Sunday, May 30, 2010

Eleventh Installment

    Now let us introduce the concept of a Mixed Strategy. All along, we have been working with two-person games whose rules allow for only a finite number of moves [with cut-off points like the rules that limit a chess game]. The Game Tree for such a game, however complex, defines a finite number of strategies for each player, even though that number may, as we have seen, be large even for very simple games. With a finite number of strategies for each player, we can convert the extensive form of the game to the normal form by constructing a payoff matrix with a finite number of rows representing A's strategies and a finite number of columns representing B's strategies. [From a mathematician's point of view, it doesn't matter how big the number of rows and columns, so long as they are finite in number]. But as we have seen, even with strictly opposed preferences, a game in which both players seek to maximize their security level may not have a stable equilibrium solution. Now von Neuman takes the final step by introducing the concept of a mixed strategy.

    When A gives her instructions to the Referee, she need not specify one of her pure strategies. Instead, she can define a probability distribution over her pure strategies and instruct the Referee to construct a Lottery that embodies that distribution. Just to be clear, this is a Lottery in which the "prizes" are strategies, not outcomes. Before leaving for her appointment, A tells the Referee to spin the wheel that has been constructed and play whichever strategy comes up. This is a real spin of the wheel. Neither A nor the referee can know which pure strategy will be played until the wheel has been spun. B can do the same thing, of course.

    Each Lottery, or probability distribution over the set of pure strategies, is a mixed strategy, and quite obviously there are an infinite number of them. With an infinite number of mixed strategies for A, and an infinite number for B, there are of course also an infinite number of mixed strategy pairs, which is to say pairs each of which consists of one mixed strategy for A and one mixed strategy for B. Notice that a pure strategy now becomes simply a mixed strategy in which all the probability weights but one are zero.

    For any mixed strategy pair, A can calculate the value to her of those mixed strategies being played against one another, although it is obviously tedious to do so. She says to herself: Suppose I play mixed strategy MA1 and B plays mixed strategy MB1. MA1 offers a .3 probability of my playing pure strategy A1, and MB1 offers B a .2 probability of his playing pure strategy B1, so I will calculate the payoff to me of A1 played against B1 and then discount that payoff, or multiply it, by (.4)(.1) = .04. Then I will do the same for A2 against B1, etc etc etc. Then I will add up all the bits of payoffs, and that is the value to me of the mixed strategy pair( MA1, MB1). I hope by now this is clear and reasonably obvious.

    However, we can no longer construct a payoff matrix, because there are infinitely many mixed strategies for each player. Instead, we need a space of points that represent the payoffs to A of each of the infinite pairs of mixed strategies. Since we are dealing now with strictly competitive zero-sum games, we do not need to represent the payoff to B. Under the normalization we have chosen, that is simply 1 minus the payoff to A.

    At this point I must do what mathematicians call "waving their hands." That is, instead of giving rigorous definitions and proofs [not that I have been doing that so far], I must simply wave my hands and describe informally what is going on, and hope that you have enough mathematical intuition to grasp what I am saying. To represent all of the mixed strategy pairs and their payoffs to A, we are going to need a space with (n-1) + (m-1) + 1 dimensions. The first (n-1) dimensions will represent the probability weights being given to A's n pure strategies. [(n-1) because the weights must add up to 1, so once you have specified (n-1) of them, the last one is implicit.] The next (m-1) dimensions represent the probability weights being given to B's m pure strategies. The last dimension, which you can think of intuitively as the height of a point off of a hyperplane, represents A's payoff. We only need to represent A's payoff because this is a zero-sum game, and B's payoff is just the negative of A's. Obviously, we are only interested in the part of the space that runs, on each axis, from 0 to 1, because both the probability weights and the payoffs all run from 0 to 1 inclusive.

    It is said that the great Russian English mathematician Besicovitch could visualize objects in n-dimensional vector space. If he wanted to know whether something was true, he would "look" at the object in the space and rotate it, examining it until it was obvious to him what its properties were. Then he would take out pen and paper and go through the tedium of proving what he already knew was true. I suspect the same must have been true of von Neuman. Well, God knows it isn't true of me, so I must just soldier on, trying to connect the dots.

    You and I can get some visual sense of what such a space would be like by thinking of the simplest case, in which A and B each have only two strategies. In that nice simple case, the number of dimensions we need is (2-1) + (2-1) + 1, or 3. And most of us can imagine a three-dimensional system of axes. Just think of a three dimensional graph with an x-axis and a y-axis forming a plane or bottom, and a z-axis sticking up. The infinity of A's mixed strategies can be represented as points along the x-axis running from the origin to plus 1. The origin represents the mixed strategy with zero weight given to A1, and therefore a weight of 1 given to A2. In other words, it represents the pure strategy A2. Any point along the line represents some mixture of A1 and A2. The point 1 on the x-axis represents the pure strategy A1. Same thing on the y-axis for B's strategies B1, B2, and the mixtures of them. We thus have a square bounded by the points (0,0), (1,0), (0,1), and (1,1). The z-axis measures the payoff to A for each point in that square, and that set of points between 0 and 1 in height that together form a surface over the square.

    Now [here goes the hand-waving], the function mapping mixed strategy pairs onto payoffs is a continuous one, because a tiny change in the assignment of probability weights results in a tiny change in the payoff. [I hope that is obvious. If it isn't, take my word for it -- which is a great thing for a philosopher to say who is trying to explain some mathematics, I know, but I have my limits!]

    OK. Got that in your mind's eye? Now, let us recall that von Neuman offered a "solution" of strictly competitive games in terms of something called security levels and equilibrium pairs of strategies. Suppose that somewhere in that space of payoffs, there is a point that represents the payoff to a pair of equilibrium mixed strategies. What would that mean and what would it look like?

    Well, what it would mean is this: First of all, if B holds to his mixed-strategy choice, any movement A makes back and forth along the x-axis is going to be worse for her. [That is what it means to maximize your security level]. Visually, that means that as she moves back and forth along the x-axis, the point in space representing the payoff to her goes down. What is more, because of continuity, it goes down smoothly. A little movement one way or the other produces a little move down of the payoff point. A bigger move one way or the other produces a bigger move down. For B, the whole situation is reversed, because B's payoffs are equal to the negative of A's payoff. [Zero-sum game]. So, if A holds to her mixed strategy choice, any movement B makes along the y-axis will push the payoff point up [up for B is bad, because the payoff point is A's payoff, and B's is the negative of that.]

    Now, what does this region of the payoff surface look like? Well, the point we are focusing on has the property that if you move back or forth in one dimension the surface falls off, and if you move back or forth in the other dimension, the surface climbs. Have you ever watched a Western movie? Have you ever ridden a horse? Can you see that the surface looks like a saddle? Especially a Western saddle, which has a pommel to grab onto and a nice high back to the seat. The point right under you on the saddle, the point you are sitting on, is a "saddle point." It has the property that if you run your finger from that point side to side [along the y-axis], your finger goes down, and if you run your finger from that point front or back, your finger goes up.

    Now we know what an equilibrium point looks like, at least in three dimensional space, for the case in which A and B each have two pure strategies. Exactly the analogous thing would be true of a hyperplane in hyperspace [you can get your light sabers at the desk before we go into warp speed]. So, we can say that If there is a saddle point in the space representing a two person zero sum mixed strategy game, then that point will occur at the intersection of an equilibrium pair of mixed strategies, and in that sense will be a Solution to the game.

    So, are there such points? Now comes the boffo ending for which all of this has been a preparation. John von Neuman proved the following theorem: Every two person zero sum mixed strategy game has a solution. That solution is represented by a saddle point in the n-dimensional vector space representing the normal form of the game. I really think von Neuman must have seen this in one exquisite flash of mathematical intuition, and then just cranked out a proof of a proposition he could just see is true. I am not going to go through the proof [I am not completely crazy], but having brought you this far, I think I owe it to you to just tell you the idea underlying it.

    In a nutshell, here it is. von Neuman defines a continuous transformation or mapping of the strategy space onto itself. He then proves that the transformation has this neat property: a point is mapped by the transformation onto itself [i.e., it remains invariant under the transformation] if and only if that point is a saddle point, and is thus a solution to the game. He then appeals to a famous theorem proved by the great Dutch mathematician L. E. J. Brower, which states that every continuous transformation of a compact space onto itself has at least one fixed point, which is to say a point that the transformation maps onto itself. [Hence, this is known as the Fixed Point Theorem.] Ta Da! [Can you believe that when I taught this stuff to my philosophy graduate students at UMass, I not only proved von Neuman's theorem, I even proved Brouwer's theorem? Ah, there were giants in the earth in those days, as the Good Book says.]

    And that is it, folks. That is the high point of formal Game Theory. There is a vast amount more to say, and I am going to say a good deal of it, but the subject never again rises to this level of formal elegance or power. Notice, before we move on, one important fact. von Neuman proves that every zero-sum two person mixed strategy game has a solution. But since Brower's theorem just tells you there exists a fixed point, and doesn't tell you how to find it, in general Game Theory cannot tell us how to solve even this limited category of games. [If there is anyone out there who has ever been involved with Linear Programming, every Linear Programming problem is equivalent to a zero-sum mixed strategy two person game, so that is why, in a certain sense of why, you also cannot be sure of solving a Linear Programming problem.] Oh yes, one final point, which we have already encountered in a simpler form. if there are two saddle points, they are equivalent, in the sense that they give the same payoffs to A and to B.


  1. Can you believe that when I taught this stuff to my philosophy graduate students at UMass, I not only proved von Neuman's theorem, I even proved Brouwer's theorem?

    Giants indeed! I've taken philosophy classes on quantum mechanics, general relativity, and large cardinals (from set theory, not ornithology) and did just fine with the math -- but I have a master's in math, and my less-mathematically-inclined fellow students always barely made it through.

    since Brower's theorem just tells you there exists a fixed point, and doesn't tell you how to find it, in general Game Theory cannot tell us how to solve even this limited category of games

    Brouwer's own version of the proof does tell you how to find the fixed point. (He was a intuitionist about mathematics, meaning, among other things, that he rejected existence proofs that didn't tell you how to find the thing in question.) But it's been too long since I studied topology to recall if his proof can be generalized to arbitrary-dimensional spaces in the way von Neumann and other game theorists would need. And the method of finding the fixed point is probably only `in principle' anyways -- it would take far too much computing power to actually calculate fixed points for the result to be useful for anything.

  2. Noumena, are you sure about that? I know Brouwer was a constructionist or intuitionist, but when I went through the proof, there was no indication of how one would construct the fixed point. It is certainly true that for various specific classes of cases one can construct the point, but in general?

    Where are you doing your work, or where did you?