2
$\begingroup$

Given the production function $q = L^\lambda + K^\gamma$, how do we determine the return to scale for different value of $\lambda$ and $\gamma$?

I know we have to determine the homogeneous degree of the function. That is:

$tf(L,K) = t(L^\lambda + K^\gamma)$

$f(tL,tK) = t^\lambda L^\lambda + t^\gamma K^\gamma$

If $\lambda = \gamma$, then $t^\lambda(L^\lambda + K^\gamma)$, then

  • Constant return to scale if $\lambda$ = 1
  • Increasing return to scale if $\lambda$ > 1
  • Decreasing return to scale if $\lambda$ < 1

This seems to be trivial. However,

  1. How would we determine the return to scales if the value for $\lambda$ and $\gamma$ is not equal?
  2. Let's assume $\lambda$ > $\gamma$, what the heck do we need to do to determine the return to scale of the production function? Is it even possible to in the general form? Perhaps we need numerical values for lambda and gamma?
$\endgroup$

1 Answer 1

2
$\begingroup$

Homogeneity of a function is a sufficient but not necessary condition to determine its returns to scale (RTS), as discussed in this post.

According to the definition of RTS,
\begin{align} \text{$F$ has increasing RTS} \quad\Leftrightarrow\quad F(aK,aL) > aF(K,L), \quad a>1 \\ \text{$F$ has decreasing RTS} \quad\Leftrightarrow\quad F(aK,aL) < aF(K,L), \quad a>1 \end{align} Applying this definition to $F(K,L)=L^\lambda + K^\gamma$, and noting that $\lambda>\gamma$, we have \begin{align} F(aK,aL) & = (aL)^\lambda + (aK)^\gamma \\ & = a^\lambda L^\lambda + a^\gamma K^\gamma &\begin{cases} > a^\gamma (L^\lambda + K^\gamma) = a^\gamma F(K,L) \\[6pt] < a^\lambda (L^\lambda + K^\gamma) = a^\lambda F(K,L) \end{cases} \end{align} Thus, \begin{align} \gamma&\ge1 \quad\Rightarrow\quad \text{$F$ has increasing RTS} \\ \lambda&\le1 \quad\Rightarrow\quad \text{$F$ has decreasing RTS} \end{align} Otherwise the returns to scale is inconclusive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.