1
$\begingroup$

I don't understand its importance in this production function.

I know that the productivity parameter is A in the function:

$F(K,L) = AK^α L^{(1-α)} $

So what could it be?

Thank you!

$\endgroup$
1
  • 2
    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Giskard
    Sep 8, 2022 at 7:08

2 Answers 2

0
$\begingroup$

The parameter $A$ typically refers to total factor productivity which is a measure of how much output is created per unit of input.

If you rearrange your equation you can see this more clearly:

$$A=\frac{F(K,L)}{K^aL^{1-a}}$$

Naturally as $A$ grows so will your output. In "words", $A$ measures the portion of your production that $K^aL^{1-a}$ does not explain.

$\endgroup$
1
  • 2
    $\begingroup$ Hi, thankyou for your answer! However, I don't think that A corresponds to the marginal productivity of a factor. Am I wrong? $\endgroup$
    – student
    Sep 8, 2022 at 14:14
0
$\begingroup$

The marginal output in all production functions explains the degree of increase in production following (1) the addition of one more unit of labor, (2) an additional unit of capital, or (3) a productivity gain of one of the production factors - the parameter A.

Generally, it is assumed that the returns are constant, so the production function can be rewritten as:

$Y=F(AK, AL) = AF(K,L) $

meaning that if you double the quantity of input, the quantity of output doubles as well.

The expression $F(AK)$ refers the efficient capital while $F(AL)$ refers to the efficient labor. You can see that if you increase the efficiency of one factor, it increases the output too. To know at which rate, you got to measure the marginal productivity :

$\frac{\partial Y}{\partial AL}$ or $\frac{\partial Y}{\partial AK}$

$\endgroup$

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.