Introduction
A minimax value is an infimum over decision rules. An infimum need not be the risk of any rule in the allowed class: the best achievable performance may be approached ever more closely without being achieved. This distinction can arise from a simple constraint on the actions, even when the observation reveals the parameter exactly.
We study a one-parameter decision problem in which the parameter and observation lie in the closed interval \([0,1]\), but actions must lie in the open interval \((0,1)\). Under squared-error loss, the open action constraint excludes the two actions needed for zero risk at the endpoints. We will show that the minimax value is zero, no rule attains it, and rules with maximum risk as small as desired do exist.
The Decision Problem and Its Minimax Value
The measurability condition ensures that the rule defines a valid action as a function of the observation. Here the observation is nonrandom under each parameter value, so the expectation simplifies:
The minimax value is the infimum of maximum risks over all admissible rules, \(R^*=\inf_{\delta}\sup_{\theta\in\Theta}R(\theta,\delta)\). A minimax rule is a rule whose maximum risk equals \(R^*\). For \(\varepsilon>0\), an \(\varepsilon\)-minimax rule has maximum risk at most \(R^*+\varepsilon\). These definitions distinguish the value of the optimization problem from whether a rule achieves that value.
Worked Example: A Rule With a Fixed Boundary Margin
Choose \(\eta=0.1\) and use the rule that moves observations below \(0.1\) up to \(0.1\), moves observations above \(0.9\) down to \(0.9\), and otherwise leaves them unchanged: \(\delta_{0.1}(x)=\max\{0.1,\min(x,0.9)\}\). Its actions all belong to \((0,1)\).
If \(0.1\leq\theta\leq0.9\), then \(\delta_{0.1}(\theta)=\theta\), so the risk is zero. If \(0\leq\theta<0.1\), the action is \(0.1\), and the risk is \((0.1-\theta)^2\leq0.1^2=0.01\). If \(0.9<\theta\leq1\), the action is \(0.9\), and the risk is \((0.9-\theta)^2\leq0.1^2=0.01\). At \(\theta=0\) and \(\theta=1\), the risk is exactly \(0.01\). Thus \(\sup_\theta R(\theta,\delta_{0.1})=0.01\).
Approaching the Infimum Without Attaining It
The general version of the preceding rule clamps the observation to an interval separated from both forbidden actions. For \(0<\eta<1/2\), define \(\delta_\eta(x)=\max\{\eta,\min(x,1-\eta)\}\). The minimum and maximum of continuous functions are continuous, so this rule is Borel-measurable. Its values lie in \([\eta,1-\eta]\), a subset of the allowed action space.
Proof of the value and the nonattainment claim. Squared loss is nonnegative, so every rule has maximum risk at least zero. For any \(0<\eta<1/2\), consider \(\delta_\eta\). When \(\eta\leq\theta\leq1-\eta\), its risk is zero. When \(0\leq\theta<\eta\), its action is \(\eta\), and \(R(\theta,\delta_\eta)=(\eta-\theta)^2\leq\eta^2\). When \(1-\eta<\theta\leq1\), its action is \(1-\eta\), and \(R(\theta,\delta_\eta)=(1-\eta-\theta)^2\leq\eta^2\). The endpoint risks equal \(\eta^2\), so the maximum risk is exactly \(\eta^2\). Taking \(\eta\) arbitrarily close to zero shows that the infimum of maximum risks is at most zero; combined with nonnegativity, this gives \(R^*=0\).
If a deterministic rule attained \(R^*=0\), its risk at \(\theta=0\) would be \(\delta(0)^2=0\), forcing \(\delta(0)=0\). But every admissible action is strictly positive. Thus no deterministic rule attains the value.
The same obstruction applies if randomized rules are allowed, with risk defined as expected loss over both the observation and the randomized action. At \(\theta=0\), the observation is \(0\), and the randomized action \(A\) must belong to \((0,1)\) with probability one. Consequently \(A^2>0\) with probability one, which implies \(E_0[A^2]>0\). To see the implication explicitly, the sets \(\{A\geq 1/m\}\), for positive integers \(m\), have union \(\{A>0\}\). At least one such set must have positive probability, and for that \(m\), \(E_0[A^2]\geq m^{-2}P_0(A\geq1/m)>0\). No randomized rule can therefore have maximum risk zero either.
Finally, given any \(\varepsilon>0\), choose \(0<\eta<1/2\) such that \(\eta^2\leq\varepsilon\). The established risk calculation gives \(\sup_\theta R(\theta,\delta_\eta)=\eta^2\leq R^*+\varepsilon\), so \(\delta_\eta\) is \(\varepsilon\)-minimax. This proves the theorem. \(\square\)
Worked Example: Choosing a Rule for a Specified Tolerance
Suppose the required excess over the minimax value is at most \(\varepsilon=0.0004\). Since \(R^*=0\), it is sufficient to choose \(\eta\) with \(\eta^2\leq0.0004\). Taking \(\eta=0.02\) gives \(\eta^2=0.0004\).
Use \(\delta_{0.02}(x)=\max\{0.02,\min(x,0.98)\}\). For \(0.02\leq\theta\leq0.98\), the risk is zero. For \(0\leq\theta<0.02\), the risk is \((0.02-\theta)^2\leq0.0004\); for \(0.98<\theta\leq1\), it is \((0.98-\theta)^2\leq0.0004\). Both endpoints have risk \(0.0004\). Hence the maximum risk is exactly the requested tolerance, and the rule is \(0.0004\)-minimax.
Why the Observation Matters
Worked Example: Constant Rules Do Not Exploit the Data
Consider the restricted class of constant rules \(\delta_c(x)=c\), where \(0<c<1\). Their risk is \(R(\theta,\delta_c)=(c-\theta)^2\). Over \(\theta\in[0,1]\), the largest squared distance from \(c\) occurs at an endpoint, so \(\sup_\theta R(\theta,\delta_c)=\max\{c^2,(1-c)^2\}\).
Since \(\max\{c,1-c\}\geq1/2\), every constant rule has maximum risk at least \(1/4\). Equality holds at \(c=1/2\), where both endpoint risks are \(1/4\) and all interior risks are no larger. Thus the best constant rule has maximum risk \(1/4\), even though the unrestricted minimax value is zero. The difference is that \(\delta_\eta\) uses the observation, which equals \(\theta\), except where it must respect the open action constraint.
This contrast also clarifies what causes the failure of attainment. It is not lack of information: the observation identifies the parameter. Nor does the minimax value remain positive because of unavoidable uncertainty. The obstacle is that, at \(\theta=0\), zero squared error requires action \(0\), and, at \(\theta=1\), it requires action \(1\). Neither action is allowed. The rules \(\delta_\eta\) get close to those endpoint actions while remaining admissible, making their maximum risks tend to zero.
Common Mistake: Treating an Infimum as a Minimum
A tempting argument is: “The minimax value is zero, so there must be a rule with maximum risk zero.” That conclusion does not follow. By definition, an infimum is a lower bound that can be approached arbitrarily closely; it is not necessarily achieved by an element of the set being optimized over. Here the rules \(\delta_\eta\) have maximum risk \(\eta^2\), which can be made arbitrarily small, while no rule can have zero risk at \(\theta=0\).
A related mistake is to say that the action constraint forces a positive minimax value. It forces positive risk at the endpoint for each individual rule, but it does not force a common positive lower bound across all rules. For every fixed \(\eta>0\), the maximum risk of \(\delta_\eta\) is positive; nevertheless, the sequence of these maximum risks can approach zero. Positive risk for every candidate and a positive infimum are different claims.
Check Your Understanding
Use the specified model and action space to answer the following questions.
- What is the risk of an arbitrary deterministic rule at parameter value \(\theta\), given that \(X=\theta\) with probability one?
- For \(0<\eta<1/2\), why is the maximum risk of \(\delta_\eta\) exactly \(\eta^2\), rather than merely bounded above by \(\eta^2\)?
- Why does the fact that every rule has positive risk at \(\theta=0\) not imply that the minimax value is positive?
- Can randomizing over actions in \((0,1)\) make the risk at \(\theta=0\) equal to zero? Justify your answer.
- Find a value of \(\eta\) that makes \(\delta_\eta\) an \(\varepsilon\)-minimax rule when \(\varepsilon=0.01\), and state its maximum risk.