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Chapter 7: Q7.5-2E (page 395)

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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Most popular questions from this chapter

In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).

4. \(Q\left( x \right) = 3x_1^2 + 3x_2^2 + 5x_3^2 + 6x_1^{}x_2^{} + 2x_1^{}x_3^{} + 2x_2^{}x_3^{}\).

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

Question: [M] The covariance matrix below was obtained from a Landsat image of the Columbia River in Washington, using data from three spectral bands. Let \({x_1},{x_2},{x_3}\) denote the spectral components of each pixel in the image. Find a new variable of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2} + {c_3}{x_3}\) that has maximum possible variance, subject to the constraint that \(c_1^2 + c_2^2 + c_3^2 = 1\). What percentage of the total variance in the data is explained by \({y_1}\)?

\[S = \left[ {\begin{array}{*{20}{c}}{29.64}&{18.38}&{5.00}\\{18.38}&{20.82}&{14.06}\\{5.00}&{14.06}&{29.21}\end{array}} \right]\]

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.

17. \(\left( {\begin{aligned}{{}}1&{ - 6}&4\\{ - 6}&2&{ - 2}\\4&{ - 2}&{ - 3}\end{aligned}} \right)\)

Question: 11. Given multivariate data \({X_1},................,{X_N}\) (in \({\mathbb{R}^p}\)) in mean deviation form, let \(P\) be a \(p \times p\) matrix, and define \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).

  1. Show that \({Y_1},................,{Y_N}\) are in mean-deviation form. (Hint: Let \(w\) be the vector in \({\mathbb{R}^N}\) with a 1 in each entry. Then \(\left( {{X_1},................,{X_N}} \right)w = 0\) (the zero vector in \({\mathbb{R}^p}\)).)
  2. Show that if the covariance matrix of \({X_1},................,{X_N}\) is \(S\), then the covariance matrix of \({Y_1},................,{Y_N}\) is \({P^T}SP\).
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