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3. 3 A Beautiful Mind John Forbes Nash Jr. What it's about: On the surface, he had everything: Good looks, a devoted wife and child, and a mathematical genius that earned the respect of some of the world's greatest minds. But by his early 30s, the brilliant eccentric had descended into mental illness, his schizophrenia leaving him unable to think either creatively or rationally. As his condition worsened, he fell into obscurity. But his work did not. After a decades-long struggle to conquer his illness, John Forbes Nash Jr. resurfaced. And in 1994, his early work won him the Nobel Prize in economics.

4. 4...and now THE MOVIE starring

5. 5 Based upon: The biography by Sylvia Nasar, “A Beautiful Mind”: A Biography of John Forbes Nash Jr. about the true story of the Nobel Prize-winning mathematical genius. Also starring: Ed Harris (Parcher), Jennifer Connelly (Alicia Nash), Christopher Plummer, Paul Bettany (Charles), Judd Hirsch (Professor), Jason Gray- Stanford (Ainsley), Anthony Rapp (Bender), Adam Goldberg (Sol) Directed by Ron Howard Screenplay by Akiva Goldsman Release date: December 2001 Imagine Entertainment / Universal Status: Currently filming Filming locations: New York and New Jersey: (New York City, Yonkers, Bayonne, Princeton, Jersey City, Belleville). Filming started March 26, 2001

6. 6 1.10 Axioms of Nash 1950s Method developed by Nash (Nobel prize winner, 1994) An attempt to provide a methodology for identifying a unique solution for 2-person cooperative games. Gave a set of axioms, which, he suggested, are reasonable conditions for a solution to satisfy.

7. 7 To introduce the axioms Nash developed we need a point that is “acceptable” by both players to initiate the negotiation. We call it the status quo point, q = (q 1, q 2 ) in C. We also have the cooperative payoff set C. This status quo point can typically be the pair of security levels, (v 1, v 2 ) But other points can be used. It is assumed that there is at least one point in C that dominates the status quo point, otherwise they would just use the status quo point.

8. 8 Set-up C q F c in C Game Situation solution

9. 9 Set - up Suppose a solution is c o C q F c o in C Game Situation solution c o = F(C,q)

10. 10 1.10.1 The Axioms of Nash Axiom 1: The point c o = (c o 1, c o 2 ) = F(C,q), is non-dominated and in particular, c o 1 ≥ q 1 and c o 2 ≥ q 2. Commentary: Makes a lot of sense! Why should a player be willing to accept less than his part of the status quo point?

11. 11 Axiom 2: The solution is invariant to linear transformations of the type L(u,v) = (a 1 u+b 1, a 2 v+b 2 ), a 1 > 0, a 2 > 0. In other words, F(L(C), L(q)) = L(F(C,q)) = L(c o ). Commentary: Is the World of Games linear? This axiom is controversial.

12. 12 Axiom 3: If D is a subset of C and both q and c o are in D, then F(D,q) = F(C,q) = c o. Commentary: This is a controversial axiom. We shall meet one of its variants later in the semester. Mathematically speaking, it assumes some “independence”.

13. 13 Axiom 4: If (c 1, c 2 ) in C implies (c 2, c 1 ) in C, (that is the cooperative payoff set is symmetrical about “y = x” line) then q 1 = q 2 implies c o 1 = c o 2. Commentary. This is the “symmetry axiom”: it says that if the game is “symmetric” with regard to the two players, then the solution is also “symmetric”.

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15. 15 Nash’s Theorem Subject to the above axioms, the game has a unique optimal solution which can be easily obtained by solving the following nonlinear optimization problem: Since the objective function is monotonic increasing with u and v as we move from (q 1, q 2 ), it follows that the optimal solution must be in the negotiation set!

16. 16 SO: NS = Negotiation set

17. 17 1.10.2 Example See lecture for details Security levels are equal to 3 (saddles). if we set q = (3,3), the game is symmetric and from Axiom 4 we have that c o 1 = c o 2.

18. 18 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678

19. 19 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 Cooperative Payoff Set

20. 20 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 Cooperative Payoff Set Status quo point

21. 21 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 Cooperative Payoff Set Status quo point Non-dominated points

22. 22 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 Status quo point Non-dominated points Security level dominance Cooperative Payoff Set

23. 23 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 Cooperative Payoff Set Status quo point Non-dominated points Security level dominance NS

24. 24 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 NS c 1 = c 2

25. 25 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 NS c 1 = c 2 Axiom 4 solution

26. 26 C1C1 C2C2 1 2 3 4 5 6 7 8 12345678 NS Axiom 4 c 0 =(4,4) c 1 = c 2

27. 27 C1C1 C2C2 1 2 3 4 5 6 7 8 8 NS Axiom 4 c 0 =(4,4) c 1 = c 2 (4,4) is a convex combination of (0, 8) and (8, 0) I.e. (4, 4) = a(0, 8) + (1 – a)(8, 0) a = 1/2 The players will use (a 1, A 1 ) and (a 2, A 2 ) each with probability 1/2.

28. 28 1.10.3 Example use security level expected payoffs for status quo point. 0 2 4 6 -2 -4 -6 C1C1 C2C2 Cooperative Payoff Set NOT symmetric A non-symmetric example:

29. 29 See lecture for more detail. 0 2 4 6 -2 -4 -6 C1C1 C2C2 V 1 =1, v 2 = –3 Negotiation Set v = 5 – (3/5)u, 1 ≤ u ≤ 5 Status quo point (1, –3) A (0, 5) B (5, 2) C (1, -5)

30. 30 Analysis We set q = (v 1,v 2 ) = (1, –3). Thus, to find the Nash solution we have to solve max {(u  1)(v+3): (u,v) in NS} (*) Since NS is a line segment, ie. NS = {(u,v): 1 ≤ u ≤ 5, v =5  (3/5)u} we can replace (*) by max{(u-1)(5  (3/5)u + 3): 1 ≤ u ≤ 5} (**) Namely max{(u-1)(40-3u)/5: 1 ≤ u ≤ 5} max {(43/5)u  (3/5)u 2  8): 1 ≤ u ≤ 5} (***)

31. 31 The following is one way of approaching it, see lectures for an alternative. The optimal solution (u*) will be either at an end point of the interval [1,5] or at a point where the derivative of the objective function is equal to zero. The later happens when (43/5)  (6/5)u = 0 namely at u* = 43/6. But this point is outside the feasible region for u, which is [1,5]. Thus, the optimal solution is either u* = 1 or u* = 5 (or both). Evaluating the objective function at these two points indicates that u* = 5 is optimal.

32. 32 optimal v is therefore v* = 5 – (3/5)u* = 5 – (3/5)5 = 2 Thus, the optimal expected payoff is (5, 2) and the optimal strategies are x* = (1, 0), y* = (0, 1). I.e. play (a 1, A 2 ) all the time.

33. 33 Remarks The Nash model can produce a satisfactory outcome only if both players agree to use it. You should be able to cope with problems in which the negotiation set consists of a number of line segments. You can simply consider each segment separately.

34. 34 c1c1 c2c2 Cooperative payoff set Negotiation set

35. 35 Example Determine the solution generated by Nash’s axioms for the game with payoff matrix below, using the security level expected payoff pair as the status quo point. The security level pair is (1/2, 17/8). Make sure you can find this using methods from earlier in the course - do it for practice. See lecture for solution.

36. 36 Max of g graphically - see 261 u v objective function: z = (u-q 1 )(v-q 2 ) Level curves: v = (z/(u-q 1 )) + q 2 Different values of z.

37. 37 Uniqueness of the solution Proof (By contradiction) Assume that there are two (distinct) optimal solutions, say (u", v") and (u', v'). Thus, z* = (u'– q 1 )(v'– q 2 ) = (u"– q 1 )(v"– q 2 ). Now, consider the point (u*, v*) defined by (u*, v*) = ( 0.5(u' + u"), 0.5(v' + v") ) namely it is the mid point of the line segment connecting (u',v') and (u",v").

38. 38 Geometry (u',v') (u",v")

39. 39 Geometry (u',v') (u",v") (u*,v*)

40. 40 Let g(u,v):= (u – q 1 )(v – q 2 ). Then, by definition, g(u*, v*):= (u*– q 1 )(v*– q 2 ) = (0.5(u'+u") – q 1 )(0.5(v'+v") – q 2 ) = 0.5((u’– q 1 )+(u”– q 1 )) x 0.5((v’– q 2 )+(v”– q 2 )) = 0.5(u'– q 1 )(v'– q 2 ) + 0.5(u"– q 1 )(v"– q 2 ) + 0.25(u'– u")(v"– v') = 0.5 z* + 0.5z* + 0.25(u'– u")(v"– v') = z* + e, say. where z* is the optimal payoff and e > 0. Convince yourself that e > 0 (see lecture).

41. 41 This means that (u*,v*) is better than the two given optimal solutions (u',v') and (u",v") unless u' = u" and v' = v" in which case we don’t have two different optimal solutions! Strictly speaking, we have to show that (u*,v*) is feasible (Hint: ?)

42. 42 Geometry (u',v') (u",v") (u*,v*) How do we know that this point is feasible?