Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 42 / 3 / i / Solution

Take logarithms entry by entry before centering. For tree $i$, let $z_i=(\log G_i,\log H_i,\log V_i{)}^T$, and put $\bar{z}=31^{-1}{\sum }_i{z}_i$. The command var(log(b)) computes the sample covariance matrix with the unbiased estimator normalizationThus entry $(j,k)$ is the sum of products of the centered log measurements divided by $30$, not the logarithm of the covariance between the original measurements. Its diagonal entries are the three log-variable sample variances, and the off-diagonal entries are their sample covariances. The positive off-diagonal entries indicate positive association of the log measurements. Taking logarithms turns a multiplicative relation such as $V\approx c{G}^{2}H$ into the approximately linear relation $\log V\approx \log c+2\log G+\log H$, which motivates investigating principal components of these transformed observations.