Why the Loss Function Matters
A procedure is not admissible in isolation. Admissibility is a comparison of risk functions, and risk is calculated using a specified loss function. If the loss changes, the statistical model and the procedures can stay exactly the same while their risk rankings—and therefore their admissibility—change.
This is more than a matter of reporting the same comparison in different units. A loss function states which consequences of a decision count, and how costly they are. A procedure that is hard to improve when one consequence matters most may be dominated when another consequence receives a different penalty.
Decision Problem and Definitions
Consider a decision problem with parameter space \(\Theta\), observation \(X\), action space \(\mathcal A\), and loss function \(L(\theta,a)\). A decision rule \(\delta\) chooses an action using the observation; randomized rules may choose an action according to a probability distribution. Its risk under loss \(L\) is the expected loss at each parameter value:
When two rules are compared, the loss function must be held fixed within that comparison. The notation \(R_L\) makes this dependence explicit.
The example below uses the same parameter space, action space, observation model, and rules in both comparisons. Only the loss changes. The losses are given as tables; there is no general requirement that an action be a numerical estimate or that the loss be zero for any particular action.
A Finite Decision Problem with Two Losses
Let \(\Theta=\{0,1\}\), and let the action space be \(\mathcal A=\{A,B,C\}\). The observation is uninformative: it takes one fixed value with probability one under either parameter. Consequently, deterministic rules choose one of the three actions, and randomized rules choose a distribution over those actions that does not depend on \(\theta\).
Specify two losses, \(L_1\) and \(L_2\), as follows. Entries are losses, with columns indexed by the parameter and rows indexed by the action.
| Action | \(L_1(0,a)\) | \(L_1(1,a)\) | \(L_2(0,a)\) | \(L_2(1,a)\) |
|---|---|---|---|---|
| A | 1 | 1 | 0 | 2 |
| B | 0 | 0 | 2 | 0 |
| C | 2 | 2 | 3 | 3 |
Worked Example: Comparing the Deterministic Rules
Let \(\delta_A,\delta_B,\delta_C\) denote the rules that always choose \(A,B,C\), respectively. Because each rule chooses a fixed action, its risk at each parameter is the corresponding table entry.
Under \(L_1\), the risk vectors, listed in parameter order \((0,1)\), are \(R_{L_1}(\delta_A)=(1,1)\), \(R_{L_1}(\delta_B)=(0,0)\), and \(R_{L_1}(\delta_C)=(2,2)\). Therefore \(B\) strictly improves on \(A\) at both parameters, and also strictly improves on \(C\) at both parameters. In particular, \(\delta_A\) is inadmissible under \(L_1\).
Under \(L_2\), the risk vectors are \(R_{L_2}(\delta_A)=(0,2)\), \(R_{L_2}(\delta_B)=(2,0)\), and \(R_{L_2}(\delta_C)=(3,3)\). Neither \(A\) nor \(B\) dominates the other: \(A\) has lower risk at parameter 0, while \(B\) has lower risk at parameter 1. Each of them does dominate \(C\). Thus the ranking of \(A\) and \(B\) changes from strict dominance under \(L_1\) to incomparable risks under \(L_2\).
Admissibility Can Change, Not Just the Ranking
Incomparability with one particular competitor does not by itself establish admissibility: another rule might still dominate. In this example we can check the entire class of randomized rules, not just the three deterministic ones.
Proof. Write a randomized rule as probabilities \((p_A,p_B,p_C)\), where each probability is nonnegative and \(p_A+p_B+p_C=1\). The action distribution is the same at both parameter values because the observation is constant. Under \(L_1\), the risk vector is
If the rule is not the pure rule \(B\), then \(p_A+2p_C>0\). The pure rule \(B\), with risk vector \((0,0)\), is consequently no worse at either parameter and strictly better at both. It is itself undominated: no risk can be below zero at either parameter, since all entries of \(L_1\) are nonnegative. This proves the claim for \(L_1\).
Under \(L_2\), the risk vector is
If \(p_C>0\), compare this rule with the randomized rule that uses \(B\) with probability \(p_B\) and \(A\) with probability \(1-p_B\). The latter rule assigns zero probability to \(C\), and its risk vector is \((2p_B,2(1-p_B))=(2p_B,2p_A+2p_C)\). Its first risk is lower by \(3p_C>0\), and its second risk is lower by \(p_C>0\). It therefore dominates the original rule.
It remains to show that every rule with \(p_C=0\) is admissible. Such a rule has risk vector \((2p_B,2p_A)\), whose two coordinates sum to 2. If any rule dominated it, that rule's two nonnegative risks would each be no greater, with at least one strictly smaller. Their sum would then be less than 2. But for any rule under \(L_2\), the sum of the risks is \(2p_B+3p_C+2p_A+3p_C=2+4p_C\geq2\). This is a contradiction. Thus precisely the rules with \(p_C=0\) are admissible under \(L_2\). In particular, \(p_A=1\) is admissible for \(L_2\) but not for \(L_1\). \(\square\)
Worked Example: Verifying the Status of Rule A
The theorem establishes the full admissible classes, but the conclusion for rule \(A\) can also be checked directly. Under \(L_1\), \(\delta_A\) has risk \((1,1)\), while \(\delta_B\) has risk \((0,0)\). Since \(0\leq1\) at both parameters and \(0<1\) at both, \(\delta_B\) dominates \(\delta_A\).
Under \(L_2\), \(\delta_A\) has risk \((0,2)\). Suppose a randomized rule with probabilities \((p_A,p_B,p_C)\) weakly improved on it. At parameter 0, its risk would have to satisfy \(2p_B+3p_C\leq0\). Since \(p_B,p_C\geq0\), this forces \(p_B=p_C=0\), and therefore \(p_A=1\). That rule has exactly the same risk as \(\delta_A\), so it cannot strictly improve at either parameter. No rule dominates \(\delta_A\), which verifies its admissibility under \(L_2\).
Worked Example: Auditing a Randomized Rule
Consider the rule \(\rho\) that chooses \(A\) with probability \(1/4\), \(B\) with probability \(1/2\), and \(C\) with probability \(1/4\). Its risks under \(L_1\) are \(R_{L_1}(0,\rho)=\tfrac14(1)+\tfrac12(0)+\tfrac14(2)=\tfrac34\) and \(R_{L_1}(1,\rho)=\tfrac14(1)+\tfrac12(0)+\tfrac14(2)=\tfrac34\). Thus \(\delta_B\), with risks \((0,0)\), dominates \(\rho\).
Under \(L_2\), the risks are \(R_{L_2}(0,\rho)=\tfrac14(0)+\tfrac12(2)+\tfrac14(3)=\tfrac74\) and \(R_{L_2}(1,\rho)=\tfrac14(2)+\tfrac12(0)+\tfrac14(3)=\tfrac54\). Now use the rule \(\eta\) that chooses \(A\) and \(B\) with probability \(1/2\) each. Its \(L_2\) risks are \((1,1)\). Since \(1<\tfrac74\) and \(1<\tfrac54\), \(\eta\) dominates \(\rho\) under \(L_2\) as well. This calculation illustrates that the same rule can be dominated under both losses, even though the identity of a useful comparison rule and the admissibility status of other procedures depend on the loss.
What the Comparison Does—and Does Not—Say
The example isolates the role of the loss: the parameter space, observation distribution, action space, and available rules did not change. Under \(L_1\), the loss table rewards \(B\) at both parameter values, so it improves on \(A\) everywhere. Under \(L_2\), the consequences of \(A\) and \(B\) trade off across parameter values. That trade-off prevents either pure rule from dominating the other and makes \(A\) admissible, even though it was inadmissible under \(L_1\).
Admissibility is a property of a rule evaluated against a specified decision problem, not an intrinsic label attached to the rule. A conclusion about admissibility should therefore identify the loss function and the class of competitors. Changing the loss may represent a substantively different goal—for example, assigning different penalties to different kinds of errors—not a correction to the original risk calculation.
Common Pitfall
- Wrong reasoning: “Rule \(A\) is inadmissible, so it is inadmissible regardless of how loss is defined.” This treats admissibility as a property of the rule alone.
- Correction: Compute risks using the loss currently under consideration, then check whether any rule has no larger risk at every parameter and strictly smaller risk somewhere. Here \(A\) is dominated under \(L_1\), but the direct check shows that it is admissible under \(L_2\).
- Another caution: Showing that two rules do not dominate each other is not enough to establish that either is admissible. A third rule—or a randomized rule—could dominate one of them. The proof above checks the full randomized class.
Check Your Understanding
Use the loss tables and decision problem above to answer the following questions.
- Under \(L_1\), what risk vector does the rule that always chooses \(C\) have, and which deterministic rule dominates it?
- Why does the fact that \(A\) and \(B\) have incomparable risks under \(L_2\) not, by itself, prove that either rule is admissible?
- For a randomized rule with probabilities \((p_A,p_B,p_C)\), write its risk vector under \(L_1\).
- Under \(L_2\), why can a rule with \(p_C>0\) be improved by a rule assigning zero probability to \(C\)?
- Which feature of the decision problem stayed fixed when the admissibility of rule \(A\) changed?