Simultaneous Estimation
[14/0%]In Example 4.2, show that an estimator is equivariant if and only if it satisfies (4.11) and (4.12).
Show that a function satisfies (4.12) if and only if it depends only on .
Verify the Bayes estimator (4.15).
Let be independent with binomial distribution . For estimating with average squared error loss (4.17), find the minimax estimator of , and determine whether it is admissible.
Establishing the admissibility of the normal mean in two dimensions is quite difficult, made so by the fact that the conjugate priors fail in the limiting Bayes method. Let
The conjugate priors are .
For this sequence of priors, verify that the limiting Bayes argument, as in Example 2.8, results in inequality (4.18), which does not establish admissibility.
Stein (in James and Stein 1961), proposed the sequence of priors that works to prove is admissible by the limiting Bayes method. A version of these priors, given by Brown and Hwang (1982), is
for . Show that a.e. as .
A special case of the very general results of Brown and Hwang (1982) state that for the prior , the limiting Bayes method (Blyth's method) will establish the admissibility of the estimator [the generalized Bayes estimator against ] if
Show that this holds for and that , so is admissible.
Stein (1956b) originally established the admissibility of in two dimensions using an argument based on the information inequality. His proof was complicated by the fact that he needed some additional invariance arguments to establish the result. See Theorem 7.19 and Problem 7.19 for more general statements of the Brown/Hwang result.
Let be independent with . The following heuristic argument, due to Stein (1956b), suggests that it should be possible, at least for large and hence large , to improve on the estimator .
Use a Taylor series argument to show
so, with high probability, the true is in the sphere . The usual estimator is approximately the same size as and will almost certainly be outside of this sphere.
Part (a) suggested to Stein an estimator of the form . Show that
Establish that , and , and, hence,
Show that the leading term in part (c) is minimized at , and since , this leads to the estimator of (4.20).
If is distributed as , use (2.2.5) to show that .
In Example 4.7, show that is nonsingular for and and singular for and .
Show that the function of Example 4.7 is convex.
In Example 4.7, show that is admissible for
and
.
In Example 4.8, show that is admissible under the assumptions (ii)(a).
Let be a family of loss functions and suppose there exists and a minimax estimator with respect to such that in the notation of (4.29),
Then, is minimax with respect to ; that is, it minimizes .
Assuming (4.25), show that is the unique unbiased estimator of the risk (5..4.25), and that is inadmissible.
The estimator is also unbiased for estimation of the loss . See Note 9.5.
A natural extension of risk domination under a particular loss is to risk domination under a class of losses. Hwang (1985) defines universal domination of by if the inequality
holds for all loss functions that are nondecreasing, with at least one loss function producing nonidentical risks.
Show that universally dominates if and only if it stochastically dominates , that is, if and only if
for all and with strict inequality for some .
For , show that the James-Stein estimator does not universally dominate x.
Hwang (1985) and Brown and Hwang (1989) explore many facets of universal domination. Hwang (1985) shows that even does not universally dominate unless the class of loss functions is restricted. We also note that although the inequality in part (a) may seem reminiscent of the "Pitman closeness" criterion, there is really no relation. The criterion of Pitman closeness suffers from a number of defects not shared by stochastic domination (see Robert et al. 1993).