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I have learned in my class that for the usual simple linear regression model $Y_i=\beta_0 + \beta_1x_i+\epsilon_i$, the estimators are $b_1 = \frac{S_{xy}}{S_{xx}}$ and $b_0 = \bar{y}-b_1\bar{x}$.

Now I am asked to find the estimator for $\beta_0$ of $Y_i = \beta_0 -X_i + \epsilon_i$. Here is what I have:

$Q = \sum (Y_i- \beta_0+X_i)^{2} \Rightarrow \frac{\partial Q}{\partial \beta_0} = -2\sum(Y_i-\beta_0+X_i)$. Setting this equal to zero, I get

$\sum Y_i - nb_0 +\sum X_i = 0 \Rightarrow b_0 = \frac{\sum Y_i +\sum X_i}{n}$.

Is there an error in my work? Thank you.

EDIT: I forgot to post the second problem.

Let $b^{*}_1 = \frac{\sum x_iy_i}{\sum x_i^{2}}$. Find $k_i$ such that $b^{*}_1 = \sum k_iy_i$.

I know that $\sum x_i^{2} - n\bar{x}^{2} = S_{xx}$. So, I rewrite $b^{*}_1 = \frac{\sum x_iy_i}{S_{xx} + n\bar{x}^{2}}$. Then, let $k_i = \frac{x_i}{S_{xx}+n\bar{x}^{2}}$, so $k_iy_i = \frac{x_iy_i}{S_{xx}+n\bar{x}^{2}}$ and so $\sum k_iy_i = \frac{\sum x_iy_i}{S_{xx}+n\bar{x}^{2}} = b^{*}_1$.

I am not sure if this second problem is correct. My thoughts are that $S_{xx}+n\bar{x}^{2}$ is a constant and so I can define my $k_i$ as such.

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    $\begingroup$ I am not sure whether conceptually the problem is reasonable, though the algebra is correct! You can rewrite your answer as $b_0=\overline{Y}+\overline{X}$. $\endgroup$
    – ssane
    Oct 12, 2019 at 14:44
  • $\begingroup$ what do you mean by conceptually reasonable? $\endgroup$
    – mXdX
    Oct 12, 2019 at 14:52
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    $\begingroup$ In practice, I didn't see a reason that one put $\beta_1=-1$, and estimate only intercept. $\endgroup$
    – ssane
    Oct 12, 2019 at 14:57
  • $\begingroup$ I see. Can I ask you about one more problem? I'll edit it into the post. $\endgroup$
    – mXdX
    Oct 12, 2019 at 15:32

1 Answer 1

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The first part of the question: Yes, your algebra is correct, and you can rewrite $b_0=\overline{Y}+\overline{X}.$

The second part of the question: By definition we have that $$b_1^*=\frac{\sum_{i=1}^{n} x_iy_i}{\sum_{i=1}^{n} x_i^2}=\frac{x_1y_1}{x_1^2+...+x_n^2}+...+\frac{x_ny_n}{x_1^2+...+x_n^2}.$$ From the explicit representation above we can easily see that each $y_i$ is multiplied by $k_i=\frac{x_i}{x_1^2+...+x_n^2}$ for $i=1,...,n$. Therefore, $$b_1^*=\sum_{i=1}^{n}k_iy_i.$$

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  • $\begingroup$ Would my answer be equivalent to yours for part 2? $\endgroup$
    – mXdX
    Oct 12, 2019 at 18:30
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    $\begingroup$ @mXdX Yes, it is! You are right that $S_{xx}+n\overline{x}^2$ is constant, but this is only denominator, and the numerator is $x_i$, thereofore the notation $k_i$ is valid. $\endgroup$
    – ssane
    Oct 12, 2019 at 18:40

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