Why Ask for a Deterministic Representative?
In a finite decision problem, randomization lets a rule mix actions after observing the data. The resulting risk vector is a convex combination of deterministic risks, but that fact alone does not ensure that a deterministic rule has risk no larger in every parameter state. The practical question is when the action and loss structure gives a deterministic replacement with the same or better coordinatewise risk.
The distinction matters when reducing a class of procedures: if a randomized rule can be replaced without increasing risk at any parameter value, randomization adds no advantage for that comparison. The replacement must be justified from the risk table, not inferred merely from the existence of a convex combination.
Finite Decision Problems and Risk Vectors
Let \(\Theta\) be a finite parameter set, \(\mathcal X\) a finite sample space, and \(\mathcal A\) a finite action set. For each \(\theta\in\Theta\), let \(p_\theta(x)=P_\theta(X=x)\), and let \(L(\theta,a)\) be the loss from action \(a\) when the parameter is \(\theta\). A deterministic rule \(\delta\) selects one action \(\delta(x)\) for each observation \(x\). A randomized rule specifies probabilities \(q(a\mid x)\) over actions at each \(x\), with \(q(a\mid x)\geq 0\) and \(\sum_{a\in\mathcal A}q(a\mid x)=1\).
The risk at state \(\theta\) is the expected loss. Collecting these risks across the finite parameter set gives the risk vector. For vectors indexed by \(\Theta\), “no larger” means no larger in every coordinate.
At a fixed observation \(x\), the randomized rule forms a convex combination of the action losses. The combination is local to \(x\); the overall risk then weights each observation by its probability under each parameter state. This makes local comparisons useful, while also warning that a single unweighted comparison need not tell the whole story when observation probabilities differ across states.
A Local Replacement Theorem
For each observation and action, define the weighted local risk vector \(v_x(a)\), whose \(\theta\)-coordinate is \(p_\theta(x)L(\theta,a)\). The randomized rule's local contribution is \(\sum_a q(a\mid x)v_x(a)\). Summing these contributions over observations gives its full risk vector.
Proof. Fix any \(\theta\). The assumed inequality holds at each \(x\); summing it over the finite sample space gives \[ \sum_xp_\theta(x)L(\theta,a_x) \leq \sum_xp_\theta(x)\sum_aq(a\mid x)L(\theta,a). \] The left side is \(R(\theta,\delta)\) and the right side is \(R(\theta,q)\). Since this holds for every \(\theta\), the deterministic rule risk-dominates the randomized rule. If each local inequality is an equality, summing gives equality at every \(\theta\). \(\square\)
A convenient stronger condition is to compare the unweighted loss vectors at each observation. If \(L(\theta,a_x)\leq\sum_aq(a\mid x)L(\theta,a)\) for every \(\theta\), then multiplying by the nonnegative \(p_\theta(x)\) gives the theorem’s condition. The weighted version also handles observations that are impossible under some parameter states: their local contribution in those coordinates is zero whatever action is chosen.
Worked Example: Local Improvements Across Two Observations
Let \(\Theta=\{\theta_1,\theta_2\}\), \(\mathcal X=\{u,v\}\), and \(\mathcal A=\{A,B,C,D\}\). The loss table lists the pair \((L(\theta_1,a),L(\theta_2,a))\):
| Action | Loss at \(\theta_1\) | Loss at \(\theta_2\) |
|---|---|---|
| A | 1 | 4 |
| B | 2 | 2 |
| C | 4 | 1 |
| D | 3 | 1.5 |
Suppose \(P_{\theta_1}(u)=0.8\), \(P_{\theta_1}(v)=0.2\), \(P_{\theta_2}(u)=0.25\), and \(P_{\theta_2}(v)=0.75\). At \(u\), let the randomized rule choose \(A\) and \(C\) with probability \(0.5\) each. Its expected loss pair there is \((2.5,2.5)\), because \(0.5(1,4)+0.5(4,1)=(2.5,2.5)\). Action \(B\), with loss pair \((2,2)\), is no larger in either coordinate.
At \(v\), let it choose \(A\) with probability \(0.25\) and \(C\) with probability \(0.75\). The expected loss pair is \((3.25,1.75)\): its first coordinate is \(0.25(1)+0.75(4)=3.25\), and its second is \(0.25(4)+0.75(1)=1.75\). Action \(D\), with pair \((3,1.5)\), is no larger in either coordinate. Thus the theorem applies to the deterministic rule choosing \(B\) at \(u\) and \(D\) at \(v\).
The randomized risk at \(\theta_1\) is \(0.8(2.5)+0.2(3.25)=2.65\). At \(\theta_2\), it is \(0.25(2.5)+0.75(1.75)=0.625+1.3125=1.9375\). The deterministic rule has risk \(0.8(2)+0.2(3)=2.2\) at \(\theta_1\), and \(0.25(2)+0.75(1.5)=0.5+1.125=1.625\) at \(\theta_2\). Both coordinates are strictly smaller.
Conditions for Risk Matching
Risk matching is a stronger demand than domination: every coordinate must agree exactly. One simple way to ensure it is for the randomized rule to mix, at each observation, only actions with the same loss in every parameter state. More generally, matching is possible when the randomized local loss vector equals that of some single action. The equality can be checked directly in a finite loss table.
This condition is sufficient, not necessary: local contributions at different observations could offset one another in the full risk vector. The local condition is useful because it guarantees equality without relying on such cross-observation compensation.
Worked Example: A Randomized Rule with an Exact Representative
Consider one observation, two parameter states, and three actions with loss vectors \(A=(0,4)\), \(B=(4,0)\), and \(C=(2,2)\). Let the randomized rule choose \(A\) and \(B\) with probability \(0.5\) each. Its risk vector is \[ 0.5(0,4)+0.5(4,0)=(2,2). \] The deterministic rule choosing \(C\) has risk vector \((2,2)\), so it exactly matches the randomized rule at both states. This is not because \(A\) and \(B\) have identical losses; rather, their particular mixture equals the loss vector of another available action.
A Useful Corollary: Ordered Loss Vectors
A particularly transparent sufficient condition is that, at each observation, the action loss vectors are ordered coordinatewise. This means that for any two actions, one has loss no greater than the other at every parameter state. A finite set with this property has a least loss vector, and choosing an action that attains it cannot be worse than any mixture.
Proof. Fix \(x\). Because \(\mathcal A\) is finite and its loss vectors are totally ordered, there is an action \(a_x\) whose loss is no larger in every coordinate than the loss of any action \(a\). For each \(\theta\), \[ L(\theta,a_x)\leq L(\theta,a). \] Multiplying by \(q(a\mid x)\geq0\) and summing over \(a\) gives \(L(\theta,a_x)\leq\sum_aq(a\mid x)L(\theta,a)\). Multiplying by \(p_\theta(x)\geq0\) yields the local inequality in the theorem. Applying that theorem gives a deterministic rule that risk-dominates \(q\). \(\square\)
The ordering hypothesis is strong. In many problems, different actions trade off risk across parameter states, so no single action is least in every coordinate. The next example shows why a deterministic representative cannot then be presumed to exist.
Worked Example: A Randomized Risk with No Deterministic Dominator
There is one possible observation, two parameter states, and two actions. Their loss vectors are \(A=(0,3)\) and \(B=(3,0)\). A randomized rule choosing each action with probability \(0.5\) has risk vector \[ 0.5(0,3)+0.5(3,0)=(1.5,1.5). \] The only deterministic risk vectors are \((0,3)\) and \((3,0)\). Action \(A\) has second-coordinate risk \(3>1.5\), while action \(B\) has first-coordinate risk \(3>1.5\). Therefore neither deterministic rule has risk no larger than the randomized rule in every state. The convex combination is attainable by randomization, but it is not a coordinatewise improvement target for either deterministic action.
Common Mistake and Practical Use
A common mistake is to reason, “The randomized risk is a convex combination of action risks, so one of those actions must have risk no larger in every coordinate.” Convex combinations do not work that way: different actions may be better in different coordinates, as the last table demonstrates. A mixture can balance those coordinates even when neither constituent action dominates the mixture.
The correct check is to state the desired comparison coordinate by coordinate. For a proposed deterministic replacement, compare its local weighted loss contribution with the randomized local contribution at each observation, or compute both complete risk vectors directly. A coordinatewise local comparison proves global domination by summation; a failure of that sufficient check does not by itself rule out domination through comparisons of the full risk vectors.
For each observation and parameter state, calculate \(p_\theta(x)L(\theta,a)\) for each action.
Average the action contributions using \(q(a\mid x)\), separately in each parameter coordinate.
Check whether an available action has local contributions no larger in every coordinate, or equal contributions if exact matching is required.
The local inequalities add to the corresponding full-risk comparison for every parameter state.
These checks identify cases in which randomization can be removed while preserving or improving the entire risk vector. When the table contains competing coordinatewise trade-offs, the comparison must be made explicitly; no deterministic replacement follows from convexity alone.
Check Your Understanding
Use coordinatewise comparisons, keeping the parameter states separate.
- State the local weighted risk vector for an action at observation \(x\).
- Why does a local coordinatewise inequality at every observation imply the same inequality for the full risk vector?
- For loss vectors \(A=(1,3)\), \(B=(2,2)\), and \(C=(3,1)\), what risk vector results from mixing \(A\) and \(C\) equally? Does \(B\) dominate it?
- Give a sufficient condition for exact risk matching by a deterministic rule.
- Why does the fact that a randomized risk is a convex combination not guarantee a deterministic rule with no larger risk in every parameter state?