From Statistical Risks to a Finite Game
A minimax calculation becomes concrete once the statistical risks are written as a payoff table. The decision maker chooses an action to keep risk low; Nature chooses a parameter value that makes risk high. When either player randomizes, the payoff is the corresponding average risk. This tutorial carries that calculation through for three actions and two parameter values, including an explicit check that the resulting strategies form a saddle point.
Consider a finite experiment with a single possible observation \(x_0\), which occurs with probability one under either parameter value. There are two possible parameter values, \(\theta_0\) and \(\theta_1\), and three available actions, \(a_1,a_2,a_3\). The loss is specified below. Because the observation is constant, a nonrandomized rule is identified with the action it chooses. A randomized rule selects among the three actions with specified probabilities.
| Action | Loss at \(\theta_0\) | Loss at \(\theta_1\) |
|---|---|---|
| \(a_1\) | 0 | 4 |
| \(a_2\) | 3 | 3 |
| \(a_3\) | 4 | 0 |
The risk \(R(\theta,a_i)\) is the expected loss when action \(a_i\) is used at parameter value \(\theta\). Here the loss is deterministic, so each risk is just the corresponding table entry. We use risk as the payoff: the decision maker minimizes it, and Nature maximizes it.
Worked Example: Find the Mixed Strategies and Game Value
Worked Example: Find the Mixed Strategies and Game Value
Use the risk table above to find a decision-maker strategy and a Nature strategy that form a saddle point. Calculate their common game value.
Step 1: Write the risks of a randomized action. If the decision maker uses probabilities \(p_1,p_2,p_3\), the risk at \(\theta_0\) is $$ R(\theta_0,p)=0p_1+3p_2+4p_3=3p_2+4p_3, $$ and the risk at \(\theta_1\) is $$ R(\theta_1,p)=4p_1+3p_2+0p_3=4p_1+3p_2. $$
A useful candidate puts no probability on \(a_2\) and balances the two risks. Setting \(p_2=0\), \(p_3=1-p_1\), and \(4p_3=4p_1\) gives \(p_1=p_3=1/2\). Thus the candidate decision-maker strategy is \(p^*=(1/2,0,1/2)\). Its risks are $$ R(\theta_0,p^*)=4(1/2)=2,\qquad R(\theta_1,p^*)=4(1/2)=2. $$ So this strategy has worst-case risk 2.
Step 2: Find a Nature strategy that prevents lower risk. Let Nature put probability \(q\) on \(\theta_0\). The expected risk of each pure action against this mixture is $$ G((1,0,0),q)=4(1-q),\qquad G((0,1,0),q)=3,\qquad G((0,0,1),q)=4q. $$ At \(q=1/2\), these three numbers are \(2,3,2\). Every pure action therefore has expected risk at least 2 against this Nature strategy. A mixture of actions also has expected risk at least 2, because its expected risk is a weighted average of the three pure-action risks.
Step 3: Check both sides of the saddle point. Against any Nature mixture, \(p^*\) has risk 2, since its risks at \(\theta_0\) and \(\theta_1\) are both 2. Against \(q^*=1/2\), every decision-maker strategy has risk at least 2. Therefore neither player can improve its position by changing strategy alone, and the game value is 2.
The equilibrium strategies are \(p^*=(1/2,0,1/2)\) and \(q^*=1/2\). In words, the decision maker chooses \(a_1\) or \(a_3\) with equal probability, and Nature chooses either parameter value with equal probability.
A Saddle-Point Test
The calculation illustrates a general verification principle. It is not enough to find a mixture that seems balanced: one must also check that it controls risk against every Nature strategy and that the proposed Nature strategy makes every decision-maker strategy pay at least the claimed value.
Proof. For any fixed \(q\), the decision maker can choose \(p^*\), so $$ \min_p G(p,q)\leq G(p^*,q)\leq v. $$ Taking the maximum over \(q\) gives \(\max_q\min_pG(p,q)\leq v\). In the other direction, for any fixed \(p\), Nature can choose \(q^*\), so $$ \max_qG(p,q)\geq G(p,q^*)\geq v. $$ Taking the minimum over \(p\) gives \(\min_p\max_qG(p,q)\geq v\). For every pair of strategies, \(\max_q\min_pG(p,q)\leq\min_p\max_qG(p,q)\); combining these inequalities shows both values equal \(v\). The displayed conditions also give the saddle-point inequalities directly. \(\square\)
For the worked game, the first condition holds because \(G(p^*,q)=2\) for every \(q\). The second holds because at \(q^*=1/2\), the pure-action risks are \(2,3,2\), so any mixture has risk at least their minimum, 2. This checks the theorem’s conditions without relying on a visual impression of the table.
Worked Example: Bayes Risk as a Function of the Prior
Worked Example: Bayes Risk as a Function of the Prior
Suppose Nature’s mixture is interpreted as a prior distribution with \(P(\theta_0)=q\). Find the Bayes risk and identify a prior that maximizes it.
There is no informative observation in this experiment, so the prior expected risk of an action is the probability-weighted average of its two losses: $$ r(a_1;q)=0q+4(1-q)=4(1-q),\qquad r(a_2;q)=3q+3(1-q)=3,\qquad r(a_3;q)=4q+0(1-q)=4q. $$
For \(0\leq q\leq 1/2\), \(4q\leq 4(1-q)\), and \(4q\leq 2<3\), so \(a_3\) is a Bayes action. For \(1/2\leq q\leq 1\), \(4(1-q)\leq4q\), and \(4(1-q)\leq2<3\), so \(a_1\) is a Bayes action. At \(q=1/2\), both have risk 2 and \(a_2\) has risk 3. Thus the minimum prior expected risk is $$ r_{\mathrm{Bayes}}(q)=\min\{4q,4(1-q),3\}=4\min\{q,1-q\}. $$ This quantity increases to 2 as \(q\) increases from 0 to \(1/2\), and decreases from 2 to 0 as \(q\) increases from \(1/2\) to 1. Its maximum is 2, attained at \(q=1/2\), exactly the Nature mixture found by the saddle-point calculation.
Worked Example: An Asymmetric Risk Table
Worked Example: An Asymmetric Risk Table
Now change the loss at \((\theta_0,a_1)\) from 0 to 0 as before, the loss at \((\theta_1,a_1)\) to 6, and keep the risks of \(a_2\) at \((3,3)\) and \(a_3\) at \((4,0)\). Find and verify the equilibrium for this modified game.
If the decision maker mixes only \(a_1\) and \(a_3\), with probability \(s\) on \(a_1\), the risks are \(4(1-s)\) at \(\theta_0\) and \(6s\) at \(\theta_1\). Equalizing them gives $$ 4(1-s)=6s,\qquad 4=10s,\qquad s=\frac{2}{5}. $$ The resulting risks are both \(4(3/5)=6(2/5)=12/5=2.4\). Thus \(p^*=(2/5,0,3/5)\) guarantees risk 2.4.
Let Nature put probability \(q\) on \(\theta_0\). The three pure-action expected risks are \(6(1-q)\), \(3\), and \(4q\). Choose \(q=3/5\). These become $$ 6(1-3/5)=12/5,\qquad 3,\qquad 4(3/5)=12/5. $$ Every decision-maker mixture therefore has risk at least \(12/5\) against this Nature strategy. The candidate decision-maker mixture has risk \(12/5\) against either parameter value, hence against every Nature mixture. Both saddle inequalities hold, so the game value is \(12/5\).
What the Verification Does—and Does Not—Say
The saddle-point check is a certificate for the particular strategies and value: one inequality bounds the risk the decision maker can guarantee, and the other bounds what Nature can force. In these finite examples, checking the inequalities against pure strategies is enough. Every mixed strategy is a weighted average of pure strategies, so an inequality that holds for each pure strategy also holds for all their mixtures.
A common mistake is to equalize the decision maker’s risks across the parameter values and immediately declare the result optimal. Equalizing can identify a useful candidate, but by itself it does not establish that Nature has a strategy under which every alternative decision rule incurs at least that risk. In the first game, the missing check is that the risks \(2,3,2\) against \(q=1/2\) are all at least 2. The correction is to verify both saddle inequalities, not just the equal-risk calculation.
Another useful check is that probabilities are valid and the reported value agrees from both directions. For the first game, the decision mixture has nonnegative probabilities summing to one, its risks are both 2, and Nature’s mixture makes the three pure-action risks \(2,3,2\). These calculations independently check the mixture, the value, and the claimed equilibrium.
Check Your Understanding
Use the risk table and the saddle-point calculations to answer the following questions.
- For the first game, calculate the risks at \(\theta_0\) and \(\theta_1\) if the decision maker uses \(p=(1/4,1/2,1/4)\).
- Against Nature’s mixture \(q=1/2\) in the first game, why is the expected risk of every mixed decision-maker strategy at least 2?
- For the first game, calculate the expected risk of \(a_1,a_2,a_3\) when Nature uses \(q=1/4\).
- In the asymmetric example, verify that the decision-maker probabilities \(2/5\) and \(3/5\) sum to one and that both parameter-specific risks equal \(12/5\).
- Explain why balancing the risks of a candidate mixture is not, on its own, a complete saddle-point verification.