What a Minimax Audit Decides
A minimax interchange replaces the order of optimization: the statistician’s worst-case risk, \(\inf_\delta\sup_\theta R(\delta,\theta)\), is compared with the largest risk after nature’s choice is averaged or fixed first. The two values need not agree automatically. The point of an audit is to check whether the particular payoff and strategy spaces satisfy a theorem that licenses the interchange.
The audit has two distinct outcomes. If all hypotheses of an applicable theorem hold, the interchange follows. If a hypothesis fails, that theorem cannot be invoked; this alone does not prove that the minimax values differ. A strict gap requires a separate argument or calculation.
A rule may be randomized: for each observation \(x\), it specifies a probability distribution over actions. Such rules can be mixed with one another. Nature’s mixed strategies are priors. These choices make the strategy spaces convex in many decision problems, but they do not by themselves establish compactness or semicontinuity.
Match the Payoff to the Minimax Theorem
Write \(f(\delta,\pi)=R(\delta,\pi)\), with the rule as the minimizing strategy and the prior as the maximizing strategy. One useful orientation of Sion’s minimax theorem is the following sufficient condition: the rule space \(D\) is compact and convex in a linear topological space, the prior space \(\Pi\) is convex in a linear topological space, and \(f:D\times\Pi\to\mathbb R\) is real-valued, lower semicontinuous and convex in \(\delta\) for each \(\pi\), and upper semicontinuous and concave in \(\pi\) for each \(\delta\). Then
There is a complementary orientation: if \(\Pi\) is compact and convex, \(D\) is convex, \(f\) is upper semicontinuous and concave in \(\pi\), and lower semicontinuous and convex in \(\delta\), then the same equality holds. In either version, a compact set is required on the side specified by that version; convexity and semicontinuity are required in the corresponding variable. The full hypotheses of the selected version matter.
For a risk game, the concavity condition in the prior variable is often easy: \(R(\delta,\pi)\) is affine in \(\pi\). The difficult checks are often compactness in the chosen topology and semicontinuity of the risk as a function of the rule or prior. For example, weak compactness of a set of priors does not by itself make \(\pi\mapsto\int R(\delta,\theta)\,\pi(d\theta)\) continuous. That continuity follows under weak convergence when the integrand is bounded and continuous in \(\theta\), not merely because the payoff is an integral.
To verify the first identity, condition on the rule mixture: with probability \(\lambda\) use \(\delta_1\), and otherwise use \(\delta_2\). Linearity of expectation gives \(R(\delta_\lambda,\theta)=\lambda R(\delta_1,\theta)+(1-\lambda)R(\delta_2,\theta)\). Integrating both sides with respect to \(\pi\) gives the first identity. For the second, linearity of integration gives
Thus the payoff is affine, hence both convex and concave, in each mixed strategy. This proves the payoff-shape part of an audit when the strategies really are mixtures and the risks are finite. It does not prove that either strategy space is compact, nor that the payoff has the required semicontinuity in the topology being used. \(\square\)
A Practical Audit
Before interchanging optimization, record the variables, their domains, and the topology on each domain. Then check the relevant conditions rather than relying on an informal claim that “the game is convex.”
State whether nature chooses a parameter, a prior, or another strategy. Define the risk as a real-valued function on the exact strategy spaces.
For the minimizing variable, check convexity; for the maximizing variable, check concavity. Affinity from mixing is a direct way to verify both conditions.
Verify lower semicontinuity in the minimizing variable and upper semicontinuity in the maximizing variable. Do not substitute pointwise continuity for a proof in the topology being used.
Identify which strategy set is compact and confirm that it is convex. If neither set meets the compactness requirement of the available version, do not claim that version applies.
If every hypothesis holds, conclude the equality. If one fails, say the theorem does not justify the equality; determine the values separately if a conclusion about a gap is needed.
Worked Audits
Worked Example: A Finite Parameter Game
Suppose the observation carries no information about \(\theta\in\{0,1\}\), the available actions are \(0\) and \(1\), and the loss is \(1\) for an incorrect action and \(0\) for a correct action. A randomized rule chooses action \(1\) with probability \(p\in[0,1]\). A prior assigns probability \(q\in[0,1]\) to \(\theta=1\).
At \(\theta=0\), the error probability is \(p\); at \(\theta=1\), it is \(1-p\). Therefore the prior-averaged risk is
Both strategy sets are the compact convex interval \([0,1]\). The payoff is affine, and hence continuous, in each coordinate. The compact-rule orientation of Sion’s theorem applies, so the minimax values agree. Direct calculation checks the conclusion: \(f(p,q)\) is linear in \(q\), so its supremum over \(q\in[0,1]\) is \(\max\{p,1-p\}\). Minimizing this over \(p\in[0,1]\) gives \(1/2\), attained at \(p=1/2\). For fixed \(q\), the minimum over \(p\) is \(\min\{q,1-q\}\); its supremum over \(q\in[0,1]\) is also \(1/2\), attained at \(q=1/2\).
Worked Example: Why Mixing Can Repair the Payoff Structure
Keep the same two parameters and zero-one loss, but restrict the statistician to a pure action \(a\in\{0,1\}\). Against a fixed parameter \(\theta\), the loss is \(\mathbf{1}\{a\ne\theta\}\). The pure action set is not convex: the average of \(0\) and \(1\) is not an allowed pure action. The pure minimax value is \(1\), because either fixed action has loss \(1\) against the other parameter. In contrast, for each fixed \(\theta\), the statistician can choose \(a=\theta\), so \(\sup_\theta\inf_a\mathbf{1}\{a\ne\theta\}=0\). These pure-strategy values differ.
Now allow randomized rules, represented by \(p=P(a=1)\), and priors represented by \(q=P(\theta=1)\). The payoff is \(p+q-2pq\), as calculated above. Its minimax and maximin values are both \(1/2\). The audit therefore identifies the issue precisely: the pure action space is nonconvex, while the randomized strategy space is convex and yields an affine payoff. The equality here is for the randomized game, not for the original pure-action game.
Worked Example: A Noncompact Rule Space
Consider an experiment with no informative observations, parameter space \(\Theta=[0,\infty)\), action space \(A=(0,1)\), and loss \(L(\theta,a)=\theta a\). A rule is just a choice \(a\in(0,1)\). For each fixed finite \(\theta\), the risk \(f(a,\theta)=\theta a\) is continuous and affine in \(a\), and for each \(a\) it is continuous and affine in \(\theta\). But \(A\) is not compact, and \(\Theta\) is not compact.
For every \(a>0\), \(\sup_{\theta\geq0}\theta a=\infty\), so \(\inf_{a\in(0,1)}\sup_{\theta\geq0}\theta a=\infty\). For each fixed \(\theta\geq0\), \(\inf_{a\in(0,1)}\theta a=0\): when \(\theta>0\), values approach zero as \(a\) decreases to zero, and when \(\theta=0\), the risk is already zero. Thus \(\sup_{\theta\geq0}\inf_{a\in(0,1)}\theta a=0\). Here the values really do differ. The calculation demonstrates why continuity and affine payoff alone are not enough; the needed compactness hypotheses are absent.
Worked Example: A Discontinuous Payoff on Compact Convex Sets
Let both strategy spaces be \([0,1]\), and define \(f(x,y)=1\) if \(x=y\), and \(f(x,y)=0\) otherwise. The spaces are compact and convex, but the payoff is discontinuous and is not convex in \(x\) or concave in \(y\). For every \(x\), nature can choose \(y=x\), so \(\inf_x\sup_y f(x,y)=1\). For each fixed \(y\), the minimizing player can choose any \(x\ne y\), giving \(\inf_x f(x,y)=0\); hence \(\sup_y\inf_x f(x,y)=0\). Compactness alone does not license interchange. This example does not isolate a single failed condition, but it shows why the payoff-shape and regularity checks must accompany the compactness check.
Interpreting a Failed Check
A common mistake is to reason: “The payoff is linear in each strategy, so the minimax theorem applies.” Linearity verifies a payoff-shape condition, but the theorem also needs the specified compactness and semicontinuity conditions. Another mistake goes in the opposite direction: “The rule space is not compact, so minimax values must differ.” That conclusion does not follow either. A failed hypothesis removes a guarantee; it is not itself a counterexample.
For statistical risks, verify the actual object and topology. If the maximizing strategies are priors, affinity in \(\pi\) is useful, but upper semicontinuity under weak convergence still depends on how \(R(\delta,\theta)\) varies with \(\theta\). If the minimizing strategies are rules, mixing rules may give convexity, but compactness of a rule space can require a particular topology and additional conditions on the experiment. Keep these claims separate: “convex,” “compact,” and “semicontinuous” are different properties.
Check Your Understanding
For each question, identify the relevant hypothesis or calculation rather than relying on a general impression.
- In the compact-rule version of Sion’s theorem, which semicontinuity condition is required in the rule variable, and which in the prior variable?
- Why does affine dependence of \(R(\delta,\pi)\) on \(\pi\) not, by itself, establish continuity under weak convergence of priors?
- In the finite zero-one-loss example, calculate the two values \(\inf_p\sup_q f(p,q)\) and \(\sup_q\inf_p f(p,q)\).
- In the noncompact example, explain why the inner infimum over \(a\in(0,1)\) is zero for every fixed finite \(\theta\), even though it is not attained when \(\theta>0\).
- If a hypothesis of the selected minimax theorem fails, what can you conclude immediately, and what requires a separate calculation?