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Q14E

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Found in: Page 331

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# 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 $$\left( {\overline x ,\overline y } \right)$$. That is, show that $$\overline x$$ and $$\overline y$$ satisfies the linear equation $$\overline y = {\hat \beta _0} + {\hat \beta _1}\overline x$$. (Hint: Derive this equation from the vector equation $${\bf{y}} = X{\bf{\hat \beta }} + \in$$. Denote the first column of $$X$$ by 1. Use the fact that the residual vector $$\in$$ is orthogonal to the column space of $$X$$ and hence is orthogonal to 1.)

It is verified that, $$\bar y$$ and  satisfies the linear equation $$\bar y = {\beta _0} + \bar x{\beta _1}$$.

See the step by step solution

## The General Linear Model

The equation of the general linear model is defined as:

$${\bf{y}} = X\beta + \in$$

Here, {\bf{y}} = \left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right) is an observational vector, X = \left( {\begin{aligned}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right) is the design matrix, \beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right) is parameter vector, and \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right) is a residual vector.

## Determine whether $$\overline x$$ and $$\overline y$$ satisfies the given linear equation

Write the design matrix as X = \left( {\begin{aligned}1&{\bf{x}}\end{aligned}} \right).

As the equation of the general linear model is given by $${\bf{y}} = X\beta + \in$$. Isolate residual vector from the equation.

$$\in = {\bf{y}} - X{\bf{\hat \beta }}$$

As the residual vector is orthogonal to columns of $$X$$, so $$1 \cdot \in = 0$$.

\begin{aligned}1 \cdot \left( {{\bf{y}} - X{\bf{\hat \beta }}} \right) = 0\\{1^T}{\bf{y}} - \left( {{1^T}X} \right){\bf{\hat \beta }} = 0\\\left( {\begin{aligned}1\\1\\ \vdots \\1\end{aligned}} \right)\left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right) - {\left( {\begin{aligned}1\\1\\ \vdots \\1\end{aligned}} \right)^T}\left( {\begin{aligned}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{aligned}} \right) \cdot {\bf{\hat \beta }} = 0\\\left( {{y_1} + {y_2} + \cdots + {y_n}} \right) - \left( {\begin{aligned}{1 + 1 + \ldots n{\rm{ times}}}&{{x_1} + {x_2} + \cdots + {x_n}}\end{aligned}} \right) \cdot {\bf{\hat \beta }} = 0\\\sum y - \left( {\begin{aligned}n&{\sum x }\end{aligned}} \right)\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right) = 0\\\sum y - \left( {n{\beta _0} + \sum x {\beta _1}} \right) = 0\end{aligned}

Write $$\sum y$$ as $$n\bar y$$ and $$\sum x$$ as $$n\bar x$$.

\begin{aligned}n\bar y - \left( {n{\beta _0} + n\bar x{\beta _1}} \right) = 0\\n\bar y - n{\beta _0} - n\bar x{\beta _1} = 0\\\bar y - {\beta _0} - \bar x{\beta _1} = 0\\\bar y = {\beta _0} + \bar x{\beta _1}\end{aligned}

This implies that $$\bar y$$ and $$\bar x$$ satisfies the linear equation $$\bar y = {\beta _0} + \bar x{\beta _1}$$.