A Risk-Based Characterization of Admissibility

Why Admissibility Is a Risk-Function Question

A decision rule is judged by its risk across the parameter space, not by how it behaves at just one parameter value. A rule may have lower risk than another at some parameter values and higher risk at others; neither then provides a pointwise improvement over the other. To determine admissibility, the comparison must include the rules that are actually available in the specified class.

This leads to a precise characterization: a rule is admissible exactly when no rule in its class has risk no greater at every parameter value and strictly smaller at even one. The word attainable matters. A risk function that would improve on the rule but cannot be produced by any rule in the class is not a counterexample to admissibility.

Definition: Fix a statistical model indexed by a nonempty parameter space $\Theta$, an action space, a loss function $L(\theta,a)$, and a specified class $\mathcal D$ of measurable decision rules. Assume the loss incurred by each $\delta\in\mathcal D$ is measurable and has a finite expectation under every parameter value. Its risk function is $$ R(\theta,\delta)=E_\theta[L(\theta,\delta(X))], \qquad \theta\in\Theta. $$ A rule $\tilde\delta\in\mathcal D$ dominates $\delta\in\mathcal D$ if $R(\theta,\tilde\delta)\le R(\theta,\delta)$ for every $\theta\in\Theta$, with strict inequality for at least one $\theta$. The rule $\delta$ is admissible within $\mathcal D$ if no rule in $\mathcal D$ dominates it.

The class $\mathcal D$ is part of the definition. It might contain all measurable rules of a specified kind, or just a restricted collection. An admissibility claim is therefore always relative to the model, loss, parameter space, and rule class being used. In what follows, risks are finite real-valued functions on $\Theta$.

The Characterization

The risk functions attainable in the class form the set $$ \mathcal R_{\mathcal D}=\{R(\,\cdot\,,\delta):\delta\in\mathcal D\}. $$ For functions $r,s$ on $\Theta$, say that $s$ is a pointwise weak improvement on $r$, strict somewhere, when $s(\theta)\le r(\theta)$ for every $\theta$, and $s(\theta_0)<r(\theta_0)$ for at least one $\theta_0$. In the admissibility question, the potential improvement must belong to $\mathcal R_{\mathcal D}$.

Theorem: A rule $\delta\in\mathcal D$ is admissible within $\mathcal D$ if and only if its risk function $R(\,\cdot\,,\delta)$ has no attainable pointwise weak improvement that is strict somewhere. Equivalently, there is no $r\in\mathcal R_{\mathcal D}$ such that $$ r(\theta)\le R(\theta,\delta)\ \text{for every }\theta\in\Theta, \qquad r(\theta_0)<R(\theta_0,\delta)\ \text{for some }\theta_0\in\Theta. $$

Proof. Suppose first that $R(\,\cdot\,,\delta)$ has an attainable pointwise weak improvement strict somewhere. By attainability, there is a rule $\tilde\delta\in\mathcal D$ whose risk is that function. The displayed inequalities say that $\tilde\delta$ has risk no greater than $\delta$'s at every parameter and strictly less at at least one. Thus $\tilde\delta$ dominates $\delta$, so $\delta$ is not admissible.

Conversely, if $\delta$ is not admissible, the definition gives a rule $\tilde\delta\in\mathcal D$ whose risk is no greater at every parameter and strictly less at some parameter. Its risk function belongs to $\mathcal R_{\mathcal D}$, so it is an attainable pointwise weak improvement strict somewhere. Therefore the risk function of $\delta$ has such an improvement. The two implications prove the characterization. $\square$

This is an exact equivalence, not a sufficient-condition test. It also does not say that a risk function is admissible merely because it looks small at selected parameter values. The comparison is over the entire parameter space and over all risk functions attainable in the class.

Worked Example: A Rule Dominated in a Three-Action Class

Let the observation space consist of a single point, let $\Theta=\{0,1\}$, and let the available actions be $a,b,c$. Consider the class containing exactly the three deterministic rules that choose one of these actions. Set the losses as follows:

Action	Loss at $\theta=0$	Loss at $\theta=1$
$a$	2	2
$b$	1	2
$c$	3	1

Because the observation space is a singleton and the class consists exactly of these three deterministic rules, the attainable risk vectors, in parameter order $(0,1)$, are precisely $$ R_a=(2,2),\qquad R_b=(1,2),\qquad R_c=(3,1). $$

For $a$, the risk of $b$ is no greater at either parameter: $1\le 2$ at $\theta=0$, and $2\le 2$ at $\theta=1$. It is strictly smaller at $\theta=0$. Thus $b$ dominates $a$, and $a$ is inadmissible. For $b$, action $a$ has risk 2 rather than 1 at $\theta=0$, while $c$ has risk 3 rather than 1 there. Neither can dominate $b$. For $c$, both $a$ and $b$ have risk 2 at $\theta=1$, which exceeds $c$'s risk 1 there. Hence neither dominates $c$. Both $b$ and $c$ are admissible within this class.

The restriction to the stated class is essential to the calculation: it tells us exactly which risk vectors are attainable. A different observation space or a larger class of rules could add other attainable risks and change the conclusion.

Weak at Every Parameter, Strict at One

The improvement condition combines two requirements. First, the alternative must be no worse at every parameter value. Second, it must be better at at least one. It need not be strictly better everywhere, and it need not be better by a fixed positive amount. A tie at some parameter values is fully consistent with domination.

Worked Example: Incomparable Risks Can Both Be Admissible

Let $\Theta=\{0,1\}$, let the observation space again be a singleton, and use squared-error loss $L(\theta,a)=(a-\theta)^2$. Restrict the class to the two constant rules $\delta_0$ and $\delta_1$, choosing actions 0 and 1, respectively.

Since there is no informative observation, each risk is just the corresponding loss at the parameter: $$ R(\,\cdot\,,\delta_0)=(0^2,(0-1)^2)=(0,1), \qquad R(\,\cdot\,,\delta_1)=((1-0)^2,(1-1)^2)=(1,0). $$

The first rule has lower risk at $\theta=0$, since $0<1$, but higher risk at $\theta=1$, since $1>0$. It does not dominate $\delta_1$. The reverse comparison also fails: $\delta_1$ has higher risk at $\theta=0$. Because these are the only two rules in the class, neither has an attainable pointwise weak improvement strict somewhere. Both are admissible within this class.

A comparison at just one parameter would give the wrong conclusion. Admissibility requires one rule to meet the weak-improvement inequality over the whole parameter space.

A Finite-Class Existence Result

The characterization also yields a useful existence fact when both the parameter space and the attainable risk set are finite. In that setting, one can find an admissible rule by minimizing the sum of its risks across the parameter values. This sum is a device for proving existence; it does not replace the pointwise definition of admissibility.

Proposition: Suppose $\Theta=\{\theta_1,\ldots,\theta_m\}$ is finite and nonempty, and $\mathcal R_{\mathcal D}$ is finite and nonempty. Then at least one rule in $\mathcal D$ is admissible within $\mathcal D$.

Proof. For each attainable risk function $r$, define $S(r)=\sum_{j=1}^m r(\theta_j)$. Since $\mathcal R_{\mathcal D}$ is finite and nonempty, at least one $r_*\in\mathcal R_{\mathcal D}$ minimizes $S(r)$. Choose a rule $\delta_*$ with risk function $r_*$, which exists by the definition of the attainable risk set.

If $\delta_*$ were dominated, there would be an attainable risk function $s$ with $s(\theta_j)\le r_*(\theta_j)$ for every $j$, and strict inequality for at least one index $k$. Summing the inequalities gives $$ S(s)=\sum_{j=1}^m s(\theta_j) < \sum_{j=1}^m r_*(\theta_j)=S(r_*). $$ This contradicts the minimizing property of $r_*$. Therefore $\delta_*$ is admissible. $\square$

The assumptions matter. The sum used in the proof is a finite sum because the parameter space is finite, and a minimizer is guaranteed here because the attainable risk set is finite. The argument makes no general claim that a minimizing risk function exists when either condition is dropped.

Worked Example: Checking Every Rule in a Two-Observation Model

Let $\Theta=\{0,1\}$, and suppose $X=\theta$ with probability one. The action space is $\{0,1\}$, the loss is squared error, and $\mathcal D$ contains all deterministic rules from $\{0,1\}$ to $\{0,1\}$. There are exactly four such rules: the identity rule $\delta_I(x)=x$, the constant rules $\delta_0(x)=0$ and $\delta_1(x)=1$, and the rule $\delta_F(x)=1-x$.

At $\theta=0$, the observation is 0; at $\theta=1$, it is 1. Evaluating each rule at those observations gives the complete risk list: $$ R(\,\cdot\,,\delta_I)=(0,0),\quad R(\,\cdot\,,\delta_0)=(0,1),\quad R(\,\cdot\,,\delta_1)=(1,0),\quad R(\,\cdot\,,\delta_F)=(1,1). $$

The identity rule has risk 0 at both parameters, so its risk is no greater than that of any other rule; it is strictly lower than each other rule's risk at one or both parameter values. Each of the other three rules is therefore dominated by $\delta_I$. The identity rule itself is admissible: all losses are nonnegative, and no rule in this complete four-rule list can have risk strictly below 0 anywhere. This conclusion depends on having checked all four rules, which is possible because the specified class contains exactly those four deterministic maps.

Common Pitfall: Comparing Only the Rules You First Notice

A frequent error is to infer that a list of a few proposed rules gives the entire attainable risk set. That inference is valid only when the model and class justify it. For instance, a random observation can allow a deterministic rule to choose different actions depending on the observed value. Such a rule may have a risk vector that is not the risk of any constant-action rule.

The correction is to state the rule class and determine which rules it contains before drawing conclusions from a risk table. In a finite model, enumerate the rules in the class or otherwise verify that the list is complete. In a general model, the theorem still applies, but the comparison is with every attainable risk function, not just the examples already calculated.

Wrong reasoning: “This rule is admissible because none of the other two rules in my table dominates it.”
Correction: “It is admissible only if the table includes all attainable risks in the stated class, or if I have separately ruled out every omitted rule as a possible dominator.”
Wrong reasoning: “A competitor ties at one parameter, so it cannot dominate.”
Correction: A tie is allowed; domination requires weak inequality everywhere and strict inequality somewhere.

Key takeaway: Admissibility within a specified class is exactly minimality of a rule's attainable risk function under pointwise weak comparison with strict improvement somewhere. Identify the class, compare across the full parameter space, and require the improving risk to be attainable.

Check Your Understanding

Use the definition and characterization to answer each question.

What two inequalities must an attainable risk function satisfy to dominate a given rule?
If two rules have risks that cross over the parameter space, can either dominate the other? Explain.
In the three-action example, why was it necessary to specify a singleton observation space and the deterministic rule class?
Can a rule dominate another if their risks are equal at some parameter values? State the complete condition.
Why does minimizing the sum of risks prove the finite-class existence proposition, but not define admissibility in general?

Action	Loss at \(\theta=0\)	Loss at \(\theta=1\)
\(a\)	2	2
\(b\)	1	2
\(c\)	3	1