Incomparable Risks and Crossing-Risk Examples

Why Risk Comparisons Can Go Both Ways

A decision rule is not uniformly preferable merely because it has lower risk in one part of the parameter space. Another rule may do better elsewhere. When this happens, their risk functions cross: neither procedure has risk no larger than the other's at every parameter value. Understanding this distinction is essential because dominance is a pointwise comparison over the entire parameter space, not a judgment based on one selected value or an overall average.

This tutorial develops a direct way to diagnose such comparisons. We will calculate risk differences, examine their signs, and use explicit models to see how continuous risk curves and finite risk tables can each show that two rules are incomparable.

Risk Differences and Incomparability

Definition: In a statistical decision problem, let $\theta\in\Theta$ be the unknown parameter, let $X$ denote the data, let $\delta$ be a decision rule, and let $L(\theta,a)$ be the loss from taking action $a$. The risk of $\delta$ at $\theta$ is $$ R(\theta,\delta)=E_\theta[L(\theta,\delta(X))]. $$ The expectation is with respect to the distribution of the data under $\theta$; if a rule uses randomization, it also averages over that randomization. We consider rules with finite risks at the parameter values being compared. A rule $\delta_1$ dominates $\delta_0$ if $R(\theta,\delta_1)\leq R(\theta,\delta_0)$ for every $\theta\in\Theta$, with strict inequality for at least one parameter value. Two rules are incomparable if neither dominates the other.

For two rules, define the signed risk difference $D(\theta)=R(\theta,\delta_1)-R(\theta,\delta_0)$. Its sign has a simple interpretation: $D(\theta)<0$ means $\delta_1$ has lower risk, $D(\theta)>0$ means $\delta_0$ has lower risk, and $D(\theta)=0$ means they have equal risk at that parameter. In particular, if $D$ is negative somewhere and positive somewhere, each rule has a strict advantage at some parameter value.

Theorem: Suppose $\delta_0$ and $\delta_1$ have finite risk on $\Theta$. If there are parameter values $\theta_-$ and $\theta_+$ such that $$ R(\theta_-,\delta_1)<R(\theta_-,\delta_0) \quad\text{and}\quad R(\theta_+,\delta_1)>R(\theta_+,\delta_0), $$ then neither rule dominates the other.

Proof. If $\delta_1$ dominated $\delta_0$, its risk would be no larger at every parameter value. That contradicts the strict inequality $R(\theta_+,\delta_1)>R(\theta_+,\delta_0)$. If $\delta_0$ dominated $\delta_1$, its risk would be no larger at every parameter value, contradicting $R(\theta_-,\delta_1)<R(\theta_-,\delta_0)$. Both possible dominance claims are ruled out, so the rules are incomparable. $\square$

Worked Examples of Crossing Risks

Worked Example: Bernoulli Estimation and the Risk-Difference Sign

Let $X\sim\operatorname{Bernoulli}(\theta)$, where $\theta\in[0,1]$, and use squared-error loss for estimating $\theta$. Compare $\delta_X(X)=X$ with the constant rule $\delta_{1/2}(X)=1/2$. Calculate both risks and determine where each rule has lower risk.

Since $E_\theta[X]=\theta$ and $\operatorname{Var}_\theta(X)=\theta(1-\theta)$, the risk of $\delta_X$ is $$ R(\theta,\delta_X) =E_\theta[(X-\theta)^2] =\operatorname{Var}_\theta(X) =\theta(1-\theta). $$ The constant rule has risk $$ R(\theta,\delta_{1/2}) =E_\theta[(1/2-\theta)^2] =(1/2-\theta)^2. $$

Set $D(\theta)=R(\theta,\delta_X)-R(\theta,\delta_{1/2})$. Expanding the square gives $$ D(\theta)=\theta(1-\theta)-(1/2-\theta)^2 =2\theta(1-\theta)-\frac14. $$ At the endpoints and midpoint, $$ D(0)=-\frac14,\qquad D(1/2)=\frac14,\qquad D(1)=-\frac14. $$ Thus the constant rule has lower risk at $\theta=1/2$, while $\delta_X$ has lower risk at $\theta=0$ and $\theta=1$. In particular, the sign at the midpoint is positive: there $R(1/2,\delta_X)=1/4$ and $R(1/2,\delta_{1/2})=0$.

For the exact crossing points, solve $D(\theta)=0$: $$ 2\theta(1-\theta)=\frac14 \quad\Longleftrightarrow\quad \theta^2-\theta+\frac18=0 \quad\Longleftrightarrow\quad \theta=\frac{1\pm 1/\sqrt{2}}{2}. $$ The quadratic $D(\theta)$ is positive between these two roots and negative outside them. Therefore, the constant rule is better between the roots, and $\delta_X$ is better outside them. Each rule has a strict advantage somewhere, so neither dominates.

Worked Example: A Normal-Mean Rule and a Constant Rule

Suppose $X\sim N(\theta,1)$, with $\theta\in\mathbb{R}$, and use squared-error loss. Compare $\delta_X(X)=X$ with the rule $\delta_0(X)=0$. Their risks are $$ R(\theta,\delta_X)=E_\theta[(X-\theta)^2]=\operatorname{Var}_\theta(X)=1 $$ and $$ R(\theta,\delta_0)=E_\theta[(0-\theta)^2]=\theta^2. $$

Taking $D(\theta)=R(\theta,\delta_X)-R(\theta,\delta_0)$ gives $D(\theta)=1-\theta^2$. At $\theta=0$, $D(0)=1$: the constant rule has risk $0$, compared with risk $1$ for $\delta_X$. At $\theta=2$, $D(2)=-3$: $\delta_X$ has risk $1$, compared with risk $4$ for the constant rule. The equality points solve $1-\theta^2=0$, so they are $\theta=-1$ and $\theta=1$. The constant rule has lower risk for $-1<\theta<1$; $\delta_X$ has lower risk for $\theta<-1$ or $\theta>1$. The risk ordering reverses, so neither rule dominates the other.

Worked Example: Incomparability in a Finite Risk Table

Let the parameter space be $\Theta=\{0,1,2\}$. Suppose the data $X$ takes values $a$ and $b$, each with probability $1/2$ at every parameter value. The loss is squared error, $L(\theta,d)=(d-\theta)^2$, and we compare the constant-action rules $\delta_0(X)=0$ and $\delta_2(X)=2$. Because their actions do not depend on $X$, their risks are the corresponding squared errors: $$ R(\theta,\delta_0)=(0-\theta)^2,\qquad R(\theta,\delta_2)=(2-\theta)^2. $$

Evaluating these expressions at each parameter gives:

$\theta$	$R(\theta,\delta_0)$	$R(\theta,\delta_2)$	Lower-risk rule
0	0	4	$\delta_0$
1	1	1	Tie
2	4	0	$\delta_2$

At $\theta=0$, $\delta_0$ has strictly lower risk; at $\theta=2$, $\delta_2$ has strictly lower risk. The tie at $\theta=1$ does not change either strict comparison. Since neither rule has risk no larger at all three parameter values, neither dominates the other. A finite table therefore establishes incomparability just as directly as a plot or an algebraic risk difference.

When Risk Curves Are Continuous

For a continuous parameter space, an opposing pair of strict risk comparisons often means the curves meet somewhere between those parameter values. Continuity is needed for that further conclusion; it is not needed to establish incomparability itself.

Result: Suppose $\Theta$ contains the interval between $\theta_a$ and $\theta_b$, and both risk functions are finite and continuous on that interval. If $D(\theta_a)<0$ and $D(\theta_b)>0$, then there is at least one $\theta_c$ strictly between $\theta_a$ and $\theta_b$ for which $R(\theta_c,\delta_1)=R(\theta_c,\delta_0)$.

Proof. The difference $D(\theta)=R(\theta,\delta_1)-R(\theta,\delta_0)$ is continuous because it is the difference of two continuous functions. The endpoint signs are opposite, so $0$ lies between $D(\theta_a)$ and $D(\theta_b)$. By the intermediate value theorem, there is a value $\theta_c$ between $\theta_a$ and $\theta_b$ such that $D(\theta_c)=0$. By the definition of $D$, the two risks are equal at $\theta_c$. $\square$

The Bernoulli and normal examples show the kinds of equality points this result describes. It does not say that there is exactly one such point: a continuous difference can have multiple zeros. Nor does a zero alone establish incomparability; the decisive evidence is that the difference also takes both signs.

What Incomparability Does—and Does Not—Tell You

A crossing comparison answers a specific question about a pair of rules: neither is uniformly at least as good as the other in risk. It does not show that either rule is admissible. A third rule could have risk no greater than one of them everywhere, with strictly lower risk somewhere. Establishing admissibility requires checking possible competitors, not just comparing two selected procedures.

It is also important not to replace pointwise comparison with a single summary. A rule can have smaller maximum risk or smaller risk averaged under a chosen distribution on $\Theta$, yet still have greater risk at some parameter value. Such a summary can be useful for a different decision criterion, but it does not erase a point where the rule loses a pointwise comparison.

Common Mistakes

Reversing the sign interpretation. If $D=R(\theta,\delta_1)-R(\theta,\delta_0)$, then a positive value means $\delta_0$ has lower risk, not $\delta_1$. Substitute the two risks at a specific parameter to check the sign.
Checking only one parameter value. A rule that wins at one value need not dominate. Dominance requires a no-larger risk at every parameter value.
Treating a tie as evidence of dominance. Equality at some parameters is compatible with incomparability. Look for strict comparisons in both directions across the parameter space.
Assuming a crossing must be unique or smooth. Opposing signs establish incomparability without either assumption. Continuity guarantees at least one equality point between opposite signs, not exactly one.
Calling an incomparable rule admissible. The comparison has considered only one opponent. Another procedure may still dominate it.

Key takeaway: To establish that two rules are incomparable, find one parameter value where each has strictly lower risk than the other. The sign of their risk difference records these reversals; a crossing point is useful for describing them, but is not itself the criterion for incomparability.

Check Your Understanding

Answer each question using pointwise risk comparisons.

If $D(\theta)=R(\theta,\delta_1)-R(\theta,\delta_0)$ is negative at one parameter and positive at another, what does each sign say about the lower-risk rule?
In the Bernoulli example, calculate $D(0)$ and $D(1/2)$. Which rule has lower risk at each value?
In the normal-mean example, at which values of $\theta$ are the risks equal, and which rule has lower risk at $\theta=2$?
In a finite risk table, one rule has strictly lower risk at one parameter and the other rule has strictly lower risk at another. Does a tie at a third parameter alter the conclusion? Explain.
Why does showing that two rules are incomparable not establish that either one is admissible?

\(\theta\)	\(R(\theta,\delta_0)\)	\(R(\theta,\delta_2)\)	Lower-risk rule
0	0	4	\(\delta_0\)
1	1	1	Tie
2	4	0	\(\delta_2\)