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Q17E

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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

# a. Rewrite the data in Example 1 with new $$x$$-coordinates in mean deviation form. Let $$X$$ be the associated design matrix. Why are the columns of $$X$$ orthogonal? b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, $$y = {\beta _0} + {\beta _1}x*$$, where $$x* = x - 5.5$$.

(a) The data in mean deviation form can be written as:

X = \left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)

As the sum of all entries of the second column is 0, so the columns of are orthogonal.

(b) The normal equation is \left( {\begin{aligned}4&0\\0&{21}\end{aligned}} \right)\left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right) = \left( {\begin{aligned}9\\{7.5}\end{aligned}} \right). The least-squares line is $$y = \frac{9}{4} + \frac{5}{{14}}\left( {x - 5.5} \right)$$.

See the step by step solution

## Find the mean of the given data

(a) The data in example 1 is: $$\left( {2,1} \right),\left( {5,2} \right),\left( {7,3} \right)$$ and $$\left( {8,3} \right)$$.

Find the mean of the data.

\begin{aligned}\bar x &= \frac{{\sum x }}{n}\\ &= \frac{{2 + 5 + 7 + 8}}{4}\\ &= \frac{{22}}{4}\\ &= 5.5\end{aligned}

So, the mean of the data is $$5.5$$.

## Find the value of $$x*$$

It is given that, $$x* = x - 5.5$$. Find $$x* = x - 5.5$$ for different values of $$x$$ in the given data set.

\begin{aligned}x* = 2 - 5.5\\ = - 3.5\end{aligned}

\begin{aligned}x* = 5 - 5.5\\ = - 0.5\end{aligned}

\begin{aligned}x* = 7 - 5.5\\ = 1.5\end{aligned}

\begin{aligned}x* = 8 - 5.5\\ = 2.5\end{aligned}

So, the data in mean deviation form can be written as,

X = \left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)

## Determine whether $$X$$ is orthogonal

On adding the elements of the second column of $$X$$, the value is 0, so the columns of $$X$$ are orthogonal.

## The General Linear Model

The equation of the general linear model is given 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.

## Find design matrix, observation vector, parameter vector for given data

Write the design matrix and observational vector for the given data points.

Design matrix: X = \left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)

Observational matrix: {\bf{y}} = \left( {\begin{aligned}1\\2\\3\\3\end{aligned}} \right)

And the parameter vector for the given equation is,

{\bf{\beta }} = \left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right)

## Normal equation

The normal equation is given by,

$${X^T}X\beta = {X^T}{\bf{y}}$$

## Find least-squares line

(b)

The normal equation is $${X^T}X\beta = {X^T}{\bf{y}}$$, find $$\beta$$ for the given data by using the normal equation.

\begin{aligned}\left( {{{\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)} \right)\beta & = {\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)^T}\left( {\begin{aligned}1\\2\\3\\3\end{aligned}} \right)\\\left( {\begin{aligned}4&0\\0&{21}\end{aligned}} \right)\beta & = \left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\\\left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right) &= {\left( {\begin{aligned}4&0\\0&{21}\end{aligned}} \right)^{ - 1}}\left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\\ &= \frac{1}{{84}}\left( {\begin{aligned}{21}&0\\0&4\end{aligned}} \right)\left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\end{aligned}

Solve further,

\begin{aligned}\left( {{{\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)} \right)\beta & = \left( {\begin{aligned}{\frac{1}{4}}&0\\0&{\frac{1}{{21}}}\end{aligned}} \right)\left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{\frac{9}{4}}\\{\frac{{7.5}}{{21}}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{\frac{9}{4}}\\{\frac{5}{{14}}}\end{aligned}} \right)\end{aligned}

Substitute the obtained values into $$y = {\beta _0} + {\beta _1}x*$$.

$$y = \frac{9}{4} + \frac{5}{{14}}\left( {x - 5.5} \right)$$

Hence, the required least-squares line is $$y = \frac{9}{4} + \frac{5}{{14}}\left( {x - 5.5} \right)$$.