Why Posterior Propriety Is Not an Admissibility Proof
An improper prior is a measure with infinite total mass, often used because it expresses a useful invariance or limiting model. It may still produce a well-defined posterior distribution after data are observed. But these are different facts: posterior propriety is a statement about a conditional distribution at observed data, while admissibility compares a rule's risk with that of every competing rule across the parameter space.
For an improper-prior rule to support an admissibility argument, one needs more than a finite posterior normalizing constant. A common route is a Blyth-type argument: approximate the improper prior by finite measures and show that the integrated risk gap between the candidate rule and the Bayes risk under those measures tends to zero. The limit must be strong enough that a genuine local improvement in risk could not disappear in the approximation.
Improper Priors and Proper Posteriors
The marginal integral is sometimes called the marginal likelihood, although with an improper prior it is not a probability density for the data. Multiplying \(\mu\) by a positive finite constant changes \(m_\mu(x)\) by that constant but leaves the posterior unchanged. This scale invariance is useful, but it does not turn \(\mu\) into a probability measure.
Proof. If \(0<m_\mu(x)<\infty\), the posterior measure is nonnegative and its total mass is \(\int_\Theta f(x\mid\theta)\mu(d\theta)/m_\mu(x)=1\). If \(m_\mu(x)=0\), there is no positive normalizing constant that produces a probability distribution from this formula. If \(m_\mu(x)=\infty\), dividing by a finite positive constant cannot give it finite total mass. Thus the stated condition is necessary and sufficient. \(\square\)
Worked Example: A Flat Prior in the Normal Location Model
Suppose \(X\mid\theta\sim N(\theta,1)\), with \(\theta\in\mathbb R\), and use the Lebesgue measure \(\mu(d\theta)=d\theta\) as a prior. This prior is improper because \(\int_{\mathbb R}d\theta=\infty\).
For an observed \(x\), the likelihood as a function of \(\theta\) is proportional to \(\exp\{-(x-\theta)^2/2\}\). Its integral against the prior is
The posterior density is therefore \((2\pi)^{-1/2}\exp\{-(\theta-x)^2/2\}\), so \(\theta\mid x\sim N(x,1)\). Under squared-error loss, the posterior expected loss is minimized by the posterior mean \(x\). This establishes posterior propriety and identifies the generalized Bayes rule; neither calculation, by itself, establishes that the rule is admissible.
Posterior propriety can even depend on the observed data. For example, take \(X\mid\theta\sim\operatorname{Poisson}(\theta)\), \(\theta>0\), and the improper measure \(d\theta/\theta\). The posterior normalizing integral is proportional to \(\int_0^\infty e^{-\theta}\theta^{x-1}\,d\theta\). For an observed integer \(x\geq1\), this equals \(\Gamma(x)=(x-1)!\), and the posterior is \(\operatorname{Gamma}(x,1)\). For \(x=0\), the integral \(\int_0^\infty e^{-\theta}\theta^{-1}\,d\theta\) diverges at zero, so the posterior is not proper. A statement that this improper prior has a proper posterior must therefore specify for which observations it holds.
The Integrated-Risk Gap
Let \(L(\theta,a)\geq0\) be a measurable loss, let \(\delta\) be a measurable decision rule, and write \(R(\theta,\delta)=E_\theta[L(\theta,\delta(X))]\). For a finite prior measure \(\nu\), define its Bayes risk by \(r(\nu)=\inf_\eta\int_\Theta R(\theta,\eta)\nu(d\theta)\), where the infimum is over measurable decision rules. The measure need not have total mass one: multiplying it by a positive constant multiplies both the integrated risks and the Bayes risk by that constant, without changing its Bayes rules.
A generalized Bayes rule for an improper prior \(\mu\) minimizes posterior expected loss wherever the posterior is proper, with a measurable choice of minimizer. This pointwise posterior minimization does not automatically make the rule Bayes for any finite prior, nor does it ensure a small integrated-risk gap under finite approximations. Those are separate requirements.
The second assumption describes how the finite measures detect a strict improvement. A standard sufficient situation is that the risk difference is continuous, is strictly positive at some point, and the measures assign a uniformly positive amount of mass to a neighborhood where that difference stays positive, while the difference is nonnegative everywhere. The measures need not be probability measures.
Proof. Suppose, for contradiction, that a measurable rule \(\delta'\) dominates \(\delta\) and strictly improves its risk somewhere. By the detection assumption, for all sufficiently large \(k\),
Since \(r(\nu_k)\) is the infimum of integrated risk over all rules, it is no greater than the integrated risk of \(\delta'\). Consequently,
This contradicts \(\Gamma(\delta;\nu_k)\to0\). Hence no rule dominates \(\delta\) with a strict improvement somewhere, which is the definition of admissibility. \(\square\)
Worked Examples of the Required Limit
Worked Example: Expanding Proper Priors for a Normal Mean
Return to \(X\mid\theta\sim N(\theta,1)\) and squared-error loss. For \(\tau>0\), let the proper prior be \(\pi_\tau=N(0,\tau^2)\). Completing the square in the product of the prior and likelihood gives the posterior variance \(\tau^2/(\tau^2+1)\) and posterior mean \(\delta_\tau(x)=\{\tau^2/(\tau^2+1)\}x\).
Under the prior \(\pi_\tau\), the integrated risk of the rule \(\delta(x)=x\) is \(E[(X-\theta)^2]=1\), since \(X-\theta\sim N(0,1)\). The posterior mean is Bayes, and its Bayes risk is the posterior variance: \(r(\pi_\tau)=\tau^2/(\tau^2+1)\).
Thus the integrated-risk gap is
The probability priors \(\pi_\tau\) spread their mass over a wider range as \(\tau\) grows. Their mass on any fixed bounded interval tends to zero, so this calculation alone does not show that they detect every local improvement with a uniform positive amount. Rescale them to finite measures \(\nu_\tau(d\theta)=\sqrt{2\pi}\tau\,\pi_\tau(d\theta)=\exp\{-\theta^2/(2\tau^2)\}\,d\theta\). The Bayes rule is unchanged by this scaling, while \(\Gamma(\delta;\nu_\tau)=\sqrt{2\pi}\tau/(\tau^2+1)\to0\). On every fixed bounded interval, the density of \(\nu_\tau\) tends uniformly to \(1\). This illustrates the two distinct ingredients: a vanishing gap and enough local mass to detect a risk improvement. Verifying all the criterion's conditions is still necessary before concluding admissibility.
Worked Example: Why Normalization Can Hide the Comparison
Use the same normal model and rule \(\delta(x)=x\). The probability priors \(\pi_\tau=N(0,\tau^2)\) have integrated-risk gap \(1/(\tau^2+1)\), which tends to zero. It is tempting to regard this limit as sufficient on its own.
Let \(K=[-1,1]\). The prior mass of \(K\) tends to zero: its density is at most \(1/(\sqrt{2\pi}\tau)\), so \(\pi_\tau(K)\leq2/(\sqrt{2\pi}\tau)\to0\). A risk improvement confined to \(K\) may therefore contribute an amount tending to zero to an integral under \(\pi_\tau\), even if the improvement is bounded away from zero on that interval. The normalization that makes each \(\pi_\tau\) a probability measure can erase the local comparison needed in the theorem.
For the rescaled measures \(\nu_\tau=\sqrt{2\pi}\tau\pi_\tau\), the density on \(K\) is at least \(\exp\{-1/(2\tau^2)\}\), which tends to \(1\). Thus \(\nu_\tau(K)\) does not vanish. The rescaled measures have the same Bayes rules, and their integrated-risk gap still tends to zero. The scale of the measures matters for the admissibility argument even though it does not matter for the posterior distribution or the Bayes rule.
What Must Be Checked in an Admissibility Argument
A proper posterior is a conditional-distribution calculation. An admissibility proof based on an improper prior is a risk-integral argument, and it must bridge the two. The following checks make the distinction explicit.
Show that \(0<m_\mu(x)<\infty\) for the relevant observations, and specify any exceptional observations where this fails.
Show that the posterior expected loss has a measurable minimizer. This identifies a generalized Bayes rule but does not establish admissibility.
For each measure, ensure the integrated risk of the candidate is finite and the Bayes risk is well defined. Scaling is allowed, but the scale affects the integrated-risk gap.
Show that the integrated-risk gap tends to zero and that a strict risk improvement would retain a positive integrated advantage under the approximating measures.
Common Pitfall
The incorrect argument is: “The prior is improper, but the posterior is proper; therefore its Bayes rule is admissible.” Posterior propriety only says the likelihood-weighted prior can be normalized at the observed data. It gives no limit for \(\int R(\theta,\delta)\nu_k(d\theta)-r(\nu_k)\), and it does not ensure that a strict risk improvement remains visible under any chosen sequence of finite measures.
The correction is to keep the two claims separate. First verify that the posterior and generalized Bayes rule exist. Then give a finite-measure approximation and prove the integrated-risk gap tends to zero at a scale that also preserves local mass where a strict improvement could occur. Without both parts, the improper-prior calculation is not an admissibility proof.
Check Your Understanding
Use the definitions and criterion above to answer the following questions.
- What condition on \(m_\mu(x)\) makes the posterior proper at an observation \(x\)?
- In the Poisson example with prior measure \(d\theta/\theta\), why is the posterior proper for \(x\geq1\) but not for \(x=0\)?
- Why does posterior propriety not imply that the generalized Bayes rule is admissible?
- In the normal-mean example, calculate the integrated-risk gap under \(\pi_\tau\) and explain why rescaling the prior does not change its Bayes rule.
- What role does local mass play in applying a vanishing integrated-risk-gap criterion?