Risk Equivalence Without Rule Equality

When Different Rules Have the Same Risk

A decision rule specifies what action to take for each possible observation. Risk, by contrast, summarizes a rule’s expected loss at a parameter value. Because an expectation can conceal differences between procedures, two distinct rules may have exactly the same risk at every parameter value. This is risk equivalence: equality of performance as measured by risk, not equality of the procedures themselves.

This distinction matters when comparing rules. A rule that differs from another only on observations that cannot occur may have the same risk for an elementary reason. Symmetric actions can also give identical losses. And sometimes different rules incur different losses at particular observations, but those differences balance in expectation. In none of these cases does equality of risk by itself establish strict domination.

Definition: Consider a statistical decision problem with parameter space $\Theta$, observation $X$ having distribution $P_\theta$ at parameter $\theta$, action space $\mathcal{A}$, and nonnegative loss $L(\theta,a)$. A deterministic decision rule $\delta$ is a measurable function from the observation space to $\mathcal{A}$. Its risk is $$ R(\theta,\delta)=E_\theta[L(\theta,\delta(X))]. $$ Assume the risks under discussion are finite. Two rules $\delta$ and $\eta$ are risk-equivalent if $R(\theta,\delta)=R(\theta,\eta)$ for every $\theta\in\Theta$. They are distinct as rules if $\delta(x)\ne\eta(x)$ for at least one observation $x$.

The quantifier “for every $\theta$” is essential: equality at one parameter value is not risk equivalence. Also, two rules can be distinct as functions even if no data generated by the model can reveal their difference. The following results make these points precise.

Changing a Rule on Impossible Observations

An observation can be part of the formally specified sample space and still have probability zero under every parameter value. A rule’s action there cannot affect its expected loss. More generally, it is enough that two rules agree outside a set that has probability zero under every distribution in the model.

Theorem: Let $N$ be a measurable subset of the observation space such that $P_\theta(X\in N)=0$ for every $\theta\in\Theta$. If two measurable rules $\delta$ and $\eta$ agree for every $x\notin N$, then $R(\theta,\delta)=R(\theta,\eta)$ for every $\theta$, provided the risks are defined.

Proof. Fix any $\theta$. Since $\delta(x)=\eta(x)$ outside $N$, the losses $L(\theta,\delta(X))$ and $L(\theta,\eta(X))$ are equal whenever $X\notin N$. The event $X\in N$ has $P_\theta$-probability zero. Thus the two loss random variables are equal almost surely under $P_\theta$, and almost surely equal integrable random variables have the same expectation. Therefore $R(\theta,\delta)=R(\theta,\eta)$. Since $\theta$ was arbitrary, the equality holds throughout $\Theta$. $\square$

Worked Example: A Rule Differs Only at an Impossible Sample Value

Let $\Theta=[0,1]$, let the observation space be $\{0,1,2\}$, and suppose $P_\theta(X=1)=\theta$, $P_\theta(X=0)=1-\theta$, and $P_\theta(X=2)=0$. Actions are real numbers and the loss is squared error, $L(\theta,a)=(a-\theta)^2$. Compare $\delta(x)=x$ with the rule $\eta$ given by $\eta(0)=0$, $\eta(1)=1$, and $\eta(2)=17$.

The rules differ at $x=2$, so they are distinct functions. But that observation has probability zero for every $\theta$. Under either rule, the action is $0$ when $X=0$ and $1$ when $X=1$. Consequently, the risk of each is

(1-\theta)(0-\theta)^2+\theta(1-\theta)^2 =\theta^2(1-\theta)+\theta(1-\theta)^2 =\theta(1-\theta).

This calculation includes the endpoints: the risk is zero at $\theta=0$ and $\theta=1$. The rules remain distinct, but the model assigns no probability to the only observation at which they differ.

Symmetric Actions and Identical Loss

A second source of risk equivalence is symmetry in the loss. Suppose a transformation of the action space changes the selected action but leaves its loss unchanged at every parameter value. Applying that transformation to a rule produces a different rule whenever it changes the rule’s action, yet leaves the risk exactly the same.

Theorem: Let $s:\mathcal{A}\to\mathcal{A}$ be a measurable transformation such that $L(\theta,s(a))=L(\theta,a)$ for every $\theta\in\Theta$ and $a\in\mathcal{A}$. For any measurable rule $\delta$, define $\eta(x)=s(\delta(x))$. Then $R(\theta,\eta)=R(\theta,\delta)$ for every $\theta$ for which the risks are defined.

Proof. For every observation $x$ and parameter $\theta$, the assumed invariance gives $L(\theta,\eta(x))=L(\theta,s(\delta(x)))=L(\theta,\delta(x))$. In particular, the losses evaluated at the random observation $X$ are equal for every possible outcome. Taking expectations under $P_\theta$ gives $R(\theta,\eta)=R(\theta,\delta)$. This holds for each $\theta$. $\square$

Worked Example: Two Sign-Symmetric Actions

Let $\Theta=\{0,1\}$, let the action space be $\{-1,1\}$, and use the loss $L(\theta,a)=(a^2-\theta)^2$. The observation may have any distribution depending on $\theta$, since the rules below do not use it. Consider the constant rules $\delta(x)=1$ and $\eta(x)=-1$.

The transformation $s(a)=-a$ switches the two actions. But $s(a)^2=a^2$, so $L(\theta,s(a))=L(\theta,a)$. Directly, both actions have square $1$, giving

L(\theta,1)=L(\theta,-1)=(1-\theta)^2.

Thus each rule has risk $(1-\theta)^2$: it is $1$ at $\theta=0$ and $0$ at $\theta=1$. The rules choose opposite actions at every observation, yet their risks agree because the loss does not distinguish the signs. The equivalence is a property of the specified loss and model, not a claim that the actions themselves are identical.

Different Outcome-by-Outcome Losses Can Balance

Risk equivalence does not require the two rules to make the same choice on impossible observations or to have identical loss for each observation. The expectation may instead average out their differences. In the next example, both sample outcomes have positive probability, and the two rules exchange which outcome produces an error.

Worked Example: Opposite Rules in an Uninformative Experiment

Let $\Theta=\{0,1\}$. The observation $X$ is a fair binary variable under either parameter: $P_\theta(X=0)=P_\theta(X=1)=1/2$. The action space is $\{0,1\}$, with loss $L(\theta,a)=1$ if $a\ne\theta$, and $0$ otherwise. Compare $\delta(x)=x$ with $\eta(x)=1-x$.

At $\theta=0$, rule $\delta$ is wrong when $X=1$, which has probability $1/2$; rule $\eta$ is wrong when $X=0$, also with probability $1/2$. Hence both risks are $1/2$.

At $\theta=1$, rule $\delta$ is wrong when $X=0$, of probability $1/2$; rule $\eta$ is wrong when $X=1$, also of probability $1/2$. Therefore

R(0,\delta)=R(0,\eta)=\frac12, \qquad R(1,\delta)=R(1,\eta)=\frac12.

The rules disagree at both possible observations, and their losses are not equal observation by observation. Rather, each rule makes one correct and one incorrect decision across the two equally likely outcomes, for either parameter value. This is another way distinct rules can have the same complete risk function.

Why Risk Equivalence Is Not Domination

In risk comparisons, domination conventionally requires a weak improvement at every parameter value and a strict improvement at least somewhere. For example, $\eta$ dominates $\delta$ if $R(\theta,\eta)\le R(\theta,\delta)$ for every $\theta$, with strict inequality for at least one $\theta$. Risk equivalence satisfies the non-strict inequalities in both directions, but supplies no strict inequality.

Result: If $\delta$ and $\eta$ are risk-equivalent, neither strictly dominates the other under the definition above.

Proof. Risk equivalence gives $R(\theta,\eta)=R(\theta,\delta)$ for every $\theta$. Therefore $R(\theta,\eta)\le R(\theta,\delta)$ holds everywhere, but strict inequality cannot hold at any parameter value. So $\eta$ does not dominate $\delta$. Interchanging the rules gives the same conclusion in the other direction. $\square$

This does not mean equality is irrelevant. It means the risk comparison alone cannot rank the two rules by strict improvement. If a convention uses “weakly no worse” to mean only the non-strict inequalities, then equivalent rules are weakly no worse than one another. That statement is weaker than domination as defined here, which requires a strict risk reduction somewhere.

A practical comparison should therefore keep two questions separate: Are the rules different? and Are their risks ordered? Differences between rules may matter for implementation, interpretation, or constraints, even when risk cannot distinguish them. Conversely, matching risks do not prove that the rules take the same actions or behave identically on every observation.

Common Pitfall

Wrong reasoning: “These rules have equal risk everywhere, so one dominates the other because its risk is never higher.” This overlooks the strict-improvement requirement in domination.
Correction: Equal risk everywhere gives no parameter value at which either rule has strictly lower risk. The rules are risk-equivalent, not strictly ordered by risk.
Another error: “The rules differ at some sample value, so their risks must differ.” The sample value may have probability zero under every parameter, or differences in loss across possible outcomes may balance in expectation. Check the risk function rather than inferring it from the rules’ pointwise differences.

Key takeaway: Risk is a function of the parameter, while a decision rule is a function of the observation. Distinct rules can map to the same risk function; equality of those functions alone provides no strict domination.

Check Your Understanding

Use the definitions and calculations above to answer the following.

What must be true of two rules’ risk functions for the rules to be risk-equivalent?
Why does changing a rule on a set that has probability zero for every parameter leave its risk unchanged?
In the sign-symmetric example, which property of the loss makes the two constant rules risk-equivalent?
In the fair-binary-observation example, are the rules’ losses equal at each observation? Explain how their risks nevertheless match.
Under the stated definition of domination, what condition fails when two distinct rules are risk-equivalent?