Prior Distributions as Mixed Strategies for Nature

From a Parameter Choice to a Mixed Strategy

In a statistical decision problem, Nature selects a parameter value $\theta$, data are generated according to the model at that value, and a decision rule chooses an action. The rule’s risk $R(\theta,\delta)$ is its expected loss when the parameter is fixed at $\theta$. In the game interpretation, the statistician chooses a rule and Nature chooses the parameter. Nature can also randomize: instead of selecting one parameter value with certainty, Nature can draw it from a probability distribution.

That distribution is a mixed strategy for Nature. Its payoff against a rule averages the rule’s risk over parameter values. The averaging does not replace the original risk, which remains a function of the fixed parameter. It defines the payoff for the game in which Nature is allowed to mix.

Definition: Let $\Theta$ be a parameter space with a specified sigma-algebra, and let $\pi$ be a probability measure on $\Theta$. In the statistical game, $\pi$ is Nature’s mixed strategy: Nature first draws $\theta$ according to $\pi$, after which data are generated under the sampling distribution $P_\theta$. Against a decision rule $\delta$, the mixed-strategy payoff is the expected loss over this entire random process.

The order of the process matters conceptually. Nature’s draw of $\theta$ is not an extra observation given to the statistician. The statistician sees the data, not Nature’s private draw. A prior can also be used without asserting that the unknown parameter is literally generated by a random mechanism: it is still a probability distribution with which to average payoffs in the game.

Risk and the Prior-Weighted Payoff

Let $\mathcal X$ be the data space and $\mathcal A$ the action space, both equipped with sigma-algebras. Assume $\theta\mapsto P_\theta(B)$ is measurable for each measurable data set $B$; that is, $P_\theta$ is a probability kernel from $\Theta$ to $\mathcal X$. A randomized rule $\delta$ is a probability kernel from $\mathcal X$ to $\mathcal A$: given data $x$, it assigns a probability distribution $\delta(da\mid x)$ over actions. A nonrandomized rule is included by taking this distribution to put all its mass on the action selected at $x$.

Let $W(\theta,a)$ be a measurable, nonnegative loss function. The risk at a fixed parameter value is

R(\theta,\delta) = \int_{\mathcal X}\int_{\mathcal A} W(\theta,a)\,\delta(da\mid x)\,P_\theta(dx).

The inner integration averages over the rule’s action randomization, and the outer integration averages over the data generated at $\theta$. The risk can be infinite if the loss is unbounded. The mixed-strategy payoff against prior $\pi$ is

G(\pi,\delta)=\int_\Theta R(\theta,\delta)\,\pi(d\theta).

This is called the integrated risk, or the prior-averaged risk. Nonnegative losses allow these integrals to be interpreted as extended nonnegative integrals, including the value $+\infty$. If losses can be negative, the same identity below applies provided the absolute expected loss is integrable under the joint distribution induced by $\pi$, the sampling model, and the rule.

Worked Example: A Binary Parameter and an Informative Observation

Let $\Theta=\{0,1\}$, let the action be $a\in\{0,1\}$, and use zero-one loss $W(\theta,a)=\mathbf 1\{a\ne\theta\}$. The data are $X\in\{0,1\}$, with $P_0(X=1)=0.2$ and $P_1(X=1)=0.8$. Consider the rule that chooses action $a=X$. Nature uses the prior $\pi(\theta=1)=0.7$, so $\pi(\theta=0)=0.3$.

At $\theta=0$, the rule errs exactly when $X=1$, so $R(0,\delta)=0.2$. At $\theta=1$, it errs exactly when $X=0$, which has probability $1-0.8=0.2$, so $R(1,\delta)=0.2$. Thus Nature’s mixed-strategy payoff is

G(\pi,\delta) =0.3R(0,\delta)+0.7R(1,\delta) =0.3(0.2)+0.7(0.2)=0.2.

For comparison, the rule that always chooses action $0$ has risk $R(0,\delta_0)=0$ and $R(1,\delta_0)=1$. Its payoff against the same prior is $0.3(0)+0.7(1)=0.7$. The payoff depends on both the rule and the distribution with which Nature mixes.

Why the Payoff Is Exactly Integrated Risk

Theorem: Suppose $\Theta,\mathcal X,\mathcal A$ are measurable spaces, $P_\theta$ is a probability kernel from $\Theta$ to $\mathcal X$, $\delta$ is a probability kernel from $\mathcal X$ to $\mathcal A$, and $W$ is a nonnegative measurable loss. For any prior probability measure $\pi$, $$ G(\pi,\delta) = \int_\Theta R(\theta,\delta)\,\pi(d\theta) = \int_\Theta\int_{\mathcal X}\int_{\mathcal A} W(\theta,a)\,\delta(da\mid x)\,P_\theta(dx)\,\pi(d\theta). $$

Proof. The kernels and the prior define the joint probability law of the parameter, data, and action by first drawing $\theta$ from $\pi$, then $x$ from $P_\theta$, then $a$ from $\delta(\cdot\mid x)$. The loss $W$ is nonnegative and measurable, so Tonelli’s theorem permits the iterated integrations in any order, whether or not the resulting expectation is finite. For each fixed $\theta$, integrating over the data and action gives $R(\theta,\delta)$ by its definition. Integrating that quantity over $\pi$ gives $G(\pi,\delta)$. Thus the game’s expected payoff and the prior-integrated risk are the same expectation, written as an iterated integral. $\square$

If $W$ is signed, Tonelli’s theorem alone does not justify rearranging the integrals. If the absolute value of the loss has finite expectation under the joint law, Fubini’s theorem gives the same identity. Stating this condition is important: a formal exchange of integrals is not automatically valid for arbitrary signed, nonintegrable losses.

Worked Example: A Randomized Rule in a Finite Game

Suppose there is no data, so a rule chooses an action directly. There are three parameter values, $\theta_1,\theta_2,\theta_3$, and two actions, $a_1,a_2$. The losses are $W(\theta_1,a_1)=0$, $W(\theta_2,a_1)=4$, $W(\theta_3,a_1)=2$, while $W(\theta_1,a_2)=3$, $W(\theta_2,a_2)=1$, and $W(\theta_3,a_2)=0$. Let Nature’s prior assign probabilities $0.2,0.5,0.3$ to the three parameter values.

The integrated risk of always choosing $a_1$ is

$$ 0.2(0)+0.5(4)+0.3(2)=2.6. $$

For always choosing $a_2$, it is

$$ 0.2(3)+0.5(1)+0.3(0)=1.1. $$

Now let the rule choose $a_1$ with probability $0.4$ and $a_2$ with probability $0.6$. Its risk at each parameter is the corresponding action-loss average:

\begin{aligned} R(\theta_1,\delta)&=0.4(0)+0.6(3)=1.8,\\ R(\theta_2,\delta)&=0.4(4)+0.6(1)=2.2,\\ R(\theta_3,\delta)&=0.4(2)+0.6(0)=0.8. \end{aligned}

Averaging these risks over Nature’s prior gives

$$ 0.2(1.8)+0.5(2.2)+0.3(0.8)=0.36+1.10+0.24=1.70. $$

Equivalently, the rule randomizes between the two pure-action payoffs: $0.4(2.6)+0.6(1.1)=1.04+0.66=1.70$. Both calculations give the same expected loss because expectations are linear.

A Prior Gives a Lower Bound on Minimax Risk

Theorem: For nonnegative loss, define the Bayes risk under $\pi$ by $$ r(\pi)=\inf_\delta \int_\Theta R(\theta,\delta)\,\pi(d\theta), $$ where the infimum is over the allowed decision rules. Define the minimax value by $$ V=\inf_\delta \sup_{\theta\in\Theta}R(\theta,\delta). $$ Then $r(\pi)\le V$ for every prior $\pi$.

Proof. Fix any rule $\delta$. Since an average of a nonnegative function under a probability measure cannot exceed its supremum,

\int_\Theta R(\theta,\delta)\,\pi(d\theta) \le \sup_{\theta\in\Theta}R(\theta,\delta).

This inequality holds for every allowed rule, including when either side is infinite. Taking the infimum over rules on both sides preserves the inequality, so

r(\pi) = \inf_\delta\int_\Theta R(\theta,\delta)\,\pi(d\theta) \le \inf_\delta\sup_{\theta\in\Theta}R(\theta,\delta) =V.

Thus a chosen prior supplies a lower bound on the best achievable worst-case risk. This argument does not assert that the bound is sharp, that an optimal rule exists, or that a minimax interchange holds. Establishing such conclusions requires additional conditions. $\square$

Worked Example: A Normal Model and a Continuous Prior

Suppose $X\mid\theta\sim N(\theta,1)$, the action is a real-valued estimate, and the loss is squared error $W(\theta,a)=(a-\theta)^2$. Let Nature use $\theta\sim N(0,4)$. Consider first the rule $\delta(X)=X$. Conditional on $\theta$, its estimation error has mean zero and variance one, so

R(\theta,\delta)=E_\theta[(X-\theta)^2]=1 \quad\text{for every }\theta.

Its integrated risk is therefore $\int 1\,\pi(d\theta)=1$. The same calculation can be made from the joint experiment: write $X=\theta+\varepsilon$, where conditionally $\varepsilon\sim N(0,1)$. Then the squared error of this rule is $\varepsilon^2$, with expectation $1$.

For the rule $\delta_0(X)=0$, the risk at a fixed $\theta$ is $\theta^2$. The integrated risk is

\int_\Theta R(\theta,\delta_0)\,\pi(d\theta) =E[\theta^2] =\operatorname{Var}(\theta)+\{E[\theta]\}^2 =4+0=4.

The prior-weighted game payoffs are thus $1$ for the rule $X$ and $4$ for the rule that always chooses zero. These values are expectations over Nature’s parameter draw as well as the data process; they are not obtained by substituting the prior mean into the risk function.

Common Pitfalls and Interpretation

Evaluating risk at the prior mean. The integrated risk is $\int R(\theta,\delta)\,\pi(d\theta)$, not $R(\int\theta\,\pi(d\theta),\delta)$. In the normal example, the zero rule has risk $\theta^2$: its risk at the prior mean $0$ is $0$, but its integrated risk is $4$.
Forgetting the data and action randomness. Nature’s prior draw is only one part of the experiment. The payoff also averages over $P_\theta$ and any randomization in $\delta$.
Treating the prior as information revealed to the rule. The rule uses the observed data. The prior specifies Nature’s mixed strategy and how payoffs are averaged; it does not mean that the realized parameter is disclosed.
Assuming a prior average equals the worst-case risk. For any fixed rule, integrated risk is at most the supremum risk, but it may be strictly smaller. A prior gives a lower bound on minimax risk, not automatically its exact value.

The useful strategic translation is precise: the pure parameter $\theta$ is Nature’s pure strategy, a prior $\pi$ is a mixed strategy, and the payoff against a rule is its integrated risk. This lets results about statistical risk be interpreted as statements about expected payoffs, while keeping clear which conclusions follow from averaging alone and which need a minimax theorem.

Check Your Understanding

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

In a statistical game, what random events are averaged when computing the payoff of a prior against a randomized rule?
Why does a nonnegative measurable loss permit the iterated-integral identity even when the integrated risk is infinite?
In the finite example, calculate the integrated risk of the rule that chooses $a_1$ with probability $0.2$ and $a_2$ with probability $0.8$.
For the normal example, what is the integrated risk of the constant-zero rule if the prior changes to $N(0,9)$?
Explain why the Bayes risk for any fixed prior cannot exceed the minimax value, and why this inequality alone does not establish equality.