1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ In your notation, $x_i(\beta + \epsilon_i) \neq x_i \beta + \epsilon_i$ $\endgroup$
    – Jon
    May 16, 2019 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 $\endgroup$ May 16, 2019 at 16:36

1 Answer 1

-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.