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Let's define elasticity, in a regression setting, as $\frac{\partial \log E(y|u)}{\partial \log x_j }$. How do we relate this definition to the percentage change interpretation, i.e., seeing this elasticity as the approximate change in $E(y|x)$, when $x_j$ increases by 1 percent? I'm reading Wooldridge.

Any help would be appreciated.

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  • $\begingroup$ So you want the percent change of elasticity (which is a percent change)? $\endgroup$
    – DornerA
    May 11, 2016 at 14:13
  • $\begingroup$ @DornerA I've edited the question. I think it now answers your question. $\endgroup$
    – An old man in the sea.
    May 11, 2016 at 16:21
  • $\begingroup$ Are they not the same huber in principle? The calculus version is an exact statement, the 1% version is an (Taylor) approximation of what a 1% change would bring given the derivatives evaluated at 0%. Does this make sense? $\endgroup$
    – Fix.B.
    May 11, 2016 at 17:44
  • $\begingroup$ @Fix.B. could you please elaborate a bit? thanks $\endgroup$
    – An old man in the sea.
    May 11, 2016 at 18:31
  • $\begingroup$ The derivative of $log x$ is $1/x$, which implies that $d log x = (dx)/x$. Therefore $e=d log y / d log x$ is the same as $e=(dy)/y / (dx)/x$. This ratio $e$ is known as the elasticity of $y$ with respect to $x$. We usually express derivatives $f'(x)$ as the change in $f$ when $x$ moves one unit, but $f(x)+ 1 * f'(x)$ is just an approximation (a Taylor approximation) of $f(x+1)$.Therefore saying that a 1% change in $x$ will lead to an $e$ change in $y$ is just an approximation. Its a rate of change that is exact only at the $x$ point, but not exact as you move away from it to $x+0.01*x$. $\endgroup$
    – Fix.B.
    May 11, 2016 at 18:56

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The derivative of $log(x)$ is $1/x$, which implies that $dlog(x)=(dx)/x$. Therefore, $e=dlog(y)/dlog(x)$ is the same as $e=(dy)/y/(dx)/x$. This ratio $e$ is known as the elasticity of $y$ with respect to $x$. We usually express derivatives $f′(x)$ as the change in $f$ when $x$ moves one unit, but $f(x)+1∗f′(x)$ is just an approximation (a Taylor approximation) of $f(x+1)$. Therefore, saying that a 1% change in $x$ will lead to an $e$ change in $y$ is just an approximation. It's a rate of change that is exact only at the $x$ point, but not exact as you move away from it to $x+0.01∗x$.

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