(a) (i) What are orthogonal polynomials ? How do you fit an orthogonal polynomial of degree 'p' ? (ii) For the model Y_(n×1) =…

(a) Orthogonal polynomials and the residual sum of squares

(i) Orthogonal polynomials are polynomials ξ₀(x) = 1, ξ₁(x), …, ξ_p(x) of degrees 0, 1, …, p that are mutually orthogonal over the observed values x₁, …, x_n: Σᵢ ξ_r(xᵢ)ξ_s(xᵢ) = 0 for r ≠ s. For equally spaced x with spacing h, put u = (x − x̄)/h: ξ₁ = λ₁u, ξ₂ = λ₂[u² − (n² − 1)/12], ξ₃ = λ₃[u³ − (3n² − 7)u/20], …, the constants λ_r being chosen to make the values integers. For n = 5: ξ₁ = (−2, −1, 0, 1, 2), ξ₂ = (2, −1, −2, −1, 2), ξ₃ = (−1, 2, 0, −2, 1), ξ₄ = (1, −4, 6, −4, 1). To fit degree p, write y = a₀ξ₀ + a₁ξ₁ + … + a_pξ_p + e. Orthogonality makes X′X diagonal, so each coefficient is found separately: â_r = Σyξ_r/Σξ_r², with sum of squares (Σyξ_r)²/Σξ_r² on 1 d.f. These sums of squares are independent and the lower-order coefficients do not change when a higher degree is added. Each term is tested against the error mean square, the degree is raised until the next term is insignificant, and the fit is converted back to powers of x.

(ii) Let M = Iₙ − X(X′X)⁻¹X′. M is symmetric and idempotent, MX = 0, and tr M = n − tr[(X′X)⁻¹X′X] = n − k. Since Y = Xβ + u, MY = Mu, so Y′MY = u′Mu. With E(u) = 0 and E(uu′) = σ²Iₙ, E(u′Mu) = tr[M E(uu′)] = σ² tr M. E[Y′(Iₙ − X(X′X)⁻¹X′)Y] = (n − k)σ², so Y′MY/(n − k) is an unbiased estimator of σ².

(b) Horvitz–Thompson estimator with replacement sampling

There are 9 ordered samples (i, j), each with probability pᵢpⱼ. The inclusion probability of unit i in two draws is πᵢ = 1 − (1 − pᵢ)²: π₁ = 0.51, π₂ = 0.36, π₃ = 0.75. The estimator is Ŷ_HT = Σyᵢ/πᵢ over the distinct units in the sample, with yᵢ/πᵢ = 21.569, 16.667, 33.333 for i = 1, 2, 3.

Sample: probability; Ŷ_HT

• (1,1): 0.09; 21.569
• (1,2): 0.06; 38.235
• (1,3): 0.15; 54.902
• (2,1): 0.06; 38.235
• (2,2): 0.04; 16.667
• (2,3): 0.10; 50.000
• (3,1): 0.15; 54.902
• (3,2): 0.10; 50.000
• (3,3): 0.25; 33.333

The probabilities add to 1, and E(Ŷ_HT) = 0.09(21.569) + 0.12(38.235) + 0.30(54.902) + 0.04(16.667) + 0.20(50.000) + 0.25(33.333) = 42.000 = 11 + 6 + 25, the population total. In general E(Ŷ_HT) = Σᵢ yᵢP(i ∈ s)/πᵢ = Σᵢ yᵢ, so the estimator is unbiased.

(c) Missing plot technique in a Latin square

If an observation is lost for reasons unrelated to treatment (e.g. a damaged plot), the missing plot technique estimates it by the value that minimises the error sum of squares (Yates) and then analyses the data as a complete design with adjusted degrees of freedom.

Derivation (m × m Latin square): let x be the missing yield, and R, C, T, G the totals of the available observations in its row, its column, its treatment and overall. Then SSE = Σy² − (ΣR_i² + ΣC_j² + ΣT_k²)/m + 2G′²/m², G′ = G + x. The part involving x is x² − (R + x)²/m − (C + x)²/m − (T + x)²/m + 2(G + x)²/m². Setting its derivative to zero and multiplying by m²/2: m²x − m(R + C + T + 3x) + 2(G + x) = 0, so x = [m(R + C + T) − 2G]/[(m − 1)(m − 2)].

Analysis:

1. Insert x and do the usual Latin-square ANOVA (rows, columns, treatments, error).
2. Reduce total and error d.f. by 1: total m² − 2, error (m − 1)(m − 2) − 1.
3. The treatment SS from the filled data is biased upwards; subtract [G − R − C − (m − 1)T]²/[(m − 1)²(m − 2)²] before the F-test.
4. The standard error of the difference between the treatment having the missing plot and any other is √{MSE[2/m + 1/((m − 1)(m − 2))]}.