All of Statistics by Larry Wasserman

Author:Larry Wasserman
Language: eng
Format: epub, pdf
Publisher: Springer-Verlag Wien 2012
Published: 2014-06-11T16:00:00+00:00

12.6 Admissibility

Minimax estimators and Bayes estimators are “good estimators” in the sense that they have small risk. It is also useful to characterize bad estimators.

12.17 Definition. An estimator is inadmissible if there exists another rule such that

Otherwise, is admissible.

12.18 Example. Let X ∼ N(θ, 1) and consider estimating θ with squared error loss. Let . We will show that is admissible. Suppose not. Then there exists a different rule with smaller risk. In particular, . Hence, . Thus, . So there is no rule that beats . Even though is admissible it is clearly a bad decision rule. ■

12.19 Theorem (Bayes Rules Are Admissible). Suppose that Θ ⊂ ℝ and that is a continuous function of θ for every Let f be a prior density with full support, meaning that, for every θ and every . Let be the Bayes’ rule. If the Bayes risk is finite then is admissible.

Proof. Suppose is inadmissible. Then there exists a better rule such that for all θ and for some θ0. Let . Since R is continuous, there is an ϵ > 0 such that for all θ ∈ (θ0 − ϵ, θ0 + ϵ). Now,

Hence, . This implies that does not minimize which contradicts the fact that is the Bayes rule. ■

12.20 Theorem. Let X1,..., Xn ∼N(μ, σ2). Under squared error loss, is admissible.

The proof of the last theorem is quite technical and is omitted but the idea is as follows: The posterior mean is admissible for any strictly positive prior. Take the prior to be N(a, b2). When b2 is very large, the posterior mean is approximately equal to .

How are minimaxity and admissibility linked? In general, a rule may be one, both, or neither. But here are some facts linking admissibility and minimaxity.

12.21 Theorem. Suppose that has constant risk and is admissible. Then it is minimax.

Proof. The risk is for some c. If were not minimax then there exists a rule such that

This would imply that is inadmissible. ■

Now we can prove a restricted version of Theorem 12.14 for squared error loss.

12.22 Theorem. Let X1,..., Xn ∼ N(θ, 1). Then, under squared error loss, is minimax.

Proof. According to Theorem 12.20, is admissible. The risk of is 1/n which is constant. The result follows from Theorem 12.21. ■

Although minimax rules are not guaranteed to be admissible they are “close to admissible.” Say that is strongly inadmissible if there exists a rule and an ϵ > 0 such that for all θ.

12.23 Theorem. If is minimax, then it is not strongly inadmissible.

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