# 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. Commented Feb 5, 2023 at 21:08
• You're welcome! I'm happy to have been useful. Commented Feb 5, 2023 at 21:24
• 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/… Commented Mar 28 at 7:43
• 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 '. Commented Mar 28 at 11:46
• 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? Commented Mar 28 at 12:05