Is any Gauss-Markov assumption violated by the simple OLS regression transformation? I am creating a simple linear model with the following form:
$$ y_i/x_i = \alpha + \beta x_i + u_i $$
The response variable has different name other than $y/x$, but it is essentially normalized by X as shown above.
So if the response, by definition includes part of the predictor can I regress this response on X, without introducing problems of bias or inefficiency? If no, can I have help with the intuition?
 A: Your model is equivalent to:
$$y_i=\alpha x_i+ \beta x_i^2+u_i x_i .$$
Note $\varepsilon_i = u_i x_i$, such that this model rewrites:
$$y_i = \alpha x_i + \beta x_i^2 + \varepsilon_i .$$
So the two differences with a usual linear regression model are:

*

*you don't have a constant term,

*the errors $\varepsilon_i$ are proportional to the covariates $x_i$.

I don't think that the absence of constant term is a big deal (maybe I'm wrong...), but the other point is.
To have Gauss-Markov theorem, you want the $\varepsilon_i$ to have:

*

*null expectation,

*constant variance,

*null covariance.

In order to have that, you need assumptions on you $x_i$ (which you don't need to have Gauss-Markov theorem for a classical linear regression).
Such assumption could be that the $x_i$ are independent from the $u_i$ and have constant mean and variance. Then you'll have that the $\varepsilon_i$ satisfying the three conditions.
So, as Nick Cox commented, the answer is: it depends.
If you have a fixed design ($x_i$ are constant), then no (the $\varepsilon_i$ won't have constant variance). If you suspect the $u_i$ not to be independent from $x_i$, then no again.
But if the $x_i$ are i.i.d. and independent from the $u_i$, then I think yes.
