Teorie her pro FJFI ČVUT řešené úlohy

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Dalibor Liška
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1 Tyto úlohy volně doplňují přednášky z kursu teorie her. Rozsah látky a použité značení odpovídá slajdům dostupným na stránce věnované výuce. Γ S S Γ 3 o = o = o 3 = vítězná o o Γ u u(o ) = u(o ) = u(o 3 ) = 0 (s, s ) S S v = u (s, s ) := u(o i ) {, 0, } o i s, s s s v = u (s, s ) = s v = u (s, s ) = s v = u (s, s ) = 0 3 j {,, 3} j {,, 3} 4 j zpětné indukce N = {,, 3}

2 3 a b a 3 c 3 b 3 a b c d e f g h c d e f (3,, 4) (0, 0, 5) (0, 0, 5) (0,, 3) (, 5, 0) (5, 4, 0) (0,, 5) (0, 0, 3) (, 3, 4) (,, 3) c d 3 a b a 3 c 3 b 3 a b c d e f g h c d e f (3,, 4) (0, 0, 5) (0, 0, 5) (0,, 3) (, 5, 0) (5, 4, 0) (0,, 5) (0, 0, 3) (, 3, 4) (,, 3) 3 3 a b a 3 f c 3 a b (3,, 4) (0,, 3) (, 5, 0) (0,, 5) (, 3, 4)

3 a b (3,, 4) (, 3, 4) 3 a b a 3 c 3 b 3 a b c d e f g h c d e f (3,, 4) (0, 0, 5) (0, 0, 5) (0,, 3) (, 5, 0) (5, 4, 0) (0,, 5) (0, 0, 3) (, 3, 4) (,, 3) i N s i V i s i (x) A(v i ) A(v i ) i x V i (a, c, f ) G = (N, (S i ) i N, (u i ) i N ) n = N β i : S i R i N β i (s i ) = max t i S i u(t i, s i ), s i S i, s i S i i N nejlepší odezvou na (n )-tici strategií s i S i u i (s i, s i ) = β i (s i ). n (s i ) i N S S n s i s i i N u i (s i, s i) u i (t i, s i) t i S i u i (s i, s i) = β i (s i)

4 Test rovnovážného řešení pomocí čistých strategií. G = (N, (S i ) i N, (u i ) i N ) G = (N, ( i ) i N, (U i ) i N ) p := n n i U i (p ) U i (p i, p i), p i i p G i S i = {s i,..., s m i i s j i S i δ s j i δ s j(s k i ) = i { k = j, 0 k j, } m i N k =,..., m i. m i i p i i m i p i = p i (s j i )δ s j. i U i ( mi ) U i (p i, p i) = U i p i (s j i )δ m i ( s j, p i = p i (s j i )U i i j= j= j= δ s j i ), p i U i (p ), p Partnerský spor. K F u (s, s ) u (s, s ) K F K, 0, 0 F 0, 0,

5 (K, K) (F, F ) x := p (K) y := p (K) p (K) K p (K) K p (F ) = p (K) = x p (F ) = p (K) = y [0, ] ({, }, ([0, ], [0, ]), (U, U )), U (x, y) = xy + ( x)( y) U (x, y) = xy + ( x)( y) x, y [0, ] P([0, ]) [0, ] BR, BR : [0, ] P([0, ]) { BR (y) := x [0, ] U (x, y) = max BR (x) := x [0,] { y [0, ] U (x, y) = max U (x, y ) y [0,] } U (x, y), y [0, ], }, x [0, ]. 0 0 y <, x <, 3 BR (y) = [0, ] y = 3, BR (x) = [0, ] x = < y,, 3 < x. 3 3 BR BR 0 y 0 x (x, y ) [0, ] x BR (y ) y BR (x ).

6 BR BR x 0 y (x, y ) [0, ] BR BR {(0, 0), (, ), (, )} 3 3 (F, F ) (K, K) (p, p ) p (K) = 3 p (K) = 3 Matching Pennies. 0 0 u (s, s ) u (s, s ) = u (s, s ) (s, s ) {0, } x := p (0) y := p (0) [0, ] G = ({, }, ([0, ], [0, ]), (U, U )), U (x, y) = 4xy x y + U (x, y) = U (x, y) x, y [0, ] x 0 x <, Λ(x) := min U (x, y) = 0 x = y [0,], x < x, y 0 y <, Λ(y) := max U (x, y) = 0 y = y [0,], y < y.

7 max Λ(x) = min Λ(y) = 0, x [0,] y [0,] G 0 (x, y ) = (, ) S = S = [0, ] u (s, s ) = + (s s ), (s, s ) [0, ]. u [0, ] dolní horní cenu Λ(s ) = min s [0,] u (s, s ), Λ(s ) = max u (s, s ). s [0,] v = max Λ(s ), s [0,] v = min s [0,] Λ(s ). { Λ(s ) = +(s 0 s ), < s +s, Λ(s ) =, s [0, ]. v = Λ( ) = 4 5 < v = Cournotův model duopolu. i =, q i 0 i c > 0 P (q, q ) = a b(q + q ) a, b > 0 a c

8 i f i : [0, ) R f i (q, q ) = P (q, q )q i cq i = (a c)q i bq i bq q. S = S = [0, ) (q, q ) [0, ) f (q, q ) f (q, q ) f (q, q ) f (q, q ) q, q [0, ) (q, q) [0, ) f (., q) f (q,.) [0, ) f q = a c bq bq, f q = a c bq bq. a c bq bq = 0 a c bq bq = 0 q = q = a c 3b f q = f q = b < 0, f (., q) f (q,.) (q, q) = ( a c, a c) 3b 3b ( a b A = c d ), a, b, c, d R. x = (x, x) T y = (x, y) T x, y [0, ] U : [0, ] R U(x, y) = x T Ay = (a + d b c)xy + (b d)x + (c d)y + d, (x, y ) U x (x, y ) = U y (x, y ) = 0.

9 (a + d b c)y + b d = (a + d b c)x + c d = 0 a + d b c 0 x = (x, x ) T = a+d b c (d c, a b)t, y = (y, y ) T = a+d b c (d b, a c)t. a + d b c = 0 U U(x, y) = (b d)x + (c d)y + d. b d c d a = b + c d a c a b A ( ) 0 4 A =. 5 x = (, 0) T y = (0, ) T b d c < d b < d c d b < d c < d 0 A =. 0 x 0 x = (x, x, x 3 ) T x 0 A T x x 0 0, 3 x i =, i= x 0.

10 y 0 y = (y, y ) T y 0 Ay y 0 0, y i =, i= y 0. x 0 x + x 3 x 0 0, x + x x 0 0, x + x + x 3 =, x, x, x 3 0, y 0 y y 0 0, y + y y 0 0, y y 0 0, y + y =, y, y 0. x = (0, 3, 3 )T, y = ( 3, 3 )T, x 0 = y 0 = 3 p U v p min p U(p, p ) = v. p U(p, p ) v

11 p v U(p, p ) = v {, } x x x C x 4 T C T x 3 C T C x 5 T x 6 x 7 x 8 x 9 x 0 b p x 3 x 8 ρ(x j, (b, b )) b b W = {x } W = {x 4, x 5 } b (α, β) [0, ] b (C, W ) = α b (C, W ) = β W W x 0 p s : {W, W } {C, T, C, T } s (W ) {C, T } s (W ) {C, T } p (s ) = b (s (W ), W ) b (s (W ), W ). ρ (x 8, b ) x 8 b ρ (x 8, b ) = b (T, W ) b (T, W ) = ( α) ( β). s S(x 8 ) x 0 x 8 s (W ) = T s (W ) = T ρ (x 8, p ) = p (s ) = b (T, W ) b (T, W ) = ( α) ( β).

12 ρ (x 3, b ) ρ (x 3, p ) ρ (x 3, b ) = b (C, W ) = α. x 3 s (W ) = C s (W ) = C s (W ) = C s (W ) = T S (x 3 ) = {s, s } ρ (x 3, p ) = p (s ) + p (s ) = = b (C, W ) b (C, W ) + b (C, W ) b (T, W ) = αβ + α( β) = α. ρ(x j, (b, b )) x j b (C, W ) = b (T, W ) = ρ(x 3, (b, b )) = 5 ρ (x 3, b )ρ (x 3, b ) = 5 α, ρ(x 7, (b, b )) = ρ 5 (x 7, b )ρ (x 7, b ) = ( α)β, 5 ρ(x 8, (b, b )) = ρ 5 (x 8, b )ρ (x 8, b ) = ( α)( β), 5 ρ(x 9, (b, b )) = 3ρ 5 (x 9, b )ρ (x 9, b ) = 3 5 ρ(x 0, (b, b )) = 3 5 ρ (x 0, b )ρ (x 0, b ) = 3 5 ρ(x 6, (b, b )) = 3 5 ρ (x 6, b )ρ (x 6, b ) = 3 5. β, ( β), S P S P K N 3 (u, u, u 3 ) W A = {x } W B = {x 3, x 4 } W C = {x }

13 0 A x x C S P P S (,, ) K x 3 N B K x 4 N (,, ) (0, 0, 0) (,, ) (0, 0, 0) (,, ) W B = S A = {S, P }, S B = {K, N}, S C = {P, S }. S A S B S C 3 K S P S (0, 0, 0) (,, ) 4 4 P (,, ) (0, 0, 0) 4 4 N S P S (0, 0, 0) (,, ) 4 4 P (,, ) (0, 0, 0) 4 4 (S, K, S ) (P, N, P ) Γ

14 x 0 A B x L R (6, 0) l r W l r (8, 0) (0, 8) (0, 8) (8, 0) Γ(x 0 ) = Γ Γ(x ) x Γ(x ) x. krok. Γ(x ) i b i β := b (l, W ) γ := b (L, x ) L R l (8, 0) (0, 8) r (0, 8) (8, 0) U (β, γ) = 6βγ 8β 8γ + 8 U (β, γ) = 6βγ + 8β + 8γ. BR BR 0 0 γ <, 0 β <, BR (γ) = [0, ] γ =, BR (β) = [0, ] β = < γ,, 0 < β. (β, γ ) [0, ] β BR (γ ) γ BR (β ).

15 (β, γ ) = (, ) Γ(x ) b (l, W ) =, b (L, x ) =. ( U (, ), U (, )) = (4, 4). krok A x 0 B (4, 4) (6, 0) B b (B, x 0 ) = Závěr: b b v N = {,, 3} { 0 A =, v(a) = A A. A B = A + B A B v v(a B)+v(A B) v(a)+v(b) C(v) x v,π π N π() = π() = 3 π(3) = x = v({}) v( ) = 0, x 3 = v({, 3}) v({}) =, x = v({,, 3}) v({, 3}) =. {(0,, ), (, 0, ), (,, 0)} x + x + x 3 =

16 v : N R N = {,, 3} 0 A =, A = {}, {}, v(a) = A = {3}, 4 A =, 5 A = N. v v v(a B) v(a)+v(b) A, B N A B = v(n) < v({, } + v({3}) v C(v) x C(v) x + x + x 3 = 5 x + x 4 x 3 5 = x + x + x 3 6 Jednoduchá hra v : N {0, } i N vetující v A N v(a\{i}) = 0 i v(n \ {i}) = 0 v W N C(v) = {x R n x(w ) =, x i 0 i W x j = 0 j N \ W }. v(n \ {i}) = 0 v(a \ {i}) = 0 A k N x R n { i = k, x i = 0 i k. v v(n) = = i N x i = x(n) A N k A x(a) = v(a) k / A x(a) = 0 = v(a) k x C(v) v v x C(v) x(n) = i N

17 x i > 0 x(n \ {i}) = x i < i v(n \ {i}) = > x(n \ {i}) x C(v) A N v(a) = A W x x(n) = x(w ) = A N v(a) = 0 x(a) 0 v(a) = A W x(a) x(w ) = = v(a), x C(v) x C(v) x i 0 i N x(n) = x i = 0 i N \ W i N \ W i x(n) = x(n \ {i}) x i = 0 = x(n) x(n \ {i}) v(n \ {i}) =, 3 50 % 5 % v N = {,, 3} { A = N, {, }, {, 3}, v(a) = A N. 0 i i φ S (v) = 3, φs (v) = φ S 3 (v) = 6. φ B (v) 3 3 φ B (v) = 3 5, φb (v) = φ B (v) = 5.

18 v N = {,..., n} C(v) = conv {x v,π π Π}, Π N φ S (v) φ S (v) = π Π c π x v,pi c π 0 π Π c π = i N φ S i (v) = π Π n! (v(aπ π (i) ) v(aπ π (i) )). x v,π x v,π i = v(a π π (i) ) v(aπ π (i) ), c π = n! π Π φs (v) C(v) C(v) ψ : Γ R n Γ ψ i (v) = v({,..., i}) v({,..., i }), i N. ψ i N ψ i (v) = i N v, w Γ (v({,..., i}) v({,..., i })) = v(n) v( ) = v(n). ψ i (v + w) = (v + w)({,..., i}) (v + w)({,..., i }) = (v({,..., i}) v({,..., i })) + (w({,..., i}) w({,..., i })) = ψ i (v) + ψ i (w). i N v(a {i}) = v(a) A N A = {,..., i } ψ i (v) = 0 ψ N = {,, 3} { A = {, 3}, N v(a) = A N. 0 ψ(v) = (0, 0, ) 3 v({, }) = v({, 3})

19 Reference An Introductory Course on Mathematical Game Theory Graduate Studies in Mathematics Game Theory Game theory

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