Paper I — Q6
(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…
(X, Y) has bivariate normal distribution BN(μ₁, μ₂, σ₁², σ₂², ρ). Show that X and Y are independent if and only if ρ = 0. 6 marks
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
If (X, Y) follows BN(0, 0, 1, 1, 0), what will be the distribution of Z = Y/X? 4 marks
State the multivariate extension of (i) when X̃ follows Nₚ(μ̃, Σ). 4 marks
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
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
हिंदी में प्रश्न पढ़ें
(X, Y) का द्विवर प्रसामान्य बंटन BN(μ₁, μ₂, σ₁², σ₂², ρ) है। दर्शाइए कि X और Y स्वतंत्र हैं, यदि और केवल यदि ρ = 0 है। (6 अंक)
यदि (X, Y) का बंटन BN(3, 1, 16, 25, 3/5) है, तो P(3 < Y < 8 | X = 7) निकालिए, दिया है Φ(2) = 0.9772 और Φ(-0.25) = 0.4017 तथा Φ(x), -∞ से x तक का मानक प्रसामान्य वक्र के अंतर्गत क्षेत्रफल दर्शाता है। (6 अंक)
यदि (X, Y) का बंटन BN(0, 0, 1, 1, 0) है, तो Z = Y/X का बंटन क्या होगा? (4 अंक)
जब X̃, Nₚ(μ̃, Σ) का अनुसरण करता है, तो (i) का बहुचर विस्तरण लिखिए। (4 अंक)
मुख्य घटकों और विहित सहसंबंध को परिभाषित कीजिए। मुख्य घटकों का उपयोग करके कोई दत्त समान्यन कैसे प्राप्त कर सकता है? यदि (X₁, X₂) का सहप्रसरण आव्यूह Σ = [[1, ρ], [ρ, 1]] है, तो मुख्य घटकों को ज्ञात कीजिए। (15 अंक)
एक साधारण रैखिक समाश्रयन निदर्श y = β₀ + β₁x + ε के लिए, जहाँ β₀ और β₁ प्राचल हैं तथा ε का माध्य 0 और प्रसरण σ² अज्ञात है, न्यूनतम वर्ग सिद्धांत और अधिकतम संभाविता विधि से β₀ और β₁ के आकलक निकालिए। जाँच कीजिए कि क्या वे एकसमान हैं। (15 अंक)
Model answer
Drafted with AI 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. How we build rubrics and answers.
(a)(i) Independence iff ρ = 0
For BN(μ₁, μ₂, σ₁², σ₂², ρ), the joint density is
f(x,y)=1/(2πσ₁σ₂√(1-ρ²)) exp[-(1/(2(1-ρ²)))(((x-μ₁)/σ₁)² - 2ρ((x-μ₁)/σ₁)((y-μ₂)/σ₂) + ((y-μ₂)/σ₂)²)].
If ρ=0, the exponent separates into a function of x plus a function of y, and the joint density factors into the product of the two marginal normal densities. Hence X and Y are independent. Conversely, if X and Y are independent, then Cov(X,Y)=0. But Cov(X,Y)=ρσ₁σ₂, so ρ=0 because σ₁,σ₂>0.
Final: X and Y are independent iff ρ=0.
(a)(ii) P(3 < Y < 8 | X = 7)
For bivariate normal, the conditional distribution is
Y|X=x ~ N(μ₂ + ρ(σ₂/σ₁)(x-μ₁), σ₂²(1-ρ²)).
Here μ₁=3, μ₂=1, σ₁=4, σ₂=5, ρ=3/5. Thus
mean = 1 + (3/5)(5/4)(7-3) = 1+3 = 4,
variance = 25(1-9/25)=16, sd=4.
So Y|X=7 ~ N(4,16). Therefore
P(3<Y<8|X=7)=P((3-4)/4<Z<(8-4)/4)=P(-0.25<Z<1)
=Φ(1)-Φ(-0.25)=0.8413-0.4017=0.4396 (Φ(1) = 0.5 + 0.3413 from the table).
Final: 0.4396. The supplied Φ(2) would correspond to the upper limit 12; if the intended upper limit was 12, the value would be 0.9772-0.4017=0.5755.
(a)(iii) Distribution of Z = Y/X
Since ρ=0 and variances are 1, X and Y are independent standard normal. Let Z=Y/X. Put z=y/x and w=x, so y=zw. The Jacobian is |w|. Hence
f_Z(z)=∫ (from -∞ to ∞) (1/(2π)) exp[-(w²+z²w²)/2] |w| dw
=1/(2π) ∫ (from -∞ to ∞) |w| exp[-w²(1+z²)/2] dw
=1/π ∫ (from 0 to ∞) w exp[-w²(1+z²)/2] dw
=1/(π(1+z²)), −∞<z<∞.
Final: Z has standard Cauchy distribution with density 1/(π(1+z²)).
(a)(iv) Multivariate Extension
Let X̃ ~ Nₚ(μ̃, Σ). Partition X̃ into X₁ and X₂ with covariance blocks Σ₁₁, Σ₂₂ and cross-covariance Σ₁₂. Then X₁ and X₂ are independent iff Σ₁₂=0. In particular, all components of X̃ are mutually independent iff Σ is diagonal.
Final: The multivariate extension is: subvectors are independent iff their cross-covariance block is zero; all components are independent iff Σ is diagonal.
(b) Principal Components, Canonical Correlation and Data Reduction
Principal components are uncorrelated linear combinations a'X of maximum variance subject to a'a=1 and zero correlation with earlier components. Their coefficients are eigenvectors of the covariance matrix Σ, and their variances are the corresponding eigenvalues. Canonical correlation between two vectors X and Y is the maximum correlation between linear combinations a'X and b'Y; successive canonical correlations are obtained subject to uncorrelatedness with earlier canonical variates.
Data reduction by principal components: sort eigenvalues λ₁≥λ₂≥...≥λₚ, retain the first k components whose cumulative variance proportion (λ₁+...+λₖ)/(λ₁+...+λₚ) is sufficiently large, and replace the original p variables by these k components. If units differ, use the correlation matrix.
For Σ=[[1,ρ],[ρ,1]],
det(Σ-λI)=(1-λ)²-ρ²=0, so λ=1+ρ, 1-ρ.
Assuming ρ≥0, the eigenvector for λ₁=1+ρ is (1,1)/√2, and for λ₂=1-ρ is (1,-1)/√2.
Final: PC₁=(X₁+X₂)/√2 with variance 1+ρ, and PC₂=(X₁-X₂)/√2 with variance 1-ρ. If ρ<0, their order is reversed.
(c) Least Squares and Maximum Likelihood Estimates of β₀ and β₁
For the model yᵢ=β₀+β₁xᵢ+εᵢ, least squares minimizes
S(β₀,β₁)=Σ(yᵢ-β₀-β₁xᵢ)².
The normal equations are
Σyᵢ=nβ₀+β₁Σxᵢ,
Σxᵢyᵢ=β₀Σxᵢ+β₁Σxᵢ².
Solving, with x̄=Σxᵢ/n, ȳ=Σyᵢ/n,
β̂₁=[Σ(xᵢ-x̄)(yᵢ-ȳ)]/[Σ(xᵢ-x̄)²],
β̂₀=ȳ-β̂₁ x̄.
For maximum likelihood, assume εᵢ are independent N(0,σ²). Then
L(β₀,β₁,σ²)=(2πσ²)^(-n/2) exp[-Σ(yᵢ-β₀-β₁xᵢ)²/(2σ²)].
Maximizing L over β₀,β₁ is equivalent to minimizing Σ(yᵢ-β₀-β₁xᵢ)². Hence the same normal equations result.
Final: β̂₀ and β̂₁ are identical by least squares and by maximum likelihood under normal errors; the MLE of σ² is Σ(yᵢ-β̂₀-β̂₁ xᵢ)²/n. Condition: Σ(xᵢ-x̄)²>0.
How this answer will be evaluated
What earns full marks: Rigorous derivations with all steps shown; correct calculations; clear interpretation of results.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a)(i) (a)(i) Show X and Y are independent iff ρ=0 for BN(μ₁,μ₂,σ₁²,σ₂²,ρ) (6 marks) 6 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Writes the bivariate normal density correctly
- With ρ=0 the density factorises into two univariate normal densities, so X and Y are independent
- Conversely independence gives Cov(X,Y)=ρσ₁σ₂=0
- Concludes ρ=0 since σ₁,σ₂ > 0
Loses marks
- Proving only one direction
- Claiming zero correlation implies independence in general
Earns more
- Stresses that zero correlation implies independence only under joint normality
- Clear two-direction structure
Extra mark
- Counter-example of uncorrelated but dependent non-normal variables
- (a)(ii) (a)(ii) P(3<Y<8 | X=7) for BN(3,1,16,25,3/5) (6 marks) 6 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Conditional distribution Y|X=x ~ N(μ₂+ρ(σ₂/σ₁)(x−μ₁), σ₂²(1−ρ²))
- Conditional mean 4 and variance 16 (sd 4)
- Standardised limits −0.25 and 1
- Probability Φ(1)−Φ(−0.25)=0.8413−0.4017=0.4396
Loses marks
- Using the marginal instead of the conditional distribution
- Taking σ₁=16, σ₂=25 as standard deviations
Earns more
- Reads Φ(1) from the supplied table
- Notes the given Φ(2) does not match the limits
Extra mark
- Contrasts with the unconditional probability
- (a)(iii) (a)(iii) Distribution of Z=Y/X when (X,Y) ~ BN(0,0,1,1,0) (4 marks) 4 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- X and Y are independent N(0,1)
- Transformation z=y/x, w=x with Jacobian |w|
- Integrates to f(z)=1/[π(1+z²)]
- Identifies the standard Cauchy distribution
Loses marks
- Dropping the absolute value of the Jacobian
- Claiming Z is normal or t
Earns more
- Correct handling of |w| by symmetry
- Range −∞ < z < ∞ stated
Extra mark
- Notes the Cauchy has no finite mean
- (a)(iv) (a)(iv) Multivariate extension of (i) when X ~ N_p(μ,Σ) (4 marks) 4 marks
explain— definition/context → points in order → small example → short close
Must cover
- Partitions X into subvectors X₁, X₂ with blocks Σ₁₁, Σ₁₂, Σ₂₂
- X₁ and X₂ independent iff Σ₁₂=0
- All components mutually independent iff Σ is diagonal
- Relies on joint multivariate normality
Loses marks
- Stating pairwise zero correlations without joint normality
- Confusing Σ₁₂=0 with Σ=I
Earns more
- Mentions factorisation of the density or MGF
- Clear notation
Extra mark
- Brief proof sketch via the characteristic function
- (b) (b) Define principal components and canonical correlation; data reduction by PCs; PCs for Σ=[[1,ρ],[ρ,1]] (15 marks) 15 marks
define— precise definition → the distinguishing feature → one example
Must cover
- PCs: uncorrelated normalised linear combinations of maximum variance given by eigenvectors of Σ
- Canonical correlation: maximum correlation between a′X and b′Y, with successive uncorrelated canonical variates
- Data reduction: retain the first k PCs explaining a large share of total variance
- Eigenvalues 1±ρ; PC₁=(X₁+X₂)/√2 and PC₂=(X₁−X₂)/√2 (order reversed if ρ<0)
Loses marks
- Not normalising the eigenvectors
- Confusing PCA with factor analysis or canonical correlation
Earns more
- Mentions scree plot or Kaiser criterion
- Advises the correlation matrix when units differ
Extra mark
- Notes the proportion of variance explained (1+ρ)/2
- (c) (c) LS and ML estimates of β₀, β₁ in y=β₀+β₁x+ε; are they identical? (15 marks) 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Normal equations Σy=nβ₀+β₁Σx and Σxy=β₀Σx+β₁Σx²
- β̂₁=Sxy/Sxx and β̂₀=ȳ−β̂₁x̄
- Normal errors assumed for ML; maximising L in β₀, β₁ equals minimising the RSS
- Concludes β estimates coincide, while σ̂²_ML=RSS/n differs from the unbiased RSS/(n−2)
Loses marks
- Claiming they are identical without assuming normal errors
- Algebraic slips in solving the normal equations
Earns more
- States Σ(xᵢ−x̄)² > 0 as the condition
- Notes ML needs a distributional assumption whereas LS does not
Extra mark
- Mentions Gauss–Markov BLUE property
What "Prove" is asking you to do
Establish that the statement holds for every case it claims, not for one representative case. The argument must be closed: each line follows from a definition, a hypothesis, or a named theorem you are entitled to use.
Structure that answers it
Given and to prove, restated → theorem or construction to be used, named → the argument line by line → conclusion stated as proved
Where marks are lost
Testing one example, which illustrates but proves nothing. On an if and only if claim, proving one direction and stopping forfeits that half outright, and degenerate cases — zero, the empty set, the equality case — have to be disposed of rather than assumed away.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
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