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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

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  • $\begingroup$ 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 at 6:11
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Given

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

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.

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

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