# These two estimators are equal?

I have a question so basically.

I have two model $$y= X_1b_1+X_2b_2+u$$ And $$X_2=y\beta_1+X_2\beta_2+v$$

I want to show $$b_1=\beta_1$$

Are these estimators are equivalent?

$$b_1= (X_1’M_1X_1)^{-1} X_1’M_2y$$

And

$$\beta_1= (X_1’M_2y)^{-1} (X_1’M_1X_1)$$

• Seems like $\beta b = I$ – Dayne Dec 7 '20 at 16:59
• Usually the dimensionality of $X_1^\top M_2 y$ would be $K \times 1$ and $X_1^\top M_1 X_1$ would be $K \times K$ so the multiplication $(X_1^\top M_2 y)^{-1}(X_1^\top M_1 X_1)$ does not make sense also the inverse $(X_1^\top M_2 y)^{-1}$ does not make sense. So you should probably try to define the "stuff" you are writing. I suggest to close this question, as it stands it is non-sensical. – Jesper Hybel Dec 7 '20 at 17:06
• As an off chance, you might be thinking of: en.wikipedia.org/wiki/… – Walrasian Auctioneer Dec 7 '20 at 20:41

They are both OLS estimators but generally $$\beta_1 \neq b_1$$.

In statistics broadly speaking estimator is just some rule, method, or criterion for arriving at an estimate of the value of a parameter. According to Stachurski: A Primer in Econometric Theory, pp 217:

An estimator is a statistic used to infer some feature $$\gamma(P)$$ of an unknown distribution $$P$$. Thus a statistic becomes an estimator when paired with and compared to feature of the distribution. Nothe that there is nothing in the definition of an estimator that implies it will be a sensible estimator of the target feature, let alone a good one.

Consequently, by definition if you use the following formulas:

$$b_1= (X_1’M_1X_1)^{-1} X_1’M_2y$$

$$\beta_1= (X_1’M_2y)^{-1} (X_1’M_1X_1)$$

for an inference they become estimators. Moreover, both of these look like OLS estimator. So in that sense they are the same type of estimator.

However, running the same estimator will not give you the same result when you change position of dependent and independent variable (in fact OLS estimator will be biased if there is reverse causality so it does not make any sense to do it in the first place). Consequently generally speaking $$\beta_1 \neq b_2$$.

In fact in your specific case:

$$y= X_1b_1+X_2b_2 + u \implies X_1=y\frac{1}{b_1}-X_2\frac{b_2}{b_1}-\frac{u}{b_1}$$

Consequently, we have:

$$\beta_1 = \frac{1}{b_1}; \beta_2 = -\frac{b_2}{b_1}; v = -\frac{u}{b_1}$$

Trivially since $$\beta_1 = \frac{1}{b_1}$$, $$\beta_1 \neq b_1$$, save for some special cases such as when $$b_1=1$$ for example.