 Welcome back to NPTEL course on game theory. In the previous session we have proved the existence of Nash equilibrium using Brouwer fixed point theorem and of course the proof is for a general class of games that means the strategy spaces can be any convex and compact subsets of Euclidean space. As I mentioned the convex compact subsets in Euclidean space that can be relaxed and you can go to any infinite dimension spaces. You can take the convex and compact subsets of any infinite dimension space. The only issue that requires to be changed is instead of using the Brouwer fixed point theorem we have to use for example Schroder fixed point theorem that is mainly the place what we need to change. Now we will consider a situation where the game is given by 2 matrices and we will again derive the proof but unlike in zero sum games in zero sum games we have we could avoid using Brouwer fixed point theorem by using the convex arguments but in a non-zero sum games we cannot do that but we will give another proof which is originally due to John Nash. So we will consider the a bimatrix game where A is the payoff matrix corresponding to player 1, B corresponds to player 2 and of course we are looking at these are M by N matrices. So let me recall the pure strategies here. Pure strategies are basically for a player 1 the pure strategies are exactly the rows. So the player 1 is going to choose one of the row that is his pure strategies and similarly choosing a column is a pure strategy for player 2. Now what are mixed strategies it is exactly like in zero sum games there is no difference you are choosing the rows with probability certain probabilities. So it is like x1, x2, xm and delta 1 is the corresponding simple x these are all xi are greater than equals to 0 and this sum to 1. So this is basically the delta 1 is all points like this that is delta 1. Similarly for this is for player 1 for player 2 y1, y2, yn in delta 2 where yj are non-negative and this sum to 1 these are the mixed strategies for player 2 and this defines delta 2. So once we have this pure mix pure and mixed strategies the payoff extension is given by pi 1 x, y which is nothing but x transpose Ay pi 2 x, y is nothing but x transpose By these are the mixed payoffs functions. The most important thing is that these are bilinear functions. These are bilinear functions therefore pi 1 is concave in x variable certainly and similarly pi 2 is concave in y variable. Therefore the min max the existence theorem proved in previous session can be applied guarantee the existence of mixed Nash equilibrium. So of course before applying it we need to realize that delta 1 and delta 2 both are convex. This is obvious we have seen it previously also delta 1 and delta 2 are convex and compact and pi 1, pi 2 satisfies the necessary assumptions. So therefore this theorem can be applied. So therefore there is always a mixed Nash equilibrium. This is the theorem but we will try to give you another proof for this result. First thing what I would like to say here is that when I want to make the following climb x star, y star is Nash equilibrium x star transpose Ay star this is greater than or equals to E i A y star this should be true for all i in 1 to m and similarly x star B y star this should be greater than or equals to x star B E E j for all j is equals to 1 to n. Remember E i and E j that we have this notation that we have been using throughout this course E i are the pure strategies for player 1 E j are the pure strategies for player 2. This is why is this true? Basically this comes from the bilinearity of pi 1 and pi 2. So with the ideas we have been using again and again so therefore this does not require any further clarification. So this is a kind of an obvious statement once you recall all the arguments that we have been using. Now we will define a function f. So we will define a function f from the delta 1 cross delta 2 to delta 1 cross delta 2. It is like the previously also we have done the same thing we have used the best response structure there but here we have used something different. So how we define? I will define f x y to be x prime y prime where x prime y prime are defined as follows. So I define for c i x comma y this is nothing but max of 0 comma E i A y minus x transpose A y this and d j x y is max of 0 comma x B E j minus x B y and x i prime the ith coordinate of x i prime is given by x i plus c i x y by 1 plus summation c i prime x comma y where i prime 1 to m. Similarly, y j prime is given by y j plus d j x y by 1 plus summation j prime goes from 1 to n to and d j prime x y. So let us try to understand what these butler terms are given. So what is this term E i prime E i transpose A y minus x transpose A y. Suppose if the player has played x player 2 has played y and player 1 has played x so he is going to get x transpose E y. Instead of x if the player 1 uses the pure strategy E i how much is he going to get that is how much extra he is going to get. Suppose if he is getting more then I would like to increase the probability of the E i the in x let us say I am playing with certain probability x i and then if by playing the pure strategy E i if I am going to get more than 0 then I would like to move towards E i. So that is exactly is captured here in this term x i originally x i is there and now I am moving towards c i x y I increase this probability because c i x y is bigger than 0 therefore x i. Now if E i A y is not is less than 0 is less than x transpose A y for example then that means it is going to be negative value so therefore this is 0 then I will not change x i. So this is the direction in which I am moving the probability with which I play i pure strategy. Now of course when I move like that I do not know whether that is going to be a probability vector or not so therefore I am normalizing so sum over all this thing and sum over x i is nothing but 1 and sum over this c i that is exactly this and similarly in for the player 2 DJ x y is giving the excess pay the player 2 will get by deviating to pure strategy E j. If he instead of playing y if he deviates to E j the jth pure strategy the excess pay of that he is going to get that is given by DJ x y and now if it is greater than 0 then the player 2 would like to move increase y j in the direction of that DJ how that probability is this y j plus DJ x y and that normalizing and this is the normalizing. Now this is so now f x y is going to x prime and y prime how is this x prime is defined as x i plus c i x y c i x y is nothing but this excess pay of function this is clearly continuous function and this is also a continuous function. So therefore x i plus c i x y is a continuous function y j plus DJ x y this is also a continuous function and the term that is there in the denominator is a non-zero term because c i's are always non-negative and 1 plus some non-negative term so therefore this is always greater than or equals to 1. So therefore these are all well defined things and continuous function and bivariate because we have normalized by these factors the sum x i prime is going to be 1 similarly some y j prime is also going to be 1. Therefore the function f that we have defined is actually takes the values of delta 1 cross delta 2 into delta 1 cross delta 2. Therefore we are we have a continuous map now and Brouwer fixed point theorem immediately implies there exists x star y star such that f x star y star is same as x star y star. Brouwer fixed point theorem now gives a fixed a point x star and y star such that f x star y star is same as x star y star. Therefore what we have is that x i star is same as x i star plus c i x star y star by 1 plus summation c i prime x star y star. So let us look at this very careful from here we would like to get some contradictions or to say that x i x star and y star are nationally how do we prove it? This immediately implies as by cross multiplication we have x i star summation i prime summation over i prime c i prime x star y star is same as c i x star y star. So now the interesting thing here in this summation for example i prime wherever c i is prime x star y star is 0 that need not be considered if for some because the whole idea here for us is to show that this c i prime x star y star has to be 0. So whenever we need to show c i x star y star is 0 whenever x i star is greater than 0. So why do we want to do this one? If x i star is strictly greater than 0 that means the player 1 is playing the pure strategy E i with a positive probability. Now look at the definition of c i if the player has played E i with a positive probability that means E i transpose A y star should be same as x star A y star. So that should happen therefore this c i x y should be 0. So this has to happen whenever x i star is greater than 0. So this is essentially the idea now. So let us look at we need to show this fact. So let us take i to be set of all i such that x i star is greater than 0. So now therefore let me also take j to be all i such that c i x star y star is greater than 0. So let me take therefore clearly from here if whenever x i star is greater than 0 look at those i's and here I am looking at i's where c i x star y star is greater than 0. So now look at this if x i star is greater than 0 because we are interested in proving this one if x i star is greater than 0 I need to show that c i x star y star should be 0. Suppose if that does not happen we are going via contradiction. So in fact the our claim is going to be summation c i prime x star y star this is going to be 0. So suppose this does not happen that means summation c i prime x star y star this is greater than 0. So therefore the summation in some sense I can take it to be i prime in j c i prime x star y star this is greater than 0 of course these two are same. So now look at look at the for all let us take i belongs to i this implies x i star is greater than 0 this implies c i x star y star is greater than 0 because this is by our contradiction c i x star y star is greater than 0 this is coming from this contradiction because we have assumed this. Therefore this c i x star y star is strictly greater than 0 what this implies by the definition of c i c i is basically the excess pay that he is getting what it says is that e i a y star is bigger than x star a y star of course x star transposes this happens. Now what is x star transpose a y star this is same as summation x i star of e i transpose a y star where i is equals to 1 to m this is certainly greater than equals to summation i prime i belongs to i of x i star e i transpose a y star because if i is not in i then x star is 0 therefore this happens and then from this previous thing we have this j is greater than j is there here so this is going to be greater than summation x i star into if I let me call this as a u 1 u 1 and this is going to be u 1. So what we have got is the u 1 is strictly bigger than u 1 so this is a contradiction this contradiction happened because we have assumed this therefore summation c i prime x star y star is going to be 0 therefore c i x star y star is 0 for each i this implies e i transpose a y star this is certainly less than equals to x star transpose a y this is true for all i running from 1 to m this implies okay it is not y star x star is optimal for player 1 in a similar fashion y star is optimal for player 2 therefore x star y star is Nash equilibrium okay so this proof is again a simple proof which requires you to construct this specific function by this way and then we show that this is a continuous function and then this gives you a fixed point and we have to finally show that this fixed point is indeed a Nash equilibrium. So this proves the existence of a Nash equilibrium our next task now is to get some optimization problem so basically how do we compute Nash equilibrium are there ways to do ways to compute the Nash equilibrium. So we will now discuss one optimization problem now and of course the solving algorithms we will postpone to next session but now we will look at a please look at a optimization problem which gives you the another reformulation of a Nash equilibrium so the problem is the following thing optimization problem let me write it first this is nothing but maximize x y alpha beta so that x transpose a plus b y minus alpha minus beta subject to a y minus alpha 1 I put a bold 1 means it is a vector of 1s less than equals to 0 similarly summation okay similarly b transpose x minus beta again a vector of 1s this is less than equals to 0 and of course x belongs to delta 1 y belongs to delta 2 okay so this is a quadratic programming. So what we have is that in this there are four decision variables x y and alpha beta and we need to choose the x y alpha beta which maximizes this the whatever maximizes this one the x star y star and then alpha star and beta star they correspond to the Nash equilibrium x star is going to be the optimal strategy for player 1 y star is for player 2 and alpha star is the value of player 1 and beta star is going to be the value of player 2. So we will try to prove this fact so first let us see let x star y star be Nash equilibrium let alpha star to be x star a y star and similarly beta star let me put it as x star transpose b y star let us take this. Now we know now in the previous itself we have seen that is that e i a y star is certainly less than or equals to alpha star this is true for all i that means e i a by star is less than equals to alpha star for every i that means every entry of a y star is going to be less than equals to alpha star this implies a y star minus alpha star the vector of ones this is less than equals to 0 this is obvious in a similar fashion from here from beta star the definition of beta star we can easily see that b transpose x star minus beta star ones less than equals to 0 this is an obvious thing from here and of course x star y star are all there and then the next thing is that for any x in delta 1 x transpose a y star is less than equals to alpha star and similarly for any y in delta 2 x star transpose a y star is less than or equals to beta star putting all these things together and combining with these facts can easily verify x star y star alpha star beta star is optimal solution for the optimization problem. So x star y star alpha star beta star will satisfy so this here when you are trying to prove from here we have to use these facts as well. This is a not a very hard thing it is a straight forward but it requires a little effort so one should try proving it here we also have to understand the following fact in this optimization problem using these facts we need to show that this is always less than or equals to 0. So because a y minus alpha is less than equals to 0 therefore x transpose a y minus alpha is less than equals to 0 that has to be used here therefore this maximum value is always non positive and at Nash equilibrium this is equal to 0 therefore x star y star alpha star beta star is going to be the optimal solution. So the details have to be furnished here but they are a simple exercise. Next let x star y star alpha star beta star be optimal solution of the optimization problem. So once we take this one we need to show that this x star y star corresponds to Nash equilibrium. So let us see how we prove first thing is because it is a solution of this optimization problem we have the following this thing conditions a y star minus alpha 1 this is less than equals to 0 this immediately tells me that x star a y star is less than equals to alpha that is there and similarly x star transpose b y star is less than equals to beta that is also there so these are there. Now if this is always true so that is the first fact that we need to show here then we will try to show that x star y star satisfies the equilibrium condition how do we prove that. So take any x in delta 1 this implies x a plus b y star minus alpha star minus beta star is less than equals to x star a plus b y star minus alpha star minus beta star as I said we have to verify that x star y star alpha star beta star is a optimal is an corresponds to the Nash equilibrium how do we do this let us look at it first thing is that we know that x star into a plus b y star minus alpha star minus beta star this is nothing but the maximum over x y alpha beta of x a plus b y minus alpha minus beta this is true. Now the what we really need to understand here is that this is always less than equals to 0 therefore this is less than equals to 0. Now here is another important thing that we need to show here is that there is we already proved the existence of a Nash equilibrium so therefore there exists a Nash equilibrium so let me call that as a x hat and y hat for the Nash equilibrium x hat and y hat the this value is going to be 0 because x hat y hat corresponding values satisfy the constraints of this optimization problem therefore this value is going to be 0 therefore this is going to be 0 that is the first important thing this once this is there now we use the constraints. So by using the constraints we can easily see that because we can write it as x star a y star minus alpha star this is one term plus another term is x star b y star minus beta star both of them the sum of these two terms is equal to 0 but by the constraints this is x star a y star minus alpha star should be less than equals to 0 this is also less than equals to 0 both of them are non negative numbers therefore what we get is that they must be 0 once these are zeros then now using the constraint this thing immediately we can prove that x star y star is Nash equilibrium so this proves the equivalence between these two problems so an existence if you have an existence of a Nash equilibrium that immediately says that the solution the Nash equilibrium is the solution corresponding to this optimization problem similarly a solution of this optimization problem is a Nash equilibrium so in this second part we have used the fact that the equilibrium existence happens with this we conclude this session we will meet again in the next session