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If we have the linear model $Y_i-\bar{Y} =b(X_i - \bar{X}) + \varepsilon_i,\ i=1,..., n$ , then we can find easily an unbiased estimator $b^*$ for $b$ with the least squares method. If $c=(Y_n - \bar{Y})/(X_n - \bar{X})$ is another unbiased estimator, then which one is better to choose? I believe that the least squares estimator is better for the linear model because it minimizes the distance of the observations $Y_i - \bar{Y}$ from regression line, which means that the SSE is smaller. Therefore the SSR is closer to SST, and that means that $R^2$ is closer to 1, which I know is not always what we want, however I believe that it could be one reason that $b^*$ is more preferable than $c$.

Is my thought correct, and if it is, is there any other reason why we choose $b^*$ than $c$, and if it is false, can someone explain why?

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  • $\begingroup$ $Yn = n $ times $Y$ or $Y_n$? $Xn$? What does "the SSR is closer to SST" mean? Assume SSE is sum of square of residual, SSR = sum of square of model (regression) and SST = total sum of square. $\endgroup$
    – user158565
    May 30, 2017 at 22:56
  • $\begingroup$ Is $Y_n$ intended to be the largest $Y$, the $Y$ that goes with the largest $X$, or simply the last $Y$ in a list -- presumably just be some random observation? I'd suggest you compare not the $R^2$ but the variance of the two estimators of the slope$ $\endgroup$
    – Glen_b
    May 31, 2017 at 5:50
  • $\begingroup$ @a_stastician By Yn I mean just the n-th observation . When I say is closer to SST , I mean while the SSE is smaller and SST is constant we have that SSR is larger so SSR is ''closer'' to SST . $\endgroup$
    – evans
    May 31, 2017 at 7:56
  • $\begingroup$ @Glen_b Y_n is the n-th observation if we think that we have n observation of a variable (maybe I would make it a little confusing now but in my question I wrote Y as the sum of Y_i divided by n, not the variable Y) . So for your answer , it would be better if the variance is smaller because if for example is significant smaller than the variance of c it would more likely reject the null hypothesis (b=0) with a t-statistical test ? Thank you both for your time ! $\endgroup$
    – evans
    May 31, 2017 at 8:03
  • $\begingroup$ I'm not sure what you're saying there; I wasn't talking about significance at all, merely precision. Conventionally one would write the random variable representing the average of the variables $Y_1$... $Y_n$ as $\bar{Y}$ $\endgroup$
    – Glen_b
    May 31, 2017 at 8:24

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