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Let $\overline{x}=\frac{1}{n}\left(x_{1}+\cdots+x_{n}\right)$ and $\overline{y}=\frac{1}{n}\left(y_{1}+\cdots+y_{n}\right) .$ Show that the least-squares line for the data $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$ must pass through $(\overline{x}, \overline{y}) .$ That is, show that $\overline{x}$ and $\overline{y}$ satisfy the linear equation $\overline{y}=\hat{\beta}_{0}+\hat{\beta}_{1} \overline{x}$ .[Hint: Derive this equation from the vector equation $\mathbf{y}=X \hat{\boldsymbol{\beta}}+\boldsymbol{\epsilon} .$ Denote the first column of $X$ by $\mathbf{1}$ . Use the fact that the residual vector $\epsilon$ is orthogonal to the column space of $X$ and hence is orthogonal to $\mathbf{1} . ]$

$\sum y-n \hat{\beta}_{0}-\hat{\beta}_{1} \sum x=n \overline{y}-n \hat{\beta}_{0}-n \hat{\beta}_{1} \overline{x}$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 6

Applications to Linear Models

Vectors

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we can write a genetic regression line problem in this way where we have on the first column the wise off the points. Then we have the big Matics X, which is the mother lode of coefficients, where we have in the first column all once and in the second columns the axis of the points that is my exact multiplies the Matics Beetle, which contains the parameters that we want to find a better one and bitter, too. And then we have the air or the residuals now to solve the problem. Let me just write a the metrics one over end. Well, all ones in a row. And we upset. Right now that is one over end. The first column of acts. Well, transport's because we put it as a raw and we're going to use this observation later. So now let's multiply from the left by a because remember that why bar and X bar are the average is off the points and well, when we multiply from the left by A were basically obtaining the means so well different from the left. The previous equation we have a wise equal to X beat up, plus a Absalon. An hour's observed on the left hand side. We know that it's just why bar So the average of the wise and on the right inside will a X is going to be the vector. Well, the first entries one because the average is awful ones. And then the second column is the average of the axis. So just x bar. And then that multiplies the metrics beetle, which is better one bitter too, and then plus a Absalom. And so basically, the point is, we want to show that a absolute zero. And if we do that, the proof is over. So we know that the residuals saw Absalon is or for gonna to the columns off X or other way of writing that is that a genetic column of X. We put it, transports and we multiply by Absalon. We should get zero. But we observed that the first column Oh X is a so a upside down. He's one over end. The first column of Ex Put as a row so transposed the multiplies Absalon and therefore that zero. Therefore from the previous page, we see that why bar is equal to be one plus beetle to x bar, which is we wanted to show

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