Separable games A polynomial game Consider a zero-sum 2-player game between players
X and
Y, with C_X = C_Y = \left [0,1 \right ] . Denote elements of C_X\, and C_Y\, as x\, and y\, respectively. Define the utility functions H(x,y) = u_x(x,y) = -u_y(x,y)\, where :H(x,y)=(x-y)^2\, . The pure strategy best response relations are: :b_X(y) = \begin{cases} 1, & \mbox{if }y \in \left [0,1/2 \right ) \\ 0\text{ or }1, & \mbox{if }y = 1/2 \\ 0, & \mbox{if } y \in \left (1/2,1 \right ] \end{cases} :b_Y(x) = x\, b_X(y)\, and b_Y(x)\, do not intersect, so there is no pure strategy Nash equilibrium. However, there should be a mixed strategy equilibrium. To find it, express the expected value, v = \mathbb{E} [H(x,y)] as a
linear combination of the first and second
moments of the probability distributions of
X and
Y: : v = \mu_{X2} - 2\mu_{X1} \mu_{Y1} + \mu_{Y2}\, (where \mu_{XN} = \mathbb{E} [x^N] and similarly for
Y). The constraints on \mu_{X1}\, and \mu_{X2} (with similar constraints for
y,) are given by
Hausdorff as: : \begin{align} \mu_{X1} \ge \mu_{X2} \\ \mu_{X1}^2 \le \mu_{X2} \end{align} \qquad \begin{align} \mu_{Y1} \ge \mu_{Y2} \\ \mu_{Y1}^2 \le \mu_{Y2} \end{align} Each pair of constraints defines a compact convex subset in the plane. Since v\, is linear, any extrema with respect to a player's first two moments will lie on the boundary of this subset. Player i's equilibrium strategy will lie on : \mu_{i1} = \mu_{i2} \text{ or } \mu_{i1}^2 = \mu_{i2} Note that the first equation only permits mixtures of 0 and 1 whereas the second equation only permits pure strategies. Moreover, if the best response at a certain point to player i lies on \mu_{i1} = \mu_{i2}\, , it will lie on the whole line, so that both 0 and 1 are a best response. b_Y(\mu_{X1},\mu_{X2})\, simply gives the pure strategy y = \mu_{X1}\, , so b_Y\, will never give both 0 and 1. However b_x\, gives both 0 and 1 when y = 1/2. A Nash equilibrium exists when: : (\mu_{X1}*, \mu_{X2}*, \mu_{Y1}*, \mu_{Y2}*) = (1/2, 1/2, 1/2, 1/4)\, This determines one unique equilibrium where Player X plays a random mixture of 0 for 1/2 of the time and 1 the other 1/2 of the time. Player Y plays the pure strategy of 1/2. The value of the game is 1/4.
Non-Separable Games A rational payoff function Consider a zero-sum 2-player game between players
X and
Y, with C_X = C_Y = \left [0,1 \right ] . Denote elements of C_X\, and C_Y\, as x\, and y\, respectively. Define the utility functions H(x,y) = u_x(x,y) = -u_y(x,y)\, where :H(x,y)=\frac{(1+x)(1+y)(1-xy)}{(1+xy)^2}. This game has no pure strategy Nash equilibrium. It can be shown that a unique mixed strategy Nash equilibrium exists with the following pair of
cumulative distribution functions: : F^*(x) = \frac{4}{\pi} \arctan{\sqrt{x}} \qquad G^*(y) = \frac{4}{\pi} \arctan{\sqrt{y}}. Or, equivalently, the following pair of
probability density functions: : f^*(x) = \frac{2}{\pi \sqrt{x} (1+x)} \qquad g^*(y) = \frac{2}{\pi \sqrt{y} (1+y)}. The value of the game is 4/\pi.
Requiring a Cantor distribution Consider a zero-sum 2-player game between players
X and
Y, with C_X = C_Y = \left [0,1 \right ] . Denote elements of C_X\, and C_Y\, as x\, and y\, respectively. Define the utility functions H(x,y) = u_x(x,y) = -u_y(x,y)\, where :H(x,y)=\sum_{n=0}^\infty \frac{1}{2^n}\left(2x^n-\left (\left(1-\frac{x}{3} \right )^n-\left (\frac{x}{3}\right)^n \right ) \right ) \left(2y^n - \left (\left(1-\frac{y}{3} \right )^n-\left (\frac{y}{3}\right)^n \right ) \right ). This game has a unique mixed strategy equilibrium where each player plays a mixed strategy with the
Cantor singular function as the
cumulative distribution function. ==Further reading==