Prior Perturbations That Expose Dominance

Why Perturb a Prior?

A decision rule may have the same integrated risk as a competitor under one prior, even though the competitor performs better for some parameter values. Adding prior mass in a region of improvement changes the integrated comparison in a controlled way. This gives a precise test: does the added improvement outweigh any advantage the candidate had under the original prior?

The calculation compares two fixed decision rules. It does not assume that the candidate remains a Bayes rule after the prior changes, nor that the competitor becomes Bayes. The point is narrower: if the competitor has lower integrated risk under the perturbed prior, the candidate cannot be Bayes for that prior.

Integrated Risk and the Perturbation Identity

Let \(R(\theta,\delta)\) denote the risk of a measurable decision rule \(\delta\) at parameter value \(\theta\). For a finite nonzero measure \(\mu\) on the parameter space, define the integrated risk by

r(\delta;\mu)=\int R(\theta,\delta)\,\mu(d\theta).

A probability prior is a special case. Allowing finite measures that are not normalized is convenient because multiplying a prior measure by a positive constant multiplies every integrated risk by that constant and does not change which rule has lower integrated risk.

Definition: For two rules \(\delta\) and \(\delta'\), define their pointwise risk difference by \(\Delta(\theta)=R(\theta,\delta')-R(\theta,\delta)\). A negative value means that \(\delta'\) improves on \(\delta\) at \(\theta\). Their integrated risk difference under \(\mu\) is \(D_\mu=\int \Delta(\theta)\,\mu(d\theta)=r(\delta';\mu)-r(\delta;\mu)\).

In the results and examples below, take \(\mu\) and \(\nu\) to be finite nonnegative measures. Assume the risks are measurable and that both rules have finite integrated risks under each measure. These conditions ensure that the displayed risk differences are well defined and integrable.

Theorem: For \(\varepsilon\geq 0\), let \(\mu_\varepsilon=\mu+\varepsilon\nu\). Then \(D_{\mu_\varepsilon}=D_\mu+\varepsilon D_\nu\). If \(\mu\) and \(\nu\) are probability measures and \(0\leq\varepsilon\leq1\), the normalized mixture \((1-\varepsilon)\mu+\varepsilon\nu\) has risk difference \((1-\varepsilon)D_\mu+\varepsilon D_\nu\).

Proof. By additivity of integration for finite measures and integrable \(\Delta\),

D_{\mu_\varepsilon} =\int \Delta(\theta)\,(\mu+\varepsilon\nu)(d\theta) =\int \Delta(\theta)\,\mu(d\theta) +\varepsilon\int \Delta(\theta)\,\nu(d\theta) =D_\mu+\varepsilon D_\nu.

For the normalized mixture, linearity gives

\int \Delta(\theta)\bigl((1-\varepsilon)\mu+\varepsilon\nu\bigr)(d\theta) =(1-\varepsilon)D_\mu+\varepsilon D_\nu.

This proves both identities. \(\square\)

The signs have a direct interpretation. If \(D_\mu=0\) and \(D_\nu<0\), every positive perturbation weight makes the competitor’s integrated risk smaller. If instead the candidate has a baseline advantage, \(D_\mu>0\), a sufficiently strong perturbation may still reverse the comparison. For probability measures, the exact condition is

(1-\varepsilon)D_\mu+\varepsilon D_\nu<0.

When \(D_\mu>0\) and \(D_\nu<0\), this is equivalent to

\varepsilon>\frac{D_\mu}{D_\mu-D_\nu}.

The threshold is between zero and one. A perturbation below it does not erase the candidate’s baseline advantage; a perturbation above it does.

Worked Examples

Worked Example: A Normal-Mean Rule and a Point-Mass Perturbation

Suppose \(X\sim N(\theta,1)\), with squared-error loss. Compare \(\delta(X)=X\) with \(\delta'(X)=X/2\). Compute their risks and determine how much prior weight on \(\theta=0\) is needed to reverse a baseline comparison.

Since \(E_\theta[X]=\theta\) and \(\operatorname{Var}_\theta(X)=1\), the risk of \(\delta\) is

R(\theta,\delta)=E_\theta[(X-\theta)^2]=1.

The rule \(\delta'\) has bias \(-\theta/2\) and variance \(1/4\). Thus

R(\theta,\delta') =\frac14+\frac{\theta^2}{4}, \qquad \Delta(\theta)=\frac{\theta^2-3}{4}.

Use the baseline probability prior \(\mu\) that puts probability \(1/2\) at each of \(\theta=-2\) and \(\theta=2\). At either support point, \(\Delta(\theta)=(4-3)/4=1/4\), so \(D_\mu=1/4\). Let \(\nu\) put all its mass at \(\theta=0\), where \(\Delta(0)=-3/4\), giving \(D_\nu=-3/4\).

For the mixture \((1-\varepsilon)\mu+\varepsilon\nu\), the difference is

(1-\varepsilon)\frac14+\varepsilon\left(-\frac34\right) =\frac14-\varepsilon.

It is negative exactly when \(\varepsilon>1/4\). For example, at \(\varepsilon=1/2\), the difference is \(-1/4\), so \(X/2\) has lower integrated risk under the mixture. At \(\varepsilon=1/4\), the rules tie; below that weight, \(X\) retains the integrated-risk advantage.

Worked Example: Perturbing a Bernoulli-Mean Comparison

Let \(X\sim\operatorname{Bernoulli}(p)\), \(0\leq p\leq1\), and use squared-error loss to estimate \(p\). Compare \(\delta(X)=X\) with the constant rule \(\delta'(X)=1/2\). Find a baseline prior and a perturbation that change which rule has lower integrated risk.

Because \(X\) is unbiased and has variance \(p(1-p)\),

R(p,\delta)=p(1-p).

The constant rule has squared-error risk \((1/2-p)^2\). Therefore,

\Delta(p)=\left(p-\frac12\right)^2-p(1-p) =2\left(p-\frac12\right)^2-\frac14.

Take the baseline prior \(\mu\) to put equal mass at \(p=0\) and \(p=1\). At either endpoint, \(\Delta(p)=1/4\), so \(D_\mu=1/4\). Perturb toward \(\nu\), a point mass at \(p=1/2\), where \(\Delta(1/2)=-1/4\). The risk difference under the mixture is

(1-\varepsilon)\frac14+\varepsilon\left(-\frac14\right) =\frac14-\frac{\varepsilon}{2}.

This expression is negative exactly when \(\varepsilon>1/2\). At \(\varepsilon=3/4\), for instance, it equals \(1/4-3/8=-1/8\). The constant rule then has lower integrated risk, despite its worse performance at the two baseline support points: the mixture assigns enough weight to the parameter value where it improves.

Worked Example: Exposing an Improvement in a Finite Decision Problem

Consider a parameter space \(\{\theta_1,\theta_2\}\), two actions \(a,b\), and the loss table below. Compare the constant rules \(\delta=a\) and \(\delta'=b\).

Parameter	Loss of \(a\)	Loss of \(b\)	\(\Delta(\theta)\)
\(\theta_1\)	1	0	-1
\(\theta_2\)	0	1	1

Let \(\mu\) put probability \(1/2\) on each state. The integrated risks of the two rules are both \(1/2\), so \(D_\mu=0\). Let \(\nu\) be a point mass at \(\theta_1\), where \(b\) has one unit less loss than \(a\). Under the mixture \((1-\varepsilon)\mu+\varepsilon\nu\), the difference is

(1-\varepsilon)D_\mu+\varepsilon D_\nu =(1-\varepsilon)0+\varepsilon(-1)=-\varepsilon.

For every \(\varepsilon>0\), rule \(b\) has lower integrated risk. This example makes the tie case especially clear: no baseline advantage needs to be overcome, so any positive perturbation in the indicated direction changes the comparison.

Using the Identity Carefully

The perturbation calculation is useful in admissibility arguments because it separates two questions that can otherwise be conflated: how the rules compare under a starting prior, and how they compare under added mass. The first is \(D_\mu\); the second is \(D_\nu\). Their weighted sum, not either quantity alone, determines the comparison after perturbation.

For example, suppose a candidate is Bayes under a probability prior \(\mu\). For any fixed competitor, Bayes optimality implies \(D_\mu\geq0\). If the competitor has \(D_\nu<0\) under a proposed perturbation, the identity shows exactly when that competitor defeats the candidate under the mixture. In the special case \(D_\mu=0\), every positive mixture weight suffices. When \(D_\mu>0\), the threshold formula is essential; merely identifying a region of improvement does not show that a small perturbation will reverse the comparison.

If the perturbed prior is \(\mu_\varepsilon\), a lower integrated risk for \(\delta'\) establishes that \(\delta\) is not Bayes under \(\mu_\varepsilon\), since a Bayes rule must minimize integrated risk among the available rules. It does not establish that \(\delta'\) is Bayes: another rule might have still lower integrated risk. Nor does it establish a pointwise dominance claim, which would require comparing risks across parameter values rather than integrating them against a prior.

Common Mistake

Pitfall: “The competitor improves somewhere, so adding a little prior mass there must make it better overall.” This overlooks the baseline difference. If \(D_\mu>0\), a small perturbation can leave that advantage intact; for example, the normal-mean calculation remains positive for every \(\varepsilon<1/4\). The correction is to compute both \(D_\mu\) and \(D_\nu\), then evaluate \((1-\varepsilon)D_\mu+\varepsilon D_\nu\). Only a negative result establishes lower integrated risk for the competitor under the perturbed prior.

A second practical check is to keep the comparison between the same two rules throughout the calculation. The perturbation identity is linear for fixed rules. If a rule is recomputed after changing the prior, its risk difference is no longer obtained merely by substituting the two original integrated differences into the identity.

Check Your Understanding

Use the risk-difference identity to answer these questions.

If \(D_\mu=0\) and \(D_\nu=-2\), what is the risk difference under \((1-\varepsilon)\mu+\varepsilon\nu\), and for which positive \(\varepsilon\) is it negative?
If \(D_\mu=3\) and \(D_\nu=-1\), calculate the exact threshold above which the mixture favors the competitor.
In the normal-mean example, why does a perturbation weight of \(1/5\) fail to make \(X/2\) better in integrated risk?
What conclusion follows if a competitor has lower integrated risk than a candidate under a proper perturbed prior? What stronger conclusion does not follow automatically?
Why must the risk comparison use the same two fixed rules before and after perturbing the prior?