First I need to minimize this just like I do with a regular L-S estimates (assuming Gauss-Markov) expect $\sum_{i=1}^n X_i=0$:

$$ \sum_i^n e_i^2=\sum_i^n(y_i-\beta_0+\beta_1x_i)^2 $$

I take the partial deriviates:

$$ -2\sum_i^n(y_i-\beta_0-\beta_1x_i) $$

$$ -2\sum_i^n(y_i-\beta_0-\beta_1x_i)x_i $$

Set equal to zero:

$$ \sum_i^n y_i-n\beta_0-\beta_1\sum_i^n x_i=0 \rightarrow \sum_i^n y_i-n\beta_0=0 \rightarrow n\beta_0=\sum_i^n y_i \rightarrow \beta_0=\bar{y} $$

$$ \sum_i^n x_iy_i-\beta_0x_i-\beta_1x_i^2=0 \rightarrow \beta_1\sum_i^nx_i^2=\sum_i^n x_iy_i \rightarrow \beta_1=\sum_i^n\frac{x_iy_i}{x_i^2}=\sum_i^n\frac{y_i}{x_i} $$

The expected values for these estimates are:

$$ E(\beta_0)=E(\bar(y))=E(\frac{\sum_i^n y_i}{n})=\frac{\sum_i^n E(y_i)}{n}=\frac{\sum_i^n \beta_0+\beta_1 \sum_i^nx_i}{n}=\beta_0+0=\beta_0 $$

$$ var(\beta_0)=var(\bar(y))=var(\frac{\sum_i^n y_i}{n})=\sum_i^n var(\frac{y_i}{n})=\sum_i^n \frac{var(y_i)}{n^2} = \frac{\sigma^2}{n} $$

I think my $\beta_0$ estimates are correct, but I feel like I did something wrong with $\beta_1$

I'd appreciate some guidance

  • 3
    $\begingroup$ The last equality for $\beta_1$ does not simplify like that. Be careful with the summation rules. $\endgroup$
    – JohnK
    Commented Sep 20, 2015 at 8:39
  • $\begingroup$ Although your question refers to a "change," what is your reference? To what are you comparing these estimates? $\endgroup$
    – whuber
    Commented Sep 20, 2015 at 13:10
  • $\begingroup$ I'm comparing to a regular L-S estimate where you don't assume $\sum_i^n X_i=0$ $\endgroup$ Commented Sep 20, 2015 at 14:04
  • 1
    $\begingroup$ I still don't understand, because to make least-squares fits one knows the values of the $x_i$. Could you explain what it would mean, then, to "assume" the $X_i$ sum to zero? Are you supposing that the $x_i$ are iid draws from some zero-mean distribution? $\endgroup$
    – whuber
    Commented Sep 21, 2015 at 14:35

1 Answer 1


Thanks I believe I got it.

$$ \sum_i^n x_iy_i-\beta_0x_i-\beta_1x_i^2=0 \rightarrow \beta_1x_i^2=\sum_i^n x_iy_i \rightarrow \beta_1=\frac{\sum_i^n x_iy_i}{\sum_i^n x_i^2} $$

$$ E(\beta_1)=\frac{\sum_i^n x_iE(y_i)}{\sum_i^n x_i^2}=\frac{\sum_i^n x_i(\beta_0+\beta_1x_i)}{\sum_i^n x_i^2}=\frac{\beta_0\sum_i^n x_i}{\sum_i^n x_i^2}+\frac{\beta_1\sum_i^n x_i^2}{\sum_i^n x_i^2}=\beta_1 $$

$$ var(\beta_1)=\sum_i^n var(c_iy_i)=\sum c_i^2\sigma^2=\sigma^2 \frac{\sum_i^n x_i^2}{(\sum_i^n x_i^2)^2}=\frac{\sigma^2}{\sum_i^n x_i^2} $$

  • 1
    $\begingroup$ Another way of stating the answer to the question in the post is that setting $x^*_i =x_i -\bar{x}$ orthogonalizes the intercept term $\beta_0$ from the remaining coefficients $\beta_i$ for $i=1,2,3 \dots, p$. $\endgroup$ Commented Jan 12, 2018 at 3:10

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