0
$\begingroup$

Following the work of Lu (1967) (Full text available here!) I got stuck trying to derive the elasticity of substitution between factors. He use the formula developed by Allen, that when the production function is linear and homogeneous is the following:

$$\sigma =\frac{\frac{\partial V}{\partial K}\frac{\partial V}{\partial L}}{V\frac{\partial^2 V}{\partial K\partial L}}$$

The partial derivatives of L and K and the cross second-order partial derivative are the following:

$$(1)\frac{\partial V}{\partial K}=\frac{dY}{dX}=\frac{1}{X}\left ( Y-\alpha X^{-\frac{c}{b}}Y^{\frac{1}{b}} \right )=\frac{1}{K}\left ( V-\frac{\partial V}{\partial L}\cdot L \right )$$ $$(2)\frac{\partial V}{\partial L}=Y-X\frac{dY}{dX}=\alpha X^{-\frac{c}{b}}Y^{\frac{1}{b}}$$ $$(3)\frac{\partial^2 V}{\partial K \partial L}=\frac{\alpha}{bL} X^{-\frac{c}{b}-1}Y^{\frac{1}{b}-1}\left ( X\frac{dY}{dX}-cY \right )$$

And the expected result (the one the author gets) is:

$$\sigma =\frac{b}{1-\frac{cf}{cf'}}$$

Is there anyone who can help me with this? Thanks.

$\endgroup$
1
$\begingroup$

First of all, I think that 'linear and homogeneous' is a typo of 'linearly homogeneous.' Indeed, it can be shown that if the production function $V$ is linearly homogeneous, Allen Elasticity of Substitution between $K$ and $L$ can be expressed as $$ \sigma = \frac{V_K V_L}{V \cdot V_{KL}} $$ by using the fact that $V_K$ and $V_L$ are homogeneous of degree 0, resulting in $$ V_{LL}L + V_{LK}K=0 \\ V_{KL}L + V_{KK}K=0 $$ .

Moreover, by the definitions in the paper (Lu, 1967), note that $$ X \equiv \frac{K}{L} \\ Y \equiv \frac{V}{L} = F(\frac{K}{L}, 1) \equiv f(X) \\ $$

Now, by substituting the results given by the author, we have $$ \sigma = \frac{\bigg[\frac{1}{K}(V-V_LL)\bigg] \bigg[\alpha X^{-\frac{c}{b}}Y^{\frac{1}{b}}\bigg]} {V \cdot \frac{\alpha}{bL} X^{-\frac{c}{b}-1}Y^{\frac{1}{b}-1}\left ( X\frac{dY}{dX}-cY \right )} $$ , or $$ \sigma = \frac{b}{V} \cdot \frac{L}{K} \cdot \frac{(V-V_LL)XY}{X\frac{dY}{dX}-cY} \\ = \frac{b}{V} \cdot \frac{(V-V_LL)Y}{X\frac{dY}{dX}-cY} \\ = \frac{b}{V} \cdot \frac{V-V_LL}{\frac{X}{Y} \frac{dY}{dX}-c} $$

Note that, since $V$ is linearly homogeneous, $$ V-V_LL = V_KK = \frac{dY}{dX}K $$ where the second equality holds for (3.18) in the paper.

Therefore, $$ \sigma = \frac{b}{V} \cdot \frac{\frac{dY}{dX}K}{\frac{X}{Y} \frac{dY}{dX}-c} \\ = b \cdot \frac{\frac{X}{Y} \frac{dY}{dX}}{\frac{X}{Y} \frac{dY}{dX}-c} \\ = \frac{b}{1-c(\frac{X}{Y} \frac{dY}{dX})^{-1}} $$

Finally, note that $$ \frac{X}{Y} \frac{dY}{dX} = \frac{X}{f(X)} \frac{dY}{dX} = \frac{X}{f(X)}f'(X) $$ , and thus we have $$ \sigma = \frac{b}{1-\frac{cf}{Xf'}} $$

(You have a typo on this expression)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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