Nonunique Procedures and Tie Handling

Why Ties Require a Full Risk Comparison

Two procedures can have the same risk at every parameter value while making different decisions. Alternatively, two procedures can tie at one parameter value and have very different risks elsewhere. These situations look similar if we inspect only a single entry in a risk table, but they have different implications for admissibility.

The central rule is simple: admissibility is determined by the entire risk function, not by whether two rules agree on their actions or tie at a particular parameter value. A tie at one parameter neither establishes nor rules out domination. By contrast, procedures with identical risk functions have the same admissibility status, although that common status might be inadmissible.

Definition: In a decision problem with parameter space $\Theta$, observation $X$, action space $\mathcal{A}$, and loss $L(\theta,a)$, a deterministic decision rule $\delta$ is a measurable function from the observation space to $\mathcal{A}$. Its risk function is $$ R(\theta,\delta)=E_\theta[L(\theta,\delta(X))], \qquad \theta\in\Theta, $$ when the expectation exists. Two rules $\delta$ and $\eta$ are risk-equivalent if $R(\theta,\delta)=R(\theta,\eta)$ for every $\theta\in\Theta$. They are distinct rules if they are not the same function on the observation space.

For this tutorial, say that $\eta$ dominates $\delta$ if $R(\theta,\eta)\leq R(\theta,\delta)$ for every $\theta\in\Theta$, with strict inequality for at least one $\theta$. A rule is admissible within a stated class if no rule in that class dominates it. The strict improvement is required at at least one parameter value, not at every value.

Identical Risk Functions, Possibly Different Rules

Risk equivalence is a statement about expected losses, not about actions. Distinct rules can be risk-equivalent because the decisions they make have equal loss, because differences in their decisions average out under every parameter value, or because they differ only on observations that have probability zero under every parameter value. These possibilities need not make the rules identical as functions.

The important admissibility consequence is that risk-equivalent rules cannot dominate one another under the definition above: neither has strictly smaller risk at any parameter. However, both might be dominated by a third rule. Thus, identifying a tie between two procedures is not by itself an admissibility proof.

Theorem: If two rules have identical risk functions, then one is admissible if and only if the other is admissible, relative to the same rule class.

Proof. Let $\delta$ and $\eta$ be risk-equivalent, and suppose first that $\delta$ is inadmissible. Then there is a rule $\gamma$ in the class such that $R(\theta,\gamma)\leq R(\theta,\delta)$ for every $\theta$, with strict inequality at some $\theta_0$. Since $R(\theta,\delta)=R(\theta,\eta)$ for every $\theta$, it follows that $R(\theta,\gamma)\leq R(\theta,\eta)$ for every $\theta$, and $R(\theta_0,\gamma)<R(\theta_0,\eta)$. Hence $\gamma$ dominates $\eta$, so $\eta$ is inadmissible. Interchanging $\delta$ and $\eta$ proves the reverse implication. $\square$

Worked Example: Different Actions, Identical Risks

Suppose there is no informative observation: the observation space contains just one possible value. The parameter space is $\Theta=\{\theta_1,\theta_2\}$, the action space is $\{a,b,c\}$, and the loss table is as follows. The rule class consists of all deterministic rules, so here there are exactly three rules, one for each action.

Action	Loss at $\theta_1$	Loss at $\theta_2$
$a$	1	1
$b$	1	1
$c$	0	2

Because there is only one observation value, each rule always takes its chosen action. Its risk is therefore just the corresponding row of the loss table: $R(\theta_1,a)=R(\theta_2,a)=1$ and $R(\theta_1,b)=R(\theta_2,b)=1$. The rules choosing $a$ and $b$ are distinct, but their risk functions are identical.

Neither of these two rules dominates the other, because their risks are equal at both parameter values and there is no strict improvement. The rule choosing $c$ has risk vector $(0,2)$. It is better than $a$ at $\theta_1$, since $0<1$, but worse at $\theta_2$, since $2>1$; the same comparison holds against $b$. Therefore $c$ dominates neither $a$ nor $b$. Since these are all three rules in the stated class, both $a$ and $b$ are admissible. They are not identical procedures; they are distinct admissible rules with the same risk function.

A Tie at One Parameter Value

When risks tie at only some parameter values, compare the procedures at every parameter value before drawing a conclusion. A single tie cannot supply the strict inequality required for domination, but it also does not prevent strict improvement elsewhere. Similarly, strict improvement at one parameter does not establish domination if the proposed dominator has higher risk at another parameter.

Worked Example: A Tie and Crossing Risks

Again let the observation space contain one value, so a deterministic rule is simply a choice of action. Let the parameter space be $\{\theta_1,\theta_2,\theta_3\}$, with exactly two available actions $A$ and $B$. Their losses, and hence their risks, are:

Action	$\theta_1$	$\theta_2$	$\theta_3$
$A$	0	1	2
$B$	0	2	1

At $\theta_1$, the two risks tie: both equal 0. At $\theta_2$, $A$ has lower risk, because $1<2$. At $\theta_3$, $B$ has lower risk, because $1<2$. Thus $A$ does not dominate $B$, since it has higher risk at $\theta_3$; $B$ does not dominate $A$, since it has higher risk at $\theta_2$.

There are exactly two deterministic rules in this problem, one for each action. Neither dominates the other, so both are admissible within this class. The tie at $\theta_1$ does not decide the comparison; the opposite inequalities at the other parameter values do.

The next example shows the complementary possibility: two rules can tie at one parameter value even though one of them is inadmissible. The deciding factor is whether a rule in the class is at least as good everywhere and strictly better somewhere.

Worked Example: A Tie Does Not Prevent Domination

Let the observation space contain one value, the parameter space be $\{\theta_1,\theta_2\}$, and the action space be $\{A,B,C\}$. As before, the rule class contains all deterministic rules, which here means exactly the three constant-action rules.

Action	$\theta_1$	$\theta_2$
$A$	1	1
$B$	1	2
$C$	0	1

Actions $A$ and $B$ tie at $\theta_1$: each has risk 1 there. At $\theta_2$, $C$ has risk 1, compared with risk 1 for $A$ and risk 2 for $B$. Therefore $C$'s risk is no greater than $A$'s at either parameter value and is strictly lower at $\theta_1$. So $C$ dominates $A$. It also has strictly lower risk than $B$ at both parameters, so it dominates $B$ as well.

Consequently, neither $A$ nor $B$ is admissible, despite their tie at $\theta_1$. The rule choosing $C$ is admissible among these three rules: $A$ has greater risk than $C$ at $\theta_1$, and $B$ has greater risk at both parameter values, so neither dominates $C$. This checks the full stated rule class rather than comparing only the tied pair.

Pointwise Minimum Risk and Tied Choices

A useful special case occurs when a rule attains the smallest possible risk at every parameter value. This is a stronger property than merely tying with another rule at selected parameter values. If the minimum is attained simultaneously across the parameter space, the rule is admissible: no competitor can improve on it anywhere while also being no worse everywhere.

Theorem: Let $\mathcal{D}$ be a rule class. If $\delta\in\mathcal{D}$ satisfies $R(\theta,\delta)\leq R(\theta,\gamma)$ for every $\theta\in\Theta$ and every $\gamma\in\mathcal{D}$, then $\delta$ is admissible in $\mathcal{D}$.

Proof. Suppose, to the contrary, that some $\eta\in\mathcal{D}$ dominates $\delta$. By the theorem's hypothesis, $R(\theta,\delta)\leq R(\theta,\eta)$ for every $\theta$. By domination, $R(\theta,\eta)\leq R(\theta,\delta)$ for every $\theta$, with strict inequality at some $\theta_0$. At that value these inequalities would give $R(\theta_0,\delta)\leq R(\theta_0,\eta)<R(\theta_0,\delta)$, which is impossible. Thus no such $\eta$ exists, and $\delta$ is admissible. $\square$

If several rules all attain this pointwise minimum, their risk functions must be identical: each has the same minimum value at every $\theta$. They are therefore risk-equivalent and, by the earlier theorem, share admissibility status. In this special case the common status is admissible. In other problems, two risk-equivalent rules may have a risk function that is not minimal; a third rule may dominate both, as the theorem on risk equivalence allows.

How to Handle Ties Reliably

When a risk table contains equal entries, do not treat the tie as a conclusion. Record the entire risk function for each procedure, then check the weak inequalities at every parameter value and the strict inequality at at least one. This procedure separates three different situations:

Identical risk functions: the rules are risk-equivalent, so they have the same admissibility status. Neither directly dominates the other.
A tie at selected parameter values, with risks crossing elsewhere: neither rule dominates the other, though another rule might dominate one or both.
A tie at selected parameter values, with one rule no worse everywhere and strictly better somewhere: the better rule dominates, and the other is inadmissible.

The rule class matters throughout. A conclusion of admissibility is always relative to the rules being considered. In finite examples, explicitly listing the available rules is a useful safeguard: a comparison among two displayed procedures alone does not establish that either is admissible if other rules are available.

Common Mistake

Wrong reasoning: “These procedures tie at one parameter value, so neither can dominate the other.” A tie at one value only means neither has a strict advantage there. It says nothing about their risks at the remaining parameter values, and it does not rule out a third procedure that improves on one or both.

Correction: Compare risks over the full parameter space. To show that $\eta$ dominates $\delta$, verify $R(\theta,\eta)\leq R(\theta,\delta)$ for every $\theta$ and strict inequality for at least one. To show admissibility, rule out every such dominator in the stated class. If two risk functions are identical, conclude that the rules share admissibility status, not that either must be admissible.

Key takeaway: A risk tie has meaning only in the context of the complete risk functions and the specified rule class. Identical risk functions preserve admissibility status; a tie at only some parameter values does not settle domination.

Check Your Understanding

Use the full risk comparison in each answer.

Two distinct rules have identical risk at every parameter value. What can you conclude about their admissibility status, and what can you not conclude?
For two rules, the risk vectors over three parameter values are $(1,2,0)$ and $(1,1,2)$. Do they dominate one another? Show the comparisons that support your answer.
A rule attains the smallest risk among all rules in its class at every parameter value. Why must it be admissible?
Why is a comparison of two displayed rules insufficient to establish admissibility if the rule class contains additional procedures?
Give the two conditions that must both hold for one rule to dominate another.

Action	Loss at \(\theta_1\)	Loss at \(\theta_2\)
\(a\)	1	1
\(b\)	1	1
\(c\)	0	2

Action	\(\theta_1\)	\(\theta_2\)	\(\theta_3\)
\(A\)	0	1	2
\(B\)	0	2	1

Action	\(\theta_1\)	\(\theta_2\)
\(A\)	1	1
\(B\)	1	2
\(C\)	0	1