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Magazine Chapter 7: Q10E (page 395)
Question:Repeat Exercise 9 with \(S = \left( {\begin{array}{*{20}{c}}5&4&2\\4&{11}&4\\2&4&5\end{array}} \right)\).
Short Answer
The variance ofobtained as \({\lambda _1} = 15\).
Step by step solution
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 matrixis:
\(S = \frac{1}{{N - 1}}B{B^T}\)
The Variance
From question, the matrix and the maximum eigenvalue we haveis:
\(\begin{array}{l}S = \left( {\begin{array}{*{20}{c}}5&4&2\\4&{11}&4\\2&4&5\end{array}} \right)\\{\lambda _1} = 15\end{array}\)
The respective unit vector is:
The new variable will be:
Hence, this is the required answer.
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