/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q7.5-2E Question: In Exercises 1 and 2, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17鈥�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. \(\left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\)

Question: 3. Let A be an \(n \times n\) symmetric matrix of rank r. Explain why the spectral decomposition of A represents A as the sum of r rank 1 matrices.

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\) .

a. \(3x_1^2 - 4{x_1}{x_2} + 5x_2^2\) b. \(3x_1^2 + 2{x_1}{x_2}\)

(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.

27. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{6}}&{ - {\bf{8}}}&{ - {\bf{4}}}&{\bf{5}}&{ - {\bf{4}}}\\{\bf{2}}&{\bf{7}}&{ - {\bf{5}}}&{ - {\bf{6}}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{1}}}&{ - {\bf{8}}}&{\bf{2}}&{\bf{2}}\\{ - {\bf{1}}}&{ - {\bf{2}}}&{\bf{4}}&{\bf{4}}&{ - {\bf{8}}}\end{array}} \right)\)

Let B be an \(n \times n\) symmetric matrix such that \({B^{\bf{2}}} = B\). Any such matrix is called a projection matrix (or an orthogonal projection matrix.) Given any y in \({\mathbb{R}^n}\), let \({\bf{\hat y}} = B{\bf{y}}\)and\({\bf{z}} = {\bf{y}} - {\bf{\hat y}}\).

a) Show that z is orthogonal to \({\bf{\hat y}}\).

b) Let W be the column space of B. Show that y is the sum of a vector in W and a vector in \({W^ \bot }\). Why does this prove that By is the orthogonal projection of y onto the column space of B?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.