Proof coefficient in log-log model is equal to coefficient of elasticity

Question

I am trying to see how we treat $\varepsilon$ in the following proof:

Suppose we have a log-log single variable regression model

$$ \ln(y) = \alpha + \beta \ln(x) + \varepsilon $$

then take partial derivative with respect to $x$ on both sides

\begin{align}&\implies \frac{\partial}{\partial x} \ln(y) = \frac{\partial}{\partial x} (\alpha + \beta \ln(x) + \varepsilon)\\&\implies\frac{\partial \ln(y)}{\partial y} \frac{\partial y}{\partial x} = 0 + \beta \frac{1}{x} + \frac{\partial \varepsilon}{\partial x}\\&\implies\frac{1}{ y} \frac{\partial y}{\partial x} = \beta \frac{1}{x} + \frac{\partial \varepsilon}{\partial x}\end{align}

We want to show that $\beta$ is equal to the elasticity of $y$ with respect to $x$

$$ \beta = \frac{\partial y}{\partial x} \frac{x}{y} $$

But how do we know the $\dfrac{\partial \varepsilon}{\partial x} $ term is zero?

Answer 1

Because $\Bbb E[\varepsilon \mid x]= 0$ is one of the key assumptions for the estimation.

Answer 2

The normal procedure is to estimate the model using sample data for $x$ and $y$, obtaining a fitted regression line:

$$\ln(\hat{y})=\hat{\alpha}+\hat{\beta}\ln(x)$$

You then have a straightforward log-linear relation between $x$ and fitted values of $y$ and can use partial differentiation as in your question, but without the $\varepsilon$ terms, to show that:

$$\hat{\beta}=\frac{\partial \hat{y}}{\partial x}\frac{x}{\hat{y}}$$

If the sample is representative of the population of interest, one can then infer that $\hat{\beta}$ is a good estimate of elasticity in the population.

Trying instead to differentiate the original stochastic model leads into the difficult issue (see here) of whether it makes any sense to differentiate a random variable.