(a) Let **X** = (X₁ X₂ X₃)' be distributed as N₃ (μ, Σ), where μ = (2 −1 3)' and Σ = [[4, 1, 0], [1, 2, 1], [0, 1, 3]]. Find…

(a)(i) Conditional distribution of (X₁, X₂)′ given X₃ = 2

Partition X into (X₁, X₂)′ and X₃. Then μ₁ = (2, −1)′, μ₂ = 3, Σ₁₁ = [[4, 1], [1, 2]], Σ₁₂ = (0, 1)′ and Σ₂₂ = 3. Step 1: Conditional mean = μ₁ + Σ₁₂Σ₂₂⁻¹(2 − 3) = (2, −1)′ + (0, 1)′(1/3)(−1) = (2, −4/3)′. Step 2: Conditional covariance = Σ₁₁ − Σ₁₂Σ₂₂⁻¹Σ₂₁ = [[4, 1], [1, 2]] − (1/3)[[0, 0], [0, 1]] = [[4, 1], [1, 5/3]]. Hence (X₁, X₂)′ | X₃ = 2 ~ N₂((2, −4/3)′, [[4, 1], [1, 5/3]]).

(a)(ii) Partial correlation ρ₁₂.₃ and multiple correlation R₁.₂₃

Step 1: ρ₁₂.₃ = (conditional covariance)/√(product of conditional variances) = 1/√(4 × 5/3) = √(3/20) ≈ 0·387. Step 2: For R₁.₂₃, take σ₁ = (σ₁₂, σ₁₃)′ = (1, 0)′ and Σ₂₃ = [[2, 1], [1, 3]], whose inverse is (1/5)[[3, −1], [−1, 2]]. Step 3: R₁.₂₃² = σ₁′Σ₂₃⁻¹σ₁/σ₁₁ = (3/5)/4 = 3/20. Hence ρ₁₂.₃ = √(3/20) ≈ 0·387 and R₁.₂₃ = √(3/20) ≈ 0·387 (they coincide because σ₁₃ = 0).

(b)(i) Two-way classified data with r equal observations per cell

Model: yᵢⱼₖ = μ + αᵢ + βⱼ + γᵢⱼ + εᵢⱼₖ, for i = 1..a, j = 1..b, k = 1..r, with Σαᵢ = Σβⱼ = 0 and Σᵢγᵢⱼ = Σⱼγᵢⱼ = 0. Assumptions: effects are additive apart from the interaction γ; errors εᵢⱼₖ are independent N(0, σ²), with constant variance; factor effects are fixed. Step 1: Total SS = Σ(yᵢⱼₖ − ȳ...)², df abr − 1. Step 2: SSA = br Σ(ȳᵢ.. − ȳ...)², df a − 1; SSB = ar Σ(ȳ.ⱼ. − ȳ...)², df b − 1. Step 3: SSAB = r ΣΣ(ȳᵢⱼ. − ȳᵢ.. − ȳ.ⱼ. + ȳ...)², df (a − 1)(b − 1). Step 4: SSE = ΣΣΣ(yᵢⱼₖ − ȳᵢⱼ.)², df ab(r − 1); MSE estimates σ². Step 5: Test interaction first by F = MSAB/MSE. If significant, interpret the cell means; if not, test main effects by F = MSA/MSE and MSB/MSE with ab(r − 1) denominator degrees of freedom. Examples: crop yield for fertiliser and irrigation levels with replicated plots; tensile strength of a product across machines and operators with repeated trials.

(b)(ii) Estimability of p₁β₁ + p₂β₂ + p₃β₃

Assumptions: the model is linear, E(Y) = Xβ, with independent errors of zero mean and common variance σ². Step 1: X = [[1, 1, 0], [1, 0, 1], [1, 1, 0]], and p′β is estimable if and only if p lies in the row space of X. Step 2: The row space is spanned by (1, 1, 0) and (1, 0, 1), so p = a(1, 1, 0) + c(1, 0, 1) = (a + c, a, c). Step 3: Thus p₂ = a, p₃ = c and p₁ = a + c = p₂ + p₃. Conversely, if p₁ = p₂ + p₃, take a = p₂ and c = p₃, so p lies in the row space and p′β is estimable (for example by p₂Y₁ + p₃Y₂). Hence p′β is estimable if and only if p₁ = p₂ + p₃.

(c)(i) Reasons for performing principal component analysis

Reason 1: Dimension reduction: many correlated variables are replaced by a few uncorrelated components that keep most of the variance. Reason 2: Remedy for multicollinearity: orthogonal components can replace collinear regressors (principal component regression). Reason 3: Visualisation and noise reduction: scores on the first two or three components display high-dimensional data and discard low-variance noise.

(c)(ii) Canonical correlations: definition, applications and procedure

Definition: for random vectors X (p × 1) and Y (q × 1), the first canonical correlation is the maximum correlation between U = a′X and V = b′Y; later pairs are maximised subject to being uncorrelated with earlier pairs. There are min(p, q) of them. Applications: relating a set of aptitude scores to a set of job-performance measures; relating economic indicators to social indicators. Procedure: Step 1: Partition the covariance matrix into Σₓₓ, Σₓᵧ, Σᵧₓ, Σᵧᵧ (sample estimates in practice; standardise variables if scales differ). Step 2: Find the eigenvalues ρ₁*² ≥ ρ₂*² ≥ … of Σₓₓ⁻¹ΣₓᵧΣᵧᵧ⁻¹Σᵧₓ; the canonical correlations are their square roots. Step 3: The eigenvector a for each eigenvalue, scaled so a′Σₓₓa = 1, gives the canonical variate U = a′X. Step 4: b = Σᵧᵧ⁻¹Σᵧₓa/ρ*, scaled so b′Σᵧᵧb = 1, gives V = b′Y.