I am trying to analyse the impact of a trade policy on exports. My dependent variable for the main specification is log of exports but as a robustness check, I want to include a linear transformation that is square root of exports (sqrtexp). On running regressing sqrtexp on a binary independent variable, I get a beta coefficient of -1.197. I have read another post on this about using the chain rule to interpret it. In my case, using the chain rule, change in y = 2(b1)^2 which equals 2.87 and implies, for X=1, the exports will change by $2.87. But still that doesn't make sense since its too small a change for exports that are for some products as high as in millions, which means I'm still interpreting the coefficient incorrectly. I'll be grateful for any insights on this!
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$\begingroup$ I am not sure what you mean by "chain rule"? But if you want to use derivatives, for $\log(y)$ the response is proportional to $y$, because $\partial{\log(y)}=\partial{y}/y$. Similarly, for $\sqrt{y}$ the response is proportional to $\sqrt{y}$, because $\partial{(\sqrt{y})}=\frac{1}{2}\partial{y}/\sqrt{y}$. The "millions" would come from multiplying the coefficient by this scale. $\endgroup$– GeoMatt22Commented Jul 28, 2020 at 1:00
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$\begingroup$ Am sorry if that was confusing, I was referring to this (stats.stackexchange.com/questions/35982/…) post that talks about using a chain rule for interpretation. I'm simply trying to figure out how to interpret the beta coefficient of a binary variable when the dependent variable Y is in square roots. $\endgroup$– MiliCommented Jul 28, 2020 at 12:15
1 Answer
If you have an equation $$ \sqrt{y} = a + bx $$ Then you can evaluate the effect of changing $x$ from $x_0$ to $x_1$ by $$ \Delta\left(\sqrt{y}\right) = \sqrt{y_1} - \sqrt{y_0} = \left(a + b x_1\right) - \left(a + b x_0\right) = b\left(x_1 - x_0\right) = b\Delta{x} $$ If the change in $y$ is small enough, you can replace the finite differences $\Delta$ with differentials $\partial$, to get an expression for the derivative $dy/dx$. This is where the "chain rule" would come up, i.e. $$ \partial\left(\sqrt{y}\right) = \frac{1}{2}\frac{\partial{y}}{\sqrt{y}} $$
However for a binary $x$, where $x_1=1$ and $x_0=0$, you cannot necessarily do this, because $\Delta{y}=y_1-y_0$ may be too large. Instead you must solve the difference equation algebraically, i.e. if we write $$ \beta = b\Delta{x} = \sqrt{y_1} - \sqrt{y_0} $$ then $$ y_1 = y_0 + 2\sqrt{y_0}\beta + \beta^2 $$ and $$ \Delta{y} = \beta\left(2\sqrt{y_0} + \beta\right) $$ If $\beta = b\Delta{x}$ is small enough, then this reduces to the "chain rule", i.e. $$\lim_{\beta\to{0}}\Delta{y} = 2\beta\sqrt{y_0}$$
For a continous $x$, you can ensure $\beta$ is small enough by considering an infinitesimal $\Delta{x}$. But for a binary $x$, where $\beta=b$, in general you must use the more general form, i.e. $$ \Delta{y} = b\left(2\sqrt{y_0} + b\right) $$ (if $b^2 \ll y_0$ this still reduces to the "chain rule" form).
Note that the scale of the response $\Delta{y}$ is set by $y_0$, which takes care of the "millions" in your case.
(Also note that your formula using $b^2$ does not allow for a negative response, which would also suggest something is wrong there. The correct formula above does not have this problem.)
