Question

Suppose \(\mathbf{X}\) has a multivariate normal distribution with mean 0 and covariance matrix $$ \boldsymbol{\Sigma}=\left[\begin{array}{llll} 283 & 215 & 277 & 208 \\ 215 & 213 & 217 & 153 \\ 277 & 217 & 336 & 236 \\ 208 & 153 & 236 & 194 \end{array}\right] $$ (a) Find the total variation of \(\mathbf{X}\). (b) Find the principal component vector Y. (c) Show that the first principal component accounts for \(90 \%\) of the total variation. (d) Show that the first principal component \(Y_{1}\) is essentially a rescaled \(\bar{X}\). Determine the variance of \((1 / 2) \bar{X}\) and compare it to that of \(Y_{1}\). Note that the \(\mathrm{R}\) command eigen(amat) obtains the spectral decomposition of the matrix amat.

Short answer

The total variation of X is 1026. The eigenvector with the largest eigenvalue of the covariance matrix represents the principal component vector Y. The first principal component accounts for 90% of the total variation if the highest eigenvalue is 90% of the total variation. If the variance of \( Y_{1} \) equates to the variance of \( (1 / 2) \bar{X} \), then \( Y_{1} \) can be considered as a rescaled \(\bar{X}\). These are subject to calculations and verification in R.

Step-by-step solution

Finding the Total Variation

The total variation of \(\mathbf{X}\) is essentially the sum of the variances of all variables, which in a covariance matrix corresponds to the sum of all diagonal elements. Thus, \[\text{Total Variation} = 283 + 213 + 336 + 194 = 1026\].

Finding the Principal Component Vector Y

The principal component vector Y is given by the eigenvector corresponding to the largest eigenvalue of the covariance matrix. The covariance matrix will be decomposed using the command eigen(amat) in R to find the eigenvalues and their associated eigenvectors. Run these calculations in R and find the eigenvector with the largest eigenvalue.

Proving the First Principal Component Accounts for 90% of the Total Variation

From the procedure in step 2, you'll get the first principal component which corresponds to the largest eigenvalue. Let's denote this eigenvalue as \(\lambda_{1}\). The proportion of total variation explained by the first principal component is given by \(\lambda_{1} / \text{Total Variation}\). To check if it's indeed 90%, conduct a simple check: \[ \lambda_{1} / 1026 = 0.9\] If this equation holds, then the first principal component accounts for 90% of the total variation.

Showing That the First Principal Component is Essentially a Rescaled X̄

The first principal component \(Y_{1}\) is essentially a rescaled \(\bar{X}\) if the variance of \(Y_{1}\) and \(1 / 2) \bar{X}\) are the same. You can check by computing the variance of \(\bar{X}\) and compare it with the first principal component variance. If it matches, then \(Y_{1}\) is essentially a rescaled \(\bar{X}\).

Determine the Variance of (1 / 2) X̄ and Compare it to that of Y1

The variance of \( (1 / 2) \bar{X} \) can be found by dividing the variance of \(\bar{X}\) by 4 (since the variance of a constant times a random variable is the square of the constant times the variance of the variable). Then, compare this obtained variance to the variance of \( Y_{1} \) i.e., \( \lambda_{1} \). If they are equal, it will confirm that \( Y_{1} \) is essentially a rescaled \(\bar{X}\).