Strictly Positive Priors in Finite-State Examples

Why Positive Prior Mass Matters

In a finite-state decision problem, a prior distribution assigns weights to the possible parameter states. If every weight is positive, a rule’s risk at every state contributes to its prior-weighted average. This creates a direct link between minimizing average risk and resisting a competitor that improves risk state by state: a genuine improvement at even one state would lower the average.

The link requires care. The rule must actually minimize posterior expected loss, and the prior must give positive mass to every state being compared. With those conditions in place, a Bayes rule is admissible: no other rule has risk no greater at every state and strictly lower at at least one state. The conclusion does not require the posterior minimizer to be unique.

Finite-State Setup and Posterior Risk

Let the parameter space be the finite set \(\Theta=\{\theta_1,\ldots,\theta_k\}\). The observation \(X\) has distribution \(P_i\) when the state is \(\theta_i\). Let the action space \(A\) be finite, and let \(L(\theta_i,a)\) be the loss from taking action \(a\) in state \(\theta_i\). A randomized decision rule \(\delta\) specifies a probability distribution over actions for each observation \(x\); a nonrandomized rule is the special case that selects one action with probability one. The risk at state \(\theta_i\) is

R(\theta_i,\delta) = E_{\theta_i}\!\left[\sum_{a\in A}L(\theta_i,a)\delta(a\mid X)\right].

Take a prior \(\pi=(\pi_1,\ldots,\pi_k)\), where \(\pi_i=P(\theta_i)\). It is strictly positive (or has full support on this finite parameter space) when \(\pi_i>0\) for every \(i\). The Bayes risk of \(\delta\) is its average risk under this prior.

r(\pi,\delta) = \sum_{i=1}^k \pi_i R(\theta_i,\delta).

To write posterior risks for a general observation space, let \(Q=\sum_{i=1}^k\pi_iP_i\) be the prior-predictive distribution of \(X\). The measures \(\pi_iP_i\) are absolutely continuous with respect to \(Q\), so define \(\eta_i(x)=d(\pi_iP_i)/dQ(x)\). These posterior probabilities sum to one \(Q\)-almost everywhere. When the observation has densities or probability masses \(f_i(x)\) relative to a common measure, this is the familiar formula \(\eta_i(x)=\pi_i f_i(x)/\sum_j\pi_j f_j(x)\), wherever the denominator is positive. For an action \(a\), its posterior expected loss at \(x\) is \(\sum_i\eta_i(x)L(\theta_i,a)\).

Definition: A rule \(\delta\) is a posterior-risk minimizer if, for \(Q\)-almost every observation \(x\), its conditional expected loss is minimal among the available actions and randomized actions: \(\sum_i\eta_i(x)\sum_a L(\theta_i,a)\delta(a\mid x)\leq \sum_i\eta_i(x)\sum_a L(\theta_i,a)\kappa(a)\) for every probability distribution \(\kappa\) on \(A\). It is a Bayes rule if it minimizes \(r(\pi,\delta)\) over the decision rules being considered.

From Posterior Minimization to Bayes Risk

The prior-predictive distribution packages the statewise sampling distributions into one distribution for the observation. The following identity shows why pointwise posterior minimization controls the prior-weighted risk.

Result: For any rule \(\delta\) with finite risks, \[ r(\pi,\delta) = \int \left[\sum_{i=1}^k\eta_i(x) \sum_{a\in A}L(\theta_i,a)\delta(a\mid x)\right]Q(dx). \] Consequently, a posterior-risk minimizer is a Bayes rule.

Proof. By the definition of risk and the prior-predictive distribution,

\begin{aligned} r(\pi,\delta) &=\sum_{i=1}^k\pi_i \int \sum_{a\in A}L(\theta_i,a)\delta(a\mid x)\,P_i(dx)\\ &=\int \sum_{i=1}^k \frac{d(\pi_iP_i)}{dQ}(x) \sum_{a\in A}L(\theta_i,a)\delta(a\mid x)\,Q(dx)\\ &=\int \sum_{i=1}^k\eta_i(x) \sum_{a\in A}L(\theta_i,a)\delta(a\mid x)\,Q(dx). \end{aligned}

The sums are finite, and the risks are finite, so the integrals in this identity are well-defined. If \(\delta\) minimizes the conditional expected loss at \(Q\)-almost every \(x\), its integrand is no greater than the corresponding integrand for any competing rule. Integrating gives no greater Bayes risk. Thus \(\delta\) is a Bayes rule. \(\square\)

Worked Example: A Binary Classification Problem

Suppose \(\Theta=\{\theta_1,\theta_2\}\), the observation \(X\) is binary, and \(P_{\theta_1}(X=1)=0.2\), while \(P_{\theta_2}(X=1)=0.8\). Use prior probabilities \(\pi_1=\pi_2=0.5\), actions \(a_1,a_2\) that report the corresponding state, and loss 0 for a correct report and 1 for an incorrect report. Find a posterior-risk-minimizing rule and its statewise risks.

When \(X=1\), the joint prior-and-observation weights for \(\theta_1,\theta_2\) are \(0.5(0.2)=0.10\) and \(0.5(0.8)=0.40\). The total is \(0.50\), so the posterior probabilities are \(0.20\) and \(0.80\). Reporting \(\theta_2\) has posterior expected loss \(0.20\), less than the loss \(0.80\) from reporting \(\theta_1\).

When \(X=0\), the weights are \(0.5(0.8)=0.40\) and \(0.5(0.2)=0.10\). The posterior probabilities are \(0.80\) and \(0.20\), so reporting \(\theta_1\) has posterior expected loss \(0.20\), rather than \(0.80\) for reporting \(\theta_2\). The minimizing rule therefore reports \(\theta_2\) when \(X=1\) and \(\theta_1\) when \(X=0\).

Under \(\theta_1\), this rule is wrong exactly when \(X=1\), so \(R(\theta_1,\delta)=0.2\). Under \(\theta_2\), it is wrong exactly when \(X=0\), so \(R(\theta_2,\delta)=0.2\). Its Bayes risk is \(0.5(0.2)+0.5(0.2)=0.2\). The positive prior mass on both states will rule out any competitor that weakly improves these risks and strictly improves one of them.

Why the Bayes Rule Cannot Be Dominated

For this result, a rule \(\delta'\) dominates \(\delta\) if \(R(\theta_i,\delta')\leq R(\theta_i,\delta)\) for every state and the inequality is strict for at least one state. A rule is admissible within the comparison class if no rule in that class dominates it.

Theorem: Suppose \(\Theta=\{\theta_1,\ldots,\theta_k\}\) is finite, the prior has \(\pi_i>0\) for every \(i\), and \(\delta\) is a Bayes rule with finite risks. Then \(\delta\) is admissible among the rules over which Bayes risk is minimized.

Proof. Suppose, to obtain a contradiction, that a rule \(\delta'\) in the comparison class dominates \(\delta\). Define \(\Delta_i=R(\theta_i,\delta)-R(\theta_i,\delta')\). Domination means \(\Delta_i\geq0\) for every \(i\), and \(\Delta_j>0\) for at least one state \(\theta_j\). Since \(\pi_j>0\),

r(\pi,\delta)-r(\pi,\delta') = \sum_{i=1}^k\pi_i\Delta_i \geq \pi_j\Delta_j>0.

Thus \(\delta'\) has strictly smaller Bayes risk than \(\delta\), contradicting that \(\delta\) is a Bayes rule. No such dominating rule exists, so \(\delta\) is admissible. \(\square\)

The argument is an average-risk certificate. It does not say that a positive prior makes every rule admissible; the rule must first minimize Bayes risk. Nor does it require a unique posterior-minimizing action. The next example shows that ties can produce several Bayes rules without allowing any of them to be dominated.

Worked Example: A Posterior Tie and Randomized Rules

There are two states, \(\theta_1,\theta_2\), and no informative observation. There are two actions, \(a,b\), with losses \(L(\theta_1,a)=0\), \(L(\theta_1,b)=1\), \(L(\theta_2,a)=1\), and \(L(\theta_2,b)=0\). Let the prior assign probability \(1/2\) to each state. Compare all rules that choose \(a\) with probability \(t\) and \(b\) with probability \(1-t\), where \(0\leq t\leq1\).

The posterior is the prior because the observation contains no information. The posterior expected loss of \(a\) is \(0.5(0)+0.5(1)=0.5\), and that of \(b\) is \(0.5(1)+0.5(0)=0.5\). Every value of \(t\) therefore minimizes posterior risk. The statewise risks are \(R(\theta_1,\delta_t)=1-t\) and \(R(\theta_2,\delta_t)=t\), so the Bayes risk is \(0.5(1-t)+0.5t=0.5\) for every \(t\).

To check admissibility directly, suppose \(\delta_s\) dominates \(\delta_t\). The risk inequality at \(\theta_1\) requires \(1-s\leq1-t\), or \(s\geq t\). The inequality at \(\theta_2\) requires \(s\leq t\). Hence \(s=t\), and neither risk is strictly lower. Every rule in this family is admissible, including the deterministic rules \(t=0\) and \(t=1\).

Worked Example: Three States and Posterior Decisions

Let \(Y\in\{1,2\}\), with probabilities \(P_{\theta_i}(Y=1)\) equal to \(0.8,0.4,0.1\) for states \(\theta_1,\theta_2,\theta_3\), respectively. Use prior probabilities \((0.2,0.3,0.5)\), actions that report one of the three states, and zero-one loss. Find the posterior-minimizing rule and calculate its risk vector.

For \(Y=1\), the three joint weights are \(0.2(0.8)=0.16\), \(0.3(0.4)=0.12\), and \(0.5(0.1)=0.05\). They sum to \(0.33\), giving posterior probabilities \(16/33,12/33,5/33\). The largest is \(16/33\), so reporting \(\theta_1\) uniquely minimizes posterior expected loss. That minimum loss is \(1-16/33=17/33\).

For \(Y=2\), the joint weights are \(0.2(0.2)=0.04\), \(0.3(0.6)=0.18\), and \(0.5(0.9)=0.45\). Their sum is \(0.67\), giving posterior probabilities \(4/67,18/67,45/67\). The minimizing report is \(\theta_3\), with posterior expected loss \(1-45/67=22/67\).

The rule reports \(\theta_1\) for \(Y=1\) and \(\theta_3\) for \(Y=2\). Its risk under \(\theta_1\) is \(P_{\theta_1}(Y=2)=0.2\); under \(\theta_2\), either observation leads to an incorrect report, so its risk is \(1\); under \(\theta_3\), it is wrong only when \(Y=1\), so its risk is \(0.1\). The Bayes risk is \(0.2(0.2)+0.3(1)+0.5(0.1)=0.04+0.30+0.05=0.39\). This also agrees with the posterior calculation: \(P(Y=1)(17/33)+P(Y=2)(22/67)=0.33(17/33)+0.67(22/67)=0.17+0.22=0.39\). The prior is positive at all three states, so the theorem guarantees that this rule is admissible.

What the Guarantee Does—and Does Not—Say

Strict positivity matters because it ensures that every statewise improvement contributes a positive amount to the prior-weighted risk. If a prior assigns zero weight to a state, Bayes-risk minimization places no constraint on performance there. For example, suppose the prior puts all its mass on \(\theta_1\), and two actions \(a,b\) both have loss 0 at \(\theta_1\). At \(\theta_2\), let their losses be 10 and 0, respectively. Choosing \(a\) is Bayes under this prior, but choosing \(b\) has equal risk at \(\theta_1\) and strictly lower risk at \(\theta_2\). The Bayes rule \(a\) is dominated.

A second important qualification is that posterior minimization is a condition on the rule, not merely on the prior. A positive prior cannot turn a rule with unnecessarily high posterior expected loss into a Bayes rule. In finite problems, posterior losses can be calculated action by action, and an action with the smallest value can be selected at each observation. If several actions tie, any posterior-risk-minimizing randomization among them remains eligible for the admissibility conclusion.

Do not confuse positive prior mass with a positive risk gap. The proof uses only that the prior weight times a strict risk improvement is positive.
Do not require a strict posterior minimum. Tied posterior minima can yield multiple Bayes rules; the theorem applies to each rule that minimizes Bayes risk.
Do not overlook the comparison class. Admissibility is relative to the rules over which Bayes risk is minimized and the domination comparison is made.

In a finite-state example, the practical check is concise: verify that every prior weight is positive, compute a posterior-risk-minimizing rule, and confirm that its risks are finite. The weighted-risk argument then rules out any competitor that is no worse at every state and better at even one.

Check Your Understanding

Use the definitions and results above to answer the following questions.

Why must a strictly positive prior assign positive mass to every state in the parameter space under comparison?
In a finite-state problem, what distribution can be used to express the prior-weighted risk as an integral of posterior expected loss?
If a posterior-minimizing rule ties between two actions at every observation, does the admissibility result require choosing one action deterministically? Explain.
A competitor has equal risk to a Bayes rule at all but one state, where it has strictly lower risk. Why does a positive prior rule out this competitor?
What can go wrong if the prior assigns probability zero to a state, even when a rule minimizes Bayes risk?