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ゲーム理論BASIC 演習51 -完全ベイジアン均衡-
- 3. શϕΠδΞϯߧۉ
શϕΠδΞϯߧۉઓུͷ ͱ৴೦ͷମܥ ͷ Ͱ
͕ ͷͱͰஞ࣍߹ཧతͰ͋Γ
͕ ʹ߹తͰ͋ΔͷΛ ऑ
શϕΠδΞϯ͏͍ͱߧۉ
ߦಈઓུͷஞ࣍߹ཧੑల։ܗήʔϜ ʹ͓͍ͯ
৴೦ͷମܥ ͕༩͑ΒΕ͍ͯΔͱ͢Δ
͜ͷͱ͖
ઓུͷ ͕
ϓϨΠϠʔ ͷใू߹ ʹ͓͍ͯҎԼΛຬͨ͢ͱ͖
৴೦ͷମܥ ͷͱͰใू߹ ʹ͓͍ͯஞ࣍߹ཧతͰ͋Δͱ͍͏
৴೦ͷମͱܥ߹ੑల։ܗήʔϜ ʹ͓͍ͯ
ઓུͷ ͕༩͑ΒΕ͍ͯΔͱ͢Δ
͜ͷͱ͖
৴೦ͷମܥ ͕
ҙͷϓϨΠϠʔ ͷҙͷใू߹ ʹ͓͍ͯ
Ͱ͋Ε
ͯ͢ͷ ͷ ʹ͍ͭͯ
ͱͳΔͱ͖
ઓུͷ ʹ͓͍ͯ߹తͰ͋Δͱ͍͏
b μ (b, μ)
b μ μ b
Γ μ
b = (b1
, ⋯, bn
) i ui
l
b μ ui
l
Hi
(ui
l, μ, (bi,ui
l, b−i,ui
l)) ≥ Hi
(ui
l, μ, (b′

i,ui
l, b−i,ui
l)), ∀b′

i,ui
l ∈ Bi,ui
l
Γ b = (b1
, ⋯, bn
)
μ i ui
l
Prob(ui
l |b) 0 ui
l x*
μ(x*) =
Prob(x*|b)
∑x∈ui
l
Prob(x|b)
μ b
ߦಈઓུʹΑͬͯ౸ୡՄೳͳ
ܦ࿏ʹ͓͚Δ߹ཧੑͷΈߟ͑Δ
- 5. ҎԼͷήʔϜʹ͓͚Δ
ߦಈઓུͰͷશϕΠδΞϯߧۉΛٻΊΑ
ղ
1
2
2
a
b
d
e
ϓϨΠϠʔͷརಘ ϓϨΠϠʔͷརಘ
ใू߹
(2, 1)
(1, − 1)
(3, − 2)
(−1, − 1)
(0, 2)
c
d
e
s
1 − s
p
q
1 − p − q
r
1 − r
ϓϨΠϠʔͷߦಈઓུΛ Ͱද͠
ϓϨΠϠʔͷߦಈઓུΛ Ͱද͢
·ͨ
৴೦ͷମܥ
ͳ͓
Λຬͨ͢
(p, q, 1 − p − q)
(r, 1 − r)
μ = (1, (s, 1 − s))
p, q ≥ 0, 0 ≤ p + q ≤ 1 0 ≤ r ≤ 1 0 ≤ s ≤ 1
1
- 11. ҎԼͷήʔϜʹ͓͚Δ
ߦಈઓུͰͷશϕΠδΞϯߧۉΛٻΊΑ
ղ
1
2
2
a
b
d
e
ϓϨΠϠʔͷརಘ ϓϨΠϠʔͷརಘ
ใू߹
(2, 1)
(1, − 1)
(3, − 2)
(−1, − 1)
(0, 2)
c
d
e
s
1 − s
p
q
1 − p − q
r
1 − r
ϓϨΠϠʔͷஞ࣍߹ཧతͳߦಈΛߟ͑Δ
ͷͱ͖
ҙͷͰ࠷ద
͜ͷ߹ͷΈ͕
͠ϓϨΠϠʔͷใू߹ʹ౸ୡ͢ΔͷͰ͋Ε
શϕΠδΞϯߧۉͷ݅ͱͳΔ
·ͨ
͍͓ͯʹߧۉ Ͱ͋Ε
ϓϨΠϠʔͷ
ʮBΛબͿʯ߹ͱʮCΛબͿʯ߹ͷظརಘҰக͢Δ
ʮBΛબͿʯ߹͕ߴ͍ظརಘΛ༩͑ΔͷͰ͋Ε
ʮCΛબͿʯʹׂΓ͍ͯͯͨ֬Λ
ʮBΛબͿʯʹׂΓͯΕظརಘΛߴ͘͢Δ͜ͱ͕Ͱ͖Δ
ͳͷͰ
ҎԼࣜΛຬͨ͢
3s − 1 = 0 ⇔ s =
1
3
r
p, q 0
2r + (1 − r) = 3r + (−1)(1 − r) ⇔ r =
2
3
1
- 14. ҎԼͷήʔϜʹ͓͚Δ
ߦಈઓུͰͷશϕΠδΞϯߧۉΛٻΊΑ
ղ
1
2
2
a
b
d
e
ϓϨΠϠʔͷརಘ ϓϨΠϠʔͷརಘ
ใू߹
(2, 1)
(1, − 1)
(3, − 2)
(−1, − 1)
(0, 2)
c
d
e
1
3
p =
1
3
0
r =
2
3
1
3
ϓϨΠϠʔͷஞ࣍߹ཧతͳߦಈΛߟ͑Δ
ͷͱ͖
ҙͷͰ࠷ద ࠷ద
ͷ݅ͱ৴೦ͷ߹ੑΑΓ
Ҏ্ΑΓ
શϕΠδΞϯߧۉ
ϓϨΠϠʔͷߦಈઓུ
ϓϨΠϠʔͷߦಈઓུ
৴೦ͷମܥ
3s − 1 = 0 ⇔ s =
1
3
r r =
2
3
p + q = 1
p
p + q
=
1
3
⇔ p =
1
3
, q =
2
3
(p, q, 1 − p − q) =
(
1
3
,
2
3
, 0
)
(r, 1 − r) =
(
2
3
,
1
3 )
μ = (1, (s, 1 − s)) =
(
1,
(
1
3
,
2
3))
1
2
3
q =
2
3