Finite Risk Vectors and Complete-Class Geometry

From Risk Functions to Risk Vectors

A decision rule has one risk for each parameter value. When the parameter space consists of finitely many values, these risks can be assembled into a single point in Euclidean space. This representation makes one question especially direct: does another rule have risk no greater at every parameter value? If so, the first rule can be discarded when the aim is to retain rules that are not uniformly worse.

The geometry is deliberately limited to coordinatewise comparisons. It does not rank every pair of rules: a rule can have lower risk at one parameter value and higher risk at another. Such trade-offs matter in decision theory and should not be mistaken for domination.

Definition: Let the parameter space be the finite set $\Theta=\{\theta_1,\ldots,\theta_m\}$. For a decision rule $\delta$ with finite risks, its risk vector is $$ r(\delta)=\bigl(R(\theta_1,\delta),\ldots,R(\theta_m,\delta)\bigr)\in\mathbb{R}^m. $$ The attainable risk set is $A=\{r(\delta):\delta\in\mathcal{D}\}$, where $\mathcal{D}$ is the collection of rules under consideration. For $x,y\in\mathbb{R}^m$, write $y\leq x$ when $y_j\leq x_j$ for every coordinate $j$. A rule $\delta'$ dominates $\delta$ if $r(\delta')\leq r(\delta)$ and the two vectors are not equal. A risk vector is undominated in $A$ if no other vector in $A$ dominates it.

The word “finite” here refers to the number of parameter values, so a risk vector has finitely many coordinates. The attainable set $A$ need not itself be finite: for example, a problem may allow a continuum of rules. The finite-set reduction proved below applies when the set of attainable risk vectors under consideration is finite.

Reading the Coordinatewise Order

In two dimensions, each attainable vector is a point in the plane, and lower coordinates mean lower risk. A point is dominated when another attainable point is at least as low in every coordinate and strictly lower in at least one. In higher dimensions, the same definition applies coordinate by coordinate. The undominated points form the Pareto frontier of the attainable risk set.

If two distinct rules have exactly the same risk vector, neither strictly dominates the other under this definition. Nevertheless, either rule can be removed if the goal is to retain one representative for each attainable risk vector: the remaining rule has exactly the same risk at every parameter value.

Worked Example: A Three-Parameter Loss Table

Suppose the observation is always $X=0$, whatever the true parameter, and the parameter space is $\{\theta_1,\theta_2,\theta_3\}$. There are four available actions, with losses as follows. Since the observation is uninformative, each action has the same loss every time it is used, so its risk vector is its row of losses.

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

Action A dominates C because $1\leq2$, $4\leq5$, and $3\leq4$, with strict inequalities in all three coordinates. Thus C can be removed. A and B are incomparable: A has lower risk at $\theta_1$, B has lower risk at $\theta_2$, and they tie at $\theta_3$. Neither can be removed on the basis of coordinatewise dominance. D has exactly the same risk vector as A, so one of A and D can be removed if only one rule per distinct risk vector is needed. After these reductions, the distinct undominated vectors are $(1,4,3)$ and $(2,3,3)$.

Worked Example: A Two-Parameter Loss Table

Let $X=0$ with probability one under each of $\theta_1$ and $\theta_2$. There are four actions whose losses, and hence risks, are shown below.

Action	Risk at $\theta_1$	Risk at $\theta_2$	Risk vector
P	0	5	$(0,5)$
Q	3	2	$(3,2)$
S	4	4	$(4,4)$
T	2	6	$(2,6)$

Q dominates S because $3\leq4$ and $2\leq4$, with strict inequalities in both coordinates. P dominates T because $0\leq2$ and $5\leq6$, again strictly. P and Q are incomparable: P is better at $\theta_1$, while Q is better at $\theta_2$. Thus, after removing S and T, both P and Q remain. This example shows why a risk-vector comparison is not a search for a single rule that is best at every parameter value.

Risk Vectors from a Sampling Model

The coordinates are expected losses, not necessarily entries from a loss table. They are computed by averaging the loss under the sampling distribution at each parameter value. The next example derives the coordinates explicitly.

Worked Example: Computing Risks in a Bernoulli Model

Let $\Theta=\{\theta_0,\theta_1\}$, with $X\sim\operatorname{Bernoulli}(0.2)$ at $\theta_0$ and $X\sim\operatorname{Bernoulli}(0.8)$ at $\theta_1$. The action is a prediction $a\in\{0,1\}$, and the loss is $L(\theta,a)=1$ if the prediction is incorrect and $0$ otherwise, where the true label is 0 at $\theta_0$ and 1 at $\theta_1$. Consider three rules: $\delta_0$ always predicts 0; $\delta_1$ always predicts 1; and $\delta_X$ predicts $X$.

For $\delta_0$, the prediction is correct with certainty at $\theta_0$, so $R(\theta_0,\delta_0)=0$. At $\theta_1$, it is always wrong, so $R(\theta_1,\delta_0)=1$. Hence $r(\delta_0)=(0,1)$.

For $\delta_1$, the prediction is always wrong at $\theta_0$, giving risk 1. At $\theta_1$, it is always correct, giving risk 0. Thus $r(\delta_1)=(1,0)$.

For $\delta_X$, the rule is wrong at $\theta_0$ exactly when $X=1$, an event with probability $0.2$. It is wrong at $\theta_1$ exactly when $X=0$, also with probability $1-0.8=0.2$. Therefore $r(\delta_X)=(0.2,0.2)$. None of these three vectors dominates another: each has at least one coordinate larger than another vector’s corresponding coordinate. In particular, the smaller maximum risk of $\delta_X$ does not make it coordinatewise dominating.

Why Dominated Rules Can Be Removed

The coordinatewise order has a simple implication for any criterion that increases when any risk coordinate increases. Formally, let $\phi:\mathbb{R}^m\to\mathbb{R}$ be coordinatewise nondecreasing, meaning $y\leq x$ implies $\phi(y)\leq\phi(x)$. If $r(\delta')\leq r(\delta)$, then $\phi(r(\delta'))\leq\phi(r(\delta))$. Thus replacing a rule by one with a no-larger risk vector cannot worsen such a criterion. It may not strictly improve every criterion: for example, the maximum coordinate can be unchanged even when one coordinate decreases.

This is the precise sense in which a dominated rule can be removed without loss. The claim concerns criteria that respect coordinatewise risk order, not every possible preference over rules. It also does not say that one of two incomparable rules can be discarded: a decision-maker’s choice between them may depend on how parameter values are weighted or on the criterion being optimized.

Proposition: If $y\leq x$, then every coordinatewise nondecreasing function $\phi$ satisfies $\phi(y)\leq\phi(x)$. If, in addition, $\phi$ is strictly increasing in at least one coordinate where $y_j<x_j$, then $\phi(y)<\phi(x)$.

Proof. The first inequality follows directly from the definition of coordinatewise nondecreasing. For the strict statement, change the coordinates of $x$ to those of $y$ one at a time. Each change is a decrease or leaves the coordinate unchanged, so the value of $\phi$ cannot increase at any step. At the specified coordinate, strict increase of $\phi$ in that coordinate means that decreasing it strictly decreases $\phi$, with the other coordinates fixed at their values in that step. The final value is therefore strictly below the initial value. $\square$

A finite attainable set has a particularly clean reduction. Repeatedly removing dominated vectors cannot continue indefinitely, and any vector removed this way has a surviving vector that dominates it. This gives a finite representative collection that retains all undominated risk possibilities.

Theorem: Let $A\subset\mathbb{R}^m$ be a finite set of distinct attainable risk vectors. Let $P$ be the set of vectors in $A$ that are undominated in $A$. Then every $x\in A\setminus P$ is dominated by some $p\in P$. Consequently, retaining one rule for each vector in $P$ preserves a rule with no greater risk vector than every original rule.

Proof. Fix $x\in A\setminus P$. Since $x$ is dominated, choose $x_1\in A$ such that $x_1\leq x$ and $x_1\ne x$. If $x_1$ is undominated, take $p=x_1$. Otherwise choose $x_2\in A$ with $x_2\leq x_1$ and $x_2\ne x_1$. Continue in this way while the current vector is dominated. The sequence cannot repeat: coordinatewise inequalities hold along the sequence, and a return to an earlier vector would force equality at every intervening step, contradicting the strict change at each step. Since $A$ is finite, the process must terminate at an undominated vector $p\in P$. By transitivity, $p\leq x$. Also $p\ne x$, because the first step was strict and the sequence cannot return to $x$. Hence $p$ dominates $x$. This argument applies to every $x\notin P$, proving the claim. $\square$

What the Frontier Does—and Does Not—Tell Us

The Pareto frontier answers a specific question: which attainable risk vectors are not uniformly worse than another attainable vector? It does not, by itself, select a unique rule. For instance, in the Bernoulli example, $(0,1)$, $(1,0)$, and $(0.2,0.2)$ are all undominated, although their maximum coordinates are $1$, $1$, and $0.2$, respectively. A minimax criterion selects $(0.2,0.2)$ among those three, but that criterion is an additional choice; the coordinatewise order alone does not make the other two dominated.

Nor does a finite number of parameter values guarantee a finite attainable set. If the rule class permits many possible rules, there may be infinitely many risk vectors in $\mathbb{R}^m$. The finite-frontier theorem applies to a finite attainable set; using it for an infinite set would require a separate argument. Similarly, a risk limit that is not attained by any rule is not an attainable point that can be used as a replacement.

Common Mistake: Treating Incomparability as Domination

A common error is to say that a rule is dominated because it has a larger risk at one parameter value, without checking the other coordinates. For example, comparing $(0,1)$ with $(0.2,0.2)$, the first vector is better in the first coordinate but worse in the second. Neither vector is coordinatewise no greater than the other. The correction is to compare every coordinate in the same direction: domination requires $y_j\leq x_j$ for all $j$, and strict inequality for at least one coordinate.

A second error is to equate “not minimax” with “dominated.” A vector can have a larger maximum risk than another while remaining better at one parameter value. Minimizing the maximum coordinate and removing coordinatewise dominated vectors are different operations, with different answers in general.

Key takeaway: For a finite parameter space, each rule is represented by a point whose coordinates are its risks. A rule can be removed on coordinatewise grounds when another attainable risk vector is no greater in every coordinate; the undominated vectors form the Pareto frontier, not necessarily a single best rule.

Check Your Understanding

Use coordinatewise comparisons, checking every parameter coordinate before deciding whether a rule can be removed.

For risk vectors $(2,1,5)$ and $(3,1,4)$, does either dominate the other? Explain by comparing all coordinates.
If two distinct rules have identical risk vectors, does either strictly dominate the other under the definition here? Can one nevertheless be removed when keeping one representative per vector?
Why does the finite-frontier theorem require the attainable risk set, not merely the parameter space, to be finite?
Give a reason why a rule with a larger maximum risk than another rule need not be dominated by it.
State the coordinatewise condition that must hold before a rule can be removed as dominated.

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

Action	Risk at \(\theta_1\)	Risk at \(\theta_2\)	Risk vector
P	0	5	\((0,5)\)
Q	3	2	\((3,2)\)
S	4	4	\((4,4)\)
T	2	6	\((2,6)\)