(a) What is the role of properties of completeness and sufficiency in Statistical Inference ? Explain. In U (0, θ), find out Uniformly Minimum Variance Unbiased Estimator (UMVUE) of θ.

(20 marks)

(b) A survey of 400 families with four children each

Question

(a) What is the role of properties of completeness and sufficiency in Statistical Inference ? Explain. In U (0, θ), find out Uniformly Minimum Variance Unbiased Estimator (UMVUE) of θ.

(20 marks)

(b) A survey of 400 families with four children each have the following distribution :

| Number of boys | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Number of families | 16 | 89 | 145 | 118 | 32 |

Is this result consistent with the hypothesis that male and female births are equally probable at 5% level of significance ?

It is given that χ²_(.05) for 4 degrees of freedom = 9·488 and χ²_(.05) for 5 degrees of freedom = 11·070.

(c) Define Likelihood Ratio Test. In N(θ, σ²), where σ² is unknown, find out LR test for testing H₀ : θ = θ₀ against H₁ : θ ∈ (Ω – θ₀), where Ω is the parametric space for θ. α is the size of the test.

UPSC Answer Check · Accepted Answer

(a) Completeness, Sufficiency and UMVUE in U(0, θ) A statistic T is sufficient for θ if the conditional distribution of the sample X, given T, is free of θ. Sufficiency means T captures all information about θ contained in the data, so no other function of the sample can add information about θ. A statistic T is complete if E_θ[g(T)] = 0 for all θ implies g(T) = 0 almost surely. Completeness guarantees that any unbiased estimator based on T is unique, and hence of minimum variance. Their combined role is central to inference. The Rao–Blackwell theorem says that if T is sufficient and W is any unbiased estimator, then E[W|T] is unbiased with variance no greater than W. The Lehmann–Scheffé theorem strengthens this: if T is complete and sufficient and h(T) is unbiased for θ, then h(T) is the UMVUE. Thus sufficiency permits reduction of data without loss of information, while completeness ensures uniqueness and minimal variance. For X₁,…,Xₙ ~ U(0, θ), the joint density is θ⁻ⁿ for 0 < xᵢ < θ. By the factorisation criterion, T = X₍ₙ₎ = max(Xᵢ) is sufficient. Its density is f(t) = n tⁿ⁻¹/θⁿ, 0 < t < θ. For completeness, suppose E[g(T)] = 0 for all θ, i.e. (n/θⁿ)∫₀^θ g(t)tⁿ⁻¹dt = 0, so ∫₀^θ g(t)tⁿ⁻¹dt = 0 for all θ. Differentiating with respect to θ gives g(θ)θⁿ⁻¹ = 0, hence g(θ) = 0. Thus T is complete and sufficient. Now E[X₍ₙ₎] = ∫₀^θ t·n tⁿ⁻¹/θⁿ dt = nθ/(n+1). Hence E[((n+1)/n)X₍ₙ₎] = θ, so it is unbiased. By Lehmann–Scheffé, since X₍ₙ₎ is complete and sufficient, the UMVUE of θ is θ̂ = ((n+1)/n)X₍ₙ₎. (b) Chi-square goodness-of-fit test Let p be the probability of a male birth. Under H₀: p = 0.5, the number of boys in a family of four follows Binomial(4, 0.5). Expected proportions are 1/16, 4/16, 6/16, 4/16, 1/16. For 400 families, expected frequencies are 25, 100, 150, 100, 25. Observed: 16, 89, 145, 118, 32. χ² = Σ(Oᵢ − Eᵢ)²/Eᵢ = (16−25)²/25 + (89−100)²/100 + (145−150)²/150 + (118−100)²/100 + (32−25)²/25 = 81/25 + 121/100 + 25/150 + 324/100 + 49/25 = 3.24 + 1.21 + 0.1667 + 3.24 + 1.96 = 9.8167. Degrees of freedom = 5 − 1 = 4 (no parameter estimated). Critical value χ²₀.₀₅,₄ = 9.488. Since 9.8167 > 9.488, we reject H₀ at the 5% level. The result is not consistent with the hypothesis that male and female births are equally probable. (c) Likelihood Ratio Test The Likelihood Ratio Test uses λ(x) = sup_θ∈Θ₀ L(θ) / sup_θ∈Θ L(θ), where Θ₀ is the null parameter space and Θ the full space. We reject H₀ when λ ≤ k, with k chosen so the test has size α. Let X₁,…,Xₙ ~ N(θ, σ²), both unknown. Test H₀: θ = θ₀ against H₁: θ ≠ θ₀. Under Θ, the MLEs are θ̂ = x̄ and σ̂² = (1/n)Σ(xᵢ − x̄)². Under Θ₀, θ = θ₀ and σ̂₀² = (1/n)Σ(xᵢ − θ₀)². The likelihood ratio simplifies to λ = (σ̂₀² / σ̂²)^(−n/2) = [Σ(xᵢ − x̄)² / Σ(xᵢ − θ₀)²]^(n/2). Now Σ(xᵢ − θ₀)² = Σ(xᵢ − x̄)² + n(x̄ − θ₀)². Hence λ^(2/n) = Σ(xᵢ − x̄)² / [Σ(xᵢ − x̄)² + n(x̄ − θ₀)²] = 1 / [1 + t²/(n−1)], where t = (x̄ − θ₀)/(s/√n) and s² = Σ(xᵢ − x̄)²/(n−1). λ ≤ k ⟺ t² ≥ constant, so the rejection region is |t| > t_α/2, n−1. Thus the LRT reduces to the familiar Student's t-test with n − 1 degrees of freedom, rejecting H₀ when |t| exceeds the two-tailed critical value at level α.

(a) What is the role of properties of completeness and sufficiency in Statistical Inference ? Explain. In U (0, θ), find out…

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