(a) Let X₁, X₂, ..., Xₙ be a random sample from Poisson distribution with mean λ > 0. Define a statistic W = (1 − 1/n)^T, T =…

(a) Poisson sample: complete sufficiency, UMVUE of e^(−λ) and the CRLB

The question is read as: (ii) W is unbiased for e^(−λ), and (iii) W is the UMVUE.

(i) The joint pmf is [e^(−nλ) λ^(Σxᵢ)] × [1/Πxᵢ!], so T = ΣXᵢ is sufficient by the factorisation theorem, and T ~ Poisson(nλ). If E_λ[φ(T)] = 0 for all λ > 0, then Σₜ [φ(t)nᵗ/t!] λᵗ = 0 for all λ > 0. A power series vanishing on an interval has all coefficients zero, so φ(t) = 0 for every t. Hence T is complete and sufficient.

(ii) Using E(sᵀ) = e^(nλ(s−1)) with s = 1 − 1/n, E(W) = e^(nλ(−1/n)) = e^(−λ). W is unbiased and a function of the complete sufficient statistic T, so by the Lehmann–Scheffé theorem it is the UMVUE of e^(−λ).

(iii) The Fisher information per observation is I(λ) = 1/λ, so for the sample it is n/λ. With g(λ) = e^(−λ), g′ = −e^(−λ), and CRLB = (g′)²/(n/λ) = λe^(−2λ)/n. Now E(W²) = e^(nλ[(1−1/n)²−1]) = e^(−2λ+λ/n), so Var(W) = e^(−2λ)(e^(λ/n) − 1) = e^(−2λ)(λ/n + λ²/(2n²) + ⋯) > λe^(−2λ)/n. For example, n = 5 and λ = 1 give Var(W) = 0.0300 against a CRLB of 0.0271. Reason: the bound is attained only when the score is linear in the estimator, ∂log L/∂λ = (n/λ)(T/n − λ), which allows efficient estimation only of linear functions of λ. Since e^(−λ) is not linear in λ, even the UMVUE exceeds the bound, which is approached only as n → ∞.

(b) Normal scale mixture: marginals, conditional and moments

(i) Using ∫e^(−yx²/2)dx = √(2π/y), f_Y(y) = [y^(3/2)e^(−y)/√(2π)] × √(2π/y) = y e^(−y), y > 0, which is Gamma(shape 2, scale 1). The conditional density is f(x | y) = √(y/2π) e^(−yx²/2), so X | Y = y ~ N(0, 1/y).

(ii) E(Y) = ∫y²e^(−y)dy = 2; E(Y²) = Γ(4) = 6, so V(Y) = 6 − 4 = 2. E(X | Y) = 0 and V(X | Y) = 1/Y.

(iii) E(X) = E[E(X | Y)] = 0. V(X) = E[V(X | Y)] + V[E(X | Y)] = E(1/Y) + 0, and E(1/Y) = ∫(1/y)·y e^(−y)dy = 1. So V(X) = 1. X is a scale mixture of normals with heavier tails than N(0, 1); √2·X follows Student's t with 4 degrees of freedom.

(c) Kruskal–Wallis test for three training methods

The three samples are independent, of unequal sizes, with no normality assumed, so the Kruskal–Wallis H test is used.

H₀: the three methods are equally effective. H₁: at least one differs in location.

Rank all N = 18 scores together (no ties):

• Method A (n = 6): ranks 9, 11, 8, 13, 16, 4, so R_A = 61.
• Method B (n = 7): ranks 10, 12, 1, 5, 14, 3, 17, so R_B = 62.
• Method C (n = 5): ranks 18, 2, 6, 7, 15, so R_C = 48. Check: 61 + 62 + 48 = 171 = 18 × 19/2.

H = [12/(N(N+1))] ΣRᵢ²/nᵢ − 3(N+1) = (12/342)(61²/6 + 62²/7 + 48²/5) − 57 = (12/342)(1630.11) − 57 = 0.197.

Under H₀, H is approximately χ² with k − 1 = 2 d.f.; the 5% critical value is 5.991 (p ≈ 0.91). Since 0.197 < 5.991, H₀ is not rejected: the data do not show a difference, and the three methods may be regarded as equally effective.