Admissible decision rule

Remarks

Admissibility depends on the decision problem: on the model, on the loss and on the class of rules allowed.

Example: let $X$ be normal with mean $\theta \in \R$ and variance $1$, with squared error loss, and let $\delta \equiv \theta_0$ be a constant. Its risk is $(\theta - \theta_0)^2$, which is $0$ at $\theta_0$. A rule $\delta'$ with $R(\theta_0, \delta') \le 0$ must equal $\theta_0$ almost surely under $P_{\theta_0}$. All $P_\theta$ have the same null sets, so $\delta' = \theta_0$ almost surely for every $\theta$, and $\delta'$ has the same risk as $\delta$. So nothing dominates $\delta$, and it is admissible.

Non-example: in the same problem, $\delta(X) = 2X$ has risk $E_\theta[(2X - \theta)^2] = 4 + \theta^2$. The rule $X$ has risk $1 < 4 + \theta^2$ for every $\theta$, so it dominates $2X$, which is inadmissible.