 Welcome back to this NPTEL course on game theory. In the last lecture, we have seen some examples of non-zero sum games, both finite as well as continuous games. Now, we will formally define the non-zero sum game and the Nash equilibrium. So, let us start defining the non-zero sum games. So, here there are 2 players. So, they have their strategy spaces S1 and S2 and the payoff functions, the player 1's payoff functions is given by pi 1, which is a function from S1 cross S2 to R. Similarly, player 2 has a payoff function which is pi 2 from S1 cross S2 to R. Now, the players are choosing their decisions in S1 which maximizes their payoff functions. Like in zero sum games, the payoff of player 1 also depends on choice of player 2. So, therefore, the issue comes. So, here what is the definition of a Nash equilibrium? So, Nash equilibrium is a pair of strategies, let us say X star, Y star, they belongs to S1 cross S2 such that when player 2 has fixed at Y star, X star should maximize player 1's payoff. This should happen for all X in S1. Similarly, when player 1 fixes X star, Y star should maximize player 2's payoff. So, these are the 2 conditions that defines a Nash equilibrium. In fact, if you look back the examples, this is exactly what we have used it and one of the player fixes his strategy the other player whatever he chooses should maximize it. So, in a sense, what you are really saying here is that when a player let us say player 2 fixes Y star for player 1 deviating from X star is not profitable. And similarly, if player 1 fixes at X star, player 1 deviating from Y star is not profitable to him. So, this unilateral deviations are not profitable to the players. So, this defines the notion of Nash equilibrium. This is introduced by John Nash and this has become a major tool in economics and many, many other fields. Once we define this Nash equilibrium, now the whole question that comes here is that does there exist a Nash equilibrium? And even before Nash equilibrium, how is this different from saddle point equilibrium? A zero sum game. Now, let us look at a zero sum game. In a zero sum game, player 1 maximizes pi XY, player 2 minimizes the same. In other words, what we have is that pi 1 of XY is nothing but pi XY for pi 2, it is simply minus of pi XY. So, any zero sum game is in that sense is a non-zero sum game with a special property that the sum of the two payoffs is 0. So, now if I write down the definition of a Nash equilibrium, what we will get here is that pi 1 XY star is less than equals to pi 1 XY star. This is nothing but pi XY star less than equals to pi XY star. This is the first inequality that we have in the definition of saddle point equilibrium. Now, look at this particular thing. If I look at that, what it says is that pi 2 XY. Now, pi 2 X star Y is nothing but minus pi of X star Y. This is less than equals to minus pi of X star Y star. This is nothing but pi X star Y star less than equals to pi X star Y. This is true for every Y in S2. So, if I combine this inequality and then the previous inequality, what we have is exactly the definition of a saddle point equilibrium. So, Nash equilibrium automatically encompasses the definition of saddle point equilibrium. So, once we know that the saddle point equilibrium is exactly same as Nash equilibrium for zero sum games. So, now we will start seeing given a game whether a saddle point equilibrium exists or not, not saddle point Nash equilibrium. Now, like in zero sum games, we do know that we need some conditions to ensure the existence of a saddle point equilibrium. Likewise, because non-zero sum games includes zero sum games, they do require some additional conditions. What are those additional conditions? We will look at them. So, let us see. So, we take the non-zero sum game now. So, the payoff functions S1, S2 are the strategies of player 1. So, we assume S1 is compact and convex, same with S2 compact and convex subsets of Euclidean space. The payoff functions pi 1 x, y, this is a function from S1 cross S2 to R and similarly pi 2, this is a function from S1 cross S2 to R, both have to be continuous. In the zero sum games, in the x variable we assume them that to be concave in y variable we assumed convex. Now, here look at the pi 1 is required only by player 1. This is only for player 1, it does not matter how it behaves with respect to y. So, what we really want is that pi 1 as a function of x should be concave for each fixed y. Similarly, pi 2 as a function for second player should be again concave for each fixed x and s, x is in S1, y is in S2. Pi 1 should be concave in x variable, pi 2 should be concave in y variable. So, these are the assumptions that we require. So, these are all the assumptions. Now, we will prove our main theorem existence under the above assumptions there exist Nash equilibrium. So, let us prove this. So, like in zero sum games, we have to use the fixed point theorem. Of course, in zero sum games we did not use fixed point theorem, we have used the convexity properties. But in a non-zero sum games, we are forced to use the fixed point theorem, the fixed point theorem that we are going to use is known as Brauer fixed point theorem, which we have discussed already in the combinatorial games. In fact, we have used the game of hex to prove the Brauer fixed point theorem in two dimensions. But we assume that result for n dimensions, we would not go into proving the n dimensional theorem. In fact, the hex can be extended to give a proof in n dimensions as well. So, let me recall Brauer fixed point theorem, let K be compact and convex subset of some Euclidean space and f from K to K is a continuous function. Then there exists x in K such that fx is a exactly x. So, such a point x is known as a fixed point theorem, fixed point and Brauer fixed point theorem is a very, very important result. And of course, this Brauer fixed point theorem assumes that the K to be a compact and convex subset of some Euclidean space. It is a finite dimensional result. In an infinite dimension, analogous results do exist. Let us not worry about them. So, now let us go to the proof. Now let us look at the payoff functions. So, fix x bar in S1 y bar in S2. Now consider pi 1 x comma y, x comma y bar plus mod x minus x bar. I can put a square here. So, now this is a continuous function. So, look at this function and I would like to look at the r max over x in S1, pi 1 x comma y bar plus mod x minus x bar square. And as a function of x, I already know that pi 1 is a concave function in x and mod x minus x bar square. Here is a small error here. It has to be minus to make it concave. This is also a concave function. This is also a concave function minus x minus x bar is a concave. So, therefore, this is a concave function. And in fact, this Butler term forces strict concavity. We have already discussed about the strict concavity convexity. So, therefore, this function as a function of x, the maximum exists and the maximizer is unique. Therefore, there exists unique x prime in S1 such that pi 1 x prime comma y bar minus mod x prime minus x bar square is nothing but maximum x in S1 pi 1 x y bar minus mod x minus x bar square. There exists a unique x prime. The uniqueness comes from the strict concavity and this is a convex function concave function and hence the maximizer do exist. So, everything is fine. So, for a fixed, I have fixed already x bar and y bar. I have picked this x prime. The next, I will similarly I look at this pi 2 x bar phi minus y minus y bar square and I look at maximizing over y in S2 of this. Once again, in y, the function pi 2 is a concave and mod y minus y bar is convex and minus of it is concave. So, therefore, this is a concave function and minus mod y minus y bar square makes it strict. Therefore, this is a strict concave function. And therefore, there exists unique y prime in S2 such that pi 2 x bar y prime minus mod y prime minus y bar square is nothing but max y in S2 of this quantity whatever is written there. Now, we have a function. Now, x bar y bar going to x prime y prime. Look at this. Let me denote this by a function phi. So, therefore, phi is a function from S1 cross S2 to S1 cross S2. We have a function from S1 cross S2 into S1 cross S2 defined by x bar y bar going to x prime y prime. What are x prime y prime? x prime is defined as the unique minimizer of this strict concave function and y prime is defined as the unique minimizer of this strict concave function. And this maximizers this for if I as I change x bar and y bar they move continuously. So, in fact, this is an exercise from analysis, real analysis it says that phi is a continuous function. Once we know that the phi is a continuous function what is going to happen? Because S1 in S2 are convex and compact therefore S1 cross S2 is also a convex and compact sets. Therefore, and they are all you need to finite dimension spaces the Brauer fixed point theorem tells you that there exists a fixed point. Brauer implies there exists x star, y star such that phi of x star, y star is nothing but x star, y star. Now, let us rewrite what exactly it says recall this if I x bar is x star, y bar is y star then x prime is also x star, y prime is also x star. So, I have to use that here in this thing. So, let us write that. So, pi x star, y star minus mod x is pi 1 x star minus x star square. So, the when I put x bar is and y bar is x star and y star. So, this is y star, this is y star, this is x star and then x star should maximize it that means pi 1 x star y star minus x star minus x star square should be same as maximum of that. This should be same as maximum x in S1 of pi 1 x comma y star minus x minus x star square. So, x star and y star satisfies the following thing pi 1 x star, y star is nothing but max x in S1 of pi 1 x comma y star minus mod x minus x star square. Similarly, pi 2 x star, y star is nothing but max y in S2 of pi 2 of x star y minus mod y minus y star square. Now, the interesting thing that would like to point out here is that since x star is maximizing this quantity, let us assume everything is nice that means the pi 1 is a smooth function if we take it, pi 1 is a continuously differentiable function if we take it, what is going to happen intuitively is that the derivative of pi 1 at x star, y star in the variable x star minus the derivative of x minus x star square that is going to be 2 into x minus x star, but at x star that is going to be 0. Therefore, this is going to be 0. This immediately tells you that when you fix y star, pi 1 also has a maximizer at x star because pi 1 is a concave function in x star, the first order condition is also sufficient, but this is all assuming several conditions. For example, x star has to be an interior point and other kind of issues, but this is an intuition. So, essence of this intuition is that x star not only maximizes this pi 1 x y star minus mod x minus x star square, it also maximizes pi 1 x y star. So, how do we prove this fact? So, let us prove this. So, goal is to prove x star maximizes pi 1 x y star. So, how do we prove it? So, let us take some point x from S1 and also take a number lambda in this open interval 0, 1 and look at lambda x plus 1 minus lambda x star. So, this is a convex combination of x and x star and both x and x star are in S1. Therefore, this belongs to S1. Now, x star is by the definition of x star, x star maximizes this entire quantity. So, we use that now to say that pi 1 lambda x plus 1 minus lambda x star comma y star minus mod lambda x plus 1 minus lambda x star square, this is less than or equals to pi 1 of x star y star minus of course mod x star minus x star square that is going to be 0, this is 0. So, this is what we have it. Now, this particular term we know that this function pi 1 is convex in the in this variable and similarly let us look at what this particular term is going to be. If we look at it this term that is nothing but lambda x plus x star minus lambda x star. So, there is a missing term here minus x star square that is a missing term there. So, using that what we have is that lambda x plus x star minus lambda x star minus x star if you use all that what we are going to get here is plus x star minus x star gets cancelled what you have is lambda square mod x minus x star square. Now, in the first term here we use the concavity and to say that lambda pi 1 x comma y star plus 1 minus lambda pi 1 x star y star this is less than or equals to pi 1 lambda x plus 1 minus lambda x star y star the concavity used. So, therefore using that concavity and this term this entire inequality can be rewritten as follows. Therefore, this lambda pi 1 x y star plus 1 minus lambda pi 1 x star y star that is smaller than this one and then the minus of that that is minus lambda square mod x minus x star square this is less than or equals to pi 1 x star y star. So, now once this equation is there so there is a pi 1 x star y star there is again pi 1 x star with 1 here. So, these 2 get cancelled. So, if we remove that Butler thing what we have is lambda pi 1 x y star minus lambda square mod x minus x star square which is less than or equals to this minus lambda pi 1 x star y star comes to the right hand side that becomes pi 1 x star y star. Now, if we look at it in all the terms there is a lambda factor. So, divide by lambda to get pi 1 x y star minus lambda mod x minus x star square this is less than or equals to pi 1 x star y star. Now, lambda in open 0 1 this is arbitrary therefore, let lambda decrease to 0 lambda go to 0 this immediately gives us that pi 1 x y star less than or equals to pi 1 x star y star. And now again x in S 1 is arbitrary. So, because x is an arbitrary this immediately tells you that x star maximizes pi 1 x y star. Therefore, x star is now maximizing pi 1 x y star. Therefore, what we have is pi 1 x star y star is greater than or equals to pi 1 x y star for all x in S 1. A similar procedure with pi 2 similar procedure with pi 2 we will actually get pi 2 x star y star is bigger than or equals to pi 2 x star y this will happen for any y in S 2. This is exactly the definition of Nash equilibrium. This implies x star y star is Nash equilibrium. This proves the existence of Nash equilibrium of this game where the payoffs are given by pi 1 pi 2 and pi 1 is concave in x pi 2 is concave in y and of course, we have to assume that they are jointly continuous. Now, I would like to point out the most common method that people use here is using what is known as best responses. So, let me introduce that. So, we have this game with strategy sets S 1, S 2, pi 1, pi 2 are the payoff functions. It is a non-zero sum game that we have. The best response of player 1 when player 2 fixes y is nothing but set of all x in S 1 such that pi 1 x y is nothing but max x prime in S 1 pi 1 x prime y. Similarly, the best response of second player when player 1 fixes x is all y in S 2 such that pi 2 x y is max y prime in S 2 of pi 2 x y prime. Now, because of the concavity assumption, all these best response sets are convex, both Br 1 y and Br 2 x are convex under the assumptions pi 1 is concave in x, pi 2 is concave in y. Therefore, this defines a map. Let me put it this thing x comma y going to Br 1 of y cross Br 2 of x. So, what kind of map? Any point x comma y pair of points are taken to a sets here. These are subsets of S 1 cross S 2. So, these are the multi-valued maps. So, here the common approach that people follow here is to show that this Butler map satisfies certain assumptions of a theorem called Kakotani fixed point theorem. We need this certain continuity properties of this set-valued map which I will not go through. So, but our proof has an advantage that we do not need to worry about a set-valued map. This is just simply a normal continuous functions and Brouwer fixed point theorem is sufficient. In fact, the same ideas can be extended to an infinite dimensional spaces where the Brouwer fixed point theorem has to be replaced with appropriate fixed point theorem infinite dimensions. Here what helps us is shorter fixed point, which we will not go into the details. So, we will only stop here. The next thing we would like to ask is what happens by matrix games? Now, in the zero sum games we have seen that the matrix games, you can actually prove the Minmax theorem, the von Neumann Minmax theorem using nice convexity ideas. Can we really prove something like that here? The proof using just fair convexity ideas is not that easy, but we will prove a proof, we will provide a proof which is due to Nash in the next session where we also discuss some other properties of this Nash equilibrium. With this we stop this session, we will continue in the next session.