Weak Topologies on Probability Kernels in Minimax Arguments

Why Put a Topology on Randomized Rules?

In a statistical decision problem, a randomized rule assigns a distribution over actions to each observation. For minimax arguments, it is useful to have more than a collection of rules: we want a topology in which limits of rules remain valid rules, and in which their risks behave continuously. Weak convergence provides such a topology when the action space is compact and all rules share the same observation distribution.

The topology does not make every risk continuous automatically. The payoff must also satisfy appropriate regularity conditions. In particular, on a noncompact observation space, continuity in the parameter need not be continuity in the supremum norm over observations. A fixed-marginal integral norm avoids that error and gives the uniform risk control needed for worst-case risk.

Definition: Let $\mathcal X$ be a Polish observation space, $\mathcal A$ a compact metric action space, and $\mu$ a fixed probability measure on $\mathcal X$. A randomized rule is a probability kernel $K(da\mid x)$ from $\mathcal X$ to $\mathcal A$. It determines a joint probability measure $$ \pi_K(dx,da)=\mu(dx)K(da\mid x) $$ on $\mathcal X\times\mathcal A$, whose $\mathcal X$-marginal is $\mu$. We use the weak topology on these joint measures: $\pi_n$ converges weakly to $\pi$ if $\int h\,d\pi_n\to\int h\,d\pi$ for every bounded continuous function $h$ on $\mathcal X\times\mathcal A$. Kernels agreeing $\mu$-almost everywhere are identified.

This is an integrated topology on rules, not pointwise convergence of $K_n(\cdot\mid x)$ at each observation. Since $\mathcal X$ is Polish and $\mathcal A$ is compact metric, both spaces are standard Borel spaces. The disintegration theorem ensures that any joint probability measure with marginal $\mu$ can be written as $\mu(dx)K(da\mid x)$ for some probability kernel $K$.

Compactness from a Fixed Marginal

The fixed observation marginal supplies tightness, while compactness of the action space prevents mass from escaping in the action coordinate. These two facts give the basic compactness result.

Theorem: Let $\mathcal X$ be Polish, $\mathcal A$ compact metric, and $\mu$ a probability measure on $\mathcal X$. The set of probability measures on $\mathcal X\times\mathcal A$ with $\mathcal X$-marginal $\mu$ is weakly compact. Equivalently, randomized rules with observation marginal $\mu$, viewed through their joint measures, form a weakly compact set.

Proof. First, this set is tight. For any $\varepsilon>0$, tightness of $\mu$ gives a compact $C\subseteq\mathcal X$ with $\mu(C)>1-\varepsilon$. For every joint measure $\pi$ in the set, $$ \pi(C\times\mathcal A)=\mu(C)>1-\varepsilon. $$ The set $C\times\mathcal A$ is compact, so the same compact set captures at least $1-\varepsilon$ of every measure in the family. Prokhorov’s theorem therefore implies relative compactness in the weak topology.

Next, the set is closed. If $\pi_n$ in the set converges weakly to $\pi$, then for every bounded continuous $q:\mathcal X\to\mathbb R$, the function $(x,a)\mapsto q(x)$ is bounded and continuous. Hence $$ \int q(x)\,\pi(dx,da) =\lim_{n\to\infty}\int q(x)\,\pi_n(dx,da) =\int q(x)\,\mu(dx). $$ Thus $\pi$ has marginal $\mu$. A closed, relatively compact set is compact. The disintegration theorem then gives the equivalent statement about kernels. $\square$

Worked Example: A Sequence of Rules and Its Weak Limit

Let $\mathcal X=\mathbb R$, $\mu=N(0,1)$, and $\mathcal A=\{0,1\}$ with its usual discrete metric. Define the deterministic rule $K_n$ to choose action 1 when $x\geq 1/n$, and action 0 otherwise. Define $K$ to choose action 1 when $x\geq0$, and action 0 otherwise.

To check weak convergence, take any bounded continuous $h:\mathbb R\times\{0,1\}\to\mathbb R$. The integrals under the joint measures are $$ \int h\,d\pi_{K_n} =\int_{\mathbb R}h\!\left(x,\mathbf 1_{\{x\geq1/n\}}\right)\mu(dx). $$ For every $x\neq0$, $\mathbf 1_{\{x\geq1/n\}}\to\mathbf 1_{\{x\geq0\}}$. The exceptional point has $\mu$-probability zero. The integrands are bounded by $\sup_{x,a}|h(x,a)|$, which is integrable under the probability measure $\mu$. The dominated convergence theorem therefore gives $$ \int h\,d\pi_{K_n}\longrightarrow \int_{\mathbb R}h\!\left(x,\mathbf 1_{\{x\geq0\}}\right)\mu(dx) =\int h\,d\pi_K. $$ So $\pi_{K_n}$ converges weakly to the joint measure of a valid rule. The example illustrates a general point: weak convergence averages across observations; it does not require action choices to converge at every observation.

Payoff Continuity and Uniform Risk Control

Let $\Theta$ be a compact metric parameter space and let $g(\theta,x,a)$ be the payoff integrand. For a rule represented by $\pi$, write $$ R_\pi(\theta)=\int_{\mathcal X\times\mathcal A}g(\theta,x,a)\,\pi(dx,da). $$ In a dominated statistical experiment, one common form is $g(\theta,x,a)=p_\theta(x)L(\theta,a)$, where $p_\theta$ is the density of the observation distribution with respect to $\mu$, and $L$ is the loss. The conditions below are stated for $g$, so they apply whenever this product has the required regularity.

Theorem: Suppose $g(\theta,\cdot,\cdot)$ is bounded and continuous on $\mathcal X\times\mathcal A$ for every $\theta$. Then weak convergence $\pi_n\to\pi$ implies $R_{\pi_n}(\theta)\to R_\pi(\theta)$ for every fixed $\theta$. In addition, suppose $g$ is jointly continuous in $(\theta,x,a)$ and there is a measurable $H:\mathcal X\to[0,\infty)$ with $\int H\,d\mu<\infty$ such that $|g(\theta,x,a)|\leq H(x)$ for all $(\theta,x,a)$. Then $$ \sup_{\theta\in\Theta}|R_{\pi_n}(\theta)-R_\pi(\theta)|\longrightarrow0. $$

Proof. For fixed $\theta$, bounded continuity of $g(\theta,\cdot,\cdot)$ is exactly what the definition of weak convergence requires, and gives convergence of the corresponding integrals.

For uniform convergence, define the fixed-marginal integral norm $$ \|u\|_{\mu,\mathcal A} =\int_{\mathcal X}\sup_{a\in\mathcal A}|u(x,a)|\,\mu(dx). $$ For each fixed $x$, joint continuity and compactness of $\mathcal A$ imply that $$ \sup_{a\in\mathcal A}|g(\theta_n,x,a)-g(\theta,x,a)|\longrightarrow0 $$ whenever $\theta_n\to\theta$. Indeed, continuity on the compact set $\{\theta,\theta_n:n\geq1\}\times\{x\}\times\mathcal A$ gives uniform convergence over $a$. The displayed supremum is bounded by $2H(x)$. Dominated convergence therefore yields $$ \|g(\theta_n,\cdot,\cdot)-g(\theta,\cdot,\cdot)\|_{\mu,\mathcal A}\longrightarrow0. $$ Thus the image $\{g(\theta,\cdot,\cdot):\theta\in\Theta\}$ is compact in this norm: it is the continuous image of compact $\Theta$.

Fix $\varepsilon>0$. Compactness gives a finite $\varepsilon$-net $g(\theta_1,\cdot,\cdot),\ldots,g(\theta_m,\cdot,\cdot)$ for this image. For every $\theta$, choose $j$ with $\|g(\theta)-g(\theta_j)\|_{\mu,\mathcal A}<\varepsilon$. Since both $\pi_n$ and $\pi$ have marginal $\mu$, $$ \left|\int (g(\theta)-g(\theta_j))\,d\pi_n\right| \leq\|g(\theta)-g(\theta_j)\|_{\mu,\mathcal A}<\varepsilon, $$ and the same bound holds with $\pi$ in place of $\pi_n$. Consequently, $$ \sup_{\theta\in\Theta}|R_{\pi_n}(\theta)-R_\pi(\theta)| \leq 2\varepsilon+ \max_{1\leq j\leq m}|R_{\pi_n}(\theta_j)-R_\pi(\theta_j)|. $$ Each term in the finite maximum tends to zero by the fixed-parameter result. Taking the limit superior gives a bound of $2\varepsilon$; since $\varepsilon$ is arbitrary, the uniform convergence follows. $\square$

Worked Example: Uniform Risk Convergence in a Bounded Experiment

Let $\mathcal X=[0,1]$, let $\mu$ be uniform, and take $\Theta=[0,1/2]$. Define $$ p_\theta(x)=1+\theta(2x-1),\qquad 0\leq x\leq1. $$ This is a positive density relative to $\mu$: its minimum is at least $1-\theta\geq1/2$, and $\int_0^1(2x-1)\,dx=0$, so $\int p_\theta\,d\mu=1$. Let $\mathcal A=[0,1]$, $L(\theta,a)=(a-\theta)^2$, and $g(\theta,x,a)=p_\theta(x)L(\theta,a)$. This payoff is jointly continuous and bounded on the compact product space.

Let $\pi_n$ be the joint measure for the constant rule $a_n=1/n$, and let $\pi$ correspond to the constant rule $a=0$. For any bounded continuous $h$, dominated convergence applied to $h(x,1/n)$ gives $\int h\,d\pi_n\to\int h(x,0)\,\mu(dx)$, so $\pi_n\to\pi$ weakly. Because $p_\theta$ integrates to one, the risks are $$ R_{\pi_n}(\theta)=(1/n-\theta)^2,\qquad R_\pi(\theta)=\theta^2. $$ Their difference is $1/n^2-2\theta/n$. Therefore, for $0\leq\theta\leq1/2$, $$ |R_{\pi_n}(\theta)-R_\pi(\theta)| \leq \frac{1}{n^2}+\frac{1}{n}, $$ which tends to zero uniformly in $\theta$, as claimed.

Why the Fixed-Marginal Norm Matters

A continuous payoff family need not be continuous in the supremum norm over the observation space. Here is a concrete example. Let $\mathcal X=\mathbb R$, $\mu=N(0,1)$, let $\mathcal A$ contain only one action, and define $$ p_\theta(x)=1+\tfrac12\sin(\theta x),\qquad 0\leq\theta\leq1. $$ Symmetry of $\mu$ gives $\int p_\theta\,d\mu=1$, and $p_\theta(x)\geq1/2$. This is a jointly continuous, bounded family of densities. Nevertheless, for $\theta_n=1/n$, $$ \sup_{x\in\mathbb R}|p_{\theta_n}(x)-p_0(x)|=\tfrac12. $$ The upper bound $1/2$ follows from $|\sin|\leq1$; equality occurs at $x=\pi n/2$. Thus supremum-norm convergence fails.

The fixed-marginal norm does converge. Using $|\sin u|\leq|u|$ and $X\sim N(0,1)$, $$ \int|p_{1/n}(x)-p_0(x)|\,\mu(dx) \leq\frac{1}{2n}\int|x|\,\mu(dx) =\frac{1}{2n}\sqrt{\frac{2}{\pi}} \longrightarrow0. $$ The relevant control for integrating against rules with marginal $\mu$ is the integral of the pointwise supremum over actions, not the supremum over all observations. On a noncompact observation space, replacing this integral by a supremum can invalidate a finite-net argument.

Worked Example: Attainment of a Worst-Case Risk

Suppose the compactness conditions for joint measures hold and the uniform risk-convergence theorem applies. Define the worst-case risk $W(\pi)=\sup_{\theta\in\Theta}R_\pi(\theta)$. For any two rules $\pi,\pi'$, $$ |W(\pi)-W(\pi')| \leq\sup_{\theta\in\Theta}|R_\pi(\theta)-R_{\pi'}(\theta)|. $$ This follows because $R_\pi(\theta)\leq R_{\pi'}(\theta)+\sup_\vartheta|R_\pi(\vartheta)-R_{\pi'}(\vartheta)|$ for every $\theta$; taking suprema gives one direction, and reversing $\pi,\pi'$ gives the other. Uniform risk continuity therefore implies continuity of $W$.

If $W$ is finite, a continuous real-valued function on the weakly compact set of rules attains its minimum. Thus there is a rule $\pi^\star$ with $W(\pi^\star)=\inf_\pi W(\pi)$. This establishes existence of a minimax rule under the stated conditions; it does not by itself establish a minimax interchange or a saddle point.

Common Pitfall and Minimax Significance

A frequent error is to argue that joint continuity and boundedness of $g(\theta,x,a)$ imply continuity of $\theta\mapsto g(\theta,\cdot,\cdot)$ in the supremum norm. This is false when $\mathcal X$ is noncompact: the oscillations in the preceding example persist at observations that move farther out as $\theta$ changes. The correction is to integrate $\sup_a|g(\theta_n,x,a)-g(\theta,x,a)|$ against the fixed marginal $\mu$, and use dominated convergence with an integrable envelope.

Weak compactness and payoff continuity are useful ingredients in minimax arguments: compactness can provide an optimizing rule, and uniform continuity controls the worst-case risk. They do not replace the other hypotheses needed for a minimax theorem, such as the relevant convexity, concavity, or semicontinuity conditions. Each condition must be checked for the particular statistical game.

Check Your Understanding

Use the assumptions and arguments above to answer the following questions.

Why does compactness of $\mathcal A$ matter in the tightness proof for joint measures?
Why is the fixed-marginal set closed under weak convergence?
Which property of $g(\theta,\cdot,\cdot)$ gives convergence of the payoff at a fixed $\theta$?
Why does the uniform risk proof use an integral norm rather than a supremum over $x$?
Does existence of a minimizer of worst-case risk alone prove a minimax interchange? Explain.