# Linear regression when dividing the dependent variable by the independent variable

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.

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

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 $$x$$

• Given that the errors are proportional to x, can this be used to model heteroscedasticity? Nov 10, 2020 at 9:00
• @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, Nov 10, 2020 at 9:23
• @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. Dec 5, 2020 at 20:55