ゲーム理論BASIC 演習66 -ダブルオークション：ベイジアンナッシュ均衡-
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1. The document presents a model of strategic interaction between two players, S and B. 2. The players choose probabilities pS and pB simultaneously to maximize their expected utilities uS and uB. 3. The optimal probabilities pS*(vS) and pB*(vB) are found to be functions of the underlying values vS and vB, where pS*(vS) = 1/3(αB + βB) + 2/3vS and pB*(vB) = 1/3αS + 2/3vB.
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ゲーム理論BASIC 演習66 -ダブルオークション：ベイジアンナッシュ均衡-
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- 3. μϒϧΦʔΫγϣϯ ചΓख Λ࣋ͭଆ ͱങ͍ख͕ಉ࣌ʹՁ֨Λఏࣔ͠ ʮങ͍खͷՁ֨ ചΓखͷՁ֨ʯͱͳͬͨ߹ʹചങ͕ߦΘΕΔऔҾ ചΓखചͬͯΑ͍࠷Ձ֨Λఏ͓ࣔͯ͠Γ ങ͍खങͬͯΑ͍࠷ߴՁ֨Λఏ͍ࣔͯ͠Δ ͱղऍ͢Δ Ϟσϧ ചΓख4 ങ͍ख#ͦΕͧΕʹରͯ͠ධՁ Λ͍ͬͯΔͨͩ͠ ͜ͷධՁࢲతใͰ͋Δ ͜ͷධՁ ֬มͰ ʹ͓͚ΔҰ༷ʹै͏ͱԾఆ͢Δ ͜ͷʹै͏͜ͱ ചΓख4 ങ͍ख#ʹ͍ޓ͍ͬͯΔͷͱ͢Δ ചΓख4Ձ֨ ങ͍ख#Ձ֨ Λఏࣔ͠ ͷ߹ʹऔҾ͕ߦΘΕ ͦͷ߹ͷՁ֨྆Ձ֨ͷਅΜதͷ ͱ͢ΔརಘؔҎԼͷΑ͏ʹද͞ΕΔɿ Ձ֨Λઓུͱ͠ ചΓख4 ങ͍ख# ͱͯ͠ ϕΠδΞϯφογϡߧۉΛٻΊΑ ≥ vi (i = S, B) vi (i = S, B) [0, 1] pS ≥ 0 pB ≥ 0 pB ≥ pS uS(pS, pB, vS, vB) = { pB + pS 2 − vS JG pB ≥ pS, 0 PUIFSXJTF uB(pS, pB, vS, vB) = { vB − pB + pS 2 JG pB ≥ pS, 0 PUIFSXJTF pS(vS) = αS + βSvS pB(vB) = αB + βBvB
- 4. ղ Ϟσϧ ചΓख4 ങ͍ख#ͦΕͧΕʹରͯ͠ධՁ Λ͍ͬͯΔͨͩ͠ ͜ͷධՁࢲతใͰ͋Δ ͜ͷධՁ ֬มͰ ʹ͓͚ΔҰ༷ʹै͏ͱԾఆ͢Δ ͜ͷʹै͏͜ͱ ചΓख4 ങ͍ख#ʹ͍ޓ͍ͬͯΔͷͱ͢Δ ചΓख4Ձ֨ ങ͍ख#Ձ֨ Λఏࣔ͠ ͷ߹ʹऔҾ͕ߦΘΕ ͦͷ߹ͷՁ֨྆Ձ֨ͷਅΜதͷ ͱ͢ΔརಘؔҎԼͷΑ͏ʹද͞ΕΔɿ Ձ֨Λઓུͱ͠ ചΓख4 ങ͍ख# ͱͯ͠ ϕΠδΞϯφογϡߧۉΛٻΊΑ ํ ചΓख4ʹ͍ͭͯ ධՁ ͷͱ͖ʹ ͖݅ظརಘΛߟ͑ ظརಘΛ࠷େԽ͢Δ ΛٻΊΔ ങ͍ख#ʹ͍ͭͯ ධՁ ͷͱ͖ʹ ͖݅ظརಘΛߟ͑ ظརಘΛ࠷େԽ͢Δ ΛٻΊΔ ͱ༩͑ΒΕͨઓུͱͷൺֱͳͲͰ ͕ఆ·Δ vi (i = S, B) vi (i = S, B) [0, 1] pS ≥ 0 pB ≥ 0 pB ≥ pS uS(pS, pB, vS, vB) = { pB + pS 2 − vS JG pB ≥ pS, 0 PUIFSXJTF uB(pS, pB, vS, vB) = { vB − pB + pS 2 JG pB ≥ pS, 0 PUIFSXJTF pS(vS) = αS + βSvS pB(vB) = αB + βBvB vS p* S (vS) vB p* B (vB) αS, βS, αB, βB
- 5. ղ Ձ֨Λઓུͱ͠ ചΓख4 ങ͍ख# ͱͯ͠ ϕΠδΞϯφογϡߧۉΛٻΊΑ ചΓख4ʹ͍ͭͯ ධՁ ͷͱ͖ʹ ͖݅ظརಘΛߟ͑ ظརಘΛ࠷େԽ͢Δ ΛٻΊΔ ʹ͓͚ΔҰ༷ʹै͏ͷͰ pS(vS) = αS + βSvS pB(vB) = αB + βBvB vS p* S (vS) EuS = EvB[ pB(vB) + pS 2 − vS pB(vB) ≥ pS] = EvB[ αB + βBvB + pS 2 − vS αB + βBvB ≥ pS] = EvB[ αB + βBvB + pS 2 − vS vB ≥ pS − αB βB ] vB [0, 1] EuS = ∫ 1 pS − αB βB ( αB + βBvB + pS 2 − vS) ⋅ 1 dvB = ∫ 1 pS − αB βB ( αB + pS − 2vS 2 + βBvB 2 ) dvB = [ αB + pS − 2vS 2 vB + βBv2 B 4 ] 1 pS − αB βB = αB + pS − 2vS 2 + βB 4 − αB + pS − 2vS 2 ⋅ pS − αB βB − βB 4 ( pS − αB βB ) 2 0 pS − αB βB 1 1 vB
- 6. ղ ͜ΕΑΓ Ұ֊͔݅Β EuS = ∫ 1 pS − αB βB ( αB + βBvB + pS 2 − vS) ⋅ 1 dvB = ∫ 1 pS − αB βB ( αB + pS − 2vS 2 + βBvB 2 ) dvB = [ αB + pS − 2vS 2 vB + βBv2 B 4 ] 1 pS − αB βB = αB + pS − 2vS 2 + βB 4 − αB + pS − 2vS 2 ⋅ pS − αB βB − βB 4 ( pS − αB βB ) 2 dEuS dpS = 1 2 − 1 2 ⋅ pS − αB βB − αB + pS − 2vS 2 ⋅ 1 βB − βB 2 ( pS − αB βB ) 1 βB = 0 pS(vS) = 1 3 (αB + βB) + 2 3 vS
- 7. ղ Ձ֨Λઓུͱ͠ ചΓख4 ങ͍ख# ͱͯ͠ ϕΠδΞϯφογϡߧۉΛٻΊΑ ങ͍ख#ʹ͍ͭͯ ධՁ ͷͱ͖ʹ ͖݅ظརಘΛߟ͑ ظརಘΛ࠷େԽ͢Δ ΛٻΊΔ ʹ͓͚ΔҰ༷ʹै͏ͷͰ pS(vS) = αS + βSvS pB(vB) = αB + βBvB vB p* B (vB) EuB = EvS[ vB − pB + pS 2 pB ≥ pS(vS) ] = EvB[ vB − pB + αS + βSvS 2 pB ≥ αS + βSvS] = = EvB[ vB − pB + αS + βSvS 2 pB − αS βS ≥ vS] vS [0, 1] EuB = ∫ pB − αS βS 0 ( vB − pB + αS + βSvS 2 ) ⋅ 1 dvS = ∫ pB − αS βS 0 ( 2vB − αS − pB 2 − βSvS 2 ) dvS = [ 2vB − αS − pB 2 vS − βSv2 S 4 ] pB − αS βS 0 = 2vB − αS − pB 2 ⋅ pB − αS βS − βS 4 ( pB − αS βS ) 2 0 pB − αS βS 1 1 vS
- 8. ղ ͜ΕΑΓ Ұ֊͔݅Β EuB = ∫ pB − αS βS 0 ( vB − pB + αS + βSvS 2 ) ⋅ 1 dvS = ∫ pB − αS βS 0 ( 2vB − αS − pB 2 − βSvS 2 ) dvS = [ 2vB − αS − pB 2 vS − βSv2 S 4 ] pB − αS βS 0 = 2vB − αS − pB 2 ⋅ pB − αS βS − βS 4 ( pB − αS βS ) 2 dEuB dpB = − 1 2 ⋅ pB − αS βS + 2vB − αS − pB 2 ⋅ 1 βS − βS 2 ( pB − αS βS ) 1 βS = 0 pB(vB) = 1 3 αS + 2 3 vB
- 9. ղ Ձ֨Λઓུͱ͠ ചΓख4 ങ͍ख# ͱͯ͠ ϕΠδΞϯφογϡߧۉΛٻΊΑ ͱ༩͑ΒΕͨઓུͱͷൺֱͳͲͰ ͕ఆ·Δ ͱ ΑΓ ͱ ΑΓ ͱ ΑΓ Ҏ্ΑΓ ϕΠδΞϯφογϡߧۉ ֤ධՁʹର͠ ҎԼͷઓུΛͱΔͷͰ͋Δ pS(vS) = αS + βSvS pB(vB) = αB + βBvB αS, βS, αB, βB pS(vS) = αS + βSvS pS(vS) = 1 3 (αB + βB) + 2 3 vS αS = 1 3 (αB + βB) βS = 2 3 pB(vB) = αB + βBvB pB(vB) = 1 3 αS + 2 3 vB αB = 1 3 αS βB = 2 3 αS = 1 3 (αB + βB) αB = 1 3 αS βB = 2 3 αS = 1 4 αB = 1 12 pS(vS) = 1 4 + 2 3 vS pB(vB) = 1 12 + 2 3 vB