I'm struggling to understand how Khan & Reinhart (1990) go from the next production function.

$$y=A f(K,L,Z)$$

Where $y$ is the production of the economy, $A$ is a variable which contains the technological change, $K$ is the capital stock of the economy, $L$ is the labor (work) of the economy, $Z$ is a vector of variables containing usual covariates.

Into the following result as an expression of growth-rates.

$$\frac{dy}{y} = [A \frac{\partial y}{\partial K}]\frac{dK}{y} + [A \frac{\partial y}{\partial L} \frac{L}{y}]\frac{dL}{L} + [A \frac{\partial y}{\partial Z}\frac{Z}{y}]\frac{dZ}{Z} + \frac{dA}{A}$$

The information is that $y,K,L,Z$ are variables (redundant I know but I want to make it clear), however, in the article, they mention $A$ as assumed to grow an exogenous rate (I assume that $A$ is also taken as a variable in the first equation). They don't make clear anymore transformations, simply "growth terms".

The last expression is mentioned by the authors to be "growth terms" of $y=Af(K,L,Z)$. but I lost it that way. Usually for expressing growth rates I simply use logarithms and differentiate over time. but this approach is not clear to me.

My intuition is that $y$ was calculated with the total differential of the form of $$dy=\frac{\partial y}{\partial K}dK + \frac{\partial y}{\partial L}dL + \frac{\partial y}{\partial A}dA$$

and then it was divided by $y$. However, I never saw the total differentials while I was studying, so I don't know if $A$ is multiplying $f$ and we may need to proceed with the product rule of the derivatives? or what is actually the process to get the result of $dy/y$ as they get.

Reference: Khan, M. S. & Reinhart, C. M. (1990) Private investment and economic growth in developing countries World Development, Volume 18, Issue 1, January 1990, Pages 19-27, DOI: https://doi.org/10.1016/0305-750X(90)90100-C

  • $\begingroup$ I think it is just a typo. All $\partial y/\partial x$ with $x \in \{K,L,Z\}$ should be $\partial f/\partial x$ otherwise the definitions given in the text of $\alpha_1,\alpha_2$ and $\alpha_3$ are also wrong. For example they say $\alpha_1$ is the marginal productivity of capital and then state $\alpha_1 = A \frac{\partial y}{\partial K}$ but clearly when $Y = A f(K,L,Z)$ the marginal productivity of capital is $\frac{\partial Y}{\partial K} = A \frac{\partial f}{\partial K}$. $\endgroup$ Mar 16, 2021 at 6:11

1 Answer 1



$$Y = Af(K,L,Z)$$

it follows that

$$\dot Y = \dot A f + A\frac{\partial f}{\partial K}\dot K + A \frac{\partial f}{\partial L} \dot L + A \frac{\partial f}{\partial Z} \dot Z,$$

where dotted expression are time derivatives and dividing with $Y$ it follows

$$\frac{\dot Y}{Y} = \frac{\dot A}{A} + \left[A\frac{\partial f}{\partial K}\right]\frac{\dot K}{Y}+ \left[A \frac{\partial f}{\partial L} \frac{1}{Y}\right] \dot L + \left[A \frac{\partial f}{\partial Z} \frac{1}{Y}\right]\dot Z ,$$

and dividing and multiplying with $L$ and $Z$ in relevant summands it follows that

$$\frac{\dot Y}{Y} = \frac{\dot A}{A} + \left[A\frac{\partial f}{\partial K}\right]\frac{\dot K}{Y}+ \left[A \frac{\partial f}{\partial L} \frac{L}{Y}\right] \frac{\dot L}{L} + \left[A \frac{\partial f}{\partial Z} \frac{Z}{Y}\right]\frac{\dot Z}{Z} ,$$

and then terms in square brackets are defined as $\alpha_1,\alpha_2$ and $\alpha_3$ being respectively

  1. The marginal productivity of capital $\alpha_1 = A \frac{\partial f}{\partial K}$
  2. The elasticity of output with respect to labor $\alpha_2 = A \frac{\partial f}{\partial L} \frac{L}{Y}$ and
  3. The elasticity of output with respect to other factors $\alpha_3 = A \frac{\partial f}{\partial Z} \frac{Z}{Y}$

In the text the authors write $\partial Y/\partial X$ with $X\in \{K,L,Z\}$ instead of $\partial f/\partial X$ which must be a typo. For example it is hard to convince yourself that with $Y = A f(K,L,Z)$ the marginal productivity of capital is $A \frac{\partial Y}{\partial K}$ and not instead $\frac{\partial Y}{\partial K} = A \frac{\partial f}{\partial K}$.

Obviously the typo is of no consequence since the terms are parameterized for the purpose of estimation.

  • $\begingroup$ Thank you, along with @callculus you both help me a lot. $\endgroup$ Mar 16, 2021 at 16:53

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.