A Worked Admissibility Audit Across Rule Classes

Why an Admissibility Audit Needs a Rule Class

Admissibility is a comparison, not a property that can be read from a procedure in isolation. The answer depends on the model, the loss, and the procedures allowed to compete. A useful audit makes each of those ingredients explicit, calculates the risk of every relevant rule, and checks domination parameter value by parameter value.

We will use one small estimation problem throughout. There are two possible parameter values and two possible observations, so every deterministic procedure can be listed. The same model will then show how to check randomized procedures and how a restriction on the allowable rules can change the conclusion.

The strict inequality in the definition matters: a procedure with exactly the same risk at every parameter does not dominate another procedure under this convention. Admissibility is always relative to the specified class; a rule can be admissible in a restricted class yet inadmissible among all rules.

The Model and Its Four Deterministic Rules

Let \(\Theta=\{0,1\}\), let the observation be \(X\in\{a,b\}\), and suppose

The action is an estimate in \(\{0,1\}\), and the loss is squared error, \(L(\theta,d)=(d-\theta)^2\). Since both actions and parameter values are 0 or 1, the loss is 0 for a correct estimate and 1 for an incorrect estimate.

A deterministic procedure assigns one action to \(a\) and one to \(b\). There are exactly \(2^2=4\) such procedures. Write each rule as the ordered pair of its actions at \((a,b)\). The risk vector lists its risks in the order \((R(0,\delta),R(1,\delta))\).

Audit of the Deterministic Class

Compare risk coordinates using the definition of domination. The rule \((0,1)\) dominates \((1,0)\), because \(0.2<0.8\) at both parameter values. Neither constant rule dominates the other: \((0,1)\) has lower risk at \(\theta=0\), while \((1,0)\) has lower risk at \(\theta=1\). The rule \((0,1)\) also does not dominate either constant rule: its risk at \(\theta=0\) is greater than 0 for \((0,0)\), and its risk at \(\theta=1\) is greater than 0 for \((1,1)\).

Nor can either constant rule dominate \((0,1)\). The constant-zero rule has risk 1 rather than 0.2 at \(\theta=1\); the constant-one rule has risk 1 rather than 0.2 at \(\theta=0\). Thus the audit identifies three admissible deterministic procedures: \((0,0)\), \((1,1)\), and \((0,1)\). The reverse rule \((1,0)\) is inadmissible.

This conclusion follows from the full list: there are only four deterministic rules, and every possible competitor appears in the table. In a larger problem, a table may not be enough to establish admissibility unless the candidate class has genuinely been exhausted or a general argument rules out all remaining competitors.

Does Randomization Change the Audit?

A randomized procedure may choose an action at random after observing the data. Describe it by \(q_a=P(\text{choose }1\mid X=a)\) and \(q_b=P(\text{choose }1\mid X=b)\), where \(0\leq q_a,q_b\leq1\). The deterministic procedures are included as the four endpoint choices with each probability equal to 0 or 1.

Under \(\theta=0\), choosing 1 is an error; under \(\theta=1\), choosing 0 is an error. Consequently the risk vector of any randomized rule is

The constant rules remain admissible in the randomized class as well. To have risk no greater than 0 at \(\theta=0\), a rule would need \(0.8q_a+0.2q_b=0\). Both coefficients are positive and both probabilities are nonnegative, so this forces \(q_a=q_b=0\), which gives the same constant-zero rule and no strict improvement. For the constant-one rule, risk no greater than 0 at \(\theta=1\) forces \(0.2(1-q_a)+0.8(1-q_b)=0\), hence \(q_a=q_b=1\), again with no strict improvement.

Finally, the randomized rule \((0,1)\) still dominates \((1,0)\), so the latter is inadmissible. Thus randomization adds no dominating procedure for the three deterministic admissible rules: the admissible rules in the randomized class are exactly the same three deterministic rules. The argument also illustrates why merely allowing mixtures does not automatically imply a change in admissibility.

What Changes Under a Restricted Class?

Now restrict attention to procedures required to report 1 after observing \(a\). Such a restriction might represent a mandated action in that data state, or a policy constraint. In the restricted deterministic class, the only procedures are \((1,0)\) and \((1,1)\). Their risk vectors are \((0.8,0.8)\) and \((1,0)\), respectively. Neither dominates the other: the first has lower risk at \(\theta=0\), and the second has lower risk at \(\theta=1\). Both are therefore admissible relative to this restricted class.

In particular, \((1,0)\) is inadmissible in the unrestricted deterministic class but admissible in this restricted class. Its unrestricted dominator, \((0,1)\), violates the restriction because it reports 0 after \(a\). The restriction has not improved the risk of \((1,0)\); it has removed its competitor.

The same conclusion holds if randomization is allowed while retaining the restriction. The restriction sets \(q_a=1\), leaving \(q_b\in[0,1]\). The risk vector is then

For a rule in this restricted randomized class to dominate \((1,0)\), it must have \(R(0)\leq0.8\), which forces \(q_b=0\). That choice is exactly \((1,0)\), with no strict improvement. For a restricted rule to dominate \((1,1)\), it must have \(R(1)\leq0\), which forces \(q_b=1\), exactly \((1,1)\). Thus both remain admissible in the restricted randomized class.

A Reliable Audit and a Common Pitfall

For a finite problem, a disciplined audit can be organized as follows:

A common mistake is to declare \((1,0)\) admissible after noticing that it has risk 0.8 at both parameters and that each constant rule is better at one parameter. That comparison omits \((0,1)\), whose risk is 0.2 at both parameters and which dominates \((1,0)\). The correction is to list the entire relevant class or give an argument covering every possible competitor. The converse mistake is to keep using that dominator after imposing the restriction: \((0,1)\) is then not an allowed competitor, so it cannot establish inadmissibility within the restricted class.

The audit also explains why a single average of the risk coordinates is not an admissibility test. Averages depend on the weights assigned to parameter values, whereas domination requires coordinatewise inequalities. For example, the two constant rules favor different parameter values; neither dominates the other even if a chosen average happens to prefer one.