Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 46 / 2 / Solution

For the sample, put $\bar{x}=n^{-1}{\sum }_i{x}_i$ and use the sample covariance matrixOrder its eigenvalues ${\hat{\lambda }}_j$ decreasingly and choose unit orthogonal eigenvectors ${\hat{\ell }}_j$. The sample principal components are the score columnsTheir sample means are zero and their sample covariance is diagonal, since ${\hat{\ell }}_j^{T}S{\hat{\ell }}_k={\hat{\lambda }}_j{1}_{\{j=k\}}$. The $j$th sample variance is ${\hat{\lambda }}_j$, and its explained variance of a principal component is ${\hat{\lambda }}_j/\mathrm{tr}(S)$. Using divisor $n$ instead changes the numerical variances by a common factor but leaves the component directions unchanged.

Standardizing means replacing a variable with positive sample variance by $(x_{ij}-{\bar{x}}_j)/s_j$. This gives principal component analysis on a correlation matrix, using $C=D^{-1/2}S{D}^{-1/2}$ with $D=\mathrm{diag}(S)$. It removes arbitrary differences of measurement units and prevents a large-variance coordinate from dominating solely through its scale. It also changes the question being optimized: small-variance and possibly noisy coordinates receive equal marginal weight. Principal component analysis based on the covariance matrix can be more appropriate when the variables have common meaningful units and absolute variability matters. Standardization is therefore a substantive choice, not an automatic improvement. A zero-variance variable must first be removed. In the word-rating example the variables share the same numerical range, but their across-word dispersions can differ; standardization gives each attribute equal initial variance.

For the standardized eight-variable analysis, total variance is $\mathrm{tr}(C)=8$. The first three fractions of explained variance of a principal component are approximately 59.6%, 19.1% and 10.1%, respectively. The first two account for about 78.8%, and the first three for 88.9%. The first direction is dominant but one direction alone loses substantial variation. Retaining two principal components is a plausible descriptive summary: their eigenvalues exceed $1$, whereas the third is below $1$. That cutoff is a heuristic, and the third principal component may still have interpretable content.

The reported coefficient rows are not unit-length vectors. The printed rounded rows have squared lengths $4.7514$, $1.5222$ and $0.8063$, consistent with their eigenvalues to the precision of the coefficient table. They are consistent with the scaled principal component loading convention $r_j=\sqrt{{\lambda }_j}{\ell }_j$. Such rows remain eigenvectors, but the score direction with unit length is ${\ell }_j=r_j/\sqrt{{\lambda }_j}$, or more accurately the row divided by its actual Euclidean norm when using rounded entries. Under this convention $r_{ij}$ is the correlation of standardized attribute $i$ with the variance-one score for component $j$:This explains both their sizes and why they must not be treated as the unit vectors from parts (i) and (ii).

The first principal component has positive coefficients on every attribute, strongest on friendliness, goodness, niceness, bravery and strength. With the favorable endpoints coded in the positive direction, it describes a broad favorable or impressive evaluation, with a potency contribution. The second contrasts movement and speed, with some size and strength, against the first three evaluative attributes; it is chiefly an activity dimension. The third is dominated by size, with smaller positive strength and negative movement/speed coefficients, distinguishing size or potency from activity. These are interpretations of directions of variation, not evidence of causal latent traits. Reversing a principal component's overall sign changes neither its meaning as an axis nor its explained variance; endpoint coding fixes only how its positive side is described.