Boundary Parameters and One-Sided Risk Comparisons

Risk Comparisons at the Edge of a Parameter Space

When the parameter is restricted to a closed interval, the endpoints are part of the statistical problem. A rule can have the same or greater risk at every interior parameter value yet tie another rule at an endpoint; that tie does not automatically prevent domination. Conversely, favorable comparisons at both endpoints do not always settle the comparison in the interior. The difference depends on the shape of the risk difference.

This tutorial develops that distinction for estimation under squared-error loss. Its central example uses constant estimators, whose risk differences are affine functions of the parameter. For those comparisons, the endpoint risks determine the ordering over the whole interval. A Bernoulli example will show why the same conclusion cannot be assumed for arbitrary estimators.

Definition: Let \(\Theta=[L,U]\), where \(L<U\), and let \(R(\theta,\delta)=E_\theta[(\delta(X)-\theta)^2]\) be the squared-error risk of an estimator \(\delta\), assumed finite for every \(\theta\in\Theta\). An estimator \(\delta_1\) weakly dominates \(\delta_2\) on \(\Theta\) if \(R(\theta,\delta_1)\leq R(\theta,\delta_2)\) for every \(\theta\in\Theta\). It dominates \(\delta_2\) if, in addition, the inequality is strict for at least one \(\theta\in\Theta\).

For any two estimators, define the signed risk difference \(g(\theta)=R(\theta,\delta_2)-R(\theta,\delta_1)\). Thus \(g(\theta)\geq 0\) throughout \(\Theta\) means that \(\delta_1\) weakly dominates \(\delta_2\). Keeping this direction explicit prevents a common sign error: a positive value of \(g\) favors \(\delta_1\), not \(\delta_2\).

Why Constant Estimators Have an Endpoint Test

A constant estimator reports a fixed number \(a\), regardless of the observation. Its risk under squared-error loss is \(R(\theta,\delta_a)=(a-\theta)^2\). For two constants \(a\) and \(b\), their risk difference is

R(\theta,\delta_b)-R(\theta,\delta_a) =(b-\theta)^2-(a-\theta)^2 =(b-a)(a+b-2\theta).

This expression is affine in \(\theta\). That property, rather than boundedness by itself, is what makes endpoint comparisons decisive.

Theorem: Suppose a risk difference \(g\) is affine on \([L,U]\). Then \(g(\theta)\geq 0\) for every \(\theta\in[L,U]\) if and only if \(g(L)\geq 0\) and \(g(U)\geq 0\). If at least one endpoint value is strictly positive, then \(g\) is strictly positive at some parameter value. If both endpoint values are zero, then \(g\) is identically zero.

Proof. Necessity follows by evaluating a nonnegative \(g\) at \(L\) and \(U\). For sufficiency, write any \(\theta\in[L,U]\) as \(\theta=(1-t)L+tU\), where \(t=(\theta-L)/(U-L)\in[0,1]\). Since \(g\) is affine,

g(\theta)=(1-t)g(L)+t g(U).

If both endpoint values are nonnegative, this convex combination is nonnegative. If at least one endpoint value is positive, \(g\) is positive at that endpoint, so it is strictly positive at some parameter value. If both values are zero, the displayed identity gives \(g(\theta)=0\) throughout the interval. \(\square\)

Applied to constant squared-error estimators, the theorem gives a complete comparison rule: calculate the two endpoint risk differences, then check their signs. It also clarifies what a boundary tie means. A zero at one endpoint is compatible with domination if the difference is nonnegative everywhere and positive somewhere else.

Worked Example: Domination With a Tie at the Boundary

Let \(\Theta=[0,1]\), and compare the constant estimators \(\delta_0(X)=0\) and \(\delta_2(X)=2\) under squared-error loss. The observation model is immaterial because neither estimator uses the data. Determine whether either estimator dominates the other.

Their risks are \(R(\theta,\delta_0)=\theta^2\) and \(R(\theta,\delta_2)=(2-\theta)^2\). Taking the risk of \(\delta_2\) minus the risk of \(\delta_0\) gives

g(\theta)=(2-\theta)^2-\theta^2 =4-4\theta =4(1-\theta).

At the endpoints, \(g(0)=4\) and \(g(1)=0\). The affine endpoint test, or the final expression directly, shows \(g(\theta)\geq0\) throughout \([0,1]\). It is strictly positive for every \(\theta<1\), but equals zero at \(\theta=1\). Therefore \(\delta_0\) dominates \(\delta_2\): its risk is never larger and is smaller, for example, at \(\theta=0\). The tie at the right boundary does not undo domination.

Worked Example: Opposite Boundary Preferences Produce Crossing Risks

Again let \(\Theta=[0,1]\), and compare the constant estimators \(\delta_0(X)=0\) and \(\delta_1(X)=1\). The risk difference in the same direction is

g(\theta)=R(\theta,\delta_1)-R(\theta,\delta_0) =(1-\theta)^2-\theta^2 =1-2\theta.

At the left endpoint, \(g(0)=1>0\), so \(\delta_0\) has lower risk there. At the right endpoint, \(g(1)=-1<0\), so \(\delta_1\) has lower risk there. The risks are equal at \(\theta=1/2\), since \(g(1/2)=0\). The endpoint signs are opposite, so neither estimator has risk no greater everywhere than the other. This is a crossing comparison: the preferred estimator changes across the parameter space.

How Restricting the Interval Can Change the Comparison

A risk comparison is always relative to the parameter space on which it is assessed. The same two estimators may be incomparable on one interval but ordered on a subinterval. In the affine case, this is visible directly from the endpoint values of the new interval.

Worked Example: A Boundary Tie After Restriction

Consider constant estimators \(\delta_0(X)=0\) and \(\delta_{0.4}(X)=0.4\) under squared-error loss. First take \(\Theta=[0,1]\). The risk difference is

g(\theta)=R(\theta,\delta_{0.4})-R(\theta,\delta_0) =(0.4-\theta)^2-\theta^2 =0.16-0.8\theta.

Here \(g(0)=0.16>0\), whereas \(g(1)=-0.64<0\). Thus the estimators cross on \([0,1]\); solving \(0.16-0.8\theta=0\) gives the crossing point \(\theta=0.2\).

Now restrict the parameter space to \([0.2,1]\). The endpoint differences are \(g(0.2)=0\) and \(g(1)=-0.64\), so \(g\leq0\) throughout this restricted interval. In this direction, that means \(R(\theta,\delta_{0.4})\leq R(\theta,\delta_0)\) everywhere: \(\delta_{0.4}\) weakly dominates \(\delta_0\) on \([0.2,1]\), with strict inequality for \(\theta>0.2\). The equality at the new left boundary is compatible with domination. On the original interval, however, \(\delta_{0.4}\) did not dominate \(\delta_0\), because its risk was higher near \(\theta=0\).

Why Endpoint Checks Do Not Always Work

The endpoint test requires an affine risk difference. When risks are nonlinear, two endpoint values can have the same sign while the difference changes sign in the interior. The following example verifies this in a model where the estimator does use the observation.

Worked Example: Endpoint Values Miss an Interior Reversal

Suppose \(X\sim\operatorname{Bernoulli}(\theta)\), with \(\Theta=[0,1]\), and use squared-error loss to estimate \(\theta\). Compare \(\delta_X(X)=X\) with the constant estimator \(\delta_{1/2}(X)=1/2\). For the sample estimator,

R(\theta,\delta_X) =(1-\theta)(0-\theta)^2+\theta(1-\theta)^2 =\theta(1-\theta).

For the constant estimator, \(R(\theta,\delta_{1/2})=(1/2-\theta)^2\). The difference is therefore

g(\theta)=R(\theta,\delta_{1/2})-R(\theta,\delta_X) =(\theta-\tfrac12)^2-\theta(1-\theta) =2\theta^2-2\theta+\tfrac14.

At both endpoints, \(g(0)=g(1)=1/4>0\), so \(\delta_X\) has lower risk at both endpoints. But at the midpoint, \(g(1/2)=-1/4<0\), so \(\delta_{1/2}\) has lower risk there. In fact, solving \(g(\theta)=0\) gives

\theta=\frac{1\pm 1/\sqrt{2}}{2}.

The quadratic is negative between these two roots and positive outside them. The risks therefore cross twice within \([0,1]\), and neither estimator dominates the other. Endpoint values alone fail here because \(g\) is quadratic, not affine.

Common Mistake and Practical Use

Common mistake: “One estimator has lower risk at both endpoints, so it must have lower risk throughout the interval.” This conclusion is justified for affine risk differences, such as comparisons between constant estimators under squared-error loss. It is not a general rule about risk functions. The Bernoulli calculation supplies a counterexample: the endpoint difference is positive at both ends but negative in the interior.

A reliable comparison begins by defining the signed difference and checking its form. If it is affine, the endpoint theorem applies. If it is not affine, endpoint signs alone do not establish its sign in the interior; the full difference must be analyzed. In either case, equality at a boundary is not itself a failure of domination. The relevant questions are whether the risk inequality holds at every parameter value and whether it is strict somewhere.

These distinctions matter when the parameter space has a genuine boundary, as it often does for probabilities, proportions, and constrained means. Restricting the parameter space can change the risk ordering because it removes parameter values where an otherwise favorable comparison fails. A conclusion about domination must therefore name the parameter space on which it holds.

Check Your Understanding

Use the sign of the risk difference carefully, and distinguish affine comparisons from general ones.

For two constant estimators \(a\) and \(b\) under squared-error loss, derive \(R(\theta,\delta_b)-R(\theta,\delta_a)\) and state why endpoint checks suffice on a closed interval.
If an affine risk difference is zero at both endpoints of an interval, what must it be in the interior? Explain.
In the comparison of \(\delta_0\) and \(\delta_2\) on \([0,1]\), why does equality at \(\theta=1\) not prevent \(\delta_0\) from dominating \(\delta_2\)?
For the constant estimators \(0\) and \(1\) on \([0,1]\), where do their risks cross, and which estimator has lower risk at each endpoint?
In the Bernoulli example, why do the endpoint risk differences fail to determine the risk ordering over the whole interval?