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Some Remarks on Sufficiency & Completeness Sufficiency Remarks 1. If X1 , . . . , Xn is a random sample (iid) from pdf/pmf f (x|θ), θ ∈ Θ, then the ˜ ˜ order statistics X(1) , . . . , X(n) are sufficient for θ. ˜ By the factorization theorem, X(1) , . . . , X(n) are sufficient for θ because we can write ˜ n n Y Y the joint pdf/pmf f (x|θ) = f (xi |θ) = f (x(i) |θ) = g(x(1) , . . . , x(n) , θ) h(x) ., ∀ x, θ ˜˜ ˜ ˜ ˜} |{z} ˜ ˜ ˜ i=1 i=1 | {z Qn f (x(i) |θ) 1 i=1 ˜ proof: 2. If S = (S1, S2, . . . , Sk ) is sufficient for real-valued θ ∈ Θ ⊂ R, then any Bayes ˜ estimator is a function of S . ˜ Example: From homework, consider X1 , . . . , Xn iid Bernoulli(θ), 0 < θ < 1; loss L(t, θ) = (t−θ)2 ; θ(1−θ) and uniform(0,1) prior π(θ). Then the Bayes estimator is T0 = X̄n , which is sufficient for θ (by factorization theorem). 3. If S = (S1, S2 , . . . , Sk ) is sufficient for θ ∈ Θ ⊂ Rp and θ̂ is the unique MLE of θ, ˜ ˜ ˜ then θ̂ is a function of S . ˜ Completeness Remarks 1. If T is complete, then T is boundedly complete; the converse is false. 2. If T is sufficient and boundedly complete, then T is minimal sufficient. 3. Suppose T is complete and h1 (T ), h2(T ) are two estimators of γ(θ) ˜ if Eθ h1 (T ) = γ(θ) = Eθ h2 (T ), ∀θ ∈ Θ ˜ ˜ ˜ ˜ ⇒ Eθ u(T ) = 0, ∀θ ∈ Θ, where u(T ) = h1 (T ) − h2 (T ) ˜ ˜ ⇒ Pθ u(T ) = 0 = 1, ∀θ ∈ Θ ˜ ˜ ⇒ Pθ h1 (T ) = h2 (T ) = 1, ∀θ ∈ Θ ˜ ˜ Hence, there can be at most one (i.e., unique) UE of a parametric function γ(θ) ˜ that is a function of a complete statistic. 1