An exercise asked to obtain properties of the lineal model $$E[y_i]=\beta x_i\qquad i=1,\cdots,n$$

where $Var[y_i]=\sigma^2$. In one of its sections, we had to calculate and estimator for $\beta$ using the least squares method, and I got

$$\hat{\beta}=\frac{\sum_{i=1}^n x_iy_i}{\sum_{i=1}^n x_i^2}$$

which is quite similar to the estimator $\hat{\beta_1}$ obtained with the same method for the linear model $E[y_i]=\beta_0 +\beta_1x_i$, that is

$$\hat{\beta_1}=\frac{\sum_{i=1}^ny_i(x_i-\bar{x})}{\sum_{i=1}^n(x_i-\bar{x})^2}$$ since is the same one with the exception that in the expression for $\hat{\beta}$,the term $\bar{x}$ does not appear. The same thing happened when I was asked to compute $Var[\hat{\beta}]$ and when finding the pivotal quantity for a confidence interval for $\beta$. My question is then the following

Is there an intuitive reason that explains why this happens? That is, could have I known that the term $\bar{x}$ that appears when there is an intercept $\beta_0$ vanishes by knowing only that the regression line is forced to pass throug the origin?

This is not homework, I already solved the exercise, but I'm trying to gain some intuition on why this things are like they are. Any help and answer will be highly appreciate.

  • $\begingroup$ Could you explain what you mean by the "term $\bar x$ ... vanishes"? I do not see the sense in which $\hat\beta$ is the same as either of $\hat\beta_0$ or $\hat\beta_1$, nor exactly how one "lets" $\bar x=0$ in any of those estimators. You seem to have in mind some particular formulas involving $\bar x$ that you haven't disclosed. $\endgroup$ – whuber Mar 31 '16 at 17:05
  • $\begingroup$ @whuber You are right, my words were not the appropiate ones. I've edited the question and I hope it is clearer now. $\endgroup$ – user314159 Mar 31 '16 at 17:17
  • $\begingroup$ I don't think it's prudent to think of the linear model without an intercept to be conceptually different from the one with an intercept. Afterall the intercept is just $\hat{\beta_0}=\overline{y}-\beta\overline{x}$, which can be made to equal 0 by shifting $y_i\rightarrow y_i-\bar{y}$ and rescaling $x_i\rightarrow x_i-\bar{x}$. $\endgroup$ – Alex R. Mar 31 '16 at 18:00
  • $\begingroup$ I have addressed this issue in various different ways, including answers at stats.stackexchange.com/a/113207 , stats.stackexchange.com/a/46508 , and stats.stackexchange.com/a/197788 . $\endgroup$ – whuber Mar 31 '16 at 18:09
  • $\begingroup$ @whuber Thank you! Those are all amazing answers. I'll check them, though somethings treated there might be beyond my knowledge. $\endgroup$ – user314159 Mar 31 '16 at 19:26

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