I have the regression $y= \beta_0 + \beta_1 \,x + e$, along with the standard error of $\beta_1$

I would like to find the standard error of the elasticity at $\bar{x},\bar{y}$, which is given by $\beta_1 \, \bar{x}/\bar{y}$

Is that simply $\text{SE}(\beta_1)\cdot (\bar{x}/\bar{y})$

  • 2
    $\begingroup$ I think it's worth clarifying whether you are regarding $(\bar{x}, \bar{y})$ as a fixed, known point that you want to estimate the elasticity at, or whether you are really interested in elasticity at $(\mu_X, \mu_Y)$ - in which case $(\bar{x}, \bar{y})$ is only an estimate, and uncertainty in that estimate should contribute to the standard error. $\endgroup$ – Silverfish Feb 19 '15 at 10:16
  • $\begingroup$ @silverfish. Good question. I actually don't know. Is it a more common research practice to evaluate elasticities at a fixed point (here the sample means) or at the estimate of the population means? $\endgroup$ – Zslice Feb 20 '15 at 4:40
  • $\begingroup$ @Zslice that Q in your comment is more of a subject-area-knowledge question rather than a stats/ML question. $\endgroup$ – Glen_b -Reinstate Monica Jun 15 '15 at 7:20
  • $\begingroup$ @Zslice: why don't you estimate a constant elasticity function: $y=\beta_1 x ^ {\beta_2}$ or equivalently $log(y)=log(\beta_1) + \beta_2 log(x)$ ? In that case $\beta_2$ is the elasticity of y with respect to x and you find the SE on it from the regression. $\endgroup$ – user83346 Aug 16 '15 at 16:20
  • $\begingroup$ @fcoppens I couldn't take the log of y and x because they both had values less than 0. $\endgroup$ – Zslice Aug 17 '15 at 19:27

Here, you can apply the Delta Method. Denote $\omega^2$ as the asymptotic variance of $\hat{\beta}$. Then, for the regression coefficients holds $\sqrt{n}(\hat{\beta} - \beta) \xrightarrow{d} N(0, \omega^2)$. The statement of the Delta Method is that if you transform an estimator by a function $g$, the following property holds:

$\sqrt{n}(g(\hat{\beta}) - g(\beta)) \xrightarrow{d} N(0, \omega^2g'(\beta)^2)$. Where $g'$ denotes the first derivative of $g$.

This implies $V[g(\hat{\beta})] = V[\hat{\beta}] \cdot g'(\hat{\beta})^2$.

In your case, $g(\hat{\beta_1}) = \hat{\beta_1} \cdot (\bar{x}/\bar{y})$ and $g'(\hat{\beta_1}) =(\bar{x}/\bar{y})$. Hence, your standard error is $SE[\hat{\beta_1} \cdot (\bar{x}/\bar{y})] = \sqrt{(\bar{x}/\bar{y})^2 \cdot V[\hat{\beta_1}]} = \sqrt{(\bar{x}/\bar{y})^2} \cdot SE[\hat{\beta_1}] $.

Note: What I defined as asymptotic variance for introducing the Delta Method is not equal to the variance in $\hat{\beta} \sim N(\beta, \sigma^2 \cdot (X'X)^{-1}$) which is the correct distribution.

  • $\begingroup$ The $\bar y$ is negative in my data. Would my standard error be $|\bar x / \bar y| * \omega/ sqrt(n)$ $\endgroup$ – Zslice Feb 19 '15 at 1:31
  • $\begingroup$ Yes, because the Delta Method finds the variance of the function. The standard error is the square root of the variance. Since the variance is positive, the square root is well defined. I also edited my answer in order to cover the case with negative values. Thanks for pointing that out. $\endgroup$ – random_guy Feb 19 '15 at 8:58
  • $\begingroup$ Please note, I edited my answer to make it more clear. I just wanted to introduce this delta method about after rereading I reckoned this could be misleading. After all, your suggestion in your question was already right. $\endgroup$ – random_guy Feb 19 '15 at 9:49

Your answer is correct: $\sigma_{\beta_1}\frac{\bar x}{\bar y}$.

Here your $\bar x$ and $\bar y$ are given (not random), unless you used bar accent to denote the sample means of $x,y$ (which is common in physics, but not economics).

To those who are wondering what's elasticity: it's the sensitivity of a percentage change over a percentage change. For instance, if $y$ is demand, and $x$ is income, then income elasticity of demand: $$\frac{\frac{\Delta y}{y}}{\frac{\Delta x}{x}}\approx\frac{x}{y}\frac{dy}{dx}$$ When you plug the the linear regression and the derivatives you get $$\beta_1\frac{x}{y}$$


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