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?


2 Answers 2


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

  • $\begingroup$ Would that not imply (merely) that $\mathbb{E}[\partial \epsilon / \partial x] = 0$? $\endgroup$
    – user17900
    Commented Oct 15, 2019 at 10:09
  • $\begingroup$ 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 $\epsilon$. $\endgroup$
    – heh
    Commented Oct 15, 2019 at 14:49

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


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.

  • $\begingroup$ 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. $\endgroup$
    – heh
    Commented Oct 15, 2019 at 14:47
  • $\begingroup$ @heh Agreed that my original version moved from $\hat{y}$ to $y$ without a justification (edited to correct). $\endgroup$ Commented Oct 16, 2019 at 13:14
  • $\begingroup$ 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. $\endgroup$
    – heh
    Commented Oct 17, 2019 at 14:33

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