Question

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 $y_1$?

Short answer

The variance of the data by $y_1$ obtained as $93.3374\mathrm{\%}$.

Step-by-step solution

Mean Deviation form and Covariance Matrix

The Mean Deviation form of any $p\times N$is given by:

$B=\left(\begin{array}{cccc}{\hat{X}}_1& {\hat{X}}_2& ........& {\hat{X}}_N\end{array}\right)$

Whose $p\times p$covariance matrix is:

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

The Change in Variance

From exercise 1, the maximum eigenvalue is:

${\lambda }_1=95.2041$

The respective unit vector is:

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

The new variable will be:

$y_1=0.946515x_1-0.322659x_2$

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

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

Hence, this is the required answer.