# CES production function: How to show that $\sigma < 1$ implies essentialness?

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?

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 😊.

• That's it. Thank you for the clear and comprehensive answer. Feb 5 at 21:08
• You're welcome! I'm happy to have been useful. Feb 5 at 21:24