Piecewise Prior Densities Near Parameter Boundaries

Why Separate Boundary Behavior?

In admissibility arguments, an integrated risk comparison can depend on more than what happens in the middle of a parameter space. A rule may have a clear risk advantage throughout the interior but little advantage, or even a disadvantage, near an endpoint. A prior density that assigns separate amounts of mass to these regions makes that trade-off explicit.

We develop the construction on $\Theta=(0,1)$. The same calculations apply to any finite interval after rescaling. Let $R(\theta,\delta)$ denote the risk of a decision rule $\delta$ at parameter $\theta$. For two rules $\delta_0$ and $\delta_1$, write $d(\theta)=R(\theta,\delta_0)-R(\theta,\delta_1)$. Thus $d(\theta)>0$ means that $\delta_1$ has lower risk at $\theta$. We will integrate this signed risk gap against priors that may emphasize the two boundary neighborhoods differently from the interior.

A Prior with Boundary Layers

Fix $0<\varepsilon<1/2$, and partition the parameter interval, up to endpoints of Lebesgue measure zero, into the left layer $A_0=(0,\varepsilon)$, the interior $A_I=[\varepsilon,1-\varepsilon]$, and the right layer $A_1=(1-\varepsilon,1)$. Assign respective masses $m_0,m_I,m_1$, where each mass is nonnegative and $m_0+m_I+m_1=1$.

Definition: The piecewise prior density with layer width $\varepsilon$ and masses $(m_0,m_I,m_1)$ is $$ q_\varepsilon(\theta) = \frac{m_0}{\varepsilon}\mathbf{1}_{A_0}(\theta) + \frac{m_I}{1-2\varepsilon}\mathbf{1}_{A_I}(\theta) + \frac{m_1}{\varepsilon}\mathbf{1}_{A_1}(\theta). $$ Here $\mathbf{1}_A$ is the indicator of the set $A$. The density is nonnegative and integrates to $m_0+m_I+m_1=1$.

This construction allows different densities on the boundary layers and the interior. A layer can even have zero assigned mass. The conditional uniform distribution on a region describes the shape within that region; the corresponding $m_j$ determines how much total prior mass the region receives.

Worked Example: A Bernoulli Model and an Asymmetric Prior

Let $X\sim\operatorname{Bernoulli}(\theta)$, with $\theta\in(0,1)$. Consider two constant decision rules, $a_0$ and $a_1$, with losses $L(\theta,a_0)=0$ and $L(\theta,a_1)=\theta$. Their risks are $R(\theta,a_0)=0$ and $R(\theta,a_1)=\theta$, so the risk advantage of $a_0$ over $a_1$, written $d(\theta)=R(\theta,a_1)-R(\theta,a_0)$, is $d(\theta)=\theta$.

Take $\varepsilon=0.1$ and masses $(m_0,m_I,m_1)=(0.2,0.5,0.3)$. The density is $2$ on the left layer, $0.5/0.8=0.625$ on the interior, and $3$ on the right layer. These pieces integrate to $0.2$, $0.5$, and $0.3$, respectively.

The regional uniform averages of $d(\theta)=\theta$ are $0.05$, $0.5$, and $0.95$: these are the midpoints of the three regions. Therefore the integrated risk advantage is

$$ 0.2(0.05)+0.5(0.5)+0.3(0.95) =0.01+0.25+0.285 =0.545. $$

Under the uniform prior on $(0,1)$, the same gap integrates to $\int_0^1\theta\,d\theta=0.5$. The increased value here results from assigning more mass to the right layer, where this particular risk advantage is larger.

Exact Decomposition of the Integrated Risk Gap

For a region of positive length, define its uniform average of $d$ by $\bar d_{j,\varepsilon}=|A_j|^{-1}\int_{A_j}d(\theta)\,d\theta$, where $j\in\{0,I,1\}$ and $|A_j|$ is its length. To use these averages, assume $d$ is Lebesgue-integrable on each of the three regions. This assumption ensures that every regional average is defined, even if the corresponding prior mass is zero.

Theorem: If $d$ is Lebesgue-integrable on each region, then its integrated value under $q_\varepsilon$ is $$ \int_0^1 d(\theta)q_\varepsilon(\theta)\,d\theta = m_0\bar d_{0,\varepsilon} + m_I\bar d_{I,\varepsilon} + m_1\bar d_{1,\varepsilon}. $$

Proof. Substitute the definition of $q_\varepsilon$ and use additivity of the integral over the three regions:

\begin{aligned} \int_0^1 d(\theta)q_\varepsilon(\theta)\,d\theta &= \frac{m_0}{\varepsilon}\int_{A_0}d(\theta)\,d\theta + \frac{m_I}{1-2\varepsilon}\int_{A_I}d(\theta)\,d\theta + \frac{m_1}{\varepsilon}\int_{A_1}d(\theta)\,d\theta\\ &= m_0\bar d_{0,\varepsilon} + m_I\bar d_{I,\varepsilon} + m_1\bar d_{1,\varepsilon}. \end{aligned}

The last equality follows because the region lengths are $\varepsilon$, $1-2\varepsilon$, and $\varepsilon$. The integrability assumption makes each average finite. In particular, this formula does not multiply zero mass by an undefined regional average. $\square$

This identity is the basic accounting rule: the integrated gap is a weighted average of the three regional gaps. When $d$ is a risk advantage, a positive integrated value favors $\delta_1$ over $\delta_0$ on average under the chosen prior. For a signed gap, the same identity records how any unfavorable boundary contribution offsets an interior advantage.

Worked Example: Boundary Mass Can Increase the Integrated Gap

Again use $X\sim\operatorname{Bernoulli}(\theta)$ on $(0,1)$, but let the two constant actions have losses $L(\theta,a_0)=0$ and $L(\theta,a_1)=(\theta-\tfrac12)^2$. The risk advantage of $a_0$ over $a_1$ is $d(\theta)=(\theta-\tfrac12)^2$, which is largest near the endpoints.

Set $\varepsilon=0.1$ and $(m_0,m_I,m_1)=(0.3,0.4,0.3)$. On the left layer, the uniform average is

\bar d_{0,\varepsilon} =\frac{1}{\varepsilon}\int_0^\varepsilon(\theta-\tfrac12)^2\,d\theta =\frac14-\frac{\varepsilon}{2}+\frac{\varepsilon^2}{3} =\frac{61}{300}.

By symmetry, the right-layer average is also $61/300$. Over the full interval, the integral of $d$ is $1/12$. Each boundary layer contributes $\varepsilon(61/300)=61/3000$, so the interior integral is $1/12-2(61/3000)=16/375$. Dividing by the interior length $0.8$ gives $\bar d_{I,\varepsilon}=4/75$. The integrated gap is therefore

0.3\left(\frac{61}{300}\right) +0.4\left(\frac{4}{75}\right) +0.3\left(\frac{61}{300}\right) = \frac{43}{300} \approx 0.1433.

With the uniform prior, the integrated gap is $1/12\approx0.0833$. Here the boundary layers receive $60\%$ of the prior mass, and the gap is larger there than in the interior. The reweighting therefore raises the integrated advantage.

How Reallocating Mass Changes the Result

The decomposition also gives an exact formula for changing the layer masses while keeping $\varepsilon$ fixed. Suppose the original masses are $m_0,m_I,m_1$, and the new masses are $m'_0,m'_I,m'_1$, with both triples nonnegative and summing to one. The conditional averages do not change because the regions do not change. Subtracting the two decompositions gives:

Result: The change in integrated gap is $$ \Delta' -\Delta = (m'_0-m_0)\bar d_{0,\varepsilon} + (m'_I-m_I)\bar d_{I,\varepsilon} + (m'_1-m_1)\bar d_{1,\varepsilon}, $$ where $\Delta$ and $\Delta'$ are the integrated gaps under the original and new masses.

Derivation. Apply the exact decomposition once with each mass triple:

\begin{aligned} \Delta' &=m'_0\bar d_{0,\varepsilon} +m'_I\bar d_{I,\varepsilon} +m'_1\bar d_{1,\varepsilon},\\ \Delta &=m_0\bar d_{0,\varepsilon} +m_I\bar d_{I,\varepsilon} +m_1\bar d_{1,\varepsilon}. \end{aligned}

Subtract the second equality from the first to obtain the stated expression. In particular, transferring mass $t$ from the interior to the left layer, with $0\le t\le m_I$, changes the integrated gap by $t(\bar d_{0,\varepsilon}-\bar d_{I,\varepsilon})$. It increases the gap exactly when the left-layer average exceeds the interior average. $\square$

Worked Example: A Singular Gap Near the Left Boundary

Let $X\sim\operatorname{Bernoulli}(\theta)$ for $\theta\in(0,1)$. Take constant actions with losses $L(\theta,a_0)=0$ and $L(\theta,a_1)=\theta^{-1/2}$. Then $d(\theta)=\theta^{-1/2}$ is the risk advantage of $a_0$ over $a_1$. It is unbounded near zero, but integrable on every interval $(0,\varepsilon)$.

Choose $\varepsilon=1/4$ and masses $(m_0,m_I,m_1)=(0.1,0.8,0.1)$. Direct integration gives the left-layer average $\bar d_{0,\varepsilon}=2/\sqrt{\varepsilon}=4$. The interior average is $2(\sqrt{1-\varepsilon}-\sqrt{\varepsilon})/(1-2\varepsilon)=2\sqrt{3}-2$. The right-layer average is $2(1-\sqrt{1-\varepsilon})/\varepsilon=8-4\sqrt{3}$. Thus

\begin{aligned} \int_0^1 d(\theta)q_\varepsilon(\theta)\,d\theta &=0.1(4)+0.8(2\sqrt{3}-2)+0.1(8-4\sqrt{3})\\ &=1.2\sqrt{3}-0.4 \approx 1.6785. \end{aligned}

The calculation is finite because $\int_0^\varepsilon \theta^{-1/2}\,d\theta=2\sqrt{\varepsilon}<\infty$. More generally, for $d(\theta)=\theta^{-\alpha}$, the left-layer integral is finite exactly when $\alpha<1$. A positive left-layer mass cannot make an infinite regional average finite.

Quantifying Boundary Contributions

A simple bound helps interpret an interior guarantee when the boundary behavior is less favorable. Suppose $d(\theta)\ge c$ throughout the interior and $d(\theta)\ge-b$ on both boundary layers, for constants $c,b\ge0$. Then $\bar d_{I,\varepsilon}\ge c$, while each boundary average is at least $-b$.

Result: Under those pointwise bounds, $$ \int_0^1 d(\theta)q_\varepsilon(\theta)\,d\theta \ge m_Ic-(m_0+m_1)b. $$

Proof. Substitute the regional lower bounds into the exact decomposition:

\begin{aligned} \int_0^1 d(\theta)q_\varepsilon(\theta)\,d\theta &=m_0\bar d_{0,\varepsilon} +m_I\bar d_{I,\varepsilon} +m_1\bar d_{1,\varepsilon}\\ &\ge -m_0b+m_Ic-m_1b\\ &=m_Ic-(m_0+m_1)b. \end{aligned}

Consequently, the interior advantage guarantees a positive integrated gap whenever $m_Ic>(m_0+m_1)b$. If $d\ge0$ everywhere, take $b=0$: then the integrated gap is at least $m_Ic$. A large boundary mass does not automatically invalidate a comparison, but the bound shows exactly how much unfavorable boundary behavior the interior contribution can absorb. $\square$

Common Pitfall and Practical Use

A common mistake is to treat a zero-mass region as if its conditional average were automatically defined. For example, if $m_0=0$, prior integrability under $q_\varepsilon$ alone says nothing about $\int_{A_0}d(\theta)\,d\theta$, because the prior density vanishes there. For $d(\theta)=1/\theta$, that left-layer integral is infinite even though the region contributes nothing to the prior integral. Writing $0\cdot\bar d_{0,\varepsilon}$ would then be undefined, not a valid zero contribution.

The correction is to state the needed integrability explicitly, as we did here by requiring Lebesgue integrability on each region, or to define the decomposition only over positive-mass regions and assign zero contribution to every zero-mass region. When gaps can be unbounded near an endpoint, check the regional integrals before using regional averages.

Piecewise priors are useful because they separate two questions: how a risk gap behaves within each region, and how much prior mass each region receives. The exact decomposition answers how those choices combine at a fixed layer width. It also makes the role of boundary behavior visible instead of hiding it inside a single prior integral.

Check Your Understanding

Use the regional decomposition and the assumptions stated in each question.

For $\varepsilon=0.1$ and masses $(0.1,0.8,0.1)$, write the three constant values of the prior density.
If the regional averages of a risk gap are $0.2$, $0.5$, and $-0.1$, respectively, what is the integrated gap for masses $(0.2,0.6,0.2)$?
How does the integrated gap change if mass $t$ is transferred from the interior to the right layer?
For which values of $\alpha$ is $\theta^{-\alpha}$ Lebesgue-integrable on the left boundary layer $(0,\varepsilon)$?
If the interior gap is at least $c$ and the boundary gaps are at least $-b$, state a sufficient inequality for the integrated gap to be positive.