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Q7.5-2E

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Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

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

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Short Answer

Question: In Exercises 1 and 2, convert the matrix of observations to mean deviation form, and construct the sample covariance matrix.

\(2.\,\,\left( {\begin{array}{*{20}{c}}1&5&2&6&7&3\\3&{11}&6&8&{15}&{11}\end{array}} \right)\)

The mean deviation form and covariance matrix are:

\(\begin{array}{l}B = \left( {\begin{array}{*{20}{c}}{ - 3}&1&{ - 2}&2&3&{ - 1}\\{ - 6}&2&{ - 3}&{ - 1}&6&2\end{array}} \right)\\\\S = \left( {\begin{array}{*{20}{c}}{5.6}&8\\8&{18}\end{array}} \right)\end{array}\)

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Mean Deviation form and Covariance Matrix

The Mean Deviation form of any \(p \times N\) is given by:

\(B = \left( {\begin{array}{*{20}{c}}{{{\hat X}_1}}&{{{\hat X}_2}}&{........}&{{{\hat X}_N}}\end{array}} \right)\)

Whose \(p \times p\) covariance matrix is:

\(S = \frac{1}{{N - 1}}B{B^T}\)

Step 2: The Mean Deviation Form

As per the question, we have a matrix:

\(\left( {\begin{array}{*{20}{c}}1&5&2&6&7&3\\3&{11}&6&8&{15}&{11}\end{array}} \right)\)

The sample mean will be:

\(\begin{array}{c}M = \frac{1}{6}\left( {\begin{array}{*{20}{c}}{24}\\{54}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4\\9\end{array}} \right)\end{array}\)

The Mean Deviation Form is given by:

\(\begin{array}{c}B = \left( {\begin{array}{*{20}{c}}{1 - 4}&{5 - 4}&{2 - 4}&{6 - 4}&{7 - 4}&{3 - 4}\\{3 - 9}&{11 - 9}&{6 - 9}&{8 - 9}&{15 - 9}&{11 - 9}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 3}&1&{ - 2}&2&3&{ - 1}\\{ - 6}&2&{ - 3}&{ - 1}&6&2\end{array}} \right)\end{array}\)

Hence, this is the required answer.

Step 3: The Covariance Matrix.

Now, the covariance matrix will be:

\(\begin{array}{c}S = \frac{1}{{6 - 1}}B{B^T}\\ = \frac{1}{5}\left( {\begin{array}{*{20}{c}}{ - 3}&1&{ - 2}&2&3&{ - 1}\\{ - 6}&2&{ - 3}&{ - 1}&6&2\end{array}} \right){\left( {\begin{array}{*{20}{c}}{ - 3}&1&{ - 2}&2&3&{ - 1}\\{ - 6}&2&{ - 3}&{ - 1}&6&2\end{array}} \right)^T}\\ = \frac{1}{5}\left( {\begin{array}{*{20}{c}}{28}&{40}\\{40}&{90}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{5.6}&8\\8&{18}\end{array}} \right)\end{array}\)

Hence, this is the required matrix.

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