Endpoint Control on Restricted Parameter Spaces

Why Finite Endpoints Need Their Own Control

On a parameter space such as \([a,b]\), there is no tail at infinity, but the endpoints still matter. A risk function may approach a different value at an endpoint than it has nearby, or may even grow without bound as the parameter approaches an excluded endpoint. Likewise, a sequence of priors that removes thin endpoint strips may have negligible effect when risks are bounded, but not when risk is singular there. Integrated-risk arguments are valid only after these endpoint contributions are controlled.

This tutorial develops two useful checks: convergence of integrated risks when prior densities change near finite endpoints, and an explicit integrability condition when risk is unbounded. It also distinguishes a parameter endpoint from a one-sided limit and explains why finite boundaries do not make boundary terms disappear automatically.

Definition: For a decision rule \(\delta\), its risk at parameter value \(\theta\) is \(R(\theta,\delta)=E_\theta[L(\theta,\delta(X))]\), when this expectation exists. Given two rules \(\delta_0\) and \(\delta_1\), define their risk gap in favor of \(\delta_0\) by \(\Delta(\theta)=R(\theta,\delta_0)-R(\theta,\delta_1)\). If \(\Delta(\theta)\geq 0\), then \(\delta_1\) has no greater risk than \(\delta_0\) at \(\theta\). For a prior probability measure \(\Pi\), the integrated risk is \(r(\Pi,\delta)=\int R(\theta,\delta)\,d\Pi(\theta)\), provided the integral exists.

If the parameter space is the open interval \((a,b)\), the endpoints are not parameter values, so a prior density on that interval assigns them no mass. Nonetheless, the one-sided limits of risk as \(\theta\downarrow a\) or \(\theta\uparrow b\) can determine whether endpoint strips contribute negligibly to integrated risk. For a closed interval \([a,b]\), the risk must also be defined at \(a\) and \(b\); its endpoint values need not equal its interior limits unless continuity is assumed.

Bounded Risks and Prior Truncation

A common perturbation on a finite interval removes small neighborhoods of the endpoints and renormalizes the remaining prior. The basic control is simple: if the risk gap is bounded, the change in its integral is bounded by the amount of prior mass changed. The prior densities need not have matching values at the endpoints. What matters is the mass of the strips removed and the normalization used afterward.

Theorem: Let \(\Pi\) have density \(\pi\), and let \(\Pi_k\) have densities \(\pi_k\) with respect to the same measure. Suppose \(\int |\pi_k-\pi|\to 0\) and \(|\Delta(\theta)|\leq M<\infty\) almost everywhere. Then \[ \left|\int \Delta(\theta)\pi_k(\theta)\,d\theta-\int \Delta(\theta)\pi(\theta)\,d\theta\right| \leq M\int|\pi_k(\theta)-\pi(\theta)|\,d\theta\longrightarrow 0. \]

Proof. Subtract the two integrals and combine them:

\left|\int \Delta(\theta)\{\pi_k(\theta)-\pi(\theta)\}\,d\theta\right| \leq \int |\Delta(\theta)|\,|\pi_k(\theta)-\pi(\theta)|\,d\theta \leq M\int|\pi_k(\theta)-\pi(\theta)|\,d\theta.

The final expression tends to zero by the assumed convergence of the densities in \(L^1\). This proves the claim. \(\square\)

Worked Example: Removing Endpoint Strips Under a Uniform Prior

Let \(\theta\in[0,1]\), let the prior \(\Pi\) be uniform, and consider the bounded risk gap \(\Delta(\theta)=\theta(1-\theta)\). For \(0<\varepsilon<1/2\), define \(\Pi_\varepsilon\) to be uniform on \([\varepsilon,1-\varepsilon]\). Compute both integrated gaps and their difference.

Under the uniform prior on \([0,1]\),

\int_0^1 \theta(1-\theta)\,d\theta =\left[\frac{\theta^2}{2}-\frac{\theta^3}{3}\right]_0^1 =\frac{1}{6}.

Under \(\Pi_\varepsilon\), the density is \(1/(1-2\varepsilon)\) on the truncated interval. Thus

\int \Delta\,d\Pi_\varepsilon =\frac{1}{1-2\varepsilon}\int_\varepsilon^{1-\varepsilon}(\theta-\theta^2)\,d\theta =\frac{\frac16-\varepsilon^2+\frac23\varepsilon^3}{1-2\varepsilon}.

Subtracting the original integrated gap gives

\frac{\frac16-\varepsilon^2+\frac23\varepsilon^3}{1-2\varepsilon}-\frac16 =\frac{\varepsilon(1-\varepsilon)}{3}.

The difference is positive and tends to zero. In this example the density changes on strips whose total length tends to zero, and the bounded gap makes their contribution negligible. The calculation also shows that renormalization matters: simply deleting endpoint mass would not leave a probability prior.

When a Risk Gap Is Strictly Positive

A second endpoint issue appears when using prior sequences to test a proposed domination. If a nonnegative risk gap is continuous and strictly positive at an interior point, it is positive on a neighborhood of that point. A limiting prior that gives positive density there therefore assigns a strictly positive integrated gap. This conclusion does not follow merely from strict positivity at an endpoint: an endpoint has zero mass under a continuous prior density, and a gap could be positive only at that single point.

Theorem: Let \(\pi\) be a probability density on \([a,b]\), and let \(\Delta\) be bounded, nonnegative, and continuous on \([a,b]\). If \(\Delta(\theta_0)>0\) for some \(\theta_0\in(a,b)\), and \(\pi\) is positive almost everywhere on a neighborhood of \(\theta_0\), then \(\int_a^b\Delta(\theta)\pi(\theta)\,d\theta>0\). Moreover, if densities \(\pi_k\) converge to \(\pi\) in \(L^1\), then their integrated gaps converge to this strictly positive value.

Proof. Put \(c=\Delta(\theta_0)>0\). By continuity, there is an open interval \(J\) containing \(\theta_0\), contained in \((a,b)\), on which \(\Delta(\theta)>c/2\). The density is positive almost everywhere on \(J\), so \(\int_J\pi(\theta)\,d\theta>0\). Nonnegativity of \(\Delta\) then yields

\int_a^b\Delta(\theta)\pi(\theta)\,d\theta \geq \int_J\Delta(\theta)\pi(\theta)\,d\theta \geq \frac{c}{2}\int_J\pi(\theta)\,d\theta >0.

The convergence of integrated gaps follows from the preceding theorem, since \(\Delta\) is bounded and \(\pi_k\to\pi\) in \(L^1\). \(\square\)

In a prior-sequence argument, this result gives a useful contradiction test. If a proposed dominating rule has a continuous nonnegative risk improvement that is strict somewhere in the interior, its limiting integrated improvement cannot be zero under a prior with positive density there. Conversely, an improvement that occurs only at an endpoint is not detected by a prior with a density unless continuity extends it to nearby interior values.

Worked Example: A Strict Endpoint Improvement Is Not an Interior Improvement

Consider the parameter space \([0,1]\) and the bounded nonnegative function \(\Delta(\theta)=\mathbf{1}_{\{0\}}(\theta)\). For every prior with a density, including the uniform prior,

\int_0^1\Delta(\theta)\pi(\theta)\,d\theta=0,

because the singleton \(\{0\}\) has Lebesgue measure zero. Thus pointwise strictness at the endpoint alone gives no positive integrated gap under such a prior. The function in this illustration is discontinuous at \(0\); if a risk gap were continuous and positive at \(0\), it would also be positive on some interval immediately to the right, and a density positive there would detect it. The example identifies exactly why continuity and the location of the strict improvement matter.

Unbounded Risk Near an Endpoint

A bound on the risk gap is convenient but not essential. If risk becomes unbounded near an endpoint, the relevant question is whether the prior density decays fast enough to make the product integrable. Suppose, near a finite endpoint \(a\), the density is at most a constant times \((\theta-a)^s\), while the absolute risk gap is at most a constant times \((\theta-a)^{-q}\). The endpoint contribution is then controlled by a power integral.

Endpoint integrability criterion: If, for \(a<\theta<a+\eta\), \(\pi(\theta)\leq C(\theta-a)^s\) and \(|\Delta(\theta)|\leq K(\theta-a)^{-q}\), where \(C,K,\eta>0\), then the endpoint integral is finite whenever \(s-q>-1\). Under the same bounds, its contribution from \((a,a+\varepsilon)\) is at most \(\frac{CK}{s-q+1}\varepsilon^{s-q+1}\).

Derivation. Multiplying the assumed bounds and integrating gives

\int_a^{a+\varepsilon}|\Delta(\theta)|\pi(\theta)\,d\theta \leq CK\int_a^{a+\varepsilon}(\theta-a)^{s-q}\,d\theta =\frac{CK}{s-q+1}\varepsilon^{s-q+1},

provided \(s-q>-1\). This exponent condition is exactly what makes the power integral finite at the endpoint. If \(s-q=-1\), the corresponding integral diverges logarithmically; if \(s-q<-1\), it diverges as a power.

Worked Example: A Prior That Controls a Singular Endpoint Risk

On \((0,1)\), take the proper prior density \(\pi(\theta)=3\theta^2\). It is normalized because \(\int_0^1 3\theta^2\,d\theta=1\). Suppose a risk gap has magnitude \(|\Delta(\theta)|=\theta^{-1}\). The endpoint contribution from \((0,\varepsilon)\) is

\int_0^\varepsilon |\Delta(\theta)|\pi(\theta)\,d\theta =\int_0^\varepsilon \theta^{-1}3\theta^2\,d\theta =3\int_0^\varepsilon\theta\,d\theta =\frac{3}{2}\varepsilon^2.

Here \(s=2\) and \(q=1\), so \(s-q=1>-1\); the integral exists and the endpoint contribution vanishes quadratically. If instead the density were uniform, the corresponding integral would be \(\int_0^\varepsilon\theta^{-1}\,d\theta\), which diverges. A prior's proximity to an endpoint is therefore not enough to assess an integrated-risk limit; its local density and the risk's rate of growth must be considered together.

Endpoint Mass and Boundary Terms

A continuous density and an atom at an endpoint behave differently. A density assigns zero mass to the endpoint itself, while an atom can retain positive mass there no matter how narrow an interior truncation becomes. A perturbation argument must therefore check whether endpoint atoms vanish, rather than infer this from the small width of an endpoint interval.

Worked Example: A Persistent Endpoint Atom Changes Integrated Risk

Let \(\theta\in[0,1]\), and consider the risk function \(R(\theta)=\theta\), which can arise, for example, from a decision problem with a single available action and loss \(\theta\). Under the uniform prior, the integrated risk is

\int_0^1\theta\,d\theta=\frac12.

Now use the prior \(\Pi_\eta=(1-\eta)\operatorname{Unif}[0,1]+\eta\delta_0\), where \(0<\eta<1\) and \(\delta_0\) is a point mass at zero. Its integrated risk is

\int R(\theta)\,d\Pi_\eta(\theta) =(1-\eta)\frac12+\eta R(0) =\frac{1-\eta}{2}.

The difference from the uniform-prior integrated risk is \(-\eta/2\), which does not vanish unless the atom weight \(\eta\) tends to zero. Endpoint mass is a separate part of the prior and must be tracked explicitly.

Finite intervals also do not eliminate boundary terms in integration by parts. The identity \(\int_a^b u'(\theta)v(\theta)\,d\theta=[u(\theta)v(\theta)]_a^b-\int_a^b u(\theta)v'(\theta)\,d\theta\) includes the endpoint expression \(u(b)v(b)-u(a)v(a)\). To discard it, one must establish that this expression is zero or tends to zero in the relevant limit. For example, if \(u(\theta)=\theta(1-\theta)\) on \([0,1]\) and \(v(\theta)=1\), then \([uv]_0^1=0\). If \(u(\theta)=1\) and \(v(\theta)=\theta\), then \([uv]_0^1=1\), not zero. A finite boundary is still a boundary.

Common Pitfalls

Assuming a thin strip always has negligible risk. Its contribution is small when risk is bounded, but a singular risk can offset the strip's small prior mass. Check the product of density and risk near the endpoint.
Ignoring renormalization after truncation. Removing prior mass changes the total mass; a proper truncated prior must be normalized, and that change affects integrated risk.
Using endpoint strictness as though it were interior strictness. A point has zero mass under a continuous prior density. To obtain a positive integrated gap, use continuity to find an interior neighborhood with positive gap.
Discarding an integration-by-parts term because the interval is finite. Evaluate the endpoint expression or prove its limit is zero; finiteness alone does neither.

Key takeaway: On a restricted parameter interval, control both the risk's one-sided behavior and the prior's mass near each endpoint. Bounded gaps with vanishing endpoint mass are straightforward; unbounded gaps require an integrability rate, and endpoint atoms or boundary terms must be handled explicitly.

Check Your Understanding

Use the endpoint conditions, not just the interval's finite length, in answering these questions.

If \(|\Delta(\theta)|\leq 5\) and a prior sequence converges in \(L^1\), what bound controls the difference between the two integrated gaps?
Why does a continuous nonnegative risk gap that is strictly positive at an interior point have strictly positive integrated gap under a density positive on a neighborhood?
Near \(a\), suppose \(\pi(\theta)\) is bounded by a constant times \((\theta-a)^3\) and \(|\Delta(\theta)|\) by a constant times \((\theta-a)^{-2}\). What is the order of the contribution from \((a,a+\varepsilon)\)?
Why can a risk improvement at a single endpoint be invisible to a prior with a density?
What additional check is needed before dropping an integration-by-parts boundary term on a finite interval?