Common Errors in Complete-Class Arguments

Why Complete-Class Arguments Need Careful Audits

Complete-class arguments compare decision rules through their risks. A proof may produce a sequence of rules, a randomized rule, or a limiting risk function, and then claim that the original rule has been replaced. The crucial question is whether the claimed replacement is an actual rule in the permitted class, with the required risk inequalities at every parameter value.

Four distinctions prevent many errors: weak risk comparison is not the same as strict domination; pointwise convergence is not uniform convergence; randomized and deterministic rules need not have the same attainable risks; and a limit of attainable risks need not itself be attainable. The examples below isolate each distinction.

Definition: For a decision rule \(\delta\), its risk at parameter value \(\theta\) is \(R_\delta(\theta)=E_\theta[L(\theta,\delta(X))]\), when this expectation exists. A rule \(\delta_1\) is a no-worse replacement for \(\delta_0\) if \(R_{\delta_1}(\theta)\le R_{\delta_0}(\theta)\) for every \(\theta\). It strictly dominates \(\delta_0\) if it is a no-worse replacement and the inequality is strict for at least one \(\theta\). Thus equality at some parameter values is allowed in strict domination, but equality at every parameter value is not.

The no-worse comparison is useful when a class is meant to provide replacements without increasing risk. Strict domination is the relevant comparison for inadmissibility: a rule is admissible if no other rule in the comparison class strictly dominates it. Keep the two claims separate. A proof that produces an equal-risk rule has established a replacement, but has not established strict domination.

Error 1: Requiring Strict Improvement Everywhere

Worked Example: Equality at One Parameter Still Allows Domination

Consider a decision problem with two parameter values, \(\Theta=\{\theta_1,\theta_2\}\), and two rules whose risk vectors are \(R_{\delta_0}=(1,1)\) and \(R_{\delta_1}=(0,1)\), where vector coordinates are ordered as \(\theta_1,\theta_2\).

At \(\theta_1\), \(R_{\delta_1}(\theta_1)=0<1=R_{\delta_0}(\theta_1)\). At \(\theta_2\), \(R_{\delta_1}(\theta_2)=1=R_{\delta_0}(\theta_2)\). Therefore \(\delta_1\) is no worse at every parameter and strictly better at one: it strictly dominates \(\delta_0\).

A claim that domination requires strict improvement at both parameter values would incorrectly classify \(\delta_0\) as undominated. The correct test is “no larger everywhere, smaller somewhere.”

There is a converse mistake too: an improvement at one parameter alone is not enough. If a proposed rule has a lower risk at \(\theta_1\) but a higher risk at \(\theta_2\), it does not dominate the original rule. One must verify the entire family of inequalities, not just find a favorable coordinate.

Error 2: Reading Too Much Into Convergence

For risk functions, pointwise convergence means that for each fixed \(\theta\), the numerical sequence \(R_{\delta_n}(\theta)\) converges. It does not say that convergence is equally accurate across all parameter values. Uniform convergence requires the largest absolute discrepancy over the parameter space to tend to zero.

Proposition: Suppose \(R_{\delta_n}(\theta)\to r(\theta)\) pointwise and \(R_{\delta_n}(\theta)\le R_{\delta_0}(\theta)\) for every \(n\) and every \(\theta\). Then \(r(\theta)\le R_{\delta_0}(\theta)\) for every \(\theta\). This conclusion concerns the limiting function \(r\); it does not assert that \(r\) is the risk of a decision rule.

Proof. Fix any \(\theta\). The real numbers \(R_{\delta_n}(\theta)\) are all at most \(R_{\delta_0}(\theta)\), and they converge to \(r(\theta)\). A limit of real numbers bounded above by a fixed number is also bounded above by that number. Hence \(r(\theta)\le R_{\delta_0}(\theta)\). Since \(\theta\) was arbitrary, the inequality holds at every parameter. \(\square\)

The hypothesis that every risk in the sequence is no larger is important. Convergence by itself does not supply an order relation with a proposed rule. Also, even when the hypothesis holds, strictness can disappear in the limit: the limiting function may equal the comparison risk everywhere.

Worked Example: Strict Improvements Can Vanish in the Limit

Let \(\Theta=\{\theta_1,\theta_2\}\), let the comparison rule have risk vector \((1,1)\), and suppose a sequence of rules has risk vectors \((1-1/n,1-1/n)\) for integers \(n\ge 1\). Each risk is nonnegative, and for every \(n\) both coordinates are strictly below 1.

The sequence converges at both parameter values to \((1,1)\), since \(1-1/n\to1\). Thus each rule strictly dominates the comparison rule, but the limiting risk function only ties it. Even if that limit were attainable, it would not strictly dominate the comparison rule.

Worked Example: Pointwise Convergence Does Not Control Maximum Risk

Let \(\Theta=(0,1)\). Consider a model with no informative observation and actions \(a_1,a_2,\ldots\), with loss \(L(\theta,a_n)=1\) when \(\theta<1/n\), and \(L(\theta,a_n)=0\) otherwise. The rule that always chooses \(a_n\) has risk \(R_n(\theta)=\mathbf{1}\{\theta<1/n\}\).

Fix any \(\theta\in(0,1)\). Once \(n\ge 1/\theta\), we have \(1/n\le\theta\), so \(R_n(\theta)=0\). Therefore \(R_n(\theta)\to0\) pointwise. But for each \(n\), choosing \(\theta=1/(2n)\in(0,1)\) gives \(R_n(\theta)=1\); since risks never exceed 1, \(\sup_{\theta\in\Theta}R_n(\theta)=1\). The pointwise limit has supremum 0, so the maximum (supremum) risks do not converge to the supremum of the limit.

Uniform convergence gives a useful quantitative control that pointwise convergence alone cannot provide. If \(f\) and \(g\) are bounded real-valued functions on \(\Theta\), then \(\left|\sup_\theta f(\theta)-\sup_\theta g(\theta)\right|\le\sup_\theta|f(\theta)-g(\theta)|\). Indeed, writing the right-hand side as \(\varepsilon\), the inequalities \(f(\theta)\le g(\theta)+\varepsilon\) and \(g(\theta)\le f(\theta)+\varepsilon\) hold for every \(\theta\). Taking suprema gives both \(\sup f\le\sup g+\varepsilon\) and \(\sup g\le\sup f+\varepsilon\), which proves the claim. Consequently, uniform convergence of bounded risks controls their supremum risks. It still does not guarantee strict improvement: a sequence of strictly smaller risks can converge to equality, as the preceding example shows.

Error 3: Treating Randomized and Deterministic Rules as Interchangeable

A randomized rule uses additional randomness to select an action, possibly conditional on the observation. Its risk averages over both the observation and the randomization. A deterministic rule selects an action without that extra randomization. A deterministic representative with the same risk may exist under particular conditions, but it is not automatic; the permitted rule class and the model’s structure matter.

Definition: A randomized rule can be described by a probability distribution over actions for each observed value. Its risk is the expected loss under that distribution as well as under the sampling model. A risk vector or risk function is attainable in a class only if some rule belonging to that class has exactly that risk.

Worked Example: Randomization Produces an Unavailable Deterministic Risk

Suppose \(\Theta=\{0,1\}\), there is no informative observation, the available actions are 0 and 1, and the loss is \(L(\theta,a)=\mathbf{1}\{a\ne\theta\}\). The deterministic rule choosing action 0 has risk vector \((0,1)\), while the deterministic rule choosing action 1 has risk vector \((1,0)\).

Now choose action 1 with probability \(p\) and action 0 with probability \(1-p\), independently of the parameter. At \(\theta=0\), an error occurs exactly when action 1 is chosen, so the risk is \(p\). At \(\theta=1\), an error occurs exactly when action 0 is chosen, so the risk is \(1-p\). For \(p=1/2\), the risk vector is \((1/2,1/2)\).

Neither deterministic risk vector is no larger than \((1/2,1/2)\) in both coordinates: each has one coordinate equal to 1. Nor does either deterministic rule reproduce this randomized risk vector. Thus replacing the randomized rule by a deterministic one would change the comparison, not merely its description.

Before using a deterministic replacement, check the conditions of the result that supplies it and confirm that the replacement preserves the risk quantities needed in the argument. If a complete-class claim is made only for deterministic rules, a randomized competitor cannot simply be inserted into the proof. Conversely, a result about a class that includes randomized rules does not automatically establish the same claim for its deterministic subclass.

Error 4: Calling a Risk Limit an Attained Replacement

A sequence of attainable risk functions may converge to a function that no allowed rule attains. The pointwise order result above then compares a limiting function, not necessarily two rules. This difference matters whenever the conclusion requires an actual member of the class—for example, an actual rule that dominates another rule.

Worked Example: A Dominating Limit With No Dominating Rule

Let \(\Theta=\{0,1\}\), let there be no informative observation, and take the action set to be \(A=\{c,a_1,a_2,\ldots\}\). Define losses by \(L(0,c)=0,\ L(1,c)=1,\ L(0,a_n)=1/n,\ L(1,a_n)=0\). Because there is no informative observation, a deterministic rule chooses one fixed action. The risk vector of \(c\) is \((0,1)\), and the risk vector of \(a_n\) is \((1/n,0)\).

As \(n\to\infty\), the risks of \(a_n\) converge coordinatewise to \((0,0)\). That limit would be strictly better than \(c\) at \(\theta=1\) and equal at \(\theta=0\), so it would dominate \(c\) if it were attainable. But no action has risk vector \((0,0)\). Each \(a_n\) has first coordinate \(1/n>0\), while \(c\) has second coordinate 1.

In fact, no randomized rule dominates \(c\) either. If it assigns positive probability to any \(a_n\), its risk at \(\theta=0\) is a sum of nonnegative terms \(\sum_n p_n/n>0\), where \(p_n\) is the probability assigned to \(a_n\). If it assigns probability zero to all \(a_n\), it chooses \(c\) with probability one and retains risk 1 at \(\theta=1\). Therefore the unattained limit does not establish an actual dominating rule.

This example separates two valid statements from an invalid conclusion. The risks of actual rules approach \((0,0)\); the limiting function is no larger than \(c\)'s risk; but no rule in the stated class has that limiting risk, and no rule in the class dominates \(c\). A proof must preserve this distinction through its final sentence.

A Practical Audit for Risk Comparisons

When checking a complete-class argument, identify the comparison class first: does it contain deterministic rules, randomized rules, or both? Then write the desired risk inequality with its quantifiers. If the conclusion is strict domination, verify strict inequality at at least one parameter and no increase at every other parameter. If a sequence is used, record whether convergence is pointwise or uniform and whether the order inequalities hold for every term. Finally, ask whether the limiting risk is attained by an admissible rule in the comparison class.

State the comparison.
Write \(R_{\delta_1}(\theta)\le R_{\delta_0}(\theta)\) for every \(\theta\), and separately identify any parameter where strict inequality is required.

Check the rule class.
Confirm whether the proposed competitor is deterministic or randomized and whether it belongs to the class named in the conclusion.

Check the limit.
Verify the stated mode of convergence, preserve only the inequalities it implies, and do not infer uniform control from pointwise convergence.

Check attainment.
Establish that an actual rule realizes the risk used as the replacement; a limiting risk function alone is not a rule.

Common Mistakes

Wrong: “The alternative ties at one parameter, so it does not dominate.” Correction: Strict domination requires no increase everywhere and a strict decrease somewhere; ties at other parameter values are allowed.
Wrong: “The risks converge pointwise, so their maximum risks converge.” Correction: Pointwise convergence controls each fixed parameter separately. Uniform convergence, or another suitable argument, is needed for such a uniform conclusion.
Wrong: “A randomized rule can always be replaced by a deterministic rule.” Correction: This requires conditions that guarantee a risk-preserving representative; without them the attainable risk sets can differ.
Wrong: “The limiting risk dominates, so the original rule is dominated.” Correction: First show that the limit is attained by a rule in the comparison class. Without attainment, the proposed dominator may not exist.

Key takeaway: A complete-class argument is only as strong as the rule and risk comparison it actually produces. Verify the inequalities, the convergence mode, the randomization status, and attainment before claiming replacement or strict domination.

Check Your Understanding

For each question, state which risk comparison or logical step must be checked.

Two risk vectors are \((2,1)\) and \((1,1)\). Does the second strictly dominate the first under the definition in this tutorial? Explain.
If every risk in a sequence is no larger than a fixed risk function and the sequence converges pointwise, what inequality holds for the limit? What does that conclusion not guarantee?
On \((0,1)\), why does \(R_n(\theta)=\mathbf{1}\{\theta<1/n\}\) converge pointwise to zero while \(\sup_\theta R_n(\theta)\) remains one?
In the two-state, two-action classification example, calculate the randomized rule’s two risks when action 1 is chosen with probability \(1/4\). Can either deterministic rule reproduce that vector?
Why does convergence of attainable risk vectors to a function that dominates another risk function fail, by itself, to prove that an actual rule dominates?