# Least squares estimator for different model

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.

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

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, $$b_1^*=\sum_{i=1}^{n}k_iy_i.$$
• @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. Oct 12, 2019 at 18:40