Why Compare Whole Risk Functions?
A decision rule does not have a single risk value unless the parameter space contains only one point or a particular averaging criterion has been chosen. Its risk is a function of the unknown parameter. To say one rule is better than another in the pointwise sense therefore requires a comparison at every parameter value, not just at a selected value or on average.
The resulting comparisons have an order structure. Risk functions can be equal, one can be no larger than another at every parameter, or they can cross so that neither is no larger everywhere. Distinguishing these possibilities is useful when assessing admissibility and when deciding whether a rule can be replaced without increasing risk anywhere.
The relation \(\delta_1\preceq_R\delta_2\) means that \(\delta_1\) has no greater risk than \(\delta_2\) anywhere in the parameter space. It allows equality at some or even all parameter values. It does not mean that the first rule is strictly better everywhere.
Terminology for “strict dominance” is not fully uniform across sources. Here, ordinary dominance means weak inequality everywhere and strict inequality somewhere; everywhere-strict means strict inequality at every parameter. These are different conditions. Since \(\Theta\) is nonempty, everywhere-strict comparison implies ordinary dominance, but ordinary dominance need not imply everywhere-strict comparison. The nonemptiness condition matters: if an empty parameter space were allowed, “strict at every parameter” would be vacuously true, while “strict at some parameter” would be false.
The Order on Risk Functions
For real-valued functions on \(\Theta\), pointwise order compares their values separately at each parameter. The following result explains why this is a partial order on risk functions, while the corresponding comparison of decision rules need not be one.
Proof. Let \(r,s,t\) be real-valued functions on \(\Theta\), and write \(r\le s\) pointwise when \(r(\theta)\le s(\theta)\) for every \(\theta\). For each \(\theta\), \(r(\theta)\le r(\theta)\), so \(r\le r\). If \(r\le s\) and \(s\le t\), then for each \(\theta\), \(r(\theta)\le s(\theta)\le t(\theta)\); hence \(r\le t\). If \(r\le s\) and \(s\le r\), then for every \(\theta\), both inequalities hold, so \(r(\theta)=s(\theta)\). Thus \(r=s\) as functions. These are the three defining properties of a partial order.
For rules, apply the same pointwise arguments to \(r(\theta)=R(\theta,\delta)\). Reflexivity and transitivity follow. If both \(\delta_1\preceq_R\delta_2\) and \(\delta_2\preceq_R\delta_1\), the risks agree at every parameter, but the rules themselves need not be the same procedure. Thus antisymmetry can fail on rules. Define \(\delta_1\sim_R\delta_2\) when their risks agree everywhere. On equivalence classes under \(\sim_R\), the induced relation is antisymmetric: mutual comparison implies equal risk functions and therefore the same equivalence class. It is consequently a partial order on those classes. \(\square\)
This distinction is not merely formal. Two procedures may use different calculations or actions but have the same risk at every parameter. Risk comparison cannot rank one above the other; it places them in the same risk-equivalence class. In contrast, crossing risk functions are incomparable under pointwise order.
Worked Comparisons
Worked Example: A Finite Risk Table
Consider a finite parameter space \(\Theta=\{\theta_1,\theta_2,\theta_3\}\) and five rules with the following risks.
| Parameter | \(\delta_A\) | \(\delta_B\) | \(\delta_C\) | \(\delta_D\) | \(\delta_E\) |
|---|---|---|---|---|---|
| \(\theta_1\) | 1 | 2 | 2 | 1 | 1 |
| \(\theta_2\) | 2 | 1 | 2 | 2 | 2 |
| \(\theta_3\) | 1 | 1 | 2 | 1 | 0 |
Compare \(\delta_A\) with \(\delta_C\). The risk differences \(R(\theta_i,\delta_C)-R(\theta_i,\delta_A)\) are \(1,0,1\), all nonnegative. Thus \(\delta_A\preceq_R\delta_C\), and the inequality is strict at \(\theta_1\) and \(\theta_3\). This is ordinary dominance, but not everywhere-strict comparison because the risks tie at \(\theta_2\).
The risks of \(\delta_A\) are \((1,2,1)\), while those of \(\delta_B\) are \((2,1,1)\). At \(\theta_1\), \(\delta_A\) has lower risk; at \(\theta_2\), \(\delta_B\) has lower risk. Neither rule is no worse at every parameter, so they are incomparable. Rules \(\delta_A\) and \(\delta_D\) have identical risk vectors, so they are risk-equivalent. Finally, \(\delta_E\) has risks \((1,2,0)\), which are no larger than \(\delta_A\)'s and strictly smaller at \(\theta_3\); hence \(\delta_E\) ordinarily dominates \(\delta_A\).
These numbers can be realized in a decision problem with no informative observation: take one available observation value and define the loss of each action at each parameter to equal the corresponding table entry. The table therefore gives a direct, valid illustration of the pointwise comparisons.
Worked Example: Risk Functions Indexed by a Continuous Parameter
Let \(\Theta=[0,1]\). Consider a decision problem with no informative observation and three available actions whose losses, and hence risks, are \(r_A(\theta)=\theta^2\), \(r_B(\theta)=\theta\), and \(r_C(\theta)=1-\theta\). All three losses are nonnegative on \([0,1]\).
For \(A\) and \(B\), the difference is \(r_B(\theta)-r_A(\theta)=\theta-\theta^2=\theta(1-\theta)\). This is nonnegative for every \(\theta\in[0,1]\), and positive when \(0<\theta<1\). Therefore \(A\) ordinarily dominates \(B\). It is not everywhere-strict: the risks tie at both endpoints, since \(r_A(0)=r_B(0)=0\) and \(r_A(1)=r_B(1)=1\).
Compare \(A\) and \(C\). At \(\theta=0\), \(r_A(0)=0<1=r_C(0)\), while at \(\theta=1\), \(r_A(1)=1>0=r_C(1)\). The inequality changes direction, so neither risk function is pointwise no larger than the other. They are incomparable, even though they may be ordered at particular parameter values.
Worked Example: Computing Risks in a Binomial Model
Let \(X\sim\operatorname{Binomial}(1,\theta)\), with \(\Theta=\{0,\tfrac12,1\}\), and use squared-error loss \(L(\theta,a)=(a-\theta)^2\). Compare the rules \(\delta_0(X)=0\), \(\delta_1(X)=1\), and \(\delta_X(X)=X\).
For \(\delta_0\), the action is always zero, so \(R(\theta,\delta_0)=(0-\theta)^2=\theta^2\). Its risks at the three parameter values are \(0,\tfrac14,1\). For \(\delta_1\), the action is always one, so \(R(\theta,\delta_1)=(1-\theta)^2\), giving risks \(1,\tfrac14,0\).
For \(\delta_X\), \(X\) is either zero or one. The squared error is \(\theta^2\) when \(X=0\), an event with probability \(1-\theta\), and \((1-\theta)^2\) when \(X=1\), an event with probability \(\theta\). Therefore \(R(\theta,\delta_X)=(1-\theta)\theta^2+\theta(1-\theta)^2=\theta(1-\theta)\). Its risks at \(0,\tfrac12,1\) are \(0,\tfrac14,0\).
At all three parameter values, the risk of \(\delta_X\) is no greater than the risk of \(\delta_0\): the differences \(R(\theta,\delta_0)-R(\theta,\delta_X)\) are \(0,0,1\). Thus \(\delta_X\) ordinarily dominates \(\delta_0\), but not everywhere-strictly. Similarly, the differences \(R(\theta,\delta_1)-R(\theta,\delta_X)\) are \(1,0,0\), so \(\delta_X\) ordinarily dominates \(\delta_1\) as well. The two constant rules are incomparable: at \(\theta=0\), \(\delta_0\) has smaller risk, and at \(\theta=1\), \(\delta_1\) has smaller risk.
Common Pitfall: Confusing “Somewhere” With “Everywhere”
A common error is to find one parameter at which a rule improves risk and conclude that it dominates another rule. For example, in the binomial calculation, \(\delta_X\) has lower risk than \(\delta_0\) at \(\theta=1\). That fact alone would not establish dominance. The required check is the inequality at every parameter; here it holds at all three values, with ties at the first two.
The reverse mistake is to demand strict improvement at every parameter before recognizing ordinary dominance. The example shows that a rule can be no worse everywhere and strictly better somewhere while tying elsewhere. A reliable comparison records both parts: first verify the weak inequality at every parameter, then check whether it is strict at least once. Only if the inequality is strict at every parameter should the comparison be called everywhere-strict here.
Check Your Understanding
Use the definitions in this tutorial to answer each question.
- Two rules have risks \((1,2,3)\) and \((1,2,4)\) on a three-point parameter space. Which pointwise comparison holds, and is it ordinary dominance?
- Can two distinct decision rules be mutually no worse than each other at every parameter? What does that imply about their risks?
- On a nonempty parameter space, why does everywhere-strict comparison imply ordinary dominance?
- For the functions \(r_1(\theta)=\theta^2\) and \(r_2(\theta)=\theta\) on \([0,1]\), identify the parameter values at which the risks tie and those at which \(r_1\) is strictly smaller.
- Why does a lower risk at one parameter value not by itself establish pointwise dominance?