From Statistical Risks to Minimax Game Payoffs

From Risk to a Two-Sided Payoff

A statistical rule is usually assessed by its risk at each parameter value. Minimax arguments reorganize those risks as payoffs: one side selects a decision rule, while the other weights parameter values. The key translation is exact. Expected loss under the sampling model, followed by averaging over parameter values, becomes a single expectation under a joint construction.

That translation also exposes the relevant geometry. Mixing two randomized rules gives another randomized rule, and mixing two parameter distributions gives another parameter distribution. Under suitable integrability and measurability conditions, the payoff is affine in each argument separately. Affinity supplies the convexity and concavity properties used in minimax theorems; it does not, by itself, supply their other hypotheses, such as compactness or semicontinuity.

Definition: Let $\Theta$ be a parameter space, $\mathcal X$ an observation space, and $\mathcal A$ an action space, each with its specified measurable sets. A statistical model is a probability kernel $P_\theta(dx)$ on $\mathcal X$, indexed measurably by $\theta\in\Theta$. A randomized decision rule $\delta(da\mid x)$ is a probability kernel from $\mathcal X$ to $\mathcal A$. Given a measurable loss $L(\theta,a)$, its risk is $$ R(\theta,\delta) = \int_{\mathcal X}\int_{\mathcal A} L(\theta,a)\,\delta(da\mid x)\,P_\theta(dx). $$ For a probability measure $\pi$ on $\Theta$, define the averaged payoff $$ G(\delta,\pi) = \int_\Theta R(\theta,\delta)\,\pi(d\theta). $$

A useful baseline assumption is that $L$ is jointly measurable and bounded. Then the integrals are finite, the risk is measurable in $\theta$ under the usual measurable-kernel assumptions, and the iterated integrals below can be rearranged by Fubini’s theorem. Nonnegative losses may instead be handled by Tonelli’s theorem, with the resulting integrals allowed to be infinite. For signed, unbounded losses, the relevant absolute integrability must be checked before interchanging integrals.

The probability measure $\pi$ is simply a way of averaging the risks across parameter values. For this tutorial, the important point is the resulting payoff’s dependence on $\delta$ and $\pi$, not a further interpretation of $\pi$.

The Payoff Representation

The distribution of the observation depends on $\theta$, and the action drawn by the rule depends on the observation. Thus, for a fixed $\pi$, the averaged loss can be evaluated by first drawing $\theta$ according to $\pi$, then drawing $X$ according to $P_\theta$, and finally drawing $A$ according to $\delta(\cdot\mid X)$. The payoff is the expected loss in this construction.

Proposition: Suppose the model and rule are measurable kernels and $L$ is jointly measurable. If $L$ is bounded, or if $L\geq 0$, then $$ G(\delta,\pi) = \int_\Theta\int_{\mathcal X}\int_{\mathcal A} L(\theta,a)\,\delta(da\mid x)\,P_\theta(dx)\,\pi(d\theta), $$ with the order of integration interchangeable. In particular, where the integrals are finite, $G(\delta,\pi)$ is the expected loss under the joint law generated by $\pi$, $P_\theta$, and $\delta$.

Proof. By the definition of risk, $$ \int_\Theta R(\theta,\delta)\,\pi(d\theta) = \int_\Theta \left[ \int_{\mathcal X}\int_{\mathcal A} L(\theta,a)\,\delta(da\mid x)\,P_\theta(dx) \right]\pi(d\theta). $$ The integrand is jointly measurable under the kernel assumptions. If $L$ is bounded, it is integrable under the resulting probability measures, so Fubini’s theorem permits the displayed iterated integral to be read in any order. If $L\geq 0$, Tonelli’s theorem gives the same conclusion, including when the value is infinite. The successive kernels define the stated joint law, so this integral is its expected loss. $\square$

The expression is linear in the loss, but its important structural feature here is its dependence on the two choices $\delta$ and $\pi$. Mixing either choice mixes the payoff in exactly the same proportions, as the next result makes precise.

Proposition: Let $\delta_0,\delta_1$ be randomized rules, $\pi_0,\pi_1$ probability measures on $\Theta$, and $t,s\in[0,1]$. Define $$ \delta_t(\cdot\mid x) = t\delta_0(\cdot\mid x)+(1-t)\delta_1(\cdot\mid x), \qquad \pi_s=s\pi_0+(1-s)\pi_1. $$ Whenever the payoffs are finite, $$ G(\delta_t,\pi)=tG(\delta_0,\pi)+(1-t)G(\delta_1,\pi) $$ and $$ G(\delta,\pi_s)=sG(\delta,\pi_0)+(1-s)G(\delta,\pi_1). $$ Thus the rule set and the set of parameter distributions are convex, and $G$ is affine in each argument separately.

Proof. A convex combination of probability measures is a probability measure, so $\delta_t(\cdot\mid x)$ is a probability measure for each $x$. Measurability is preserved by taking a fixed convex combination of kernels. Substituting this kernel into the risk integral and using linearity of integration gives $$ R(\theta,\delta_t) = tR(\theta,\delta_0)+(1-t)R(\theta,\delta_1). $$ Integrating both sides against $\pi$ proves the first identity. For the second identity, linearity of integration against the mixture measure gives $$ \int_\Theta R(\theta,\delta)\,\pi_s(d\theta) = s\int_\Theta R(\theta,\delta)\,\pi_0(d\theta) +(1-s)\int_\Theta R(\theta,\delta)\,\pi_1(d\theta). $$ These are the claimed equalities. $\square$

Worked Examples of the Translation

Worked Example: A Randomized Classifier and a Two-Point Parameter Distribution

Let $\Theta=\{0,1\}$, and observe $X\in\{0,1\}$, with $P_0(X=1)=0.2$ and $P_1(X=1)=0.8$. The action is a classification, $A\in\{0,1\}$, and the loss is $L(\theta,a)=1$ if $a\ne\theta$, and $0$ otherwise. A randomized rule is described by $q_x=P(A=1\mid X=x)$. Take $q_0=0.25$ and $q_1=0.75$.

At $\theta=0$, action $1$ is the error. Therefore, $$ R(0,\delta) = P_0(X=0)q_0+P_0(X=1)q_1 = 0.8(0.25)+0.2(0.75) = 0.35. $$ At $\theta=1$, action $0$ is the error, so $$ R(1,\delta) = P_1(X=0)(1-q_0)+P_1(X=1)(1-q_1) = 0.2(0.75)+0.8(0.25) = 0.35. $$ If $\pi(\{1\})=0.4$ and $\pi(\{0\})=0.6$, then the payoff is $$ G(\delta,\pi) = 0.6R(0,\delta)+0.4R(1,\delta) = 0.6(0.35)+0.4(0.35) = 0.35. $$ The two risks happen to be equal, so this particular payoff does not depend on the chosen weights. The calculation still shows the translation: first compute risk at each parameter value, then average those risks according to $\pi$.

Worked Example: Verifying Affinity by Mixing Rules

Use the same observation model and loss. Consider the rule $\delta_0$ that always chooses action $0$, and the rule $\delta_1$ that always chooses action $1$. For $\delta_0$, $$ R(0,\delta_0)=0, \qquad R(1,\delta_0)=1. $$ For $\delta_1$, $$ R(0,\delta_1)=1, \qquad R(1,\delta_1)=0. $$ Now mix these rules equally, using an independent randomization that selects each rule with probability $1/2$. The resulting rule chooses each action with probability $1/2$, regardless of $X$. Its risks are $$ R(0,\delta_{1/2})=\tfrac12(0)+\tfrac12(1)=\tfrac12, \qquad R(1,\delta_{1/2})=\tfrac12(1)+\tfrac12(0)=\tfrac12. $$ For the parameter distribution in the previous example, $$ G(\delta_{1/2},\pi) = 0.6(\tfrac12)+0.4(\tfrac12) = \tfrac12. $$ Mixing the payoffs gives the same answer: $$ \tfrac12G(\delta_0,\pi)+\tfrac12G(\delta_1,\pi) = \tfrac12[0.6(0)+0.4(1)] + \tfrac12[0.6(1)+0.4(0)] = \tfrac12. $$ This verifies affinity directly in a finite model.

Worked Example: Mixing Parameter Distributions

Return to the rule $\delta_0$, which always chooses action $0$, and let $\pi_0$ put all its mass on $\theta=0$, while $\pi_1$ puts all its mass on $\theta=1$. The risks already calculated are $R(0,\delta_0)=0$ and $R(1,\delta_0)=1$. For the mixture $\pi_s=s\pi_0+(1-s)\pi_1$, $$ G(\delta_0,\pi_s) = sR(0,\delta_0)+(1-s)R(1,\delta_0) = s(0)+(1-s)(1) = 1-s. $$ For example, if $s=0.3$, then $\pi_s$ assigns mass $0.3$ to $0$ and $0.7$ to $1$, and the payoff is $0.7$. The payoff changes linearly with the weights because the risk at each parameter value is fixed while the averaging measure changes.

Which Convexity Conditions Are Supplied?

A real-valued function $g$ on a convex set is convex if $g(tx+(1-t)y)\leq tg(x)+(1-t)g(y)$, and concave if the reverse inequality holds. An affine function satisfies equality, so it is both convex and concave. It is also both quasiconvex and quasiconcave: its value at a mixture lies between the two endpoint values.

For a minimax formulation with the rule $\delta$ as the minimizing argument and the parameter distribution $\pi$ as the maximizing argument, the payoff is therefore convex (indeed affine) in $\delta$, and concave (indeed affine) in $\pi$. These are the payoff-shape conditions relevant to minimax interchange. For example, Sion’s minimax theorem asks for lower semicontinuity and quasiconvexity in the minimizing argument, and upper semicontinuity and quasiconcavity in the maximizing argument, together with its specified convexity and compactness assumptions on the sets. The affine calculation supplies the quasiconvexity and quasiconcavity, but it does not establish semicontinuity or compactness. Those are separate checks.

Randomization matters because the collection of randomized rules is closed under mixtures. Deterministic rules alone generally are not. Also, mixing rules means mixing probability kernels, not necessarily averaging the actions themselves.

Worked Example: Why Averaging Actions May Not Be Mixing Rules

Suppose there is no informative observation, the action space is $\mathcal A=\{0,1\}$, the parameter is $\theta=0.4$, and the loss is squared error $L(\theta,a)=(a-\theta)^2$. The deterministic action $0$ has loss $$ (0-0.4)^2=0.16, $$ while action $1$ has loss $$ (1-0.4)^2=0.36. $$ A randomized rule that chooses each action with probability $1/2$ has expected loss $$ \tfrac12(0.16)+\tfrac12(0.36)=0.26. $$ The arithmetic average of the actions is $0.5$, but $0.5\notin\mathcal A$, so it is not an available deterministic decision in this problem. Even in a different problem where the action space includes $0.5$, taking the loss at the averaged action would give $(0.5-0.4)^2=0.01$, not the randomized rule’s expected loss $0.26$. A kernel mixture averages losses according to the probabilities of the selected actions; it is not generally equivalent to taking a loss at an averaged action.

Common Pitfall and Practical Check

Wrong reasoning: “The payoff is an average of risks, so it is bilinear and minimax interchange follows.” This skips two distinct issues. First, the required measurability and integrability must justify defining the risk and rearranging the integrals. Second, affinity gives the appropriate convexity and concavity, but not the topological hypotheses of a minimax theorem.

Correction: Build the payoff from the risk integral, verify that the relevant integrals are well defined, and prove its affine dependence on each argument by mixing kernels and measures. Then check the chosen minimax theorem’s remaining hypotheses separately. In particular, do not infer compactness or semicontinuity merely from the fact that a payoff is affine.

The practical translation can be summarized as follows: a statistical model and loss determine $R(\theta,\delta)$; a distribution $\pi$ averages those risks; and randomized rules make the rule side convex. This is the bridge from pointwise statistical risk to a two-sided minimax payoff.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

Write the iterated-integral expression for $G(\delta,\pi)$ in terms of $L$, $\delta$, $P_\theta$, and $\pi$.
Why does a fixed convex mixture of two randomized rules remain a randomized rule?
If $R(\theta_0,\delta)=0.2$ and $R(\theta_1,\delta)=0.8$, what is the payoff when $\pi$ assigns masses $0.25$ and $0.75$ to $\theta_0$ and $\theta_1$, respectively?
In a minimax formulation minimizing over rules and maximizing over parameter distributions, which argument is convex and which is concave when the payoff is affine?
Name one minimax hypothesis that affinity alone does not establish.