Why the Hypotheses Must Be Checked Separately
Blyth’s method turns a sequence of integrated-risk comparisons into an admissibility argument. The sequence may make the candidate rule nearly Bayes in an integrated sense, but that fact alone does not rule out a competitor that improves risk at a particular parameter value. To draw that conclusion, one must also know that the candidate has finite risk, that the prior measures retain enough mass near each parameter value, and that a strict pointwise improvement persists on a neighborhood.
The purpose of this tutorial is to audit those requirements explicitly. The result below is a useful sufficient version of the argument. Its conditions are deliberately stated separately: failure of any one of them blocks this particular proof, even if the others hold.
A Sufficient Hypothesis Checklist
Let \(\pi_n\) be finite nonnegative measures on \(\Theta\). For a rule \(\eta\), define its integrated risk by \(r_n(\eta)=\int_\Theta R(\theta,\eta)\,d\pi_n(\theta)\). Suppose the following conditions hold:
- Finite candidate risk: \(R(\theta,\delta_0)<\infty\) for every \(\theta\), and the integrated risks used below are defined and finite.
- Vanishing integrated gap: \(b_n=r_n(\delta_0)-\inf_\eta r_n(\eta)\) tends to zero, where the infimum is over the decision rules being considered.
- Local mass: for every nonempty open neighborhood \(U\) of every point of \(\Theta\), \(\liminf_{n\to\infty}\pi_n(U)>0\).
- Continuity: for every rule \(\delta\) with risk no greater than that of \(\delta_0\), the function \(\Delta_\delta\) is continuous on \(\Theta\).
Under these conditions, \(\delta_0\) is admissible within the stated class of rules. This formulation uses finite measures that need not be probability measures. Multiplying a prior measure by a positive constant changes both its integrated risks and its gap by that constant; for the criterion, the resulting measures must still have gaps tending to zero and satisfy the local-mass condition.
Here is the full argument. Suppose, for contradiction, that \(\delta\) dominates \(\delta_0\). Then \(\Delta_\delta(\theta)\geq0\) everywhere. Since the candidate risk is finite and the competitor has no larger risk, their integrated risks are finite under each finite \(\pi_n\) in the hypothesis. By the definition of the infimum,
Thus the nonnegative integrals of the improvement tend to zero. If dominance were strict at \(\theta_0\), then \(\Delta_\delta(\theta_0)>0\). Continuity gives an open neighborhood \(U\) of \(\theta_0\) and an \(\varepsilon>0\) such that \(\Delta_\delta(\theta)\geq\varepsilon\) for every \(\theta\in U\). Consequently,
The local-mass condition makes the right-hand side bounded away from zero for all sufficiently large \(n\), contradicting \(b_n\to0\). Therefore strict dominance is impossible. \(\square\)
Worked Example: Auditing Blyth’s Conditions for a Normal Mean
Worked Example: The One-Dimensional Normal Location Model
Observe \(X\sim N(\theta,1)\), where \(\theta\in\mathbb R\), use squared-error loss \(L(\theta,a)=(a-\theta)^2\), and consider \(\delta_0(X)=X\). We check the conditions and calculate the integrated gap.
Candidate risk. Since \(X-\theta\sim N(0,1)\),
\(R(\theta,\delta_0)=E_\theta[(X-\theta)^2]=1\) for every \(\theta\). Thus candidate risk is finite everywhere.
Prior measures and Bayes rules. For \(\tau>0\), let the probability prior be \(N(0,\tau^2)\), and define the finite measure \(\pi_\tau\) to be \(\tau\) times that probability measure. Its density with respect to Lebesgue measure is
Completing the square in the normal likelihood and prior gives the posterior mean \(\delta_\tau(X)=\frac{\tau^2}{1+\tau^2}X\). Under squared-error loss, the posterior mean minimizes posterior expected loss. Its integrated risk is therefore the minimum integrated risk. Under the probability prior, the integrated risk of \(X\) is 1, while that of the posterior mean is the posterior variance \(\frac{\tau^2}{1+\tau^2}\). Scaling the prior by \(\tau\) scales both integrated risks, so
Local mass. If \(U\) is any nonempty open interval, it contains a bounded interval \((a,b)\) with \(a<b\). On \((a,b)\), \(w_\tau(\theta)\) converges uniformly to \(1/\sqrt{2\pi}\) as \(\tau\to\infty\). Hence \(\liminf_{\tau\to\infty}\pi_\tau(U)\geq(b-a)/\sqrt{2\pi}>0\).
Continuity check. Consider any rule \(\delta\) that dominates \(X\). Its risk is at most 1 at every \(\theta\). To verify continuity, fix \(M>0\) and restrict \(\theta\) to \([-M,M]\). The normal density \(\phi(x-\theta)\) is bounded above, uniformly for those \(\theta\), by a constant times \(\phi(x-(M+1))+\phi(x+(M+1))\). For example, on \(x\geq0\) its ratio to \(\phi(x-(M+1))\) is at most \(\exp(-(x-M-\tfrac12))\leq e^{M+1/2}\); the negative half-line is handled by the other endpoint density. Finite risk at \(\theta=\pm(M+1)\) implies that \(\delta(x)^2\) is integrable against each of these endpoint densities, since \(\delta^2\leq2(\delta-\theta)^2+2\theta^2\). Also \((\delta(x)-\theta)^2\leq2\delta(x)^2+2M^2\) on the interval. These bounds give an integrable dominating function. The integrand \((\delta(x)-\theta)^2\phi(x-\theta)\) is pointwise continuous in \(\theta\), so dominated convergence proves continuity of \(R(\theta,\delta)\), and therefore of \(\Delta_\delta\).
All four conditions hold. The theorem consequently proves that \(X\) is admissible in this model. Notice that the proof needs both the vanishing gap and nonvanishing local mass: neither calculation replaces the other.
Worked Example: Why Normalized Diffuse Priors Are Not Enough
Worked Example: Vanishing Mass Near a Point
Keep the normal location model, but use the probability priors \(Q_\tau=N(0,\tau^2)\) without multiplying by \(\tau\). Their Bayes-risk gaps for the rule \(X\) are \(1/(1+\tau^2)\), which tend to zero. However, for any bounded interval \(U=(a,b)\),
For a direct illustration of the logical gap, the continuous nonnegative function \(g(\theta)=e^{-\theta^2}\) is strictly positive at zero, yet
Thus a vanishing integral under these normalized priors does not by itself rule out a strict improvement at a fixed parameter. The illustration is not a claim that \(g\) is a risk difference; it shows why the local-mass hypothesis in the proof cannot simply be omitted. Rescaling the priors in the preceding example repairs that particular hypothesis, while retaining a vanishing integrated gap.
Worked Example: A Candidate with Infinite Risk
Worked Example: A Finite Integrated Comparison Cannot Be Assumed
Again let \(X\sim N(\theta,1)\) with squared-error loss, but consider \(\widetilde\delta(X)=e^{X^2}\). At any fixed \(\theta\), its risk includes the term \(E_\theta[e^{2X^2}]\). The integrand in this expectation, apart from a positive constant, is
As \(|x|\to\infty\), the positive quadratic term dominates the linear term \(\theta x\). The integrand does not have a finite integral over \(\mathbb R\), so \(E_\theta[e^{2X^2}]=\infty\) and \(R(\theta,\widetilde\delta)=\infty\). A proposed Blyth argument that starts with this rule cannot use a finite candidate-risk comparison. This calculation does not show that every rule used in an admissibility proof must have finite risk under every conceivable formulation; it shows that finiteness is a specific hypothesis of the sufficient criterion stated here and must be verified before applying it.
What Each Check Prevents
The proof uses each condition at a distinct point. Candidate finiteness and finite integrated comparisons make the risk differences meaningful. The vanishing gap bounds the integrated improvement of any proposed dominator. Local mass prevents the prior sequence from avoiding the region where that improvement occurs. Continuity ensures that a strict improvement at one parameter is not confined to a set that the measures can miss.
A common mistake is to reason, “the integrated gap tends to zero, so no rule can improve strictly anywhere.” This omits the change from an integrated statement to a pointwise one. The correction is to assume a dominator, use continuity to find a neighborhood of positive improvement, and then use the prior sequence’s local mass to force the integrated improvement to stay positive. Without either ingredient, the contradiction does not follow.
Verify that the candidate’s risk is finite at every parameter and that the integrated comparisons in the argument are defined.
Show that the candidate’s integrated risk minus the minimum integrated risk tends to zero.
Confirm that neighborhoods retain positive prior mass and that strict risk improvement persists locally.
A dominator would produce both a vanishing integrated improvement and a positive lower bound on that improvement.
Check Your Understanding
For each question, identify which hypothesis or proof step is involved.
- In the normal location example, why does multiplying each probability prior by \(\tau\) not change its Bayes rule?
- What lower bound contradicts a vanishing integrated gap if a continuous risk improvement is strict at some \(\theta_0\)?
- Why does convergence of integrated improvements to zero under \(N(0,\tau^2)\) probability priors not alone exclude a pointwise improvement?
- In the infinite-risk example, which part of the integrand shows that the risk is infinite?
- If a proposed competitor has risk at most that of a finite-risk candidate, why is its integrated risk finite under each finite prior measure used in the criterion?