MT368, Continuous Games Begun

MT369, Continuous games 1

MT369, Part 6 Continuous games on the unit square, begun Notation. Some of this is non-standard. IX denotes the unit interval 0  x  1 and IY the interval 0  y  1 . The unit square written  is the square IX  IY in the X – Y plane: I will write u   to mean that u is a point in the square. The interior of  ( written  ) is the set of points 0 < x < 1 and 0 < y < 1 . A continuous game on the unit square is played by 2 persons A and B . A secretly chooses a number x  IX and B secretly chooses a number y  IY . Then they reveal their choices, and A wins an amount z from B where z is some predetermined function of x and y . The payoff function is written P( x , y ) and is usually called the payoff kernel . The traditional example is Example. A duel . A and B each have a pistol with one bullet. At time t = 0 they face each other & start walking towards each other. If nobody gets shot, they meet at time t = 1. If A fires at time t=x , he hits B (payoff 1) with probability x . Likewise if B fires at time t=y he hits B (payoff 1) with probability y . A player who is hit cannot then shoot. If neither is hit, or if both hit, the payoff is 0 . There are 2 versions of this: 1. With silencers. Neither player knows if the other has fired unless he is hit. 2. No silencers. If A shoots first & misses, then B can wait until t = 1. Ex. Find the payoff functions under the 2 rules. Done in class.

Definition. A point ( u , v )   is a saddle point if P( x , v )  P( u , v )  P ( u , y ) (5.1) for all other x  IX and y  IY . If so, then ( u , v ) are optimum strategies for A and B . Note that this is not the normal definition of a S.P . Normally P has a SP if P has a local max in any direction and a local min in any other direction, but in Game Theory P must have local max in the X direction and min in the Y direction. Hence Theorem. Let P be twice differentiable for all points in  with a saddle point at ( u , v ) . Then either u = 0 and P/x < 0 or u = 1 and P/x > 0 or P/x = 0 and 2P/x2  0 and similarly for y with < and > signs switched.

1 Example: P = xy – x/3 – y/2 has a SP at ( ½ , /3 ) Example: P = xy + x + y . SP at ( 1 , 0 ) .

Given the function P we try to solve the game as follows: first look for a SP in  . If that fails, we may need to look at the boundaries of  : there are 4 sides and 4 corners. If that fails, we must try to find a solution in mixed strategies, once we have decided what a mixed strategy is. MT369, Continuous games 2

5.2 Distributions Definition. A mixed strategy for A is a decision by A to choose numbers x from IX by some random process according to a selected probability distribution (  , say) . (examples) Definition. A probability distribution ( P.D. for short) is an increasing function  on IX such that (0) = 0 and (1) = 1 . Given a random process that produces numbers x in IX , the distribution function of that process is given by  prob( x   ) . For the moment, assume that  is differentiable. Then the density function of  is d/dx . A value of x is active if d/dx > 0 and non-active if d/dx = 0 . The support or spectrum of  is the set of active values of x . A distribution might be discontinuous. For example if A decided to choose the exact value x = ¼ with prob. ½ , then the corresponding  jumps up by ½ at that point.  is a step point of  if  jumps in this way. Because   it can only have simple jumps. All the step points are active also. So when A is playing a mixed strategy, he first chooses the function  . Then each time the game is played, he chooses x according to that distribution. Imagine  as mapping IX into IZ . The easiest way for A to choose numbers x according to the distribution  is: first choose   IZ (all numbers equally likely) and then x = sup{ u  IX: (u)   } . (examples) Suppose that A has chosen a strategy represented by the dist.  ( a  map from IX into IZ ) and B chooses to play y . What is A's expected payoff? If IX was a finite set, then the expected payoff would be EP( y )= Prob( x = a ) P ( a , y ) a IX (5.2) Since IX is actually a slice of the real line, we must replace this sum by an integral. Now for any given  , the probability that a random x is actually equal to  is 0 . So we must replace Prob( x =  ) by the probability that   x   + x . This prob. is equal to ( x )  . If  is differentiable, this is approximately x'() . So 1 EP(q , y )= P ( x , y ) dq dx (5.3) x=0 dx If  jumps by an amount  at x =  , that means that A chooses the exact value x =  with probability  . Then we must modify the RHS of (5.3) by adding a term P(  , x )  ; and we need one such term for each step point. Then the integral on the RHS of (5.2) has to be replaced by: 1 EP( y ,q )= P ( x , y )dq dy + P ( a , y ) dq ( a ) x=0 dy (5.4) step points x=a

1 We write this mixture of sum and integral thus: EP( y ,q )= P ( x , y ) d q . (5.5) x=0 Definition. Let  be a distribution on IX and u(x) a function defined on IX . If A chooses 1 points in IX with P.D.  , then the mean value of u is written u= u( x ) dq , interpreted as x=0 follows: 1 1 u= u( x ) dq = u ( x )dq dx + u ( a ) dq ( a ) 蝌x=0 x = 0 dx (5.6) step points x=a So (5.6) means that we integrate when  is smooth and sum where it jumps. MT369, Continuous games 3

Now suppose that A chooses strategy  and B chooses  . For each y the expected payoff is given by (5.5). Then the overall mean payoff will be got by a 2nd integrate-or-sum operation: 1 1 EP(q , f )= P ( x , y ) d q d f (5.7) 蝌y=0 x = 0

Definition. Let  and  be a pair of strategies for the two players A , B , respectively. Then * and * are an equilibrium pair of strategies if EP(l , f *)＃ EP ( q *, f *) EP ( q *, m ) (5.8) for every other strategy for A and  for B . Neither player can improve her payoff by deviating. Ville’s Theorem. (not proved, beyond scope) Let P( x , y ) be continuous on  . Then the corresponding game has a solution in mixed strategies; that is, there exist equilibrium strategies  and  as defined above.

The opt. strategies are guaranteed to exist, but there is no known method to find them. There is one special case.

Definition. Let f(x) be a function defined on an interval I  R . Then f is a convex function if f(a p+ (1 - a ) q )� a f ( p ) - (1 a ) f ( q ) (5.9) for all p , q  I and all  in the range 0 <  < 1 . f is strictly convex if (5.9) holds with < in place of  .

This means that if we take 2 points on the graph of f , namely ( p , f(p) ) and ( q , f(q) ) and join by a line, the graph of u sags below the line. If f is twice differentiable, then f convex implies that f "  0 in I . Roughly speaking: a convex function either goes downhill (getting flatter) , or uphill (getting steeper) or downhill then flat , then uphill . Examples: ex , ex , x2 1/x . Non-example: (x-1)(x-2)(x-3) (draw graph & observe bends). Definition. f(x) is (strictly ) concave if f(x) is (strict) convex . Non-standard notation: from a book on another subject: convex  or concave  makes it easier to remember which is which. Definition. A game P( x , y ) is convex if P( x , y ) is a strictly convex  function of y for each fixed x  IX . Note: non –symmetrical definition.

Theorem. (proof will be omitted unless there is time, in any case the proof is not examinable) Let f(x) be convex  and bounded in an interval I : a  x  b . Then there exist numbers c and d such that a  c  d  b and 1 if a < c then f(x) is strictly decreasing in [ a , c ] 2 if c < d then f(x) is constant ( = , say) in ( c , d ) . If c = d, put  = f(c) . 3 if d < b then f(x) is strictly increasing in [ d , b ] and  is the minimum value of f(x) in I . 4 If f is strict convex, then c = d i.e. the min at c is unique. MT369, Continuous games 4

Proof. (outline only, omitting the difficult parts). Let  = min value of f in I and c and d the upper and lower bounds of points where f(x) =  . Then f(c) = f(d) =  . If c < x < d , then x = c + ( 1 )d for some  . Then (5.9) says that f(x)   . But  = min value  f(x) =  for all x in ( c , d ) . This proves (2). But if f is strict convex, then (5.9) says that f(x) <  , contradiction. Proves (4) . To prove (1). Let a  p < q  c , to prove that f(p) > f(q) . Since  = min (f) and c = least x for which f =  , f(p) >  , say f(p) =  . Now q = p + ( 1 )c for some  . Then (5.9) says that f(q)   f(p) Proves 1 . (3) is similar.

Definition. A game P( x , y ) is convex if P( x , y ) is a strictly convex  function of y for each fixed x  IX . Theorem. Assuming that optimum strategies exist, if P( x , y ) is convex then B has a unique opt. strategy which is a pure strategy. i.e. B should play y=u (fixed) every time. Proof. For each fixed x , P( x , y) is a strict-convex function of y . Let p < q < r be any 3 values of y in IY , say q = p + r . By (5.9): P( x , q ) < P( x , p ) + P( x , r )  P( x , p ) + P( x , r )  P( x , q ) > 0 (5.10) Now suppose that  is an optimum strategy for A . Multiply (5.10) by d and integrate. Recall 1 that EP( , y ) P ( x , y ) d  is the expected payoff to A if A plays  and B x0 plays y . Applying this to (5.10) we get: P(  , p ) + P(  , r )  EP(  , q ) > 0  EP(  , q ) < P(  , p ) + P(  , r ) So the expected payoff EP(  , y ) is a strictly convex function of y  it has a unique minimum, say at y = u . Since this min. is unique, we have EP(  , u ) < P(  , y ) (5.11) for all y  IY different from u . Since  is opt. for A we have also EP(  , u )  P(  , u ) (5.12) where  is any deviant strategy for A . But now A might have two opt strategies, say  and  . By the above argument, EP(  , y ) is strict convex,  it has a unique min, say at y = v . Required to prove that u = v . Suppose not. Since  and y=v are a second pair of opt. strategies for A and B , they must satisfy the equivalent of (5.11) and (5.12) which are: EP(  , v )  P(  , v ) < P(  , y ) (5.13, 5.14) where  is any deviant strategy for A and y  v. But now EP(  , u ) < P(  , v ) (by 5.11 with y = v )  P(  , v ) ( 5.13 with  < P(  , u ) (5.14 with y = u )  EP(  , u ) (5.12 with 

So EP(  , u ) < EP(  , u ) Contradiction, so u = v . 