Equal Values Do Not Guarantee Optimal Strategies
A minimax value is a number describing the best guaranteed payoff a player can approach. It need not be a payoff that any strategy actually achieves. This distinction matters when strategy sets are noncompact: a sequence of strategies can improve indefinitely while escaping every point in the set. The two minimax values may still agree, even though neither player has an optimal strategy and no saddle point exists.
Consider a zero-sum game with payoff \(f(x,y)\), where the maximizing player chooses \(x\in X\) and the minimizing player chooses \(y\in Y\). The payoff is received by the maximizing player and paid by the minimizing player. The lower and upper minimax values are, respectively, the best guaranteed payoff the maximizing player can secure and the largest payoff the minimizing player must allow.
The inequality \(v_-\leq v_+\) follows because \(f(x,y)\leq\sup_{z\in X}f(z,y)\) for every \(x,y\). Taking the infimum over \(y\), then the supremum over \(x\), gives the inequality. Equality of these numbers says nothing by itself about whether either outer supremum or infimum is achieved.
A Continuous Game on a Closed Noncompact Set
Worked Example: A Rational Payoff with No Optimal Strategies
Let \(X=Y=[0,\infty)\), a closed, convex, noncompact subset of the real line, and define \(g(t)=t/(1+t)\) and \(f(x,y)=g(x)-g(y)\). The function \(g\) is increasing, \(g(0)=0\), and \(g(t)<1\) for every finite \(t\), while \(g(t)\to1\) as \(t\to\infty\).
For fixed \(x\), the infimum over \(y\) is \(\inf_{y\geq0}\{g(x)-g(y)\}=g(x)-1\). This is an infimum, not a minimum: it would require \(g(y)=1\), which no finite \(y\) satisfies. Therefore \(v_-=\sup_{x\geq0}\{g(x)-1\}=0\), since \(g(x)-1\to0\) as \(x\to\infty\) and is strictly negative for each finite \(x\).
For fixed \(y\), the supremum over \(x\) is \(\sup_{x\geq0}\{g(x)-g(y)\}=1-g(y)\), also not attained at any finite \(x\). Hence \(v_+=\inf_{y\geq0}\{1-g(y)\}=0\): these quantities are positive for finite \(y\) and tend to zero as \(y\to\infty\). Thus \(v_-=v_+=0\), but neither outer extremum is attained.
In particular, for every fixed \(x\), its guaranteed payoff is \(g(x)-1<0=v_-\). No maximizing strategy is optimal. For every fixed \(y\), the payoff the minimizing player must allow is \(1-g(y)>0=v_+\), so no minimizing strategy is optimal either. A saddle point would require both players to be optimal, so there is no saddle point.
The payoff is continuous and has the usual concavity and convexity properties: \(g'(t)=1/(1+t)^2>0\) and \(g''(t)=-2/(1+t)^3<0\). Thus \(f\) is concave in the maximizing player's choice \(x\) and convex in the minimizing player's choice \(y\). The missing ingredient for a general attainment guarantee is compactness: continuous functions on noncompact sets need not attain their suprema or infima.
A General Separable-Payoff Result
The example's calculation follows a general pattern. Suppose the payoff separates into one term controlled by each player. The theorem below gives exact conditions under which the values agree and explains precisely when strategies attain them.
Proof. Boundedness ensures \(A\) and \(B\) are finite. For each fixed \(x\), subtracting \(b(y)\) from the constant \(a(x)\) gives \(\inf_{y\in Y}f(x,y)=a(x)-B\). Taking the supremum over \(x\) yields \(v_-=A-B\). Likewise, for each fixed \(y\), \(\sup_{x\in X}f(x,y)=A-b(y)\). Taking the infimum over \(y\) yields \(v_+=A-B\). This proves equality.
For fixed \(x^*\), its guaranteed payoff is \(a(x^*)-B\), which equals \(v_-=A-B\) exactly when \(a(x^*)=A\). For fixed \(y^*\), its upper guarantee is \(A-b(y^*)\), which equals \(v_+=A-B\) exactly when \(b(y^*)=B\). Thus both strategies are optimal precisely when both suprema are attained. If they are attained, at a pair of maximizers, \(a(x)-b(y^*)\leq a(x^*)-b(y^*)\leq a(x^*)-b(y)\), since \(a(x)\leq A=a(x^*)\) and \(b(y)\leq B=b(y^*)\). These are the saddle inequalities. Conversely, a saddle point gives optimal strategies for both players, so both suprema must be attained. \(\square\)
Worked Example: Linear Payoff on Open Strategy Sets
Take \(X=Y=(0,1)\) and \(f(x,y)=x-y\). These strategy sets are noncompact because they are not closed subsets of the real line. For every fixed \(x\), \(\inf_{0<y<1}(x-y)=x-1\): values approach \(x-1\) as \(y\) approaches \(1\), but no allowed \(y\) equals \(1\). Therefore \(v_-=\sup_{0<x<1}(x-1)=0\), approached as \(x\) approaches \(1\) but not attained.
For fixed \(y\), \(\sup_{0<x<1}(x-y)=1-y\), with the supremum approached but not attained. Consequently \(v_+=\inf_{0<y<1}(1-y)=0\), approached as \(y\) approaches \(1\) but not attained. The values agree, but every \(x\) has guaranteed payoff \(x-1<0\), and every \(y\) allows an upper payoff \(1-y>0\). Neither player has an optimal strategy, and there is no saddle point.
This example isolates the role of the missing endpoints: the relevant limits are finite, but the strategies that would realize them are excluded. In the earlier example the strategy sets were closed, but the limiting strategies escaped to infinity. Noncompactness can therefore obstruct attainment in different ways.
Why the Saddle Inequalities Require Attainment
For any saddle point \((x^*,y^*)\), the inequalities in its definition imply \(\inf_y f(x^*,y)=f(x^*,y^*)=\sup_x f(x,y^*)\). In particular, the maximizing player attains the lower value and the minimizing player attains the upper value; moreover, the two values are equal. The converse does not follow from equality alone: equality supplies a common number, not strategies that realize it.
This distinction is useful when applying minimax theorems. A theorem that establishes an interchange of \(\sup\) and \(\inf\) may identify a common value without asserting that either extremum is achieved. Compactness and semicontinuity conditions can provide attainment in suitable settings; without the relevant hypotheses, one must check attainment separately.
Randomization Does Not Necessarily Restore Attainment
Worked Example: Mixed Strategies Still Escape to Infinity
Return to \(X=Y=[0,\infty)\) and \(f(x,y)=g(x)-g(y)\), where \(g(t)=t/(1+t)\). Allow each player to use a probability distribution on \([0,\infty)\), with the players' randomizations independent. Since \(0\leq g(t)<1\), all expected payoffs exist and the payoff between distributions \(P\) and \(Q\) is \(E_P[g(X)]-E_Q[g(Y)]\).
For any distribution \(P\), \(E_P[g(X)]<1\). Indeed, \(1-g(X)=1/(1+X)>0\) with probability one, so its expectation is strictly positive. To see the strictness, if a nonnegative random variable \(Z\) is positive with probability one, then some event \(\{Z\geq1/k\}\) has positive probability (the union of these events over positive integers \(k\) has probability one); hence \(E[Z]>0\).
The maximizing player's guaranteed payoff under \(P\) is \(\inf_Q\{E_P[g(X)]-E_Q[g(Y)]\}=E_P[g(X)]-1<0\). The infimum \(1\) of the opponent's expected \(g(Y)\) is approached by choosing \(Y=N\) deterministically and letting \(N\to\infty\). Taking the supremum over \(P\) gives lower value \(0\), approached by \(X=N\) deterministically, but no distribution \(P\) attains it.
Similarly, for every \(Q\), the payoff the minimizing player must allow is \(\sup_P\{E_P[g(X)]-E_Q[g(Y)]\}=1-E_Q[g(Y)]>0\). Its infimum over \(Q\) is \(0\), approached by taking \(Y=N\) deterministically, but is not attained by any probability distribution on finite values. Thus even allowing randomization, the two values agree at zero and neither side has an optimal strategy.
Common Pitfall and Practical Check
Wrong reasoning: “The lower and upper minimax values are equal, so the players must have a saddle point.” This confuses equality of two real numbers with attainment of the extrema used to define them. In the rational-payoff example, both values are zero, yet every finite strategy of the maximizing player guarantees a strictly negative payoff, and every finite strategy of the minimizing player allows a strictly positive one.
Correction: After establishing \(v_-=v_+\), separately test whether a strategy achieves each player's defining supremum or infimum. A sequence whose payoffs converge to the value establishes approximation, not attainment. In noncompact games, explicitly identify whether the limiting strategy belongs to the strategy set.
For a fixed strategy of one player, evaluate the opponent's infimum or supremum, keeping track of whether it is attained.
Take the remaining supremum or infimum and determine whether an admissible strategy realizes it.
Verify the saddle inequalities, or establish that both players' outer extrema are attained. Value equality alone is not enough.
Check Your Understanding
For each question, distinguish the value of an extremum from whether it is attained.
- In the game \(f(x,y)=x/(1+x)-y/(1+y)\) on \([0,\infty)^2\), what sequence of strategies approaches the common minimax value?
- Why does a finite \(x\) fail to be an optimal maximizing strategy in that game?
- For \(f(x,y)=a(x)-b(y)\), state the conditions for a saddle point in terms of \(a\) and \(b\).
- In the game \(f(x,y)=x-y\) on \((0,1)^2\), which excluded endpoint is approached in each outer optimization?
- Why does equality \(v_-=v_+\) alone not imply that the common value is attained?