Elasticity of the factor share

Question

My question concerns exercise 2.4 from Hal Varian's Microeconomic Analysis.

Please note that I am not requesting a solution to the exercise, and only asking how to understand what I am being asked to do here.

2.4. Let $f(x_1, x_2)$ be a production function with two factors and let $w_1$ and $w_2$ be their respective prices. Show that the elasticity of the factor share $(w_2 x_2 / w_1 x_1)$ with respect to $(x_1 / x_2)$ is given by $1/\sigma - 1$.

Here $\sigma$ is defined earlier as the elasticity of substitution $$\frac{\text{TRS}}{x_2 / x_1}\frac{d(x_2 / x_1)}{d \text{TRS}}$$

where TRS is the technical rate of substitution $\frac{-\partial f / \partial x_1}{\partial f / \partial x_2}$.

The text has not yet defined the factor share or its elasticity yet, but in general, I interpret the phrase "the elasticity of $A$ with respect to $B$" as referring to the quantity $\frac{B}{A}\frac{dA}{dB}$. It is not hard to show using the chain rule that this quantity is equal to $d (\ln A) / d (\ln B)$.

Using the first definition, I take $A = w_2 x_2 / w_1 x_1$ and $B = x_1 / x_2$. Then $A = \frac{w_2}{w_1}\frac{1}{B}$, and $$\frac{B}{A}\frac{dA}{dB} = \frac{B}{A}\frac{w_2}{w_1} \frac{-1}{B^2}= -1$$

Using the second definition, I get the same result. We know $$\ln A = \ln (w_2 x_2 / w_1 x_1 ) = \ln (w_2 / w_1) - \ln (x_1 / x_2) = \text{const.} - \ln B $$ Differentiating both sides with respect to the symbol $\ln B$ yields $d (\ln A) / d (\ln B) = -1$.

Is my math wrong, or have I misunderstood the question?

Answer 1

Your math is not wrong.

To get to the correct solution, you should not treat $(w_2/w_1)$ as a constant. If $x_1/x_2$ changes then the prices will have to change to keep the factor shares optimal for a given level of output (I know that the question is not clear in this respect).

Use the first order conditions $\dfrac{w_2}{w_1} = \dfrac{\partial f(x_1,x_2)/\partial x_2}{\partial f(x_1, x_2)/\partial x_1}= -\dfrac{1}{TRS}$ to subsitute out.