In game theory, trembling hand perfect equilibrium, or simply perfect equilibrium, is a type of refinement of a Nash equilibrium that was first proposed by Reinhard Selten. A trembling hand perfect equilibrium is an equilibrium that takes the possibility of off-the-equilibrium play into account by assuming that the players, through a "slip of the hand" or tremble, may choose unintended strategies, albeit with negligible probability.
Definition First define a perturbed game. A perturbed game is a copy of a base game, with the restriction that only totally mixed strategies are allowed to be played. A totally mixed strategy is a mixed strategy in an n {\displaystyle n} -player strategic game where every pure strategy is played with positive probability. This is the "trembling hands" of the players; they sometimes play a different strategy, other than the one they intended to play. Then define a mixed strategy profile σ = ( σ 1 , … , σ n ) {\displaystyle \sigma =(\sigma _{1},\ldots ,\sigma _{n})} as being trembling hand perfect if there is a sequence of perturbed games strategy profiles { σ k } k = 1 , 2 , … {\displaystyle \{\sigma ^{k}\}_{k=1,2,\ldots }} that converges to σ {\displaystyle \sigma } such that for every k {\displaystyle k} and every player 1 ≤ i ≤ n {\displaystyle 1\leq i\leq n} the strategy σ i {\displaystyle \sigma _{i}} is the best reply to σ − i k {\displaystyle \sigma _{-i}^{k}} . Note: All completely mixed Nash equilibria are perfect. Note 2: The mixed strategy extension of any finite normal-form game has at least one perfect equilibrium.
Example The game represented in the following normal form matrix has two pure strategies Nash equilibria, namely ⟨ Up , Left ⟩ {\displaystyle \langle {\text{Up}},{\text{Left}}\rangle } and ⟨ Down , Right ⟩ {\displaystyle \langle {\text{Down}},{\text{Right}}\rangle } . However, only ⟨ U , L ⟩ {\displaystyle \langle {\text{U}},{\text{L}}\rangle } is trembling-hand perfect.
Assume player 1 (the row player) is playing a mixed strategy ( 1 − ε , ε ) {\displaystyle (1-\varepsilon ,\varepsilon )} , for 0 < ε < 1 {\displaystyle 0<\varepsilon <1} . Player 2's expected payoff from playing L is:
1 ( 1 − ε ) + 2 ε = 1 + ε {\displaystyle 1(1-\varepsilon )+2\varepsilon =1+\varepsilon }
Player 2's expected payoff from playing the strategy R is:
0 ( 1 − ε ) + 2 ε = 2 ε {\displaystyle 0(1-\varepsilon )+2\varepsilon =2\varepsilon }
For small values of ε {\displaystyle \varepsilon } , player 2 maximizes his expected payoff by placing a minimal weight on R and a maximal weight on L. By symmetry, player 1 should place a minimal weight on D and a maximal weight on U if player 2 is playing the mixed strategy ( 1 − ε , ε ) {\displaystyle (1-\varepsilon ,\varepsilon )} . Hence ⟨ U , L ⟩ {\displaystyle \langle {\text{U}},{\text{L}}\rangle } is trembling-hand perfect. However, a similar analysis fails for the strategy profile ⟨ D , R ⟩ {\displaystyle \langle {\text{D}},{\text{R}}\rangle } . Assume player 2 is playing a mixed strategy ( ε , 1 − ε ) {\displaystyle (\varepsilon ,1-\varepsilon )} . Player 1's expected payoff from playing U is:
1 ε + 2 ( 1 − ε ) = 2 − ε {\displaystyle 1\varepsilon +2(1-\varepsilon )=2-\varepsilon }
Player 1's expected payoff from playing D is:
0 ε + 2 ( 1 − ε ) = 2 − 2 ε {\displaystyle 0\varepsilon +2(1-\varepsilon )=2-2\varepsilon }
For all positive values of ε {\displaystyle \varepsilon } , player 1 maximizes his expected payoff by placing a minimal weight on D and maximal weight on U. Hence ⟨ D , R ⟩ {\displaystyle \langle {\text{D}},{\text{R}}\rangle } is not trembling-hand perfect because player 2 (and, by symmetry, player 1) maximizes his expected payoff by deviating most often to L if there is a small chance of error in the behavior of player 1.
… excerpt ends here. Continue reading the full article.
