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# Correspondence Systems.

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Limitedly Repeated Game: Assume amusement is rehashed N times (N > 1). Prize of every player is ... Unendingly Repeated Game: Game is rehashed until the end of time. Prize of each ...
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﻿Correspondence Networks A Second Course Jean Walrand Department of EECS University of California at Berkeley

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Repeated, Bargaining, Dynamic Motivation Repeated Games Bargaining Dynamic Games

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Motivation So far: One-shot (static) diversions Many amusements are rehashed or element: Repeated connections (market, system utilization) Multiple stages (chess, card recreations) Negotiations (bartering) New impacts: Past activities influence data Reputation of players Threats Learning

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Repeated Games Example 1: Prisoners\' Dilemma Example 2: Battle of the Sexes Folk Theorem

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Repeated Prisoners\' Dilemma Model: One-Shot: (B, R) is the remarkable NE. Instinct: If players know the diversion is rehashed, they might need to work together and play (T, L). Limitedly Repeated Game: Assume diversion is rehashed N times (N > 1). Prize of every player is the whole of the progression rewards. Players see and recall the past activities. Both players play all the while at every progression. Both players know the network of prizes.

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Repeated Prisoners\' Dilemma Definitions: Strategy: Specifies what to do at every progression, given accessible data. Subgame Perfect Equilibrium: Pair of methodologies from which no player has a motivator to digress singularly. Hypothesis: The exceptional SPE for the N-time rehashed PD is as per the following: Player 1 plays B at each progression; Player 2 plays R at each progression. Verification: Backward prompting (clear finally step, ...). 

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Repeated Prisoners\' Dilemma Infinitely Repeated Game: Game is rehashed perpetually Reward of every player is the whole of the marked down stride rewards: R i = (1 – b ) S 0  b n R i (n) where 0 < b < 1. Players see and recall the past activities. Both players play at the same time at every progression. Both players know the network of prizes.

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Repeated Prisoners\' Dilemma Definition: r is in C  r = raised brush. of prize sets and r overwhelms a NE Theorem: Consider the boundlessly rehashed PD. For any r in C, there is some b 0 < 1 to such an extent that there is a SPE that implements r if b > b 0 .

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Repeated Prisoners\' Dilemma Proof: Pick r is in C. At that point r  prizes of playing the pair (a k ,b k ) of activities a part p k  N k/N of the times, k = 1, … , 4. Procedure of player 1 [2, resp.]: We play (a 1 , b 1 ) for N 1 stages, then … (a 4 , b 4 ) for N 4 stages, and we rehash until the end of time. On the off chance that you go astray whenever, I play B [R, resp.] everlastingly from there on.

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Repeated Prisoners\' Dilemma Proof (proceeded with): Reward of P2 at steps n, n +1 , … in the event that he goes astray at time n: < (1 – b ) b n 5 + b n+1 2 + e =: v if b is sufficiently expansive. Prize of P2 at steps n, n +1 , … on the off chance that he doesn\'t go astray: > (1 – b ) b n 1 + b n+1 r 2 – e =: w if b is sufficiently extensive. Note that w > v for b sufficiently huge since r 2 > 2. The system is a SPE: Say P2 digresses at time n. At that point P1 plays B everlastingly and, knowing this, P2 must play R perpetually and P1 should as needs be play B always since (B, R) is NE. 

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Repeated Prisoners\' Dilemma Comments: The SPE that authorizes the prizes r is a "risk methodology." The key stride in the contention is to demonstrate that the danger technique is a SPE. At the end of the day, the danger is "tenable." This is the situation on the grounds that the risk is to return to playing a NE until the end of time.

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Repeated Battle of the Sexes Theorem: Consider the boundlessly rehashed PD. For any r in C, there is some b 0 < 1 with the end goal that there is a SPE that authorizes r if b > b 0 .

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Folk Theorem

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Bargaining Question: What is a sensible understanding between two gatherings if arrangements are costly? Model: Alice and Bob deal on the most proficient method to separate a pie of worth 1. After one stage, the estimation of the pie gets increased by d 1 < 1 for Alice and by d 2 for Bob. Alice makes the underlying offer which Bob acknowledges or denies and makes a substitute offer, etc. Hypothesis: (Rubinstein-Stahl, 1972, 1982) SPE: Alice dependably requests the portion x := (1 – d 2 )/(1 – d 1 d 2 ) of the pie and Bob acknowledges an offer iff it is in any event z := d 2 (1 – d 1 )/(1 – d 1 d 2 ).

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Bargaining Theorem: (Rubinstein-Stahl, 1972, 1982) SPE: Alice dependably requests the part x 1 := (1 – d 2 )/(1 – d 1 d 2 ) of the pie and Bob acknowledges an offer iff it is at any rate z 2 := d 2 (1 – d 1 )/(1 – d 1 d 2 ). Remarks: z 2 = 1 – x 1 is the littlest offer Bob acknowledges. Alice can\'t pick up by proposing a bigger 1 – x 1 to Bob. On the off chance that Alice offers under 1 – x 1 , Bob denies and offers x 2 = (1 – d 1 )/(1 – d 1 d 2 ) with the goal that Alice would get just d 1 ( 1 – x 2 ) = d 1 2 x 1 < x 1 .

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Bargaining Theorem: SPE: Alice dependably requests the part x 1 := (1 – d 2 )/(1 – d 1 d 2 ) of the pie and Bob acknowledges an offer iff it is at any rate z 2 := d 2 (1 – d 1 )/(1 – d 1 d 2 ). Verification:

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Bargaining Theorem: SPE: Alice dependably requests the division x 1 := (1 – d 2 )/(1 – d 1 d 2 ) of the pie and Bob acknowledges an offer iff it is in any event z 2 := d 2 (1 – d 1 )/(1 – d 1 d 2 ). Evidence (proceeded):

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Bargaining Theorem: SPE: Alice dependably requests the portion x 1 := (1 – d 2 )/(1 – d 1 d 2 ) of the pie and Bob acknowledges an offer iff it is at any rate z 2 := d 2 (1 – d 1 )/(1 – d 1 d 2 ). Verification (proceeded):

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Dynamic Games Example 1 Example 2 Example 3

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Dynamic Game: Example 1 Claim: 2 NEs = (L, L) and (R, R) If P1 does not pick L: Fact: Only one SPE: (R, R)

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Dynamic Game: Example 2 SPE: P1: L P2: R P1: R if P2 = L generally

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Dynamic Game: Example 3 Equivalent lattice amusement Cannot explain.

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