(a) For a multiple linear regression model with three covariates X₁, X₂ and X₃, let rᵢⱼ denote the correlation coefficient…

(a) Consistency of correlations and limits for r₁₃

(i) For three correlations, consistency requires the correlation matrix R = [[1, r₁₂, r₁₃], [r₁₂, 1, r₂₃], [r₁₃, r₂₃, 1]] to be positive semidefinite. Since all diagonal entries are 1 and the given correlations lie in (-1, 1), it is enough to check det(R) ≥ 0.

det(R) = 1 + 2r₁₂r₂₃r₁₃ − r₁₂² − r₂₃² − r₁₃² = 1 + 2(0·77)(0·52)(0·72) − (0·77)² − (0·52)² − (0·72)² = 1 + 0·576576 − 0·5929 − 0·2704 − 0·5184 = 0·194876 = 48719/250000 > 0.

Hence R is positive definite, so the given data are consistent.

(ii) Let x = r₁₃. Positive semidefiniteness requires 1 + 2(0·77)(0·52)x − (0·77)² − (0·52)² − x² ≥ 0. So x² − 2(0·77)(0·52)x + [(0·77)² + (0·52)² − 1] ≤ 0.

Solving the quadratic, x = (0·77)(0·52) ± √[(1 − (0·77)²)(1 − (0·52)²)] = 0·4004 ± √[(0·4071)(0·7296)] = 0·4004 ± √0·29702016 = 0·4004 ± 0·54499556.

Therefore −0·14459556 ≤ r₁₃ ≤ 0·94539556, or exactly 1001/2500 − √(1856376/6250000) ≤ r₁₃ ≤ 1001/2500 + √(1856376/6250000). The given value r₁₃ = 0·72 lies in this interval.

(b) Cluster sampling with equal-sized clusters

Let the population have N clusters, each of equal size M, with elements yᵢⱼ, i = 1,…,N, j = 1,…,M. Let Ȳᵢ = (1/M)Σⱼ yᵢⱼ, Ȳ = (1/N)Σᵢ Ȳᵢ. A simple random sample without replacement of n clusters is selected. Define ȳ̄ = (1/n)Σᵢ∈s Ȳᵢ. Equivalently, ȳ̄ = (1/(nM))Σᵢ∈sΣⱼ yᵢⱼ. Since each cluster has inclusion probability n/N, E(ȳ̄) = (1/n)(n/N)Σᵢ Ȳᵢ = (1/N)Σᵢ Ȳᵢ = Ȳ. Thus ȳ̄ is an unbiased estimator of the population mean Ȳ.

Now let S_b² = (1/(N−1))Σᵢ(Ȳᵢ − Ȳ)² and f = n/N. Since the sample is SRSWOR of n cluster means from N clusters, V(ȳ̄) = (1−f)S_b²/n.

Write dᵢⱼ = yᵢⱼ − Ȳ. Then Ȳᵢ − Ȳ = (1/M)Σⱼ dᵢⱼ, so Σᵢ(Ȳᵢ − Ȳ)² = (1/M²)[ΣᵢΣⱼ dᵢⱼ² + ΣᵢΣⱼ≠ₖ dᵢⱼdᵢₖ].

Let S² = (1/(NM−1))ΣᵢΣⱼ dᵢⱼ² and define the intracluster correlation ρcl = [ΣᵢΣⱼ≠ₖ dᵢⱼdᵢₖ]/[(M−1)ΣᵢΣⱼ dᵢⱼ²]. Then ΣᵢΣⱼ≠ₖ dᵢⱼdᵢₖ = (M−1)(NM−1)S²ρcl. Hence Σᵢ(Ȳᵢ − Ȳ)² = (NM−1)S²[1 + (M−1)ρcl]/M².

Substituting into V(ȳ̄), V(ȳ̄) = (1−f)(NM−1)S²[1 + (M−1)ρcl]/[M²(N−1)n]. This is valid for equal-size clusters; if M = 1, the formula reduces to the usual SRSWOR result.

(c) Conditional independence in a trivariate normal

Partition Z = (X, Y₁, Y₂)ᵀ, where X is scalar and Y = (Y₁, Y₂)ᵀ. The mean and covariance are μ = (0, 0, 1)ᵀ, Σ = [[1, 2, 1], [2, 5, 2], [1, 2, 2]].

Thus μ_X = 0, μ_Y = (0, 1)ᵀ, Σ_XX = 1, Σ_XY = (2, 1), Σ_YX = (2, 1)ᵀ, Σ_YY = [[5, 2], [2, 2]].

For a multivariate normal, Y | X = x ~ N(μ_Y + Σ_YXΣ_XX⁻¹(x − μ_X), Σ_YY − Σ_YXΣ_XX⁻¹Σ_XY). Since Σ_XX = 1, conditional mean = (0, 1)ᵀ + (2, 1)ᵀx = (2x, 1 + x)ᵀ, conditional covariance = [[5, 2], [2, 2]] − [[4, 2], [2, 1]] = [[1, 0], [0, 1]].

So conditional on X, the covariance between Y₁ and Y₂ is 0. For a bivariate normal, zero covariance implies independence. Therefore, conditional on X, Y₁ and Y₂ are independent.

Marginally, Y = (Y₁, Y₂)ᵀ ~ N((0, 1)ᵀ, [[5, 2], [2, 2]]). Here Cov(Y₁, Y₂) = 2 ≠ 0, and Corr(Y₁, Y₂) = 2/√10 = √(2/5) ≠ 0. Hence Y₁ and Y₂ are marginally dependent, not independent. Thus conditional independence holds, but marginal independence does not.