Question

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.