Solution

The set

E = Δ_{2} \times Δ_{2}

is a compact convex square. Define

u_{i} (p, q) = max {0, A (e_{i}, q) - A (p, q)}, p^{'} = \frac{p + u ( p , q )}{1 + u _{1} ( p , q ) + u _{2} ( p , q )} .

Similarly, set

v_{j} (p, q) = max {0, B (p, e_{j}) - B (p, q)}, q^{'} = \frac{q + v ( p , q )}{1 + v _{1} ( p , q ) + v _{2} ( p , q )} .

These formulae define a continuous self-map

(p, q) \mapsto (p^{'}, q^{'})

E

. The square is homeomorphic to a closed disc, so the Brouwer fixed-point theorem gives a fixed point

(p^{*}, q^{*})

Let

U = u_{1} + u_{2}

at that fixed point. From

p^{'} = p

u_{i} = p_{i} U .

U > 0

, every

i

with

p_{i} > 0

has

A (e_{i}, q) - A (p, q) > 0

. This contradicts bilinearity, because

\sum_{i} p_{i} (A (e_{i}, q) - A (p, q)) = 0.

Hence

U = 0

, all

u_{i} = 0

, and

A (p^{*}, q^{*}) \geq A (e_{i}, q^{*})

for both pure strategies. By linearity this holds for every mixed strategy

p

. The same argument with

v_{j}

gives

B (p^{*}, q^{*}) \geq B (p^{*}, q)

for every

q

. Therefore

\forall (p, q) \in E, A (p^{*}, q^{*}) \geq A (p, q^{*}), B (p^{*}, q^{*}) \geq B (p^{*}, q) .

In game terms,

p

and

q

are the players' mixed strategies, and

A, B

are their expected payoffs in a two-by-two bimatrix game. The inequalities say that neither player can improve unilaterally, so

(p^{*}, q^{*})

Solved by gpt-5.6-sol high.

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