Paper I — Q5
(a) Define general linear model with usual assumptions. If y₁ = β₁ + u₁, y₂ = –β₁ + β₂ + u₂, y₃ = –β₂ + u₃, where u₁, u₂, u₃ are…
Define general linear model with usual assumptions. If y₁ = β₁ + u₁, y₂ = –β₁ + β₂ + u₂, y₃ = –β₂ + u₃, where u₁, u₂, u₃ are mutually independent random variables with mean zero and variance σ², then find the least square estimators of β₁ and β₂. 10 marks
Given X ~ N₃(μ, Σ), where μ = (2, 4, 3)' and Σ = ⎛8 2 3⎞ ⎜2 4 1⎟ ⎝3 1 3⎠ find the regression function of X₁ on X₂ and X₃, and (ii) compute the conditional variance of X₁ given X₂ and X₃. 10 marks
What is a uniformity trial ? Explain how it can be used to determine optimum shape and size. 10 marks
In a 2⁶ – factorial experiment, the key block is given as : (1), ab, cd, ef, ace, abef, abcd, bce, cdef, acf, ade, abcdef, bde, bcf, adf, bdf. Identify the confounded effects. 10 marks
If the coefficients of variation of x and y are equal and the correlation coefficient between x and y is ρ = 2/3, compute the efficiency of ratio estimator relative to the mean of a simple random sample. 10 marks
हिंदी में प्रश्न पढ़ें
सामान्य रैखिक निदर्श को प्रचलित कल्पनाओं सहित परिभाषित कीजिए । यदि y₁ = β₁ + u₁, y₂ = –β₁ + β₂ + u₂, y₃ = –β₂ + u₃, जहाँ u₁, u₂, u₃ परस्पर स्वतंत्र यादृच्छिक चर हैं जिनका माध्य शून्य तथा प्रसरण σ² है, तो β₁ और β₂ के न्यूनतम वर्ग आकलकों को ज्ञात कीजिए । (10 अंक)
दिया गया है कि X ~ N₃(μ, Σ), जहाँ μ = (2, 4, 3)' और Σ = ⎛8 2 3⎞ ⎜2 4 1⎟ ⎝3 1 3⎠ X₁ का X₂ और X₃ पर समाश्रयण फलन ज्ञात कीजिए, और (ii) X₂ और X₃ के दिए होने पर X₁ के सप्रतिबंध प्रसरण की गणना कीजिए । (10 अंक)
एकसमानता परीक्षण क्या है ? व्याख्या कीजिए कि इसका उपयोग इष्टतम आकृति और आकार ज्ञात करने के लिए कैसे किया जा सकता है । (10 अंक)
एक 2⁶ – बहु-उपादानी प्रयोग में, मुख्य खंडक इस प्रकार दिया गया है : (1), ab, cd, ef, ace, abef, abcd, bce, cdef, acf, ade, abcdef, bde, bcf, adf, bdf. संकरित प्रभावों की पहचान कीजिए । (10 अंक)
यदि x और y के विचरण गुणांक समान हैं और x और y के बीच सहसंबंध गुणांक ρ = 2/3 है, तो अनुपात आकलक की दक्षता सरल यादृच्छिक प्रतिदर्श के माध्य के सापेक्ष परिकलित कीजिए । (10 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) General linear model and least squares estimators
The general linear model is y = Xβ + u, where y is an n × 1 vector of observations, X an n × p matrix of known constants of rank p, β a p × 1 vector of unknown parameters and u an n × 1 vector of errors. Usual assumptions: E(u) = 0 and Var(u) = σ²Iₙ, i.e. the errors are uncorrelated with a common variance (normality is needed only for inference).
Minimising (y − Xβ)′(y − Xβ) gives the normal equations X′Xβ̂ = X′y, so β̂ = (X′X)⁻¹X′y.
Here X = [[1, 0], [−1, 1], [0, −1]], so X′X = [[2, −1], [−1, 2]], |X′X| = 3, (X′X)⁻¹ = (1/3)[[2, 1], [1, 2]] and X′y = (y₁ − y₂, y₂ − y₃)′. Hence β̂₁ = [2(y₁ − y₂) + (y₂ − y₃)]/3 = (2y₁ − y₂ − y₃)/3 β̂₂ = [(y₁ − y₂) + 2(y₂ − y₃)]/3 = (y₁ + y₂ − 2y₃)/3
(b) Regression function and conditional variance
Let a = (2, 3) be the covariances of X₁ with (X₂, X₃)′, and T = [[4, 1], [1, 3]] the covariance matrix of (X₂, X₃)′. For a normal vector, E(X₁ | X₂, X₃) = μ₁ + aT⁻¹[(X₂, X₃)′ − (4, 3)′], Var(X₁ | X₂, X₃) = σ₁₁ − aT⁻¹a′.
det T = 12 − 1 = 11, T⁻¹ = (1/11)[[3, −1], [−1, 4]], so aT⁻¹ = (1/11)(6 − 3, −2 + 12) = (3/11, 10/11).
(i) Regression function: E(X₁ | X₂, X₃) = 2 + (3/11)(X₂ − 4) + (10/11)(X₃ − 3) = (3X₂ + 10X₃ − 20)/11.
(ii) aT⁻¹a′ = (3/11)(2) + (10/11)(3) = 36/11, so Var(X₁ | X₂, X₃) = 8 − 36/11 = 52/11 ≈ 4.73, smaller than Var(X₁) = 8 and free of the values of X₂, X₃.
(c) Uniformity trial for plot size and shape
A uniformity trial grows one crop variety under identical management on a field and records the yield of each small basic unit (e.g. 1 m²). It measures only soil variability, not treatment effects.
Size: adjacent units are pooled into plots of x units and the variance per unit area V(x) is found for each x. Smith's law is V(x) = V₁/xᵇ, where b (0 < b < 1) is the negative of the slope of log V(x) on log x: b near 1 means adjacent units are almost uncorrelated, so larger plots cut variance sharply; b near 0 means strongly correlated units, so they help little. With cost K₁ per plot and K₂ per unit area, the variance of a treatment mean for a given total cost is least at x = bK₁/[(1 − b)K₂].
Shape: plots of equal area but different shapes (1×4, 2×2, 4×1 units) are compared and the shape with least variance is chosen.
(d) Confounded effects
The key block has 16 = 2⁴ treatments, so there are 4 blocks and 3 effects are confounded (two independent effects and their generalised interaction).
An effect is confounded if each key-block treatment contains an even number of its letters, so that the effect has the same sign throughout the block. From ab, cd, ef: the effect contains a and b together or neither, likewise c, d and e, f. It is therefore ab, cd, ef or a union of these pairs. From ace: the count of a, c, e in the effect must be even. ab, cd, ef each give 1 (odd), abcdef gives 3 (odd); abcd, abef, cdef each give 2 (even). The remaining 12 treatments also contain an even number of letters of each of these three effects.
Confounded effects: abcd, abef, cdef (abcd × abef = cdef). All are four-factor interactions, so main effects and two-factor interactions remain estimable.
(e) Efficiency of the ratio estimator
With sampling fraction f, V(ȳ) = (1 − f)S_y²/n and, to first order, V(ȳ_R) = [(1 − f)/n][S_y² + R²S_x² − 2RρS_xS_y], where R = Ȳ/X̄.
Equal coefficients of variation: S_x/X̄ = S_y/Ȳ, so RS_x = S_y. Then V(ȳ_R) = [(1 − f)S_y²/n](1 + 1 − 2ρ) = 2(1 − ρ)V(ȳ) = (2/3)V(ȳ) for ρ = 2/3.
Efficiency = V(ȳ)/V(ȳ_R) = 1/[2(1 − ρ)] = 3/2 = 1.5, i.e. 150%.
The ratio estimator has two-thirds of the variance of the sample mean; with equal CVs it is better whenever ρ > 1/2.
How this answer will be evaluated
What earns full marks: All parts fully answered with correct derivations, calculations, and interpretations.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Define the general linear model with its usual assumptions and find the least squares estimators of β₁ and β₂ for the given three-equation model. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Model y = Xβ + u with E(u) = 0 and Var(u) = σ²I
- Normal equations X′Xβ̂ = X′y with X = [[1, 0], [−1, 1], [0, −1]] and X′X = [[2, −1], [−1, 2]]
- β̂₁ = (2y₁ − y₂ − y₃)/3 and β̂₂ = (y₁ + y₂ − 2y₃)/3
Loses marks
- Assumptions omitted
- Arithmetic slip in X′X, its inverse or X′y
Earns more
- Rank condition for (X′X)⁻¹ and |X′X| = 3
- Remark that normality is needed only for inference
Extra mark
- Residuals orthogonal to the columns of X
- (b) Find the regression function of X₁ on X₂ and X₃ and the conditional variance of X₁ given X₂ and X₃ for the given N₃(μ, Σ). 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Conditional mean μ₁ + aT⁻¹(x − μ) with a = (2, 3) and T = [[4, 1], [1, 3]]
- Regression function (3X₂ + 10X₃ − 20)/11
- Conditional variance 8 − 36/11 = 52/11
Loses marks
- Inverting the full Σ instead of T
- Not subtracting the means
Earns more
- T⁻¹ = (1/11)[[3, −1], [−1, 4]] with |T| = 11
- Noting the conditional variance does not depend on X₂, X₃
Extra mark
- Squared multiple correlation 36/88 = 9/22
- (c) Define a uniformity trial and explain how it is used to determine the optimum shape and size of plots. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Definition: uniform crop and management on the whole field, yield recorded on small basic units, measuring soil variability only
- Pooling units into plots of different sizes and shapes and computing the variance per unit area V(x)
- Smith's law V(x) = V₁/xᵇ with the meaning of b, and optimum size from costs; shape by least variance at equal area
Loses marks
- Confusing the uniformity trial with the main treatment experiment
- Taking b near 1 to mean strongly correlated adjacent plots
Earns more
- Optimum size x = bK₁/[(1 − b)K₂]
- Comparison of shapes of equal area
Extra mark
- Maximum-curvature method as an alternative for plot size
- (d) Identify the confounded effects from the given key block of 16 treatments of the 2⁶ factorial experiment. 10 marks
analyse— intro → causes → effects → stakeholders/linkages → way forward
Must cover
- 16 = 2⁴ treatments, so 4 blocks and 3 confounded effects
- Confounded effects have an even number of letters in common with every key-block treatment
- abcd, abef and cdef, with abcd × abef = cdef
Loses marks
- Listing fewer or more than three effects
- Naming an effect with an odd letter count for some key-block treatment
Earns more
- Using ab, cd, ef and ace to narrow the candidates
- Modulo 2 multiplication to obtain the generalised interaction
Extra mark
- Noting that all three are four-factor interactions
- (e) Compute the efficiency of the ratio estimator relative to the simple random sample mean when the coefficients of variation of x and y are equal and ρ = 2/3. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- V(ȳ) = (1 − f)S_y²/n and V(ȳ_R) ≈ [(1 − f)/n][S_y² + R²S_x² − 2RρS_xS_y]
- Equal CVs give RS_x = S_y, so V(ȳ_R) = 2(1 − ρ)V(ȳ)
- Efficiency 1/[2(1 − ρ)] = 3/2 = 150%
Loses marks
- Efficiency inverted to 2/3
- Omitting R = Ȳ/X̄ in the variance formula
Earns more
- First-order (large-sample) form of the ratio-estimator variance
- Interpretation: variance two-thirds of the sample mean's; ratio estimator better when ρ > 1/2
Extra mark
- Ratio estimator is biased but the bias is negligible for large n
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