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

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

Short Answer

Expert verified

The mean deviation form and covariance matrix are:

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

Step by step solution

01

Mean Deviation form and Covariance Matrix

The Mean Deviation formof 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:

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

02

The Mean Deviation Form

As per the question, we have a matrix:

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

Thesample mean will be:

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

TheMean 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.

03

The Covariance Matrix.

Now, thecovariance 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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