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Expert-verified Found in: Page 395 ### 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 # 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:

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

See the step by step solution

## 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. ### Want to see more solutions like these? 