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I have the linear regression model:

$Y_i= \alpha_0 + \alpha_1(X_i - \overline{X})$

Anyway I got through the method for find the least square estimator for $\alpha_0$ and end up with

$\sum_{i=1}^n \frac{\alpha_1(X_i-\overline{X})}{n}$

as part of my answer for $\alpha_0$ and I'm not sure how to compute this.

Ian
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John Meighan
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  • Unless $\alpha_1$ is given, what you've written is not an estimator at all, since you can't compute it by knowing only the sample. Perhaps you can write the details of what you've done, so that we can explain where you made a mistake and/or how you need to continue from where you left off. – Ian Nov 01 '15 at 17:51
  • @ian your right I meant to right $\alpha_1\hat$ also this isn't my whole formula for my $\alpha_0\hat$ just the part I'm stuck on – John Meighan Nov 01 '15 at 17:56
  • The thing is, $\hat{\alpha}_0$ and $\hat{\alpha}_1$ are inextricably linked; you essentially cannot compute one without computing the other. You might think about the simpler case where $Y_i=\alpha_0 + \alpha_1 X_i$ where the $X_i$ are independent and identically distributed. In this case you need to solve the normal equations, which look like $X^T X \begin{bmatrix} \alpha_0 \ \alpha_1 \end{bmatrix} = X^T Y$, where $X$ has a column of all $1$s and a column containing the sample $x_i$ and $Y$ is a column vector containing the sample $y_i$. – Ian Nov 01 '15 at 17:57
  • I think your case is slightly more complicated because $X_i-\overline{X}$ are not independent variables. – Ian Nov 01 '15 at 18:00
  • @Ian I used the method of least squares to estimate $\alpha_0$ starting from $\sum_i=(Yi-Yi\hat)^2$ and got to $\alpha_0$ = $Y\bar$ - $\sum_{i=1}^n \frac{\alpha_1(X_i-\overline{X})}{n}$ – John Meighan Nov 01 '15 at 18:02
  • I understand, you want to minimize $\sum_{i=1}^n (y_i-\hat{Y}i)^2=\sum{i=1}^n (y_i - \alpha_0 - \alpha_1 (x_i-\overline{x}))^2$. But when you do that, you should get something along the lines of the normal equations, possibly with a correction to account for the fact that the $\hat{Y}_i$ are not independent. – Ian Nov 01 '15 at 18:03
  • Well I have gotten essentially the same thing but instead of $\sum_{i=1}^n \frac{\alpha_1(X_i})}{n}$ = $X\bar$ I have $\sum_{i=1}^n \frac{\alpha_1(X_i-\overline{X})}{n}$ – John Meighan Nov 01 '15 at 18:08

1 Answers1

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It's the usual:

$\begin{array}\\ \sum_{i=1}^n \frac{\alpha_1(X_i-\overline{X})}{n} &=\sum_{i=1}^n \frac{\alpha_1(X_i)}{n}-\sum_{i=1}^n \frac{\alpha_1(\overline{X})}{n}\\ &=\alpha_1\sum_{i=1}^n \frac{X_i}{n}-\alpha_1\sum_{i=1}^n \frac{\overline{X}}{n}\\ &=\alpha_1\overline{X}-\alpha_1\overline{X}\\ &=0 \end{array} $

marty cohen
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