Local Risk Improvements and Positive Prior Mass

From a Pointwise Advantage to a Local One

In an admissibility argument, a competing procedure may have strictly smaller risk at one parameter value. By itself, that is only a pointwise fact. The useful next step is to ask whether the advantage persists for nearby parameter values, and whether a prior assigns positive probability to those values. Continuity supplies the local persistence; positive prior mass makes the local advantage contribute to an integrated-risk comparison.

Let $\Theta$ be a topological parameter space, and let $\pi$ be a prior probability measure on its Borel sets. For two procedures $d$ and $d'$, define their risk difference, with the sign chosen so that positive values favor $d'$, by

g(\theta)=R(\theta,d)-R(\theta,d').

Thus $g(\theta)>0$ means that $d'$ has strictly smaller risk than $d$ at $\theta$. This tutorial concerns how that strict inequality can be transferred from one parameter point to a neighborhood and then to an integral. A local improvement alone does not say that $d'$ has lower integrated risk over the entire parameter space: the risk difference could be negative elsewhere.

Definition: A prior gives positive mass to every neighborhood of $\theta_0$ if $\pi(U)>0$ for every open set $U$ containing $\theta_0$. A prior with this property at every $\theta_0\in\Theta$ has full support on $\Theta$.

The Local Improvement Lemma

The topological condition in the next result is continuity of $g$ at the particular point $\theta_0$; continuity everywhere is not needed. Measurability is a separate requirement: a function continuous at one point need not be measurable throughout a neighborhood. We therefore require measurability on the neighborhood that will be integrated.

Theorem: Suppose $g$ is real-valued, measurable on $\Theta$, and continuous at $\theta_0\in\Theta$, with $g(\theta_0)>0$. Then there is an open neighborhood $U$ of $\theta_0$ such that $$ g(\theta)>\frac{g(\theta_0)}{2} \qquad\text{for every }\theta\in U. $$ If, in addition, $\pi(U)>0$, then $$ 0<\frac{g(\theta_0)}{2}\,\pi(U) \leq \int_U g(\theta)\,\pi(d\theta) \leq \frac{3g(\theta_0)}{2}\,\pi(U)<\infty. $$

Proof. Put $\varepsilon=g(\theta_0)/2>0$. By continuity at $\theta_0$, there is an open neighborhood $U$ such that $|g(\theta)-g(\theta_0)|<\varepsilon$ for every $\theta\in U$. Consequently,

\frac{g(\theta_0)}{2}<g(\theta)<\frac{3g(\theta_0)}{2} \qquad(\theta\in U).

Since $g$ is measurable on $U$, its integral there is defined. The displayed bounds show that it is bounded and positive on $U$. Integrating the lower and upper bounds with respect to the probability measure $\pi$ gives the claimed inequalities. The strict positivity follows from $\pi(U)>0$, and finiteness follows from $\pi(U)\leq 1$. $\square$

The conclusion is local: it identifies a measurable region on which the risk difference is uniformly bounded away from zero. Positive prior mass is what converts this bound into a strictly positive integral over that region. If $g=R(\cdot,d)-R(\cdot,d')$, this local integral is a positive contribution favoring $d'$. Other regions could still contribute in the opposite direction.

Worked Examples

Worked Example: A Strict Local Contribution Without Global Dominance

Let $\Theta=[0,1]$, and suppose the observation has any distribution because both procedures ignore the data and report constants: $d=0$ and $d'=1$. Use squared-error loss $L(\theta,a)=(a-\theta)^2$, and take the uniform prior on $[0,1]$.

The risks are the losses themselves:

R(\theta,d)=\theta^2,\qquad R(\theta,d')=(1-\theta)^2,\qquad g(\theta)=\theta^2-(1-\theta)^2=2\theta-1.

At $\theta_0=7/8$, the advantage is $g(7/8)=3/4>0$. The relative neighborhood $U=(3/4,1]$ contains $\theta_0$, has prior mass $1/4$, and $g(\theta)>1/2$ throughout it. Its exact integrated contribution is

\int_{3/4}^{1}(2\theta-1)\,d\theta =\left[\theta^2-\theta\right]_{3/4}^{1} =0-\left(\frac{9}{16}-\frac{12}{16}\right) =\frac{3}{16}>0.

But $g$ is negative below $1/2$, and the full integrated difference is $\int_0^1(2\theta-1)\,d\theta=0$. Thus a positive local contribution does not, by itself, establish strict improvement in total integrated risk.

Worked Example: Global Domination Makes the Total Comparison Strict

Use the same procedures and squared-error risks, but let $\Theta=[1/2,1]$ and let $\pi$ be uniform on that interval. For every $\theta\in\Theta$,

g(\theta)=2\theta-1\geq 0.

Moreover, $g(3/4)=1/2>0$, and every relative neighborhood of $3/4$ has positive uniform-prior mass. The risks are nonnegative and integrable under this prior. Their integrated values are

\int_{1/2}^{1}\theta^2\,\pi(d\theta) =2\int_{1/2}^{1}\theta^2\,d\theta =2\left[\frac{\theta^3}{3}\right]_{1/2}^{1} =\frac{7}{12},

\int_{1/2}^{1}(1-\theta)^2\,\pi(d\theta) =2\int_{1/2}^{1}(1-\theta)^2\,d\theta =2\left[\frac{(1-\theta)^3}{-3}\right]_{1/2}^{1} =\frac{1}{12}.

Therefore the integrated risk difference is $7/12-1/12=1/2>0$. Here the local strict advantage combines with nonnegative risk difference everywhere, so no region can cancel the positive contribution.

Worked Example: A Neighborhood With Zero Prior Mass

Let $\Theta=\mathbb{R}$, and consider a decision problem with no informative observation and two actions $d,d'$. Set their deterministic losses to $L(\theta,d)=h(\theta)$ and $L(\theta,d')=0$, where

h(\theta)=\max\{1-2|\theta-1|,0\}.

The risk difference is $g(\theta)=h(\theta)$, a continuous, nonnegative function with $g(1)=1$. Hence $d'$ strictly improves on $d$ near $\theta_0=1$. Now put all prior mass at zero: $\pi=\delta_0$. For $U=(3/4,5/4)$, which contains $1$, $\pi(U)=0$; indeed $h(0)=0$, so

\int_{\mathbb{R}}g(\theta)\,\pi(d\theta)=g(0)=0.

Continuity still gives a local risk improvement, but this prior does not see it. Positive prior mass on the improving neighborhood is a substantive condition, not a consequence of continuity.

From a Local Integral to Strict Integrated-Risk Improvement

To compare total integrated risks, a positive contribution on one neighborhood must not be offset by a negative contribution elsewhere. A standard sufficient condition is global domination, $R(\theta,d')\leq R(\theta,d)$ at every parameter value. The next result also states conditions that justify the integrated comparison rather than presuming that finite risk for one procedure automatically implies finite risk for the other.

Theorem: Let $\pi$ be a probability measure. Suppose $R(\theta,d)$ and $R(\theta,d')$ are measurable, nonnegative, finite for every $\theta$, and $$ R(\theta,d')\leq R(\theta,d)\qquad\text{for every }\theta\in\Theta. $$ Let $g=R(\cdot,d)-R(\cdot,d')$ be continuous at $\theta_0$, with $g(\theta_0)>0$. Suppose every neighborhood of $\theta_0$ has positive $\pi$-mass, and $$ \int_\Theta R(\theta,d)\,\pi(d\theta)<\infty. $$ Then $d'$ also has finite integrated risk, and $$ \int_\Theta R(\theta,d')\,\pi(d\theta) < \int_\Theta R(\theta,d)\,\pi(d\theta). $$

Proof. Since both risks are measurable and $R(\theta,d')\leq R(\theta,d)$, nonnegativity gives

0\leq\int_\Theta R(\theta,d')\,\pi(d\theta) \leq\int_\Theta R(\theta,d)\,\pi(d\theta)<\infty.

Thus both integrated risks are finite. The local improvement theorem supplies a neighborhood $U$ of $\theta_0$ and a constant $c=g(\theta_0)/2>0$ such that $g(\theta)>c$ on $U$. By assumption $\pi(U)>0$, so $\int_U g\,d\pi>0$. Global domination gives $g\geq0$ on $\Theta$, and therefore

\int_\Theta g(\theta)\,\pi(d\theta) \geq \int_U g(\theta)\,\pi(d\theta)>0.

Because the two risks are integrable, linearity of the integral yields

\int_\Theta g(\theta)\,\pi(d\theta) = \int_\Theta R(\theta,d)\,\pi(d\theta) - \int_\Theta R(\theta,d')\,\pi(d\theta).

The positive difference proves the asserted strict inequality. $\square$

Assumptions That Cannot Be Dropped Casually

The local lemma and the integrated-risk theorem rely on distinct conditions. Keeping them separate prevents a common chain of invalid inferences.

Continuity is local, measurability is for integration. Continuity at $\theta_0$ gives a neighborhood bound, but does not ensure $g$ is measurable throughout that neighborhood. To write its integral there, require measurability on the neighborhood (or, as above, on all of $\Theta$).
Positive prior mass is not implied by a positive value. A prior concentrated away from the improving neighborhood assigns it zero mass, as the third example shows. Full support is a convenient sufficient condition for positive mass around every parameter point.
A local advantage is not global domination. Negative risk differences elsewhere can cancel the positive local contribution. The first example has a positive local integral but a zero total integral.
Finite integrated risk for $d$ does not alone control $d'$. Under nonnegative risks and global domination, $R(\theta,d')\leq R(\theta,d)$ supplies the required integrability. If losses may be signed, nonnegativity is unavailable; one must instead impose conditions ensuring both risks are integrable, such as absolute integrability under $\pi$, before subtracting their integrals.

These distinctions are especially useful in admissibility arguments: continuity can turn a strict inequality at one point into a robust local inequality, while support and domination determine whether that fact certifies a strict integrated comparison. The continuity step itself does not establish either global risk ordering or a prior's support.

Key takeaway: If $g(\theta_0)>0$ and $g$ is continuous at $\theta_0$, then $g$ is bounded away from zero on some neighborhood. If that neighborhood is measurable and has positive prior mass, it contributes a strictly positive, finite amount to the integrated risk difference. A strict total integrated-risk improvement additionally requires control of the difference elsewhere and conditions that make both risk integrals legitimate.

Check Your Understanding

Answer each question by identifying which condition supplies the needed step.

Suppose $g(\theta_0)=0.8$ and $g$ is continuous at $\theta_0$. What lower bound can be guaranteed on some neighborhood of $\theta_0$?
Why does measurability on the neighborhood matter even when continuity at its central point is known?
If $g$ is positive on a neighborhood but the prior assigns that neighborhood probability zero, what can be concluded about its contribution to the prior integral?
Give the extra condition in the strict integrated-risk theorem that prevents negative contributions away from the improving neighborhood.
Why does nonnegativity of the risks, together with global domination and finite integrated risk for $d$, ensure finite integrated risk for $d'$?