: $a \frac{1}{K^{-\rho}}$... But then... $Y(K,0) \neq 0$, but is some positive number instead, right? So, in case of substitutes, you will never have essentiality? Or do you need to use another trick?  
> **username_1**: I hope to understand correctly your observation. Yes, I answered the question, that concerned negative exponents only, the question  assumed $\sigma<1$, my answer concerned this case only. Of course, in $\sigma \geq 1$ it could be different, it is another matter. I think you are right, in the latter  case, there is no essentiality.  

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