Endpoint-Prior Candidates for Bounded Normal Means

Why an Endpoint Prior Is a Candidate, Not a Conclusion

For a normal mean restricted to a bounded interval, it is tempting to put prior mass at the two endpoints: they are the most separated parameter values, and their competition produces a tractable estimator. But being Bayes for that prior does not by itself make a rule minimax on the whole interval. Interior parameter values still contribute to the maximum risk. The essential check is whether the candidate’s risk anywhere in the interval exceeds its risk at the prior-supported endpoints.

We consider one observation $X\sim N(\theta,\sigma^2)$, with known $\sigma^2>0$, parameter space $\Theta=[-a,a]$ for $a>0$, and squared-error loss $L(\theta,d)=(d-\theta)^2$. A decision rule $\delta$ is a measurable function of $X$, and its risk is $R(\theta,\delta)=E_\theta[(\delta(X)-\theta)^2]$.

Definition: An endpoint prior assigns prior probability $p$ to $\theta=a$ and $1-p$ to $\theta=-a$, where $0<p<1$. Its Bayes rule minimizes prior-averaged risk. A rule is minimax if it minimizes the maximum risk $\sup_{\theta\in[-a,a]}R(\theta,\delta)$ over all measurable rules.

Deriving the Endpoint-Prior Bayes Rule

For an observed value $x$, the posterior probability of the upper endpoint follows from Bayes’ formula. The likelihood ratio of the two endpoint models is $f(x\mid a)/f(x\mid -a)=\exp(2ax/\sigma^2)$. Thus the posterior odds equal the prior odds multiplied by this likelihood ratio.

q(x):=P(\theta=a\mid X=x) =\frac{p\exp(2ax/\sigma^2)} {(1-p)+p\exp(2ax/\sigma^2)} =\frac{1}{1+\exp\!\left[-\left(2ax/\sigma^2+\log\frac{p}{1-p}\right)\right]}.

Under squared-error loss, the posterior expected loss at action $d$ is minimized by the posterior mean. Indeed, for any posterior distribution with mean $m$, $E[(d-\theta)^2\mid x]=(d-m)^2+\operatorname{Var}(\theta\mid x)$; the second term does not depend on $d$. Here the posterior is supported on $\{-a,a\}$, so its mean is $a q(x)-a(1-q(x))$.

Result: The Bayes rule for the endpoint prior is $$ \delta_p(x)=a\{2q(x)-1\} =a\tanh\!\left(\frac{ax}{\sigma^2}+\frac12\log\frac{p}{1-p}\right). $$ For equal endpoint weights, $p=1/2$, this simplifies to the odd rule $\delta_{1/2}(x)=a\tanh(ax/\sigma^2)$, which takes values strictly between $-a$ and $a$.

Worked Example: Computing the Equal-Weight Posterior Estimate

Take $\Theta=[-1,1]$, $\sigma^2=1$, and equal prior probabilities at $-1$ and $1$. If the observation is $x=0.5$, the log likelihood ratio is $2ax/\sigma^2=1$. The posterior probability of $1$ is therefore $q(0.5)=1/(1+e^{-1})\approx0.7311$.

The posterior mean, and hence the Bayes estimate, is $1(0.7311)-1(1-0.7311)=0.4622$, or equivalently $\tanh(0.5)\approx0.4621$ (the difference in the last decimal is rounding). This is an estimate between the endpoints, but that fact alone says nothing about whether its maximum risk over $[-1,1]$ is minimized.

Worked Example: How Unequal Endpoint Weights Shift the Rule

Let $a=1$, $\sigma^2=1$, and put prior probability $p=0.8$ on $1$. At $x=0$, the likelihoods at the two endpoints are equal, so the posterior probability of $1$ remains $0.8$. The posterior mean is $1(0.8)-1(0.2)=0.6$.

The formula gives the same result: $\delta_{0.8}(0)=\tanh\{\tfrac12\log(0.8/0.2)\}=\tanh(\log 2)\approx0.6$. Thus unequal weights shift the rule toward the more probable endpoint, even when the observation itself provides no evidence favoring either endpoint.

Using the Endpoint Bayes Risk as a Minimax Bound

For the equal-weight prior, the candidate rule is odd: $\delta_{1/2}(-x)=-\delta_{1/2}(x)$. The normal model is also symmetric. If $X\sim N(\theta,\sigma^2)$, then $-X\sim N(-\theta,\sigma^2)$; substituting these sign changes in the squared-error loss gives $R(-\theta,\delta_{1/2})=R(\theta,\delta_{1/2})$. In particular, the risks at $-a$ and $a$ are equal. Write their common value as $r_a$.

The prior-averaged risk of this rule is exactly $r_a$, because the prior puts equal mass on those two endpoints. For any other rule, its maximum risk over the full interval is at least its prior-averaged risk, and that average cannot be below the Bayes risk of the prior. This gives the certificate below.

Theorem (Endpoint-prior minimax certificate): Let $\delta_{1/2}$ be the equal-weight endpoint-prior Bayes rule for $X\sim N(\theta,\sigma^2)$, $\theta\in[-a,a]$, under squared-error loss. If $$ R(\theta,\delta_{1/2})\le r_a \quad\text{for every }\theta\in[-a,a], \qquad r_a=R(a,\delta_{1/2})=R(-a,\delta_{1/2}), $$ then $\delta_{1/2}$ is minimax and the minimax risk is $r_a$.

Proof. For any measurable rule $\delta$, let its maximum risk be $M(\delta)=\sup_{\theta\in[-a,a]}R(\theta,\delta)$. Since a supremum is at least any average, $M(\delta)\ge\{R(a,\delta)+R(-a,\delta)\}/2$. The right side is the risk averaged under the equal-weight endpoint prior. By the definition of a Bayes rule, it is at least the Bayes risk $r_a$. Therefore every rule has maximum risk at least $r_a$. Under the theorem’s condition, $\delta_{1/2}$ has maximum risk at most $r_a$; at the endpoints its risk equals $r_a$, so its maximum risk is exactly $r_a$. It attains the lower bound and is minimax. $\square$

The rule’s risk can be written in a form suitable for checking:

R(\theta,\delta_{1/2}) =\int_{-\infty}^{\infty} \left[a\tanh\!\left(\frac{ax}{\sigma^2}\right)-\theta\right]^2 \frac{1}{\sigma}\phi\!\left(\frac{x-\theta}{\sigma}\right)\,dx,

where $\phi$ is the standard normal density. Equivalently, with $Z\sim N(0,1)$, substitute $X=\theta+\sigma Z$ in the expectation. The rule is bounded and continuous, so this risk is continuous in $\theta$: for any convergent sequence of parameter values, the integrand in the $Z$-representation converges pointwise and is bounded by $4a^2$. Dominated convergence applies. Consequently the risk attains a maximum on the compact interval. Symmetry reduces the check to $[0,a]$, but does not show that the maximum occurs at an endpoint. That inequality must still be established.

A Risk Check Can Reject the Candidate

The next example shows why the interior check matters. We use a rigorous lower bound on risk at the midpoint and compare it with a different rule’s maximum-risk bound. The bound is sufficient to reject minimaxity; it is not an approximation to the candidate’s exact risk.

Worked Example: An Endpoint Bayes Rule That Is Not Minimax

Take $\Theta=[-3,3]$, $\sigma^2=1$, and equal prior mass at $-3$ and $3$. The endpoint-prior rule is $\delta(x)=3\tanh(3x)$. At the interior point $\theta=0$, $X=Z\sim N(0,1)$, so $R(0,\delta)=9E[\tanh^2(3Z)]$. On the event $\{|Z|\ge1\}$, monotonicity of $\tanh(t)$ for $t\ge0$ gives $\tanh^2(3Z)\ge\tanh^2(3)$. Hence

R(0,\delta)\ge 9\tanh^2(3)P(|Z|\ge1) \approx 9(0.990134)(0.317311) \approx 2.8276.

In particular, this establishes $R(0,\delta)>2.82$; it does not establish $R(0,\delta)>2.83$. Thus the maximum risk of $\delta$ exceeds $2.82$.

Now consider the projection rule $\delta_c(x)=\min(3,\max(-3,x))$, which clips $x$ to the parameter interval. For every $\theta\in[-3,3]$, projection onto that interval cannot increase distance to $\theta$, so $|\delta_c(X)-\theta|\le|X-\theta|$. Squaring and taking expectations yields $R(\theta,\delta_c)\le E_\theta[(X-\theta)^2]=1$ for every $\theta$. Its maximum risk is at most $1$, strictly below the endpoint rule’s maximum risk, which is greater than $2.82$. Therefore the endpoint-prior Bayes rule is not minimax in this example.

How to Make the Minimaxity Check

For a symmetric endpoint prior, equal endpoint risks give the natural candidate value $r_a$. The sufficient certificate requires checking that every interior risk is no larger than that value. A complete argument may derive a bound or monotonicity property for the risk; a numerical plot can suggest where to look, but does not prove the inequality over a continuum. If the risk exceeds $r_a$ anywhere, the certificate fails. That failure alone does not prove the rule is not minimax; comparison with another rule or another valid argument is needed, as in the example above.

A common mistake is to reason that a Bayes rule supported on the endpoints must have its worst risk at those endpoints. The prior’s support determines which risks are averaged in its Bayes criterion, not where the rule’s maximum risk over a larger parameter space occurs. The correction is to separate the two tasks: first derive the Bayes rule and its prior-averaged risk; then evaluate or bound its risk for all allowed parameter values.

Derive the candidate.
Compute the posterior probabilities at the two endpoints and take the posterior mean under squared-error loss.

Identify the lower bound.
For equal weights, the average of the two endpoint risks is the Bayes risk and is a lower bound on every rule’s maximum risk.

Check the full interval.
Establish that the candidate’s risk never exceeds the endpoint value. If it does, the certificate cannot certify minimaxity.

Key takeaway: An endpoint prior supplies both a tractable Bayes rule and a lower bound on minimax risk. Minimaxity follows only when the candidate’s risk across the entire bounded parameter space is controlled by that lower bound.

Check Your Understanding

Use the definitions and risk comparison above to answer the following.

For prior mass $p$ at $a$, what is the posterior odds of $a$ versus $-a$ after observing $x$?
Why is the posterior mean the Bayes action under squared-error loss?
For the equal-weight endpoint rule, why are the risks at $-a$ and $a$ equal?
State the risk condition in the endpoint-prior minimax certificate. Why must it cover interior parameter values as well?
In the interval $[-3,3]$ example, what does the event $\{|Z|\ge1\}$ establish about the midpoint risk, and what numerical inequality should not be claimed from that bound?