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?


1 Answer 1


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.

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

  • $\begingroup$ That's it. Thank you for the clear and comprehensive answer. $\endgroup$ Commented Feb 5, 2023 at 21:08
  • $\begingroup$ You're welcome! I'm happy to have been useful. $\endgroup$ Commented Feb 5, 2023 at 21:24
  • $\begingroup$ 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: economics.stackexchange.com/questions/57714/… $\endgroup$
    – Athaeneus
    Commented Mar 28 at 7:43
  • $\begingroup$ 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 '. $\endgroup$ Commented Mar 28 at 11:46
  • $\begingroup$ 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? $\endgroup$
    – Athaeneus
    Commented Mar 28 at 12:05

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