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Consider a regression model:$$y=x_{1}\beta_{1}+x_{2}\beta_{2}+u$$

Now, consider a different regression model:$$y=\frac{x_{1}}{x_{2}}\gamma_{1}+x_{2}\gamma_{2}+v$$

Of course, in the second model, the coefficients are identified because we have not induced linear dependency (the transformation is nonlinear). I have two questions:

1) What does the second model even mean? I mean whenever I think of linear regression, I always think of it as holding the value of an included regressor constant. For instance, in the first model, I would interpret $\beta_{2}$ as the marginal effect on the conditional mean of y by increasing $x_{1}$ by one unit, but holding $x_{2}$ constant. What will this even mean in the second case? If we are holding $x_{2}$ constant, considering changes in the ratio $\frac{x_{1}}{x_{_{2}}}$ is equivalent to considering changes in levels of $x_{1}$ .

2) What is the relationship between these two models?

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    $\begingroup$ I would say that the marginal effect of $x_1$ on the conditional expected value of $y$ in the second model is a function of $x_2$. In the first model, it is not. $\endgroup$ – Dimitriy V. Masterov Dec 2 '15 at 23:55
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I got it after thinking it through. Of course, in both cases, the marginal effect of $x_{1}$ is the same for a given value of $x_{1}$ and $x_{2.}$ The difference is that the second model includes something akin to an interaction term, whereby the marginal effect of $x_{1}$ is allowed to change by value of $x_{2}$.

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    $\begingroup$ If you think of $x_3=x_1/x_2$ then you measure the marginal effect of $x_3$. Note that you can disentangle linear transformations; it's quite feasible to go from say $y=(x_{1}+x_{2})\gamma_{1}+x_{2}\gamma_{2}+v$ back to the first model. The transformations just need to be invertible and two models identifiable and everything works. $\endgroup$ – Glen_b -Reinstate Monica Dec 2 '15 at 23:54
  • $\begingroup$ Kind of strange- we can affect the R2 and fit of the model without any new information... $\endgroup$ – ChinG Dec 3 '15 at 0:02
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    $\begingroup$ You do have new information -- the information about model structure. $\endgroup$ – Glen_b -Reinstate Monica Dec 3 '15 at 0:03

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