Non-unique Bayes rules admissibility I have a question about non-unique Bayes rule. Are they in general inadmissible or admissible and why? Thanks in advance!
 A: Bayes rules are admissible if they are unique in the sense of equivalence defined as the equality of the risk function.
Proof.
Proof by contradiction. Let's assume that the Bayes rule $d_0$ is not admissible and that it is unique (in the sense written above). It means that there is a rule $d_1$, which is better than $d_0$: $R_{d_1}(\theta) \leq R_{d_0}(\theta)$ for all $\theta \in \Theta$ and 
$R_{d_1}(\theta) < R_{d_0}(\theta)$ for some $\theta _0 \in \Theta$. $\theta _0$ is Bayes rule, so: $$\int_\Theta R_{d_0}(\theta)d\tau (\theta) \geq \int_\Theta R_{d_1}(\theta)d\tau (\theta)$$
Then $d_1$ is also a Bayes rule, so by uniqueness of the Bayes rule $d_0$:
$R_{d_0}(\theta)=R_{d_1}(\theta)$ for all $\theta \in \Theta$, so $d_1$ isn't better than $d_1$. A contradiction.
Note that a non-unique Bayes rule doesn't have to be admissible. Let X be a random sample from experiment:  $\Theta= \{ \theta_0, \theta_1 \}$, where $P_{\theta_0}$~ $N(0,1)$ and $P_{\theta_1}$~ $\delta(0)$. Let $\tau(\theta)= \mathbb{1}_{ \{ \theta=\theta_1 \} }$. Now let's compare two estimators of variance $d_0(X)=0$ and $d_1(X)= \mathbb{1}_{ \{ X \neq 0 \}}$. Clearly both estimators are Bayes estimator but the first one is inadmissible (under quadratic Error Function).
