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Q7.5-7E

Expert-verifiedFound in: Page 395

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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 \({y_1}\)?**

The variance of the data by \({y_1}\) obtained as \(93.3374\% \).

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 matrix** is:

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

From exercise 1, the maximum **eigenvalue** is:

\({\lambda _1} = 95.2041\)

The respective **unit vector** is:

\({u_1} = \left( {\begin{array}{*{20}{c}}{0.946515}\\{ - 0.322659}\end{array}} \right)\)

The new variable will be:

\({y_1} = 0.946515{x_1} - 0.322659{x_2}\)

Now, the percentage of **change in variance** can be obtained as:

\(\begin{array}{c}\Delta = \frac{{{\lambda _1}}}{{tr\left( S \right)}} \times 100\\ = \frac{{95.2041}}{{86 + 16}} \times 100\\ = 93.3374\% \end{array}\)

Hence, this is the required answer.

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