Common Logical Errors in Admissibility Arguments

What an Admissibility Argument Must Establish

An admissibility claim is a statement about a comparison across all parameter values and against a specified class of procedures. A calculation at a few convenient points, or an improvement under an average of risks, may be informative, but neither alone establishes that one procedure dominates another. The details of the comparison matter: weak improvement is required everywhere, and strict improvement is required somewhere.

Definition: Let \(\theta\) range over a parameter space \(\Theta\), let \(\delta\) be a procedure, and let \(L(\theta,a)\) be the loss from taking action \(a\) when the parameter is \(\theta\). The risk of \(\delta\) is \(R(\theta,\delta)=E_\theta[L(\theta,\delta(X))]\), when this expectation is defined. Within a specified class \(C\), \(\delta'\) dominates \(\delta\) if \(R(\theta,\delta')\leq R(\theta,\delta)\) for every \(\theta\in\Theta\), with strict inequality for at least one \(\theta\). The procedure \(\delta\) is admissible in \(C\) if no procedure in \(C\) dominates it.

This definition gives four separate questions for an argument to answer: What are the risks? Which parameter values are included? What class \(C\) is being considered? Is at least one comparison strictly better? Confusing any two of these questions can turn a valid calculation into an invalid conclusion.

Checking Only Selected Parameter Values

Domination is a pointwise condition over the full parameter space. An improvement at some values is not enough: the candidate procedure must also have risk no greater at every value left unexamined. A single counterexample parameter value disproves a proposed domination claim.

Worked Example: An Omitted Parameter Value Reverses the Conclusion

Suppose there is no informative data, the parameter space is \(\Theta=\{0,\tfrac12,2\}\), actions are real numbers, and loss is squared error, \(L(\theta,a)=(a-\theta)^2\). Compare the constant procedures \(\delta=1\) and \(\delta'=0\).

Because the procedures are constant, their risks are just their squared errors. At \(\theta=0\), \(R(0,\delta)=1\) and \(R(0,\delta')=0\). At \(\theta=\tfrac12\), both risks are \(\tfrac14\). Thus, on the two values checked, \(\delta'\) is no worse and is strictly better at one.

\begin{array}{c|ccc} \theta & 0 & \tfrac12 & 2\\ \hline R(\theta,\delta) & 1 & \tfrac14 & 1\\ R(\theta,\delta') & 0 & \tfrac14 & 4 \end{array}

At the omitted value \(\theta=2\), however, \(R(2,\delta')=4>1=R(2,\delta)\). Therefore \(\delta'\) does not dominate \(\delta\). The two-point calculation was correct, but it did not establish the all-parameter condition.

Result: To establish domination, it is necessary to verify \(R(\theta,\delta')-R(\theta,\delta)\leq 0\) for every \(\theta\in\Theta\), and to verify that the difference is strictly negative for at least one \(\theta\). If a single value has a positive difference, domination fails.

This is a direct consequence of the definition, but it is a useful discipline in practice: write the risk difference as a function of \(\theta\), identify its sign on the full parameter space, and separately identify where it is strictly negative. In a continuous parameter space, checking a finite grid is not a substitute for proving the sign between grid points.

An Average Risk Is Not a Pointwise Risk Comparison

A weighted average of risks summarizes performance under a chosen weighting of the parameter space. It can favor one procedure even when that procedure is worse at some parameter values. Domination instead requires the risk comparison to hold separately at each parameter value.

Result: For a finite parameter space \(\Theta\), let \(w(\theta)>0\) for every \(\theta\), with \(\sum_{\theta\in\Theta}w(\theta)=1\). If \(\delta'\) dominates \(\delta\), then \(\sum_{\theta\in\Theta}w(\theta)R(\theta,\delta')<\sum_{\theta\in\Theta}w(\theta)R(\theta,\delta)\). The converse need not hold.

To prove the forward implication, define \(d(\theta)=R(\theta,\delta')-R(\theta,\delta)\). Domination gives \(d(\theta)\leq0\) for all \(\theta\), and at some \(\theta_0\), \(d(\theta_0)<0\). Since \(w(\theta_0)>0\), the term \(w(\theta_0)d(\theta_0)\) is strictly negative, while every other term \(w(\theta)d(\theta)\) is nonpositive. Consequently,

\sum_{\theta\in\Theta}w(\theta)R(\theta,\delta') - \sum_{\theta\in\Theta}w(\theta)R(\theta,\delta) = \sum_{\theta\in\Theta}w(\theta)d(\theta)<0.

The reverse implication fails because a low weighted average can hide a large risk at a parameter value that receives little weight.

Worked Example: A Smaller Prior Average, but No Domination

Let \(\Theta=\{0,1\}\), with no informative data, actions \(0\) and \(1\), and zero-one loss: \(L(\theta,a)=0\) if \(a=\theta\), and \(1\) otherwise. Let \(\delta_0\) always choose \(0\), and \(\delta_1\) always choose \(1\). Their risks are

\begin{array}{c|cc} & \theta=0 & \theta=1\\ \hline \delta_0 & 0 & 1\\ \delta_1 & 1 & 0 \end{array}

Suppose the weights are \(w(0)=0.9\) and \(w(1)=0.1\). The weighted risk of \(\delta_0\) is \(0.9(0)+0.1(1)=0.1\), while that of \(\delta_1\) is \(0.9(1)+0.1(0)=0.9\). Thus \(\delta_0\) has the smaller weighted risk. But at \(\theta=1\), its risk is \(1\), greater than \(\delta_1\)'s risk of \(0\). Neither procedure dominates the other.

The calculation supports a statement about this particular weighting, not a claim of pointwise superiority. In particular, low average risk alone does not establish admissibility or inadmissibility.

The Procedure Class Is Part of the Claim

Admissibility is always relative to a specified class \(C\). A procedure may be admissible when only a restricted set of competitors is allowed, yet dominated when the class is enlarged. The logical implication goes one way: if a procedure is admissible in a larger class, it is admissible in any subclass containing it. The converse is not guaranteed.

Result: Let \(C_1\subseteq C_2\), and let \(\delta\in C_1\). If \(\delta\) is admissible in \(C_2\), then it is admissible in \(C_1\). Indeed, any procedure in \(C_1\) that dominated \(\delta\) would also belong to \(C_2\), contradicting admissibility there. The reverse implication need not hold: a dominator may belong to \(C_2\) but not to \(C_1\).

Worked Example: Admissible Among Constant Rules, Dominated Among All Rules

Let \(X\sim\operatorname{Bernoulli}(\theta)\), where \(\Theta=\{0,1\}\). Actions are \(0\) and \(1\), and loss is zero-one loss. First restrict \(C_1\) to the two constant rules, \(\delta_0(X)=0\) and \(\delta_1(X)=1\). Their risk vectors, in the order \((\theta=0,\theta=1)\), are \((0,1)\) and \((1,0)\). Neither dominates the other: \(\delta_1\) is worse at \(\theta=0\), while \(\delta_0\) is worse at \(\theta=1\). Thus both are admissible within \(C_1\).

Now enlarge the class to include all deterministic rules based on \(X\). The rule \(\delta_X(X)=X\) has risk \(0\) at \(\theta=0\), since then \(X=0\) with probability one, and risk \(0\) at \(\theta=1\), since then \(X=1\) with probability one. Its risk vector is \((0,0)\), so it dominates \(\delta_0\): its risk is equal at \(\theta=0\) and strictly smaller at \(\theta=1\). Therefore \(\delta_0\) is admissible in the restricted class but inadmissible in the enlarged class.

A conclusion that says only “\(\delta_0\) is admissible” leaves out an essential qualification. The class of permitted procedures must be stated.

Equal Risks Do Not Mean Strict Domination

The risk function does not always identify a procedure uniquely. Two distinct procedures can have exactly the same risk at every parameter value. Such procedures are risk-equivalent, but neither strictly dominates the other under the definition above: neither has a parameter value at which its risk is strictly lower.

Worked Example: Distinct Rules with Identical Risk Functions

Let \(\Theta=\{0,1\}\), let \(X\sim\operatorname{Bernoulli}(\tfrac12)\) under either parameter value, and use actions \(0\) and \(1\) with zero-one loss. Consider \(\delta(X)=X\) and \(\delta'(X)=1-X\). These rules differ for each possible observation.

At either parameter value, \(X\) is \(0\) with probability \(\tfrac12\) and \(1\) with probability \(\tfrac12\). For \(\delta\), the action is correct with probability \(\tfrac12\), so \(R(\theta,\delta)=\tfrac12\). For \(\delta'\), the action is also correct with probability \(\tfrac12\), so \(R(\theta,\delta')=\tfrac12\). Hence

R(0,\delta)=R(0,\delta')=\tfrac12, \qquad R(1,\delta)=R(1,\delta')=\tfrac12.

Neither rule has a strictly smaller risk anywhere. Thus neither dominates the other, despite their distinct actions for each observation. Calling one a strict improvement would confuse being a different procedure with having strictly smaller risk somewhere.

A Practical Audit

Before accepting an admissibility argument, separate the claims and check each against the definition. The following sequence keeps the logical requirements visible.

Name the comparison.
Write the candidate procedure, its proposed competitor, the loss, and the procedure class.

Compare risks pointwise.
Check the risk inequality for every parameter value, not only selected values or a weighted average.

Find strict improvement.
Identify at least one parameter value where the competitor's risk is strictly lower. Equality everywhere is not strict domination.

Check membership in the class.
Confirm that the proposed competitor is actually an allowed procedure in the class relative to which admissibility is claimed.

A common mistake is to reason, “The competitor has lower risk on the parameter values I checked, and its average risk is lower, so it dominates.” This reasoning does not establish the required pointwise inequality; the competitor could be worse elsewhere, as the examples show. The correction is to treat an average comparison as evidence about that average only, then establish the all-parameter inequalities and strictness separately. Likewise, equal risks may establish risk-equivalence, but not strict domination.

Key takeaway: A valid domination claim requires a competitor in the specified class, no greater risk at every parameter value, and strictly smaller risk at at least one value. Each condition does separate logical work.

Check Your Understanding

For each question, distinguish what the calculation establishes from what it does not establish.

A proposed competitor has lower risk at five parameter values, but the parameter space contains other values. What additional condition must be checked before concluding domination?
For a finite parameter space with strictly positive weights, why does pointwise domination imply a strictly lower weighted risk?
Two procedures have equal risk at every parameter value but differ as functions of the data. Does either strictly dominate the other? Explain.
A rule is admissible in a restricted class of constant procedures. What must be investigated before claiming it is admissible among all data-dependent procedures?
Can a rule have lower weighted average risk than a competitor and still fail to dominate it? State the pointwise feature that permits this.