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If have a simple bivariate regression model:

$ Y_i= x_i \beta + \epsilon_i $

where $i$ are the number of observations.

How do I test for the hypothesis that the OLS coefficient $\beta$ does not change across observations??

Another Question is, if instead i have the model: $ Y_i= x_i (\beta + \epsilon_i) $

then, the OLS estimator will still be consistent right?

My logic:

$\hat\beta$ = $(X'X)^{-1}X'Y$

$\hat\beta$ = $(X'X)^{-1}X'(X (\beta + \epsilon)) $

$\hat\beta$ = $\beta + (X'X)^{-1}X'(X\epsilon)) $

Now, if $E(\epsilon \vert X)$ = 0;

taking probability limits, can we say that,

$E(X'X\epsilon)$ = $E_x(E(X'X\epsilon \vert X))$

$E(X'X\epsilon)$ = $E_x(X'XE(\epsilon \vert X))$

$E(X'X\epsilon)$ = $0$

Are we allowed to do that?

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    $\begingroup$ In your notation, $x_i(\beta + \epsilon_i) \neq x_i \beta + \epsilon_i$ $\endgroup$ – Jon May 16 '19 at 16:27
  • $\begingroup$ @Jon Yes, that's why I have conditional expectations with regard to $X'X \epsilon$ and not $X' \epsilon$ Is that wrong way? How should I proceed? Also, there are two parts to the question, the first part has a normal OLS bivariate model $\endgroup$ – Raghav Goyal May 16 '19 at 16:36
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You can search for random slope (or random coeficient) models, but this is normaly done with panel data, with a $X$ variable that varies through time for each individual. Otherwise, in your simple model, if you don't put any constraint on your coefficients, you will just end up saying that each $Y$ is 100% determined by $X$.

This applies to your second model as well. It assumes that there is no error term beside the error on the coefficient. Therefore, it just assumes that $X(\beta+\epsilon_i)$ completely determines $Y$, so nobody would ever use this to try to estimate individual-specific coefficients.

But if this is still the model you want to assume, then yes OLS gives a consistent estimate of the average effect $\beta$. It just means you have heterosckedasticity, so you should take this into account for inference.

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