(a) (i) If **X** = (X₁ X₂ X₃)' is distributed as N₃ (μ, Σ), find the distribution of [(X₁ – X₂) (X₂ – X₃)]'. (5 marks) (ii)…

(a)(i) Distribution of (X₁ − X₂, X₂ − X₃)′

Let Y = AX with A = [[1, −1, 0], [0, 1, −1]]. Since X ~ N₃(μ, Σ), a linear transformation of a normal vector is normal: Y ~ N₂(Aμ, AΣA′). Mean: Aμ = (μ₁ − μ₂, μ₂ − μ₃)′. Variances: Var(X₁ − X₂) = σ₁₁ + σ₂₂ − 2σ₁₂ and Var(X₂ − X₃) = σ₂₂ + σ₃₃ − 2σ₂₃. Covariance: Cov(X₁ − X₂, X₂ − X₃) = σ₁₂ − σ₁₃ − σ₂₂ + σ₂₃. These entries form the 2 × 2 covariance matrix AΣA′ of the bivariate normal distribution.

(a)(ii) Independence of X₁ + X₂ + X₃ and X₁ − X₂ − X₃

Let U = X₁ + X₂ + X₃ and V = X₁ − X₂ − X₃. They are jointly normal, so they are independent if and only if Cov(U, V) = 0. Step 1: Σ(1, −1, −1)′ = (1 − ρ, −1, −1 − ρ)′. Step 2: Cov(U, V) = (1, 1, 1)Σ(1, −1, −1)′ = (1 − ρ) − 1 − (1 + ρ) = −1 − 2ρ. Step 3: Setting this to zero gives ρ = −1/2. Step 4: Σ is a valid covariance matrix at ρ = −1/2 since det Σ = 1 − 2ρ² = 1/2 > 0. Yes, U and V are independent exactly when ρ = −1/2.

(b) Multivariate normality and linear combinations

Necessity: if X ~ Nₚ(μ, Σ), then for any fixed l, l′X is a linear transformation of X and is N(l′μ, l′Σl), which is univariate normal (degenerate if l′Σl = 0). Sufficiency: suppose l′X is univariate normal for every l. Step 1: Let μ = E(X) and Σ = Cov(X). For any t, t′X is normal with mean t′μ and variance t′Σt. Step 2: The characteristic function of X at t is E exp(i t′X), which is the characteristic function of t′X at 1, namely exp(i t′μ − t′Σt/2). Step 3: This is the characteristic function of Nₚ(μ, Σ). By the uniqueness theorem, X ~ Nₚ(μ, Σ). Hence X is p-variate normal if and only if every linear combination of its components is univariate normal.

(c) Test of H₀: β₀ = β₁ = 0 in the simple linear model

Let Sₓₓ = Σ(xᵢ − x̄)², Sₓᵧ = Σ(xᵢ − x̄)(Yᵢ − Ȳ) and Sᵧᵧ = Σ(Yᵢ − Ȳ)². Step 1: Full model: β̂₁ = Sₓᵧ/Sₓₓ, β̂₀ = Ȳ − β̂₁x̄, and SSE_F = Sᵧᵧ − Sₓᵧ²/Sₓₓ with n − 2 degrees of freedom. Step 2: Under H₀ the fitted values are zero, so SSE_R = ΣYᵢ² = Sᵧᵧ + nȲ², with 2 restrictions. Step 3: F = [(SSE_R − SSE_F)/2]/[SSE_F/(n − 2)] = [(nȲ² + Sₓᵧ²/Sₓₓ)/2]/[(Sᵧᵧ − Sₓᵧ²/Sₓₓ)/(n − 2)]. Step 4: Under H₀, F ~ F(2, n − 2). Reject H₀ at level α if F > F(α; 2, n − 2).

(d) Completing the ANOVA table and testing treatments

Step 1: Blocks: df = SS/MSS = 21/4·2 = 5, so there are 6 blocks. Step 2: Error MSS = 12/15 = 0·8. Step 3: Error df = (b − 1)(t − 1) = 5(t − 1) = 15, so t = 4 and treatments df = 3. Step 4: Treatments SS = 5·0 × 3 = 15. Total df = 5 + 3 + 15 = 23 and total SS = 21 + 15 + 12 = 48. Step 5: Variance ratios: treatments F = 5·0/0·8 = 6·25; blocks F = 4·2/0·8 = 5·25. Step 6: With the tabulated F(3, 15) = 8·70 given, 6·25 < 8·70, so H₀ of equal treatment effects is not rejected. There is no significant difference between treatments at the 5% level. For blocks, 5·25 > F(5, 15) = 4·62, so blocks differ significantly.

(e) Regression estimator of the population mean: bias and MSE

Definition: with an auxiliary variable x whose population mean X̄ is known, the regression estimator under SRSWOR is ȳ_lr = ȳ + b(X̄ − x̄), where b = sₓᵧ/sₓ² is the sample regression coefficient. Notation: f = n/N, β = μ₁₁/μ₂₀, with μ₂₀, μ₀₂, μ₁₁, μ₂₁, μ₃₀ the population central moments. Step 1: E(ȳ_lr) − Ȳ = −Cov(b, x̄). Step 2: To first order, Cov(b, x̄) = (1 − f)(μ₂₁ − βμ₃₀)/(n μ₂₀). Step 3: Bias B(ȳ_lr) = −(1 − f)(μ₂₁ − βμ₃₀)/(n μ₂₀), which is of order 1/n. Step 4: To first order b is replaced by β, so ȳ_lr − Ȳ ≈ (ȳ − Ȳ) − β(x̄ − X̄), and MSE(ȳ_lr) = (1 − f)/n × [μ₀₂ + β²μ₂₀ − 2βμ₁₁] = (1 − f)μ₀₂(1 − ρ²)/n, where ρ is the population correlation of x and y. The squared bias is of order 1/n² and is ignored.