(a) (X, Y) has bivariate normal distribution BN(μ₁, μ₂, σ₁², σ₂², ρ).
(i) Show that X and Y are independent if and only if ρ = 0. (6 marks)
(ii) If (X, Y) follows BN(3, 1, 16, 25, 3/5), obtain P(3

Question

(a) (X, Y) has bivariate normal distribution BN(μ₁, μ₂, σ₁², σ₂², ρ).
(i) Show that X and Y are independent if and only if ρ = 0. (6 marks)
(ii) If (X, Y) follows BN(3, 1, 16, 25, 3/5), obtain P(3 < Y < 8 | X = 7), given Φ(2) = 0.9772 and Φ(-0.25) = 0.4017, and Φ(x) represents the area under the standard normal curve from -∞ to x. (6 marks)
(iii) If (X, Y) follows BN(0, 0, 1, 1, 0), what will be the distribution of Z = Y/X? (4 marks)
(iv) State the multivariate extension of (i) when X̃ follows Nₚ(μ̃, Σ). (4 marks)

(b) Define principal components and canonical correlation. How can one attain data reduction using principal components? If (X₁, X₂) has covariance matrix Σ = [[1, ρ], [ρ, 1]], then find the principal components. (15 marks)

(c) For the simple linear regression model y = β₀ + β₁x + ε, where β₀ and β₁ are parameters and ε has zero mean and an unknown variance σ², find the estimates of β₀ and β₁ by the principle of least squares as well as the method of maximum likelihood. Examine whether they are identical. (15 marks)

UPSC Answer Check · Accepted Answer

Prove the independence condition in (a)(i) using factorization of joint density; for (a)(ii)-(iv), calculate conditional distributions and identify the Cauchy distribution; for (b), define concepts then derive eigenvalues/eigenvectors for PC extraction; for (c), derive both estimators and compare. Allocate ~40% time to part (a) [20 marks], ~30% each to (b) and (c) [15 marks each], with explicit theorem statements and step-by-step derivations throughout.
- (a)(i) Prove ρ=0 ⇔ independence by showing joint density factorizes into marginal densities, using the bivariate normal PDF structure
- (a)(ii) Compute conditional distribution Y|X=7 ~ N(μ₂ + ρ(σ₂/σ₁)(x-μ₁), σ₂²(1-ρ²)), then standardize and use Φ values
- (a)(iii) Identify Z=Y/X as ratio of independent N(0,1) variables, hence standard Cauchy distribution
- (a)(iv) State multivariate extension: X̃ ~ Nₚ(μ̃, Σ) has independent components iff Σ is diagonal
- (b) Define PCs as uncorrelated linear combinations maximizing variance; define canonical correlation as correlation between linear combinations of two variable sets; data reduction by retaining top k PCs; derive eigenvalues (1±ρ) and eigenvectors for given Σ
- (c) Derive LSE by minimizing Σ(yᵢ-β₀-β₁xᵢ)²; derive MLE using normal error assumption; show identical estimators but different variance estimators
- Compare LSE (distribution-free) vs MLE (requires normality) and note σ²_MLE = SSE/n vs σ²_LSE = SSE/(n-2)

Dimension	Weight	Max marks	Excellent	Average	Poor
Setup correctness	20%	10	Correctly writes full bivariate normal PDF for (a)(i); properly identifies conditional normal parameters in (a)(ii); recognizes independence of X,Y in (a)(iii); states multivariate normal definition with precision in (a)(iv); sets up eigenvalue problem correctly in (b); specifies all regression assumptions in (c)	Writes most formulas correctly but misses constants or conditions; partial setup for eigenvalue problem; vague on regression assumptions	Incorrect PDF or missing key components; wrong conditional distribution formula; fails to identify distribution of ratio; omits multivariate extension; incorrect eigenvalue setup
Method choice	20%	10	Uses factorization criterion for independence proof; applies standard conditional normal theory with proper standardization; recognizes Cauchy via ratio of normals; uses spectral decomposition for PCs; applies calculus minimization for LSE and likelihood maximization for MLE with clear distinction	Correct general methods but inefficient or partially justified; some mixing of approaches; incomplete optimization steps	Wrong method for independence (e.g., correlation only); incorrect standardization; fails to identify distribution type; wrong PC extraction method; confused LSE/MLE derivation
Computation accuracy	20%	10	Precise calculation: conditional mean = 1 + (3/5)(5/4)(4) = 4, conditional SD = 4, Z-scores (3-4)/4=-0.25 and (8-4)/4=1, probability = 0.9772-0.4017 = 0.5755; eigenvalues 1±ρ with orthogonal eigenvectors [1,1]/√2 and [1,-1]/√2; correct normal equations and MLE solutions β̂₁=Sxy/Sxx, β̂₀=ȳ-β̂₁x̄	Minor arithmetic errors in probability or eigenvectors; correct formulas but calculation mistakes; partial derivation of estimators	Major computational errors in conditional parameters; wrong probability value; incorrect eigenvalues/eigenvectors; wrong estimator formulas
Interpretation	20%	10	Explains why zero correlation implies independence only for normal distributions; interprets conditional probability in context; explains why Cauchy has no moments; clarifies that PCs are uncorrelated and capture maximum variance sequentially; explains geometric meaning of LSE (projection) vs MLE (likelihood maximization) and conditions for equivalence	Some interpretation present but shallow or partially correct; misses key insights about normality requirement or PC variance maximization	No interpretation of results; fails to explain why results matter; purely mechanical computation without insight
Final answer & units	20%	10	Clear boxed answers: (a)(ii) P = 0.5755 or 57.55%; (a)(iii) Z ~ Cauchy(0,1); (a)(iv) Σ diagonal ⇔ independence; (b) PC1 = (X₁+X₂)/√2, PC2 = (X₁-X₂)/√2 with variances 1+ρ and 1-ρ; (c) explicit estimator formulas with note on identical β̂ but different σ² estimators; proper notation throughout	Answers present but poorly formatted or missing some parts; inconsistent notation; unclear final statements	Missing final answers; wrong conclusions; no clear presentation of results; confused notation

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