Algorithmic game theory - cslog.uni-bremen.de · Algorithmic game theory Ruben Hoeksma October 16,...
Transcript of Algorithmic game theory - cslog.uni-bremen.de · Algorithmic game theory Ruben Hoeksma October 16,...
Algorithmic game theoryRuben Hoeksma
October 16, 2018
IntroductionCourse basics
Course basics
Ruben HoeksmaMZH 3320Webpage:https://www.cslog.uni-bremen.de/teaching/winter18/agt/
Lectures: Monday 14:00 – 16:00 14:15 – 15:45Tuesday 12:00 – 14:00 12:15 – 13:45
Exercises:I approx. 1 set per weekI 1 week to finish eachI 50%: +0.3 90%: +0.6I All exercises treated during the lectures are part of the exam
Course basics
Weeks 3+4:I No lectures (Jan. 14, 15, 21, 22)I Reading assignment + exercises - part of examination
Examination:I Oral exam (around 30 minutes)I First question: Say something about your favorite topic/game
from the course.I Questions will include proofs and intuıtionI Anything spoken about during lectures + any material on webpage
IntroductionGames, selfish behavior, and equilibria
What is a game?
What is a game?
Situations with multiple actors who make their own decisions.I Situations of conflicting interestsI Situations of mutual interests
I Actors are called playersI Each player has some objectiveI Each player has choices that influence both their own objective
and that of othersI Each player is rational, i.e., they optimize for their objective
Simultaneous game
In a simultaneous game, all players, at the same time, choose astrategy from their own strategy space without knowledge about whatthe other players have done.
Definition (Simultaneous game)A simultaneous game is defined byN: Set of n playersSi : Set of strategies for each player i ∈ N
S = S1 × S2 × . . .× Sn: set of strategy vectorsui : S → R Utility function for each player i ∈ N
Example: a routing game
o d
x
10
10
x
0
I Given this directed graph with origin o and sink dI 10 players want to go from o to dI Cost, c(x), for each arc depends on number of players that use itI Cost of each player is the sum of cost of arcs they chose
Example: a routing game
o d
x
10
10
x
0
What do the players do?I There are three routes {U, L, Z}I If there are n ≤ 10 players:
ci (s) =
10 + #U(s) + #Z (s) if si = U,10 + #L(s) + #Z (s) if si = L,#L(s) + #U(s) + 2#Z (s) if si = Z .
Equilibrium
Definition (Equilibrium)An equilibrium is a state in which no player has an incentive to changetheir strategy.
Definition (Dominant strategy equilibrium (DSE))A strategy vector s ∈ S is a DSE if for each player i ∈ N, all alternativestrategies xi ∈ Si , and all strategies of the other players x−i ∈ S−i , wehave
ui (si , x−i ) ≥ ui (xi , x−i ) .
The strategy si is called a dominant strategy for player i .
s−i is the strategy vector s with player i ’s strategy omitted.S−i = S1 × . . .× Si−1 × Si+1 × . . .× Sn.
Example: Battle of the sexes
I Two players N = {Man, Woman}I Two strategies: F : go to the football match; T : go to the theaterI Woman prefers going to football and Man prefers going to theaterI Both prefer to go anywhere together over going anywhere alone
WomanF T
Man F (5, 6) (1, 1)T (2, 2) (6, 5)
I Normal form: explicitdescription of utility for allstrategy combinations
I For two-player game: matrixI Row/column player
Question: Does this game have a dominant strategy equilibrium?
Pure Nash equilibrium
Answer: The battle of the sexes game does not have a DSE.Proof. Man and Woman have two strategies F and T . If Woman playsF , Man prefers to play F . If Woman plays T , Man prefers to play T .So neither strategies is dominant for Man and no DSE exists.
Definition ((Pure) Nash equilibrium (NE))A strategy vector s ∈ S is a NE if for each player i ∈ N and allalternative strategies of that player xi ∈ Si , we have
ui (si , s−i ) ≥ ui (xi , s−i ) .
si is a best response of player i to s−i .
Rock-paper-scissors
Definition (Zero-sum game)A zero-sum game is a game where for any strategy vector s ∈ S thesum of the utilities of the players for that strategy vector is zero.∑
i∈Nui (s) = 0
Rock-paper-scissors
R P SR (0, 0) (−1, 1) (1,−1)P (1,−1) (0, 0) (−1, 1)S (−1, 1) (1,−1) (0, 0)
Q: Does RPS have a NE?A: NoProof. For any strategy of the rowplayer, there is a strategy for thecolumn player that wins. Same theother way around, so no pair ofstrategies are each a best responseto each other.
Mixed strategies
Definition (Mixed strategy)A mixed strategy of player i ∈ N is a probability distribution over theirstrategy space Si . A mixed strategy vector is a vector of mixedstrategies.
Definition (Mixed Nash equilibrium (MNE))A mixed strategy vector s ∈ S is a MNE if for each player i ∈ N and allalternative strategies of that player xi ∈ Si , we have
E[ui (si , s−i )] ≥ E[ui (xi , s−i )] .
Example: Rock-paper-scissors (cont.)
R P SR (0, 0) (−1, 1) (1,−1)P (1,−1) (0, 0) (−1, 1)S (−1, 1) (1,−1) (0, 0)
Claim. Both players playing each strategy with probability 13 is a MNE.
Proof. Let s be the mixed strategy
E[ur (sr , sc)] = 19(3 · −1 + 3 · 1 + 3 · 0) = 0
E[ur (xr , sc)] = 13(−1 + 1 + 0) = 0 ∀ xr ∈ Sr
Two more. . .
Definition (Correlated equilibrium (CorEq))Let p be a probability distribution over S. p is a CorEq if for eachplayer i ∈ N and all strategies of that player si , xi ∈ Si , we have∑
s−i∈S−i
p(si , s−i )ui (si , s−i ) ≥∑
s−i∈S−i
p(si , s−i )ui (xi , s−i ) .
Definition (Coarse correlated equilibrium (CCE))Let p be a probability distribution over S. p is a CCE if for each playeri ∈ N and all alternative strategies of that player xi ∈ Si , we have∑
s∈Sp(s)ui (s) ≥
∑s∈S
p(s)ui (xi , s−i ) .
CorEq and CCE
CorEq: ∑s−i∈S−i
p(si , s−i )ui (si , s−i ) ≥∑
s−i∈S−i
p(si , s−i )ui (xi , s−i ) .
Let player i be the row player in the following representation
s1i
s2i
...
s1−i s2
−i · · ·
CorEq and CCE
CCE: ∑s∈S
p(s)ui (s) ≥∑s∈S
p(s)ui (xi , s−i ) .
Let player i be the row player in the following representation
s1i
s2i
...
s1−i s2
−i · · ·
Example: Game of chicken (the traffic light)
D SD (−10,−10) (1,−1)S (−1, 1) (0, 0)
DSE? NoNE? (D, S) or (S, D)MNE? pi (D) = 1
10 , pi (S) = 910 for i ∈ {1, 2}
CorEq? Traffic light {(S, S), (D, S), (S, D)}
Example: Rock-paper-scissors (again)
R P SR (0, 0) (−1, 1) (1,−1)P (1,−1) (0, 0) (−1, 1)S (−1, 1) (1,−1) (0, 0)
CCE: (R, P), (P, R), (R, S), (S, R), (P, S), (S, P) all withprobability 1
6 .Claim: The above probability distribution is not a CorEq.Proof. We consider the row player playing Rock. Given that the rowplayer plays Rock, the column player plays Paper and Scissors withprobability 1
2 each and expected utility equal to 0. If the row playerplays Scissors instead their expected utility is 1
2 .
Next time
I Existence of equilibria (Nash’s theorem)
Exercise set 1 available today. Deadline 22.10.