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

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?

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

• Would that not imply (merely) that $\mathbb{E}[\partial \epsilon / \partial x] = 0$? – user17900 Oct 15 '19 at 10:09
• When it comes to the error term, the moments are the only things you can reason meaningfully with. In the context of regression analysis, once you're here: $$\implies\frac{1}{ y}\frac{\partial{y}}{\partial{x}} = \beta \frac{1}{x} + \frac{\partial{\varepsilon}}{\partial{x}}$$, take the expected value of both sides and rely on your assumptions about the distribution of $\epsilon$. – heh Oct 15 '19 at 14:49

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.

• But those "difficulties" are precisely why the question is of any practical interest, so I'm not sure dropping the error term is the right approach. Your model should have $\hat{\epsilon}$ in it. And I think there's an abuse of notation here, as you moved from $\hat{y}$ to $y$ without taking an expected value - which is the key operation, as given a well-behaved error term, that is what makes the offending derivative vanish. – heh Oct 15 '19 at 14:47
• @heh Agreed that my original version moved from $\hat{y}$ to $y$ without a justification (edited to correct). – Adam Bailey Oct 16 '19 at 13:14
• Your estimator for the error really needs to stick around, though. The true model would be what you've presented without the "hats", but again - doing this analysis on the true model is just algebra. There is no need to get philosophical about differentiation of a random variable provided one uses the necessary assumptions about the error's distribution. – heh Oct 17 '19 at 14:33