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 kevel 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. $\endgroup$ Feb 5 at 21:08
  • $\begingroup$ You're welcome! I'm happy to have been useful. $\endgroup$ Feb 5 at 21:24

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