Jianfei Shen. School of Economics, The University of New South Wales, Sydney 2052, Australia

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1 . Zero-sum games Jianfei Shen School of Economics, he University of New South Wales, Sydney, Australia emember that in a zerosum game, u.s ; s / C u.s ; s / D, s ; s. Exercise. Step efer Matrix A, we know that for Player, his minmax pure strategy is G. his is because Player guarantees ; if s D F and s D E ; if s D G and s D D ; if s D H and s D D ; if s D I and s D A or C ; if s D J and s D A: Step Similarly, you can check that Player s minmax pure strategy is E: ; if.s ; s / D.H; A/ ; if.s ; s / D u.s ; s / D ; if.s ; s / D ; if.s ; s / D ; if.s ; s / D : Exercise. efer Matrix A, and we find the unique pure Nash equilibrium is.g; E/. Exercise. emark. Find the set of Nash equilibria by yourself. Please refer Notes in the event that you forget how to solve the game of Matching Pennies. et us denote Player i s payoff in Matrix C- as eu i.s ; s /, and his payoff in Matching Pennies as u i.s ; s /, where i D ;, s f; g, and s f; g. Jianfei Shen: jianfei.shen@unsw.edu.au April,

2 A C D E F G H I J Matrix A Player Player Player Player Matrix C- Matching Pennies Step Consider Player s affine transformation eu.s ; s / D a u.s ; s / C b; where a > : [Note that by definition of an affine transformation, a must strictly greater than zero.] hus, there are four equations every equation corresponds to a pair of payoffs: D a C b ˆ< D a C b D a C b ˆ: D a C b: Note that the first equation is the same as the last one, and the second equation is the same as the third one, so we can write the above system of equations as follows: ( D a C b D a C b: Solve this system and we get ( a D b D :

3 hen we have eu.s ; s / D u.s ; s / C ; for any s ; s : [See Matrix C-]. Hence, we know that Player s payoff in Matrix C- is the affine transformation from the game of Matching Pennies, that is, his preference does not change. D C D C D C D C Matrix C- Step Now it is your turn to prove that Player s payoff in Matrix C- is the affine transformation from the game of Matching Pennies. First, write Player s payoff eu.s ; s / in Matrix C- as [you can use any parameters that you like]: eu.s ; s / D where the first parameter should satisfy the following condition by the definition of an affine transformation: ist all of the four equations and reduce the system of equations: ( () [Note that we need two equations since there are two unknown variables.] Solve the reduced system of equation, and you can get he final result is: [See Matrix C-]. eu.s ; s / D u.s ; s / ; for any s ; s : Exercise. Find the set of Nash equilibria by yourself. Please refer Notes in the event that you forget how to solve the game of OS.

4 D D D D Matrix C- Player Player Player Player Matrix D- OS Step We keep using the notation as in Exercise. Write Player s payoffs in Matrix D- as eu.s ; s / D a u.s ; s / C b; a > : hus, D a C b ˆ< D a C b D a C b ˆ: D a C b () ˆ< D a C b D b ˆ: D a C b () ( a D b D ; that is, eu.s ; s / D u.s ; s / C : Step Since Player s payoffs in Matrix D- are the affine transformations from the game of OS, we know that his preference does not change. Obviously, Player s preference does not change. Player M Player Matrix E-

5 Exercise. We first introduce a simple method, where Player s payoffs are unchanged. I will introduce a more general method in Section. Step Consider an affine transformation for Player : Step eu.s ; s / D a u.s ; s / C b; a > : o transfer the game in Matrix E- into a zerosum game, we need the following outcome: u.s ; s / C eu.s ; s / D ; s ; s : Of course, there are six equations: C. a C b/ D ; if.s ; s / D.; / C. a C b/ D ; if.s ; s / D.; M / ˆ< ; if.s ; s / D.; / C. a C b/ D ; if.s ; s / D ; if.s ; s / D ˆ: ; if.s ; s / D We pick any two district equations from the above system, that is, we suppose this system is compatible [it has a solution] and check ex post that it is true. For example, I choose the first two equations: ( C. a C b/ D C. a C b/ D ; which solves for a D ; and b D : Step Hence, Player s payoffs are transformed as eu.s ; s / D u.s ; s / ; s ; s ; and the game is now as in Matrix E-. As you can see, it is really a zerosum game.

6 Player M Player Matrix E- emark (General Method). An alternative more general argument goes as follows. et Player s affine transformation be eu.s ; s / D a u.s ; s / C b ; a > I let Player s affine transformation be eu.s ; s / D a u.s ; s / C b ; a > : Since we want to get a zerosum game, eu.s ; s / and eu.s ; s / should satisfy the following requirement: eu.s ; s / C eu.s ; s / D ; s ; s ; that is, a u.s ; s / C b C a u.s ; s / C b D a u.s ; s / C a u.s ; s / C Œb C b D : If we let a D a D, then the above equation becomes as u.s ; s / C u.s ; s / C Œb C b D C.b C b / D : [Note that u.s ; s / C u.s ; s / D throughout the game in Matrix E-.] Hence, we can choose any b and b satisfying b C b D : For example, we can let b D and b D b D. / D ; and so eu.s ; s / D u.s ; s / ; eu.s ; s / D u.s ; s / : Note that if we let b D and b D, then we get the same result as in the former way. Of course you can also let, for example, a D a D, and obtain the results without difficulty [in this case, b C b D ]. he more general case is let a D a D a >. hen