Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 1 / 3

The vector of column sample means is $\bar{x}=(1,3,2)$. Subtract it from each observation. The centred data matrix, with cell types in the original order, isEach column now has sample mean zero.

Using divisor $n=4$, the covariance matrix isFor example, the first variance is $(1+1+1+9)/4=3$, and the first-second covariance is $(3-1-1+3)/4=1$. If “sample covariance matrix” means the unbiased convention, use divisor $n-1=3$ instead: $S=(4/3)\Sigma$. This multiplies every eigenvalue by $4/3$ and leaves all principal component loadings and explained-variance fractions unchanged.

The total variance is the trace $\mathrm{tr}\Sigma =10$. Retaining two original genes preserves the sum of their marginal variances; the largest choice includes the third gene and either of the first two, giving $4+3=7$. Thus the largest proportion is 70%. This refers to retaining measured coordinates themselves, rather than replacing them by linear combinations as in principal component analysis.

The eigenvector $(1,-1,0{)}^{\mathsf{T}}$ has eigenvalue $2$. On its orthogonal complement, write vectors as $(a,a,c{)}^{\mathsf{T}}$. The eigenvalue equation becomesHence the ordered eigenvalues of the covariance matrix are $4+2\sqrt{2}$, $2$, and $4-2\sqrt{2}$. The first two principal components explainof the total variance. The denominator is the sum of all eigenvalues, which equals the trace.

The second-largest eigenvalue is $2$, with eigenvector $(1,-1,0{)}^{\mathsf{T}}$. A unit principal component loading is thereforeIt is orthogonal to the first loading $u_1=(1,1,\sqrt{2}{)}^{\mathsf{T}}/2$. Either sign of a loading is equivalent, provided its principal component scores use the same sign convention.

The centred observation for D is $d=(3,1,2{)}^{\mathsf{T}}$. Taking inner products with the unit principal component loadings gives its principal component scores:Normalizing the loadings matters: a direction alone does not fix the scale of the coordinates.

After subtracting the same training sample mean, E has vector $e=(1,-1,0{)}^{\mathsf{T}}$ and principal component scores $(0,\sqrt{2})$. The difference from D lies entirely along the first principal component. Its Euclidean distance in the retained plane isThe new point must use the original centring and loadings; recomputing them with E included would change the question.