Finite Decision Problems as Risk-Vector Geometry

From Risk Functions to Vectors

A risk function records a rule’s performance across parameter values. When the parameter space is finite, that function can be written as a finite list of numbers. This changes the form of a comparison, not its meaning: a rule is no worse at every parameter value precisely when each coordinate of its risk vector is no larger. The geometric picture helps locate dominated rules and possible representatives of a complete class.

Let the parameter space be \(\Theta=\{\theta_1,\ldots,\theta_m\}\). For a rule \(\delta\), write \(R(\theta_i,\delta)\) for its risk at \(\theta_i\), the expected loss under that parameter value. The vector records these \(m\) risks in the same fixed order for every rule.

Definition: The risk vector of a rule \(\delta\) on the finite parameter space \(\Theta\) is \[ r(\delta)=\bigl(R(\theta_1,\delta),\ldots,R(\theta_m,\delta)\bigr)\in\mathbb{R}^m. \] For two vectors \(u,v\in\mathbb{R}^m\), write \(u\leq v\) when \(u_i\leq v_i\) for every coordinate \(i\). A risk vector \(u\) dominates \(v\) if \(u\leq v\) and \(u\ne v\). An attainable risk vector is minimal if no distinct attainable vector is less than or equal to it coordinatewise.

Minimal is not the same as smallest in every coordinate, or as minimizing the sum of coordinates. It means that there is no attainable vector that improves at least one coordinate without worsening any other. Minimal vectors are sometimes called Pareto-minimal vectors. Several such vectors may exist and be incomparable: one may have lower risk at some parameter values and higher risk at others.

Result: For a finite parameter space, a rule \(\delta_1\) has risk no larger than a rule \(\delta_2\) at every parameter value if and only if \(r(\delta_1)\leq r(\delta_2)\) coordinatewise. It strictly dominates \(\delta_2\) if and only if \(r(\delta_1)\) dominates \(r(\delta_2)\).

Proof. By definition, the \(i\)th coordinate of \(r(\delta_j)\) is \(R(\theta_i,\delta_j)\). Thus \(r(\delta_1)\leq r(\delta_2)\) means exactly that \(R(\theta_i,\delta_1)\leq R(\theta_i,\delta_2)\) for every \(i=1,\ldots,m\). Since the list contains every parameter value, this is pointwise risk comparison over \(\Theta\). The vectors are distinct exactly when at least one coordinate inequality is strict. \(\square\)

Worked Risk-Vector Comparisons

Worked Example: Four Rules in a Two-State Model

Let \(X\sim\operatorname{Bernoulli}(\theta)\), with \(\theta\in\{0,1\}\). The action is a prediction \(a\in\{0,1\}\), and the loss is \(L(\theta,a)=1\) when \(a\ne\theta\), and \(0\) otherwise. Consider four deterministic rules: always predict 0, always predict 1, predict \(X\), and predict \(1-X\). Find their risk vectors, in the coordinate order \((\theta=0,\theta=1)\).

When \(\theta=0\), \(X=0\) with probability 1; when \(\theta=1\), \(X=1\) with probability 1. The always-0 rule is correct at \(\theta=0\) and incorrect at \(\theta=1\), so its vector is \((0,1)\). The always-1 rule has vector \((1,0)\).

The rule that predicts \(X\) is correct with probability 1 in both states, giving vector \((0,0)\). The rule that predicts \(1-X\) is incorrect with probability 1 in both states, giving vector \((1,1)\). Thus the attainable vectors are \((0,1),(1,0),(0,0),(1,1)\). The vector \((0,0)\) dominates each of the other three, so its rule is a representative that is no worse than all four rules.

Worked Example: Two Incomparable Minimal Vectors

Consider a decision problem with three parameter values and no informative observation. There are four available actions, \(A,B,C,D\); each rule chooses one action. The table gives the loss at each parameter value, so each row is also the action’s risk vector.

Action	\(\theta_1\)	\(\theta_2\)	\(\theta_3\)
A	1	4	2
B	2	2	1
C	3	5	3
D	2	4	2

The vectors are \(r(A)=(1,4,2)\), \(r(B)=(2,2,1)\), \(r(C)=(3,5,3)\), and \(r(D)=(2,4,2)\). Since \(1\leq 3\), \(4\leq 5\), and \(2\leq 3\), action \(A\) dominates \(C\). Also, \(r(A)\leq r(D)\), with a strict inequality in the first coordinate, so \(A\) dominates \(D\).

Neither \(A\) nor \(B\) dominates the other: \(A\) has lower risk at \(\theta_1\), but \(B\) has lower risk at \(\theta_2\) and \(\theta_3\). Neither is dominated by \(C\) or \(D\), because those vectors do not have risks no larger in every coordinate. Therefore \(A\) and \(B\) are exactly the minimal vectors among the four. The class containing both rules is complete for this four-rule problem: \(C\) and \(D\) are dominated by \(A\), while \(A\) and \(B\) are themselves in the class. Keeping only \(A\) would not be complete, since it does not dominate \(B\).

When Minimal Representatives Give a Complete Class

A class of rules is complete under pointwise risk comparison if every rule in the full class of procedures has some rule in the proposed class whose risk is no larger at every parameter value. Equality of risks is allowed. In a finite parameter problem, the vector order expresses this definition directly; but finiteness of the parameter space alone does not guarantee that minimal attainable vectors exist.

Theorem: Suppose the set \(S\) of attainable risk vectors is finite. Every vector in \(S\) is coordinatewise greater than or equal to some minimal vector in \(S\). Consequently, choosing one rule for each distinct minimal vector gives a complete class. Moreover, every complete class must contain a rule whose risk vector equals each minimal vector.

Proof. Fix \(v\in S\), and consider the nonempty finite subset \(S_v=\{u\in S:u\leq v\}\). Choose \(w\in S_v\) that minimizes the sum of its coordinates, \(\sum_{i=1}^{m}w_i\). If some \(z\in S\) satisfied \(z\leq w\) and \(z\ne w\), then \(z\in S_v\), and at least one coordinate would be strictly smaller while none would be larger. Therefore \(\sum_i z_i<\sum_i w_i\), contradicting the choice of \(w\). Hence \(w\) is minimal and \(w\leq v\). This proves the first claim.

Choose one rule for each distinct minimal vector. For any rule in the full class, its vector \(v\) belongs to \(S\). The first claim supplies a minimal vector \(w\leq v\), and the chosen representative of \(w\) therefore has no larger risk at any parameter value. The representatives form a complete class.

Finally, let \(v\) be a minimal attainable vector and suppose a complete class contains a rule with vector \(u\) no larger than \(v\). Minimality forces \(u=v\): otherwise \(u\) would be a distinct attainable vector below \(v\). Thus the class must include a rule with risk vector \(v\). \(\square\)

The theorem’s finiteness assumption is on the set of attainable risk vectors, not just on the number of parameter values. A finite parameter space puts vectors in a finite-dimensional space, but that space may contain infinitely many attainable points. For an infinite set, repeatedly finding a strict improvement need not end at a minimal point. The next example makes this distinction explicit.

Worked Example: Finite Dimension Without a Minimal Vector

Let \(\Theta=\{\theta_1,\theta_2\}\), suppose there is no informative observation, and let the available actions be indexed by \(t\in(0,1]\). Define the loss by \(L(\theta_1,t)=t\) and \(L(\theta_2,t)=1\). The rule choosing action \(t\) has risk vector \(r(t)=(t,1)\).

For every available \(t\), the action \(t/2\) is also available and has vector \((t/2,1)\). Since \(t/2<t\) and the second coordinates agree, this vector dominates \((t,1)\). Thus no attainable vector is minimal, even though there are only two parameter values.

This does not mean complete classes are impossible. For instance, the actions \(t=1/n\), for positive integers \(n\), form a complete subclass: for any \(t\in(0,1]\), choose an integer \(n\geq 1/t\). Then \(1/n\leq t\), so \((1/n,1)\leq(t,1)\). The example shows why the finite-vector theorem cannot be applied merely because the risk vectors have finitely many coordinates.

Reading the Geometry Carefully

For any attainable vector \(v\), the vectors no worse than \(v\) lie in the coordinatewise lower orthant \(\{u:u_i\leq v_i\text{ for every }i\}\). A vector is dominated when this lower orthant contains a distinct attainable point. The minimal attainable vectors are the points for which no such improvement is available. This picture is useful for organizing comparisons across parameters without collapsing them into one number.

That last qualification matters. A sum of risks or a maximum risk can be useful for a different decision question, but neither by itself represents coordinatewise dominance. A vector with a smaller sum may have a larger risk in one coordinate; it therefore need not dominate another vector. Likewise, the set of risk vectors for deterministic rules need not be convex. The coordinatewise comparisons here require neither a scalar summary nor convexity.

Common Pitfalls

Calling a vector minimal because it has a small sum. A low sum does not rule out a coordinatewise improvement. Check every coordinate of a proposed dominating vector.
Assuming minimal means uniquely best. The vectors \((1,4,2)\) and \((2,2,1)\) are both minimal in the worked table, although neither is uniformly better.
Inferring a minimal vector from a finite parameter space. Finite dimension does not imply finitely many attainable vectors or guarantee that an improvement sequence terminates.
Discarding a rule just because another has lower risk somewhere. To establish domination, the replacement must be no worse at every parameter value.

Key takeaway: With finitely many parameter values, encode each rule by its ordered list of risks. Pointwise domination becomes coordinatewise order. If the attainable vector set is finite, its minimal vectors have rule representatives that form a complete class; with infinitely many attainable vectors, that conclusion needs additional justification.

Check Your Understanding

Use coordinatewise comparisons, not a scalar summary of the risks.

For risk vectors \(u=(2,1,4)\) and \(v=(2,3,4)\), does either vector dominate the other? Explain.
In the four-action example, why does action \(A\) dominate \(D\) even though their risks agree at two parameter values?
Can two distinct minimal risk vectors be incomparable? Give the pair from the worked table and identify the coordinates that establish incomparability.
What finiteness condition in the theorem ensures that every attainable vector lies above a minimal one?
In the \(r(t)=(t,1)\) example, verify that no individual available action has a minimal risk vector.