Convex Action Spaces and Averaged Decisions

Why Average a Randomized Decision?

A randomized rule can select different actions even when it observes the same data. If the possible actions are vectors in a convex set, it is natural to ask whether replacing that random choice by its average gives a simpler rule. The answer has two parts: the average must still be feasible, and its risk must not be larger. Closed convex action spaces ensure the first property; convex loss ensures the second.

This is a comparison of decision rules, not an assertion that randomization is always unnecessary. The conclusion applies when actions can be averaged within the feasible set and the loss has the appropriate convexity. It does not apply automatically to discrete action sets or nonconvex losses.

Randomized Rules and Their Averaged Actions

Let the observation \(X\) take values in a measurable space \(\mathcal X\), with distribution \(P_\theta\) under parameter \(\theta\). Let the action set \(C\) be a nonempty subset of \(\mathbb R^d\). A randomized rule is a probability kernel \(K(da\mid x)\): after observing \(x\), it specifies a probability distribution over actions. A deterministic rule instead assigns one action \(\delta(x)\) to each observation.

For the results below, suppose \(C\) is closed and convex, and \(K(C\mid x)=1\) for every \(x\). Also suppose the conditional first moment is finite at every \(x\): \(\int_C \|a\|K(da\mid x)<\infty\). The averaged action, or barycenter, of the randomized rule at observation \(x\) is

where the vector integral is taken coordinate by coordinate. The averaged rule is deterministic: it uses \(\bar a(x)\) rather than drawing an action from \(K(\cdot\mid x)\).

Feasibility of the Average

The first issue is whether averaging can produce an action that was not allowed. In a finite action set, for example, the numerical average of two feasible actions may not itself be an available action. A convex action set is designed to rule out this problem: it contains every finite convex combination of its elements. Closedness ensures that limits of such combinations remain feasible.

Proof. Write \(b=E[A]\). Suppose, for contradiction, that \(b\notin C\). The strict separation theorem for a point and a nonempty closed convex set in finite-dimensional Euclidean space gives a vector \(u\in\mathbb R^d\) and a real number \(\alpha\) such that

Since \(P(A\in C)=1\), it follows that \(u^\mathsf{T}A\leq\alpha\) almost surely. The first-moment assumption makes \(u^\mathsf{T}A\) integrable, so taking expectations gives \(u^\mathsf{T}b=E[u^\mathsf{T}A]\leq\alpha\). This contradicts \(u^\mathsf{T}b>\alpha\). Therefore \(b\in C\). \(\square\)

Applying the theorem to the conditional distribution \(K(\cdot\mid x)\) shows that \(\bar a(x)\in C\) for every \(x\). The averaged rule is also measurable: each coordinate of \(\bar a(x)\) is the integral of a measurable coordinate function with respect to the measurable kernel, and the finite first-moment assumption ensures these coordinate integrals are finite. Thus the average is a feasible deterministic rule, not merely a pointwise formal expression.

Convex Loss and the Risk Comparison

For a rule \(K\), define its risk at parameter \(\theta\) by averaging its loss over both the observation and the action drawn by the kernel:

For a deterministic rule \(\delta\), the risk is \(R(\theta,\delta)=\int_{\mathcal X}L(\theta,\delta(x))P_\theta(dx)\). Assume the loss \(L(\theta,\cdot)\) is nonnegative, Borel measurable, and convex on \(C\) for every \(\theta\). Risks may be infinite; the nonnegativity assumption makes the iterated integrals well-defined without requiring an interchange of signed integrals.

Proof. Feasibility and measurability follow from the preceding result and the kernel measurability argument. Fix \(\theta\). For every observation \(x\), the conditional action distribution has a finite first moment, so Jensen’s inequality for the convex function \(L(\theta,\cdot)\) gives

Both sides are nonnegative measurable functions of \(x\). Integrating this pointwise inequality with respect to \(P_\theta\), using the definition of risk, yields

The inequality holds for every \(\theta\), including when the randomized rule has infinite risk. \(\square\)

The important feature is that the same averaged rule is constructed from \(K\), independently of \(\theta\). The pointwise Jensen inequality holds at each observation and for each parameter, so the comparison is simultaneous across the parameter space. In particular, taking a supremum over \(\theta\) preserves the inequality: \(\sup_\theta R(\theta,\bar a)\leq\sup_\theta R(\theta,K)\).

What the Result Does—and Does Not—Say

The comparison is useful because it provides a direct way to remove randomization when the action space and loss meet the stated conditions. A randomized rule may be convenient to describe or analyze, but its conditional mean can then supply a deterministic rule with no larger risk at any parameter. This is a risk comparison; it does not claim that every aspect of the randomized rule is preserved, or that the two rules have the same distribution of actions.

A common mistake is to reverse Jensen’s inequality and claim that a convex loss makes randomization safer. For a convex function, Jensen says the loss at the mean is no greater than the mean loss:

It is the deterministic averaged action that has the lower-or-equal loss, not the random action. Another mistake is to average actions without checking feasibility. If the allowed actions are only \(\{0,1\}\), their average can be \(1/2\), which is not an allowed action. Convexity of the action set is therefore a separate requirement from convexity of the loss.