 Welcome back to this NPTEL course on game theory. In the previous session, we introduced the Nash bargaining problem and the 5 axioms that Nash has introduced and we concluded with the statement of the Nash bargaining solution. Now we will prove this one. So before we proceed for the proof, we will just give a small example and then we go for the proof. So let us consider the following simple situation. Let us assume that the F is given by, so let us say this is a 4, 0 and this is 0, 4 and let us take this as 11. So let us consider this convex hull to be F. So F is this and let us take this V to be V is 11. So the disagreement vector is 11. Now when you take this one, what we are really looking at is x1 minus 1 into x2 minus 1 and then you are trying to maximize over all x1 greater than equals to 1, x2 greater than equals to 1 and of course x1, x2 in F. We really look at this one. In fact, we can show is that this 0.22 is going to maximize this. The solution is going to be 22 here. One can just verify that F is going to be 22 here. So this is in fact, this is a symmetric set and therefore both the values will be same. In fact, that itself will also give you the reason why it should be 22. In fact, whatever it is that will be in this vector, in this line. And as you go further, you can see that this will be maximized and this is also Pareto efficient because from this thing, either directions I cannot improve. I cannot move this. If I move this side, I am going out of it. This side also, you cannot improve without this. So therefore, this is also Pareto efficient symmetry and in fact, the scale covariance you can try that if we take some everything is scaled by some lambda and we can see that the solution will also be scaled and of course, individually rationality and all the other. So now, let us go to the theorem thing. So there exist unique solution satisfying 5 axioms. So recall, the 5 axioms are basic are the Pareto efficiencies, the strong efficiencies, then individual rationality, then scale covariance, then independence of irrelevant alternatives, then symmetry. So these are the 5 axioms and in fact, the solution is given by f of fv, this maximizes the product x1 minus v1 into x2 minus v2 such that x1 x2 is in f such that x1 greater than equals to v1, x2 greater than equals to v2. So whatever maximizes this, this f should be inside that. So now let us try to prove it. So the proof, we first prove for class of problems called essential bargaining problems. And then we generalize, we start with this thing. So let me define what is this thing. fv is called essential bargaining problem. If there exist at least one allocation y in f, that is strictly better for both the players than the disagreement allocation. That is y1 is strictly bigger than v1, similarly y2 is strictly bigger than v2. If there is a strictly better allocation for both players, then you are calling that bargaining problem as essential. So we start with this problem and then we see. So let us say fv is essential. So therefore there exist some y in f such that y1 is bigger than v1, y2 is bigger than v2. So this is automatic. Now let us consider this optimization problem. Max x1 x2 in f with x1 greater than equals to v1, x2 greater than equals to v2 of this x1 minus v1 into x2 minus v2. Let us look at this one. So this is sometimes called Nash product. So what we would like to now look at is that the function x1 x2 going to x1 minus v1 into x2 minus v2. So this is basically f2 or this function is strictly quasi concave. So let me define, we have introduced convex functions and concave functions but not quasi concave. So let me define what is quasi concave. A function f from some set S to R is quasi concave. Of course SIM is in convex sets. Quasi concave, if f of lambda x plus 1 minus lambda y is greater than equals to minimum of fx. This should be true for every xy in S and lambda in 01. If this condition holds true then you will call it as a quasi concave function and instead of less than or equals to, if I make it strict then it becomes a strict quasi concave. In fact I would like to point out the difference between concave function. In the concave function this would not be minimum of fx fy. It is simply lambda fx plus 1 minus lambda fy. And then in fact you can easily verify that every concave function is quasi concave but not otherwise. So now in fact what a simple exercise is that the Nash product is strict quasi concave for essential bargaining problem. So this is first thing. Then the second thing I would like to say is that quasi concave function will have a unique optimal solution of course here maximum. So again under the natural assumptions basically we need to assume this compactness of this thing. So we can say that this is basically there will be a unique maximum. And this is again not very hard to prove it. In fact if we go back to the convex functions that we have introduced use the same ideas and to show that this quasi concave function will automatically have a maximum. This I will leave it for you to fill these details. So these are not very hard. So now using this result Nash product has unique maximizer. So let me call this as x1 star x2 star. So remember we are always living in the compactness thing. All our solutions and everything we are making them to be bounded. If they are not bounded the solution becomes meaningless because we are maximizing them and if it is not bounded then one person can keep on increasing this thing. So therefore there is no solution. So therefore we are always in the boundedness set. So that is also necessary in this thing when I am talking about existence of an optimal solution. Okay so let us call for the Nash product that becomes a strict quasi concave for the essential bargaining problem and let us call that is maximizer to be x1 star x2 star. So now let us look at so let us take f in R2 let us take this is convex and close. Suppose then f intersection x1, x2 such that x1 greater than equals to v1, x2 greater than equals to v2 this is non-empty bounded. So now suppose f, fv satisfies the 5 axioms. Suppose fv satisfies the 5 axioms that is the strong efficiency and scale covariance independent individually rational the symmetry independent of irrelevant alternators all the 5 axioms if it is satisfied now clearly we can see that f this has to happen. Okay so if f satisfies this then f has to be this solution. So that is very easy to verify this fact so let us now slowly get into the other parts of it. Okay so let us we start using the notation n x1 x2 to be x1 minus v1 for the Nash product because we use this again and again. Okay so here of course we have to verify here all the 5 axioms satisfies that we will do it but first what we will do is that define part one in the part one define fv to be x1 star x2 star we show that f satisfies all 5 axioms. Okay that is the first part in the part 2. Okay so suppose f satisfies all axioms then fv is nothing but x1 star x2 star these are the 2 ways we have to prove it and we will prove all the things so let us start with this thing. Okay now in the part one x1 star x2 star satisfies all the assumptions this is what we need to prove so let us do it. Now the strong efficiency okay let us say if x1 x2 is less than equals to y1 y2 then n x1 x2 is certainly less than equals to n y1 y2 this is a very easy fact to observe that means if you take any vector and if you increase at least one of the coordinate then the Nash product is going to increase therefore what we can say here is that if x1 star x2 star is maximizing the Nash product there cannot be any vector which is higher than x1 star x2 star it follows from this thing therefore x1 star x2 star is strongly efficient so because if there is some vector which is higher than this that corresponding Nash product will be higher that follows from this fact therefore the strong efficiency automatically holds. Now individually rationality next assumption next axiom now if you look at the way we have defined x1 star x2 star we know that that is x1 star x2 star maximizes Nash product over all x1 greater than equals to v1 x2 greater than equals to v2 and by definition x1 star and x2 star should be bigger than v1 and v2 respectively so therefore individual rationality automatically holds okay then the scale covariance so let us take lambda 1 greater than 0 lambda 2 greater than 0 and then mu 1 mu 2 then look at the g2 v lambda 1 x1 plus mu 1 lambda 2 x2 plus mu 2 such that x1 x2 is in F. Now corresponding this thing max y1 y2 in G of y1 minus lambda 1 v1 plus mu 1 into y2 minus lambda 2 v2 plus mu 2. Now any y1 is in this fashion so therefore this Butler thing can be written as lambda 1 into x1 minus v1 plus mu 1 and minus mu 1 that gets cancelled here into lambda 2 into x2 minus v2 where x1 x2 belongs to F. Now lambda 1 lambda 2 are positive therefore maximizing this quantity is same as maximizing this into this over x1 x2 in F and therefore whatever maximizes this the corresponding lambda 1 x1 star plus lambda 2 x star maximizes in G the same Nash product. Therefore what we have proved here is that F g lambda v1 plus mu 1 lambda 2 v2 plus mu 2 this is the decrement of this one this is same as lambda 1 F 1 F F v plus mu 1 lambda 2 F 2 F v plus mu 2 so this is automatically follows. Next we need to look at the independence of irrelevant alternatives. So G is contained in F G is close convex now x1 star x2 star optimal to F v. Let y1 star y2 star optimal to let us say Gv. Now what we have is that y1 star y2 star is in G which is also in F therefore what we have is n x1 star x2 star the Nash product correspond to x1 star x2 star should be bigger than or equals to n y1 star y2 star that is the first. Now but y1 star y2 star is optimal to Gv therefore Nash product correspond to G the y1 star y2 star is the maximum therefore n y1 star y2 star must be maximum over all Nash products inside G but x1 star x2 star is in G by the independent irrelevant axiom therefore this is greater than equals to n x1 star x2 star so therefore both of them are one and the same thing therefore x1 star x2 star and y1 star y2 star they must be one and the same the reason why this is coming is that F is essential and the Nash product is quasi strictly quasi concave and therefore it has a unique optimal solution. So therefore x1 star x2 star should be same as y1 star y2 star which is coming because strict quasi concave this is necessary. So therefore independent of irrelevant alternatives also holds then the final thing that is symmetry so what we have is that if x1 x2 belongs to F implies x2 x1 is also in F in fact that is essentially this thing now and we also have v1 is equals to v2 therefore x1 star x2 star maximizes x1 minus v1 into x2 minus v1 because v1 and v2 are same this thing is there. So x1 star x2 star maximizes this not only that x2 star x1 star also will maximize this because of this symmetric nature v1 is same as this thing these two are same therefore if x1 star x2 star maximizes this x2 star x1 star also will maximize now we have the unique maximum quasi strict concave gives that there is a unique maximizer therefore x1 star x2 star should be same as x2 star x1 star by quasi concave and the strict therefore x1 star is same as x2 star that is the symmetric. So therefore the symmetric condition axiom is also satisfied and with in this thing. Therefore if I define this solution like this then I know that all these 5 axioms are satisfied now the next part is part 2. So what we have to show is that suppose F the solution satisfies all 5 axioms then we need to show that FFy maximizes the Nash product. So let us say need to show FV is nothing but x1 star x2 star where x1 star x2 star maximizes Nash product. So I am not writing this notation again here x1 star x2 star is the one which maximizing Nash product and then I need to show that if F satisfies all these axioms this must be true. Now first thing that we would like to say is that note that x1 star is strictly bigger than v1 similarly x2 star is strictly bigger than v2. This is happening because F is essential because if F is essential there is at least one vector which is higher than v1 and v2 both therefore the Nash product should be bigger than those things and hence this happens. Now consider the following thing consider lx1 x2 is equals to lambda1 x1 plus mu1 lambda2 x2 plus mu2 where lambda1 I will choose it to be 1 by x1 star minus v1 lambda2 is 1 by x2 star minus v2 mu1 is equals to minus v1 by x1 star minus v1 mu2 is minus v2 by x2 star minus v2. I consider this. Now in other words what we have is that lx1 x2 is nothing but x1 minus v1 by x1 star minus v1 x2 minus v2 by x2 star minus v2. This is transformation that we are considering it. Now once you look at it first thing I would like to notice is that lv1 v2 is 0 0 lx1 star x2 star is 1 1. Now define g2 be set of all lx1 x2 such that x1 x2 is in f therefore now the problem fv is now transformed to g0 0 by the scale this thing by scaling this thing. Now so therefore what should we now say when you do this one all these terms if we really look at it okay so this particular thing satisfies at x1 x1 star 1 1 is the maximum that can come. In fact 1 it is easy to verify that 1 1 is strong Pareto efficient point of g. This is by looking at this Butler thing one can see this one by the Nash product and other thing. Now therefore g0 0 the solution of this has to be 1 1 okay so in fact here the Nash product is going to be x1 minus 0 into x2 minus 0 which is simply x1 x2. So in fact we can show that x1 plus x2 is always less than equals to 2 that is that comes from here if I take the this plus this we can actually try to show that this is always less than or equals to 2. In fact we can prove it via contradiction so let us to show this let us assume x1 plus x2 is strictly greater than 2 okay. Suppose alpha is a number in 0 1 and take t to be 1 minus alpha into 1 1 plus alpha into x1 x2. So this is nothing but 1 minus alpha plus alpha x1 1 minus alpha plus alpha x2 okay. So g is convex 1 1 is in g therefore t is in g okay what is now t1 t2 that is nothing but 1 minus alpha plus alpha x1 into 1 minus alpha plus alpha x2. So if I calculate this one this is going to be 1 minus alpha square plus alpha square x1 x2 plus 1 minus alpha into alpha into x1 plus x2 okay. Therefore because we have assumed that x1 plus x2 is greater than 2 if x1 plus x2 this is greater than 2 then what we can say that t1 t2 this is greater than 2 I am just putting that one here is that then t1 t2 is bigger than 1 minus alpha square plus alpha square x1 x2 plus 1 minus alpha into alpha because alpha is a number between 0 and 1. So therefore this should be strictly smaller this is same as 1 minus alpha into plus alpha square x1 x2. Now we can choose alpha sufficiently small such that we can choose alpha and make t1 t2 bigger than 1 because if alpha is sufficiently small minus alpha plus alpha square x1 x2 I can make it strictly greater than 0 therefore this becomes at least bigger than 1 therefore t1 greater than t2 this is contradiction because we know that 1 1 is the t1 t2 cannot be 1 this is basically this thing okay. So the G is bounded that we have so we can always find rectangle H which is symmetric about line x1 x2 we can always find rectangle H which is symmetric which sits inside which is outside G that means G is contained in H H is convex bounded this such a thing is always possible. Further choose H such that 1 1 in G is on the boundary of H we can also choose this rectangle in such a way that this 1 1 is on the boundary of H. Now strong efficiency implies F H 0 0 should be 1 1 and now using the independence so for this thing this is same as F G 0 0. Now F 0 G 0 this is nothing but L F F V because of the scale the invariance and this thing. Now this implies L F F V is nothing but our 1 1. Now if I write down everything this implies F F V has to be x1 star x2 star star. So this becomes the this thing. Now this becomes a it proves the fact that if F satisfies all these axioms then the solution is nothing but the solution corresponds to the Nash bargain. Of course so far we have considered the assumption that the bargaining problem is essential. So now we need to look at the non-essential. Now consider F V which is inessential. So therefore this immediately implies F is convex implies there exists at least one player i such that y1 greater than equals to v1 and y2 greater than equals to v2 implies yi is equals to vi for all y1 y2 in F. That means any point if you take it which is bigger than v1 and v2 one of the player there must be a player i such that yi and vi must be same. So that is that happens. So if this does not happen that means y1 greater than v1 and y2 greater than v2. So both will happen. So therefore that is not we are assuming F V to be inessential. Without loss of generality we can take that y1 greater than equals to v1 y2 greater than equals to v2 implies y1 is equals to v1 for all y1 y2 is equals to F. Basically we are considering this normalizing this player one of the player has equality always. So now suppose x star is an allocation in F that is best for player 2 subject to constraint x1 is equals to v1. So what is the best when x1 is equals to v1 and look at the best allocation for the player 2. So this is there. In fact under this constraint note that the Nash bargaining product will be 0. Note that under the inessential thing Nash bargaining Nash product is always 0. So therefore there is no whatever maximizes that means everything should give 0. So therefore that is one point which I will come back to this again. This implies x star is Pareto efficient. So in fact strongly Pareto efficient that is automatic and individually rational all the things come axiom 1 and 2 are already satisfied. Therefore we can easily say that f of fv is going to be x1 star x2 star. Now in fact this achieves even the maximum of the Nash product. Now the remaining thing is we have to verify all the other axioms and in fact they are not very hard to show it because once we assume that one of the player has this thing and in every whenever you scale it the same condition holds true and everything can be proved. So this proves even the inessential part. So the details here has to be fixed which I will leave it to you. And one point I would like to say that under this inessential the Nash product is always 0 therefore everything maximizes. So therefore the condition whatever we have said in the theorem that is automatically true. So with this we will conclude this session. We will continue in the next session. Thank you.