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Q7.5-1E

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.$$1.\,\,\left( {\begin{array}{*{20}{c}}{19}&{22}&6&3&2&{20}\\{12}&6&9&{15}&{13}&5\end{array}} \right)$$

The mean deviation form and covariance matrix are:

$$\begin{array}{l}B = \left( {\begin{array}{*{20}{c}}7&{10}&{ - 6}&{ - 9}&{ - 10}&8\\2&{ - 4}&{ - 1}&5&3&{ - 5}\end{array}} \right)\\\\S = \left( {\begin{array}{*{20}{c}}{86}&{ - 27}\\{ - 27}&{16}\end{array}} \right)\end{array}$$

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:

$$\left( {\begin{array}{*{20}{c}}{19}&{22}&6&3&2&{20}\\{12}&6&9&{15}&{13}&5\end{array}} \right)$$

The sample mean will be:

$$\begin{array}{c}M = \frac{1}{6}\left( {\begin{array}{*{20}{c}}{72}\\{60}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{12}\\{10}\end{array}} \right)\end{array}$$

The Mean Deviation Form is given by:

$$\begin{array}{c}B = \left( {\begin{array}{*{20}{c}}{19 - 12}&{22 - 12}&{6 - 12}&{3 - 12}&{2 - 12}&{20 - 12}\\{12 - 10}&{6 - 10}&{9 - 10}&{15 - 10}&{13 - 10}&{5 - 10}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}7&{10}&{ - 6}&{ - 9}&{ - 10}&8\\2&{ - 4}&{ - 1}&5&3&{ - 5}\end{array}} \right)\end{array}$$

Hence, this is the required answer.

## Step 3: The Covariance Matrix

Now, the covariance matrix will be:

$$\begin{array}{c}S = \frac{1}{{6 - 1}}B{B^T}\\ = \frac{1}{5}\left( {\begin{array}{*{20}{c}}7&{10}&{ - 6}&{ - 9}&{ - 10}&8\\2&{ - 4}&{ - 1}&5&3&{ - 5}\end{array}} \right){\left( {\begin{array}{*{20}{c}}7&{10}&{ - 6}&{ - 9}&{ - 10}&8\\2&{ - 4}&{ - 1}&5&3&{ - 5}\end{array}} \right)^T}\\ = \frac{1}{5}\left( {\begin{array}{*{20}{c}}{430}&{ - 135}\\{ - 135}&{80}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{86}&{ - 27}\\{ - 27}&{16}\end{array}} \right)\end{array}$$

Hence, this is the required matrix.