Boundary Terms in Integration-by-Parts Proofs

Why Boundary Terms Matter

In an admissibility argument, an integrated risk comparison can turn a pointwise risk question into an integral identity. Integration by parts is often what makes that identity useful: it moves a derivative from the sampling density onto a weight, or replaces a factor involving the observation by a derivative of the estimator. But integration by parts also creates endpoint terms. A sign or vanishing claim about the resulting risk difference is justified only when those terms have been retained and controlled.

We develop two related identities in the normal location model. The first integrates risk differences over a bounded parameter interval and makes the parameter-endpoint terms explicit. The second integrates by parts in the observation and shows the boundary condition needed for the familiar normal identity.

Definition: Let \(X\sim N(\theta,1)\), with density \(\phi(x-\theta)=(2\pi)^{-1/2}\exp\{-(x-\theta)^2/2\}\). Under squared-error loss, the risk of a decision rule \(\delta\) is \(R(\theta,\delta)=E_\theta[(\delta(X)-\theta)^2]\). We compare the rule \(\delta_0(x)=x\) with a perturbed rule \(\delta(x)=x+g(x)\), where \(g\) is a measurable function.

Integrating Risk Differences Over a Parameter Interval

Let \([A,B]\) be a finite interval and let \(w\) be a nonnegative continuously differentiable weight on it. The weight need not integrate to one; normalizing it would multiply all terms in the identity by the same constant. Write

m_w(x)=\int_A^B w(\theta)\phi(x-\theta)\,d\theta, \qquad q_w(x)=\int_A^B w'(\theta)\phi(x-\theta)\,d\theta.

Here \(m_w\) is the mixture density weighted by \(w\), while \(q_w\) records how the weight changes across the interval. Assume the integrals below are absolutely integrable, including the terms involving \(g^2m_w\), \(gq_w\), and the endpoint contributions. These conditions justify changing the order of integration and applying Fubini's theorem.

Theorem: Under the conditions above, the integrated risk difference is

\int_A^B w(\theta)\{R(\theta,\delta)-R(\theta,\delta_0)\}\,d\theta = \int_{\mathbb R} g(x)^2m_w(x)\,dx -2\int_{\mathbb R}g(x)q_w(x)\,dx +2\int_{\mathbb R}g(x)\{w(B)\phi(x-B)-w(A)\phi(x-A)\}\,dx.

Proof. Expanding the two squared-error losses gives, for every \(\theta\),

R(\theta,\delta)-R(\theta,\delta_0) =E_\theta\!\left[g(X)^2+2g(X)(X-\theta)\right].

After multiplying by \(w(\theta)\) and integrating, the \(g^2\) term is \(\int g(x)^2m_w(x)\,dx\). For the cross term, the normal density satisfies

\frac{\partial}{\partial\theta}\phi(x-\theta)=(x-\theta)\phi(x-\theta).

Thus, at each \(x\), integration by parts in \(\theta\) gives

\int_A^B w(\theta)(x-\theta)\phi(x-\theta)\,d\theta = w(B)\phi(x-B)-w(A)\phi(x-A) -\int_A^B w'(\theta)\phi(x-\theta)\,d\theta.

Multiplying this equality by \(2g(x)\), integrating over \(x\), and adding the \(g^2\) term yields the stated identity. The absolute-integrability assumptions justify these integrations and the change in order of integration. \(\square\)

The endpoint contribution is not an optional correction: it is part of the identity. If \(w(A)=w(B)=0\), it disappears. If either endpoint value is nonzero, it must be included, even when the derivative \(w'\) is zero.

Worked Example: A Constant Perturbation With a Uniform Weight

Take \([A,B]=[0,1]\), \(w(\theta)=1\), and \(g(x)=c\), where \(c\) is any real constant. The competing rule is \(\delta(x)=x+c\). Since \(w'=0\), we have

m_w(x)=\int_0^1\phi(x-\theta)\,d\theta,\qquad q_w(x)=0,

and the endpoint contribution in the theorem is

2c\int_{\mathbb R}\{\phi(x-1)-\phi(x)\}\,dx =2c(1-1)=0.

Also,

\int_{\mathbb R}m_w(x)\,dx =\int_0^1\left\{\int_{\mathbb R}\phi(x-\theta)\,dx\right\}d\theta =\int_0^1 1\,d\theta=1.

The integrated risk difference is therefore \(c^2\). Directly, at each \(\theta\), the risk difference is \(E_\theta[c^2+2c(X-\theta)]=c^2\), so its integral over an interval of length one is also \(c^2\). The boundary term is not zero as a function of \(x\); its integral happens to cancel in this example. Dropping it from the identity pointwise would still be incorrect.

When the Weight Tapers to Zero

A weight that vanishes at both ends removes the explicit parameter-endpoint term. The derivative term remains and can carry the entire cross-term contribution. The following calculation checks that distinction using a nonconstant weight.

Worked Example: A Tapered Weight and a Linear Perturbation

Let \(w(\theta)=6\theta(1-\theta)\) on \([0,1]\), the Beta\((2,2)\) density, and take \(g(x)=x\), so \(\delta(x)=2x\). The endpoints satisfy \(w(0)=w(1)=0\), and \(w'(\theta)=6-12\theta\). The endpoint term in the theorem is zero.

At fixed \(\theta\), \(E_\theta[X]=\theta\) and \(E_\theta[X^2]=\theta^2+1\). Hence

R(\theta,\delta)-R(\theta,\delta_0) =E_\theta[2X(X-\theta)+X^2] =3(\theta^2+1)-2\theta^2 =\theta^2+3.

For a Beta\((2,2)\) random variable, \(E[\theta^2]=3/10\). Thus the integrated difference is \(3+3/10=33/10\). Now verify this with the identity. The first term is

\int_{\mathbb R}x^2m_w(x)\,dx =\int_0^1w(\theta)E_\theta[X^2]\,d\theta =1+\frac{3}{10}=\frac{13}{10}.

For the second term, changing the order of integration gives

\int_{\mathbb R}xq_w(x)\,dx =\int_0^1w'(\theta)\left\{\int_{\mathbb R}x\phi(x-\theta)\,dx\right\}d\theta =\int_0^1\theta w'(\theta)\,d\theta.

A further integration by parts, using \(\theta w(\theta)=0\) at both endpoints, yields

\int_0^1\theta w'(\theta)\,d\theta =[\theta w(\theta)]_0^1-\int_0^1w(\theta)\,d\theta=-1.

Consequently, the identity gives \(13/10-2(-1)=33/10\), agreeing with the direct risk calculation. The zero endpoint values removed the first boundary term, but the derivative of the weight still contributed.

Integration by Parts in the Observation

There is another useful route: fix \(\theta\) and integrate with respect to the observation \(x\). It gives an identity for a single risk difference, rather than one integrated over a parameter interval. The condition at infinity is essential.

Theorem (normal Stein identity for a risk difference): Fix \(\theta\). Suppose \(g\) is locally absolutely continuous, \(E_\theta[g(X)^2]<\infty\), \(E_\theta[|g'(X)|]<\infty\), and \(\lim_{x\to-\infty}g(x)\phi(x-\theta)=\lim_{x\to\infty}g(x)\phi(x-\theta)=0\). Then

R(\theta,\delta)-R(\theta,\delta_0) =E_\theta[g(X)^2+2g'(X)].

Proof. From the loss expansion,

R(\theta,\delta)-R(\theta,\delta_0) =\int_{\mathbb R}\{g(x)^2+2g(x)(x-\theta)\}\phi(x-\theta)\,dx.

Since \(\phi'(x-\theta)=-(x-\theta)\phi(x-\theta)\), integrate the cross term first over \([-M,M]\). Absolute continuity gives

\int_{-M}^M g(x)(x-\theta)\phi(x-\theta)\,dx =-[g(x)\phi(x-\theta)]_{-M}^{M} +\int_{-M}^M g'(x)\phi(x-\theta)\,dx.

As \(M\) tends to infinity, the displayed boundary term tends to zero by hypothesis. The derivative integral converges absolutely because \(E_\theta[|g'(X)|]<\infty\). The cross-term integral also converges absolutely: by Cauchy–Schwarz,

E_\theta[|g(X)(X-\theta)|] \leq \{E_\theta[g(X)^2]\}^{1/2}\{E_\theta[(X-\theta)^2]\}^{1/2} =\{E_\theta[g(X)^2]\}^{1/2}<\infty.

Taking limits in the finite-interval identity therefore gives \(E_\theta[g(X)(X-\theta)]=E_\theta[g'(X)]\). Substituting this into the risk expansion proves the result. \(\square\)

The assumption on \(g\phi\) cannot be replaced by a silent omission of the boundary term. On a finite observation interval, that term is explicitly \(-[g(x)\phi(x-\theta)]\); the theorem uses a condition that makes its limit zero.

Worked Example: Checking the Observation-Space Identity

Let \(g(x)=\lambda x\), so \(\delta(x)=(1+\lambda)x\). This function is locally absolutely continuous, \(g'(x)=\lambda\), and \(g(x)\phi(x-\theta)\) tends to zero at both ends because a linear function times a normal density tends to zero. The identity predicts

R(\theta,\delta)-R(\theta,\delta_0) =\lambda^2E_\theta[X^2]+2\lambda =\lambda^2(\theta^2+1)+2\lambda.

For a numerical check, take \(\theta=1\) and \(\lambda=-1/2\). The formula gives \(2(1/4)-1=-1/2\). Directly, \(\delta(X)=X/2\), so

R(1,\delta)=E_1[(X/2-1)^2] =\frac14\operatorname{Var}_1(X)+\left(\frac12E_1[X]-1\right)^2 =\frac14+\frac14=\frac12.

The rule \(\delta_0(X)=X\) has risk \(1\), making the difference \(1/2-1=-1/2\), as required.

How These Identities Support Admissibility Arguments

In an integrated-risk comparison, the weight and its endpoints are part of the mathematical argument, not just a device for averaging. An identity with a negative or vanishing derivative term may be useful, but it supports a conclusion only after the endpoint contribution has also been evaluated. Likewise, the observation-space identity can simplify a risk difference, but only for perturbations satisfying its integrability and boundary hypotheses.

These calculations are useful as certificates or intermediate steps in admissibility proofs: they expose precisely which terms must be bounded, shown to have a sign, or shown to vanish. They do not by themselves establish admissibility. That conclusion requires the relevant comparison to hold for the decision problem under consideration.

Common Mistakes

Discarding a parameter endpoint because the weight is constant. A constant weight has \(w'=0\), but its endpoint contribution is \(w(B)\phi(x-B)-w(A)\phi(x-A)\), which need not vanish.
Using the observation-space identity without checking its boundary condition. Integration by parts first produces \(-[g(x)\phi(x-\theta)]\) on a finite interval. It disappears at infinity only when the stated limits are zero.
Confusing a zero endpoint term with a zero derivative term. A tapered weight can have \(w(A)=w(B)=0\) while \(w'\) is nonzero throughout the interval, as in the Beta\((2,2)\) example.
Reading an integrated comparison as a pointwise comparison. A weighted integral of risk differences does not, on its own, show that the risk difference has the same sign at every parameter value.

Key takeaway: Write the boundary term before simplifying an integration-by-parts identity. Then verify its sign or limiting value under explicit assumptions; only after that can the remaining risk terms support an admissibility argument.

Check Your Understanding

For each question, keep track of which variable is being integrated and which endpoint conditions are relevant.

In the parameter-interval identity, what expression forms the endpoint contribution when \(w\) is continuously differentiable on \([A,B]\)?
For the uniform weight on \([0,1]\), why is the endpoint contribution not identically zero as a function of \(x\), even though its integral is zero in the constant-perturbation example?
Why does \(w(0)=w(1)=0\) remove the endpoint term in the tapered-weight example but not remove the term involving \(w'\)?
State the boundary condition used in the observation-space normal Stein identity and explain where it enters the proof.
For \(g(x)=\lambda x\), derive the risk difference from the identity and verify it by direct calculation.