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I came across an interview question that asks:

Compare 2 univariate regressions:

  1. $y = \beta'x + \epsilon$
  2. $\frac{y}{x} = \beta'x + \epsilon$.

In which setting do you expect to see a better fit (not looking for exact numbers but some mathematical intuition comparing the fit across both settings)?

I am not sure how to approach this question. Any help will go a long way.

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    $\begingroup$ Any fit is between a model and data. This thus seems like a non-sensical question without saying anything about the data. $\endgroup$ Commented Nov 10, 2020 at 20:14

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The first equation: $$ y = \beta'x + \epsilon $$

represents a linear regression where there is a linear association between $x$ and $y$ with some error $\epsilon$

Taking the 2nd equation:

$$ \frac{y}{x} = \beta'x + \epsilon $$

and multiplying through by $x$ we have:

$$ y = \beta'x^2 + \epsilon x $$

So we can interpret this as a linear regression where the functional form is quadratic and the errors are proportional to $x$

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  • $\begingroup$ Given that the errors are proportional to x, can this be used to model heteroscedasticity? $\endgroup$ Commented Nov 10, 2020 at 9:00
  • $\begingroup$ @user2974951 Yes. For example, you could exctract the residuals from a regression of $y$ on $x^2$, and then regress those, devided by $x$, on $x$. The residuals from the 2nd regression should have constant variance equal to that of $\epsilon$. I'm thinking here about normally distributed $\epsilon$. Also, this may not work so well when $x$ has values close to or equal to zero. Of course, in that situation the initial premise of $\frac{y}{x} = \dots $ also runs into trouble, $\endgroup$ Commented Nov 10, 2020 at 9:23
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    $\begingroup$ @vpy Does this answer your question ? If so please consider marking it as the accepted answer, and if not please let us know why, so it can be improved. $\endgroup$ Commented Dec 5, 2020 at 20:55

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