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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.

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

Short Answer

Expert verified

The mean deviation form and covariance matrix are:

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

Step by step solution

01

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

02

The Mean Deviation Form

As per the question, we have a matrix:

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

Thesample mean will be:

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

TheMean 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.

03

The Covariance Matrix

Now, thecovariance matrix will be:

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

Hence, this is the required matrix.

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Most popular questions from this chapter

10.Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.

10. \(\left( {\begin{aligned}{{}}{1/3}&{\,\,2/3}&{\,\,2/3}\\{2/3}&{\,\,1/3}&{ - 2/3}\\{2/3}&{ - 2/3}&{\,\,1/3}\end{aligned}} \right)\)

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

22. Show that if \(A\) is an \(n \times n\) positive definite matrix, then an orthogonal diagonalization \(A = PD{P^T}\) is a singular value decomposition of \(A\).

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

22. \(\left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\)

(M) Compute an SVD of each matrix in Exercises 26 and 27. Report the final matrix entries accurate to two decimal places. Use the method of Examples 3 and 4.

26. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{18}}}&{{\bf{13}}}&{ - {\bf{4}}}&{\bf{4}}\\{\bf{2}}&{{\bf{19}}}&{ - {\bf{4}}}&{{\bf{12}}}\\{ - {\bf{14}}}&{{\bf{11}}}&{ - {\bf{12}}}&{\bf{8}}\\{ - {\bf{2}}}&{{\bf{21}}}&{\bf{4}}&{\bf{8}}\end{array}} \right)\)

Question: Let \({x_1}\,,{x_2}\) denote the variables for the two-dimensional data in Exercise 1. Find a new variable \({y_1}\) of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2}\), with\(c_1^2 + c_2^2 = 1\), such that \({y_1}\) has maximum possible variance over the given data. How much of the variance in the data is explained by \({y_1}\)?

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