Question

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 \({y_1}\)?

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

Short answer

The variance of\(y = 0.615525{x_1} + 0.599424{x_2} + 0.511683{x_3}\)obtained as:\({\lambda _1} = 51.6957\).

The required percentage is: \(64.8872\% \).

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 matrix is:

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

The Variance

From question, the matrix and themaximum eigenvalue we haveis:

\(\begin{array}{l}S = \left[ {\begin{array}{*{20}{c}}{29.64}&{18.38}&{5.00}\\{18.38}&{20.82}&{14.06}\\{5.00}&{14.06}&{29.21}\end{array}} \right]\\{\lambda _1} = 51.6957\end{array}\)

The respective unit vector is:

\({u_1} = \left[ {\begin{array}{*{20}{c}}{0.615525}\\{0.599424}\\{0.511683}\end{array}} \right]\)

The new variable will be:

\({y_1} = 0.615525{x_1} + 0.599424{x_2} + 0.511683{x_3}\)

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

\[\begin{array}{c}\Delta = \frac{{{\lambda _1}}}{{tr\left( S \right)}} \times 100\\ = \frac{{51.6957}}{{29.64 + 20.82 + 29.21}} \times 100\\ = 64.8872\% \end{array}\]

Hence, this is the required answer.