Paper I — Q6
(a)(i) If (X, Y) follows bivariate normal BN(μ₁, μ₂, σ₁², σ₂², ρ), then obtain (A) E(e^X) (B) E(e^(X+Y)) (C) Var(e^X) and (D)…
If (X, Y) follows bivariate normal BN(μ₁, μ₂, σ₁², σ₂², ρ), then obtain (A) E(e^X) (B) E(e^(X+Y)) (C) Var(e^X) and (D) Correlation between e^X and e^Y. 3+3+3+3=12 marks
If (X, Y) have the joint probability density function g(x,y) = y e^(-y(x+1)), for x ≥ 0, y ≥ 0; 0 elsewhere, then find the regression curve of X on Y and comment on the nature of the curve. 8 marks
Let X = (X₁, X₂, X₃)' ~ N₃(μ, Σ), in which μ = (2 1 3)' and Σ = (9 2 -2 / 2 2 -3 / -2 -3 9). Obtain (i) E{X₁ | X₂ = x₂, X₃ = x₃} and (ii) Var{X₁ | X₂ = x₂, X₃ = x₃}. 15 marks
Consider the model: Y = X θ + ε, where ε is an n×1 vector of unobservable random variables such that E(ε) = 0 and D(ε) = σ²Ω, σ>0 unknown, Ω is a positive definite matrix of known constants and rank(X) = k<n. Then (i) Derive least square estimator of θ and (ii) Derive an unbiased estimator of σ². 9+6=15 marks
हिंदी में प्रश्न पढ़ें
यदि (X, Y) द्विचर प्रसामान्य BN(μ₁, μ₂, σ₁², σ₂², ρ) का अनुसरण करता है, तो (A) E(e^X) (B) E(e^(X+Y)) (C) Var(e^X) तथा (D) e^X और e^Y के बीच सहसंबंध ज्ञात कीजिए। 3+3+3+3=12 अंक
यदि (X, Y) का संयुक्त प्रायिकता घनत्व फलन निम्नवत है: g(x,y) = y e^(-y(x+1)), x ≥ 0, y ≥ 0; 0 अन्यथा, तो X का Y पर समाश्रयन वक्र ज्ञात कीजिए तथा वक्र की प्रकृति पर टिप्पणी कीजिए। 8 अंक
मान लीजिए कि X = (X₁, X₂, X₃)' ~ N₃(μ, Σ), जिसमें μ = (2 1 3)' तथा Σ = (9 2 -2 / 2 2 -3 / -2 -3 9) है। ज्ञात कीजिए (i) E{X₁ | X₂ = x₂, X₃ = x₃} और (ii) Var{X₁ | X₂ = x₂, X₃ = x₃}। 15 अंक
निदर्श पर विचार कीजिए: Y = X θ + ε, जहाँ ε अलक्ष्य यादृच्छिक चरों का एक n×1 सदिश इस प्रकार है कि E(ε) = 0 और D(ε) = σ²Ω, σ > 0 अज्ञात है, Ω ज्ञात स्थिरांकों का एक धनात्मक निश्चित आव्यूह है तथा कोटि (X) = k < n है। तब: (i) θ का न्यूनतम वर्ग आकलक व्युत्पन्न कीजिए और (ii) σ² का एक अनभिनत आकलक व्युत्पन्न कीजिए। 9+6=15 अंक
Model answer
Written by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key.
(a)(i) Since (X, Y) is bivariate normal, X ~ N(μ₁, σ₁²) and X+Y ~ N(μ₁+μ₂, σ₁²+σ₂²+2ρσ₁σ₂).
The mgf of a normal variable is E(e^(tZ)) = exp(μt + σ²t²/2).
(A) Put Z = X and t = 1: E(e^X) = exp(μ₁ + σ₁²/2).
(B) Put Z = X+Y and t = 1: E(e^(X+Y)) = exp(μ₁+μ₂ + (σ₁²+σ₂²+2ρσ₁σ₂)/2).
(C) Var(e^X) = E(e^(2X)) − (E(e^X))². Now E(e^(2X)) = exp(2μ₁ + 2σ₁²), and (E(e^X))² = exp(2μ₁ + σ₁²). Hence Var(e^X) = exp(2μ₁+σ₁²)(exp(σ₁²)−1).
(D) Cov(e^X, e^Y) = E(e^(X+Y)) − E(e^X)E(e^Y). Since E(e^Y) = exp(μ₂ + σ₂²/2), Cov(e^X, e^Y) = exp(μ₁+μ₂+(σ₁²+σ₂²)/2)(exp(ρσ₁σ₂)−1). Also Var(e^Y) = exp(2μ₂+σ₂²)(exp(σ₂²)−1). Therefore Corr(e^X, e^Y) = (exp(ρσ₁σ₂)−1) / sqrt((exp(σ₁²)−1)(exp(σ₂²)−1)), provided σ₁ > 0 and σ₂ > 0.
(a)(ii) The joint pdf is g(x,y) = y e^(−y(x+1)), x ≥ 0, y ≥ 0.
Marginal of Y: f_Y(y) = ∫₀^∞ y e^(−y(x+1)) dx = y e^(−y) ∫₀^∞ e^(−yx) dx = y e^(−y) · 1/y = e^(−y), y ≥ 0. So Y ~ Exp(1).
Conditional pdf of X given Y = y: f_X|Y(x | y) = g(x,y)/f_Y(y) = y e^(−yx), x ≥ 0. Thus X | Y = y ~ Exp(rate y), whose mean is 1/y. Hence the regression curve of X on Y is E(X | Y = y) = 1/y, y > 0. It is a rectangular hyperbola: nonlinear, decreasing and convex. As y → 0⁺, E(X | Y=y) → ∞; as y → ∞, E(X | Y=y) → 0.
(b) Partition X₁ and X₂* = (X₂, X₃)′. Here μ₁ = 2, μ₂* = (1, 3)′, Σ₁₁ = 9, Σ₁₂ = [2, −2], Σ₂₂ = [[2, −3], [−3, 9]].
For multivariate normal, E(X₁ | X₂*=x₂*) = μ₁ + Σ₁₂ Σ₂₂⁻¹ (x₂* − μ₂*), Var(X₁ | X₂*=x₂*) = Σ₁₁ − Σ₁₂ Σ₂₂⁻¹ Σ₂₁.
Now det(Σ₂₂) = 2·9 − (−3)(−3) = 9, so Σ₂₂⁻¹ = (1/9) [[9, 3], [3, 2]]. Then Σ₁₂ Σ₂₂⁻¹ = [2, −2] (1/9)[[9, 3], [3, 2]] = [4/3, 2/9]. Hence E(X₁ | X₂=x₂, X₃=x₃) = 2 + (4/3)(x₂−1) + (2/9)(x₃−3). Simplifying, E(X₁ | X₂=x₂, X₃=x₃) = (4/3)x₂ + (2/9)x₃.
Also Σ₁₂ Σ₂₂⁻¹ Σ₂₁ = [4/3, 2/9] [2, −2]′ = 8/3 − 4/9 = 20/9. Therefore Var(X₁ | X₂=x₂, X₃=x₃) = 9 − 20/9 = 61/9.
(c) The model is Y = Xθ + ε, E(ε)=0, D(ε)=σ²Ω, with Ω positive definite and rank(X)=k<n.
(i) Since Ω is positive definite, write Ω⁻¹ = Ω^(−1/2)Ω^(−1/2). Transform Z = Ω^(−1/2)Y, A = Ω^(−1/2)X, η = Ω^(−1/2)ε. Then Z = Aθ + η, with E(η)=0 and D(η)=σ²I_n. Apply ordinary least squares to the transformed model. Minimize (Z − Aθ)′(Z − Aθ) = (Y − Xθ)′Ω⁻¹(Y − Xθ). The normal equation is A′Aθ = A′Z, i.e. X′Ω⁻¹X θ = X′Ω⁻¹Y. Since rank(X)=k and Ω⁻¹ is positive definite, X′Ω⁻¹X is nonsingular. Hence the generalized least squares estimator is θ̂ = (X′Ω⁻¹X)⁻¹X′Ω⁻¹Y.
(ii) The transformed residual is ê = Z − Aθ̂ = (I − H)η, where H = A(A′A)⁻¹A′ is the projection matrix of rank k. Therefore E(ê′ê) = σ² tr(I − H) = σ²(n − k). But ê′ê = (Y − Xθ̂)′Ω⁻¹(Y − Xθ̂). Thus an unbiased estimator of σ² is σ̂² = (Y − Xθ̂)′Ω⁻¹(Y − Xθ̂)/(n − k). This is valid because n > k and X′Ω⁻¹X is nonsingular.
What "Derive" is asking you to do
Reach the stated expression from a starting relation, justifying every step. The destination is printed in the question, so only the route earns marks, and the assumptions you work under are part of that route.
Structure that answers it
Assumptions and notation defined → starting relation or governing equation → each step with its justification → the required expression → limiting case or boundary check
Where marks are lost
Writing the standard result first and fitting three lines to it, which an examiner reads at a glance. Marks also go on assumptions left unstated — lossless medium, small amplitude, errors independent with zero mean — and on symbols used before they are defined, even when the question says usual notations.
How this answer will be evaluated
Approach
Framework: Generalized Least Squares (GLS) and Multivariate Normal Theory. (a) calculate: given > formula > substitution > result | (b) calculate: given > formula > substitution > result | (c) calculate: given > formula > substitution > result Full marks: Rigorous derivations with all matrix steps shown and correct final forms.
Key points expected
- MGF of bivariate normal
- Log-normal moments
- Marginal density integration
- Partitioned covariance matrix
- Conditional expectation formula
- Conditional variance formula
- GLS estimator derivation
- Unbiased variance estimator
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Derive moments of log-normal variables and regression curve from joint PDF. 20 marks
calculate— given → formula → substitution → result
Must cover
- State MGF of bivariate normal
- Derive E(e^X) and Var(e^X)
- Compute marginal f_Y(y) for regression
- Identify regression curve as linear
Loses marks
- Skipping MGF derivation steps
- Assuming independence without proof
Earns more
- Correct correlation formula for e^X, e^Y
- Explicit integration for marginal density
- Comment on linearity of regression
Extra mark
- Verification of joint PDF normalization
- (b) Compute conditional mean and variance using partitioned covariance matrix. 15 marks
calculate— given → formula → substitution → result
Must cover
- Partition Σ into Σ11, Σ12, Σ22
- Apply conditional mean formula μ1 + Σ12Σ22^-1(x2-μ2)
- Apply conditional variance formula Σ11 - Σ12Σ22^-1Σ21
- Compute inverse of 2x2 submatrix Σ22
Loses marks
- Incorrect matrix partitioning
- Arithmetic errors in inverse
Earns more
- Correct matrix multiplication steps
- Simplified final linear expression
- Explicit scalar variance value
Extra mark
- Verification of positive definiteness
- (c) Derive GLS estimator and unbiased variance estimator for heteroscedastic model. 15 marks
calculate— given → formula → substitution → result
Must cover
- State GLS estimator θ_hat = (X'Ω^-1X)^-1X'Ω^-1Y
- Derive via minimization of (Y-Xθ)'Ω^-1(Y-Xθ)
- State unbiased estimator σ_hat^2 = SSE/(n-k)
- Define SSE using Ω^-1 weighting
Loses marks
- Using OLS formula instead of GLS
- Missing degrees of freedom in variance
Earns more
- Proof of unbiasedness E(σ_hat^2) = σ^2
- Explicit trace calculation for expectation
- Distinction between OLS and GLS
Extra mark
- Mention of Aitken's theorem
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