(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) Let Y = [(X₁ − X₂), (X₂ − X₃)]′. Then Y = AX, where A = [[1, −1, 0], [0, 1, −1]]. Since X ∼ N₃(μ, Σ), by the linear-transformation theorem for multivariate normal vectors, Y ∼ N₂(Aμ, AΣA′).

Here Aμ = [μ₁ − μ₂, μ₂ − μ₃]′. Also, if Σ = (σᵢⱼ), then Var(X₁ − X₂) = σ₁₁ + σ₂₂ − 2σ₁₂, Var(X₂ − X₃) = σ₂₂ + σ₃₃ − 2σ₂₃, Cov(X₁ − X₂, X₂ − X₃) = σ₁₂ − σ₁₃ − σ₂₂ + σ₂₃.

Hence Y ∼ N₂([μ₁ − μ₂, μ₂ − μ₃]′, [[σ₁₁ + σ₂₂ − 2σ₁₂, σ₁₂ − σ₁₃ − σ₂₂ + σ₂₃], [σ₁₂ − σ₁₃ − σ₂₂ + σ₂₃, σ₂₂ + σ₃₃ − 2σ₂₃]]).

(a)(ii) Let U = X₁ + X₂ + X₃, V = X₁ − X₂ − X₃. Since U and V are linear combinations of a multivariate normal vector, (U, V) is jointly normal. Therefore U and V are independent if and only if Cov(U, V) = 0.

Using Σ = [[1, ρ, 0], [ρ, 1, ρ], [0, ρ, 1]], Cov(U, V) = (1, 1, 1) Σ (1, −1, −1)′. Compute: = 1 − ρ + ρ − 1 − ρ − ρ − 1 = −1 − 2ρ.

Set this equal to zero: −1 − 2ρ = 0 ⇒ ρ = −1/2.

For ρ = −1/2, Σ is positive definite because its leading minors are positive and det(Σ) = 1 − 2ρ² = 1 − 1/2 = 1/2 > 0. Yes, for ρ = −1/2, U and V are independent.

(b) Let X = (X₁, X₂, ..., Xₚ)′.

First suppose X ∼ Nₚ(μ, Σ). For any fixed vector l = (l₁, l₂, ..., lₚ)′, the linear combination L = l₁X₁ + l₂X₂ + ... + lₚXₚ = l′X is a linear transformation of X. By the linear-transformation property of multivariate normal distribution, L ∼ N(l′μ, l′Σl). If l′Σl = 0, L is a degenerate univariate normal random variable. Thus every linear combination is univariate normal.

Conversely, suppose every linear combination l′X is univariate normal. Define μ = (E X₁, E X₂, ..., E Xₚ)′, Σ = (Cov(Xᵢ, Xⱼ)). For any t = (t₁, ..., tₚ)′, the random variable t′X is normal. Its mean is E(t′X) = t′μ, and its variance is Var(t′X) = t′Σt. Therefore its characteristic function is E exp(i t′X) = exp(i t′μ − (1/2)t′Σt).

But this is exactly the characteristic function of Nₚ(μ, Σ). By uniqueness of characteristic functions, X ∼ Nₚ(μ, Σ). Hence X has a p-variate normal distribution if and only if every linear combination of its components is univariate normal.

(c) The model is Yᵢ = β₀ + β₁xᵢ + eᵢ, eᵢ ∼ N(0, σ²), independently, i = 1, ..., n. We test H₀ : β₀ = β₁ = 0 against H₁ : not both zero.

Let x̄ = (1/n)Σxᵢ, Ȳ = (1/n)ΣYᵢ, Sₓₓ = Σ(xᵢ − x̄)², Sₓᵧ = Σ(xᵢ − x̄)(Yᵢ − Ȳ), Sᵧᵧ = Σ(Yᵢ − Ȳ)².

Under the full model, the least-squares estimators are β̂₁ = Sₓᵧ / Sₓₓ, β̂₀ = Ȳ − β̂₁x̄. The full-model residual sum of squares is SSE_F = Σ(Yᵢ − β̂₀ − β̂₁xᵢ)² = Sᵧᵧ − Sₓᵧ² / Sₓₓ, with degrees of freedom n − 2.

Under H₀ : β₀ = β₁ = 0, the fitted value is 0, so SSE_R = ΣYᵢ² = Sᵧᵧ + nȲ². The number of restrictions is 2. Therefore the F statistic is F = [(SSE_R − SSE_F)/2] / [SSE_F/(n − 2)] = [(nȲ² + Sₓᵧ²/Sₓₓ)/2] / [(Sᵧᵧ − Sₓᵧ²/Sₓₓ)/(n − 2)].

Under H₀, F ∼ F₂,ₙ₋₂. Reject H₀ at level α if F > F_α;2,n−2. This requires n > 2 and Sₓₓ > 0.

(d) From the block row: MSS = SS/df ⇒ 4.2 = 21/df ⇒ df for blocks = 21/4.2 = 5.

Error MSS = 12/15 = 0.8.

Blocks df = 5 ⇒ number of blocks b = 6. Error df = (b − 1)(t − 1) = 15 ⇒ 5(t − 1) = 15 ⇒ t = 4. Thus treatments df = t − 1 = 3.

Treatments MSS = 5.0, so Treatments SS = 5.0 × 3 = 15.

Total df = 5 + 3 + 15 = 23. Total SS = 21 + 15 + 12 = 48.

Completed ANOVA entries:

• Blocks: df = 5, SS = 21, MSS = 4.2, VR = 4.2/0.8 = 5.25.
• Treatments: df = 3, SS = 15, MSS = 5.0, VR = 5.0/0.8 = 6.25.
• Error: df = 15, SS = 12, MSS = 0.8, VR = —.
• Total: df = 23, SS = 48, MSS = —, VR = —.

For treatments, the critical value is F₀.₀₅(3,15) = 8.70. Since observed F = 6.25 < 8.70, we fail to reject H₀. There is no significant difference between treatments at the 5% level. For completeness, blocks show F = 5.25 > F₀.₀₅(5,15) = 4.62, so blocks differ significantly.

(e) Let Y be the study variable and X an auxiliary variable whose population mean X̄ is known. In simple random sampling without replacement of size n, let ȳ and x̄ be sample means. The regression estimator of the population mean Ȳ is ȳ_lr = ȳ + b(X̄ − x̄), where b = sₓᵧ/sₓ² = Σ(xᵢ − x̄)(yᵢ − ȳ) / Σ(xᵢ − x̄)². If the population regression coefficient β is known, the estimator is ȳ_reg = ȳ + β(X̄ − x̄).

For the usual case where b is estimated, define population moments about means: μ₂₀ = (1/N)Σ(Xᵢ − X̄)², μ₀₂ = (1/N)Σ(Yᵢ − Ȳ)², μ₁₁ = (1/N)Σ(Xᵢ − X̄)(Yᵢ − Ȳ), μ₂₁ = (1/N)Σ(Xᵢ − X̄)²(Yᵢ − Ȳ), μ₃₀ = (1/N)Σ(Xᵢ − X̄)³, and β = μ₁₁/μ₂₀. Let f = n/N.

To the first order of approximation, the bias of ȳ_lr is B(ȳ_lr) = − (1 − f)/(n μ₂₀) (μ₂₁ − β μ₃₀) = (1 − f)/(n μ₂₀) (β μ₃₀ − μ₂₁).

Its mean square error to the first order is MSE(ȳ_lr) ≈ (1 − f)/n [μ₀₂ + β² μ₂₀ − 2β μ₁₁] = (1 − f)/n [μ₀₂ − μ₁₁²/μ₂₀] = (1 − f)/n S_Y²(1 − ρ²), where ρ = μ₁₁/√(μ₂₀ μ₀₂) is the population correlation coefficient between X and Y. The bias squared is of order n⁻² and is omitted. These results require large n, SRSWOR, μ₂₀ > 0, and known population mean X̄.