 Welcome back to this NPTEL course on game theory. In the previous session, we proved that Shapley value is the unique allocation rule satisfying certain axioms. Now, in this session, we will discuss another class, another notion of a solution to cooperative games. Known as a nuclear loss, the concept of nucleus depends on what is called excess of a correlation. So let us S is a correlation of N, the excess is defined as the following thing. So E X S is V of S that is the, V S is basically the worth of the correlation S minus how much the people in the correlation S are allocated in this imputation X, here X is a imputation. Recall imputation is defined in the previous lectures, it is basically X is an allocation vector which is both individually rational and collectively rational. So this, of course this we are defining it for any subset S of N. So therefore, there are 2 power N minus 1 X S. So now what is X S? So basically in the allocation vector and their total worth and the deviation from the total worth of that correlation S and this is what. So now this also measures, this measures dissatisfaction with an offered imputation X in a correlation. So in other words what I am trying to say that if S is the correlation that you are doing this V E X S, the X S function here is basically telling you because X is the imputation that is offered to us. So how much dissatisfaction is there? So if V S is minus this thing, if the worth of V S is more that means they are not happy. So you want it as E X S should be as small as possible. So that is basically the idea of this thing. Here the following vector, E X this is basically nothing but E 1 X E 2 X E M X where M is of course nothing but 2 power N minus 1. What is E 1 X? E 1 X is basically E of X X S function corresponding to some correlation S 1 like this E M X is also X S function correspond to some correlation. How are these chosen? S 1, S M are indexed according to the following. The E 1 X is largest X S function E 2 X basically the various correlations look at the maximum excess over all correlations. So that I am writing it as E X S 1 that is E 1 X and the least dissatisfaction is given by the correlation S M for the particular imputation X so that is E M X. So now I have ordered all of them and then I have written this vector E X vector of excesses is in the decreasing order here E 1 X E 2 X E M X this is in a decreasing order. Now to define the concept of this thing I need to introduce some this. So let us take 2 vectors X Y in R M I will say that X is lexicographically smaller than Y if the following thing happens basically this is denoted according to X less than equals to E Y is the notation. What it says is that of course E X the excess vector so there is a small error here this is R N we are taking the 2 allocations X and Y and then I am saying that X is lexicographically smaller than Y this I am using the excess vectors if both are same then of course X is less than equals to Y and Y less than equals to X or there exists some K such that 1 less than equals to K less than equals to M and E I X is same as E I Y for I that is up to K minus 1 and E K X is strictly less than E K Y it is a dictionary ordering so you are looking at it if the first E 1 X E 2 X this thing E 1 X is smaller than E 1 Y then I will say that X is less than equals to Y similarly if X first 2 coordinates are same E 1 X and E 2 X E 1 X and E 1 Y are same look at the next coordinate which is smaller and like that go on. So this is exactly the dictionary order. So now once you define this dictionary order on this E X's that is influencing an order on the allocation vectors X and Y then we take that as this thing. So what now we see is that the lexicographic minimum the lexicographic minimum is called is called the nucleus of the game. So with respect to this ordering if there is an allocation vector X which has the minimum which is the minimum that means every other allocation vector has a bigger access vector than this then I will call that particular allocation such an allocation vector is the nucleus of the game of course does this exist. So in fact the there is a very important result here let me put it this thing there exists a unique nucleus for each cooperative game. So the theorem actually says that given any cooperative game and we there is always a nucleus in fact it is unique. So now let us compare with the three notions that we have seen so far. The first we have seen the core we know that core can be multiple the size of the core can be as large as possible. The second is Shapley value which is unique which is actually a very famous concept that is used in the cooperative games Shapley value is uniquely characterized and now we are seeing the nucleus once again this is a unique this. So nucleus and cooperative game enjoys the uniqueness whereas the core need not be unique. So now first we will work out some examples and then we will go to the other parts. So let us look at an example. So let us say a small company goes bankrupt owing money 3 credits. So for a let us say some 10,000 not specific units for B 20,000 and for C let us say 30,000 but company has 36,000 only basically to cover this. So how should the money be divided? So let us look at this particular example. So for A they have to give 10,000 for B 20 is for C 30 and but the company has only 36,000. Now how this money be divided? So there are several ways you can do it for example you can say the pro rata why can do it but now we will actually work out how this nuclear less and for example one Shapley value help here. So let us write down the characteristic function there are 3 players. So these are the 3 players and then characteristic function V of empty set of course is 0 and V of ABC is 36. So measured in 1000 so let me not worry about that and so let us say if V of A what will be V of A? If V of A is 0 the reason is that B and 6 gets 20 and 30. So if A alone is coming that B and C can together climb the money and then they can try to split therefore V of A becomes 0. Similarly V of B will also be 0 because A and C their climbs is bigger than 36. So therefore V of B will also be 0. So V of C if you look at it if A and B form a coalition if I divide the if they take their money then 30,000 the remaining 6. So therefore this will be 6. Similarly we can find that if we write down in the same fashion V of AB will be 6, V of AC will be 16, V of BC will be 26. So the same way we can argue and then this thing. So now this is basically the coalition form game now let us take an allocation X to be X1, X2, X3. Let us say this is a efficient allocation. Recall what is an efficiency X1 plus X2 plus X3 has to be 36. Now we will look at the X's functions. So we have to write down here S V S E X S. So if I take A as this thing V of S is 0 if A is the coalition S is equals to A then V of A is 0 and E of X comma A because V of S is 0 this will be simply minus X1. So B if I take it 0 this is minus X2. So like that I can actually calculate all the things C for C it is 6 it is 6 minus X3 AB this is 6. Therefore this will be 6 minus X1 minus X2 AC is 16 this is going to be 16 minus X1 minus X3 BC is 26 and then this is going to be 26 minus X2 minus X3. Then these are the excess functions that we have defined. Now what we really need to now look at is for different X's allocation vectors what you have to calculate these values for example if I take X to be the 6, 12, 18 we can calculate this to be minus 6 minus 12 minus 12 this is also minus 12 this will be minus 8 minus 4. So like that we can do it for for example 5, 12, 19 if I take it we can calculate this as minus 5 minus 12 minus 13 minus 11 minus 8 minus 5. Like that we can do it for many this thing. In fact by looking at these things and by little work we can say that the nucleolus here is going to be 5, 10.5, 20.5. So it requires one to calculate this because it you remember we have to for each X we have to look at the order them and then try it. So here it is worked out for two examples but one has to be little careful to see what exactly is the nucleolus. So it requires a little work but we can do it. Now how does the Shapley value here? In fact we can say here is that Shapley value is going to be 6 here, Shapley value for the B is going to be 11, Shapley value for C is going to be 19. Just apply the formula and then they use this thing and what you have here is the allocation vector here is 6, 11, 19. So in fact if you try to compare these two examples we can easily see that these two are not same. So this is a very important point to observe here. Now we will try to give some ideas of the proof of this nucleolus in existence of nucleolus. Let us say look at the proof. Of course we will give only some ideas of existence. So let us first remember the following thing. So the components what is E1X? E1X is basically the maximum of all the excess functions. So it is a maximum of all I1 to M recall M is 2 power N minus 1 of the excess functions, maximum of all those things. Now what is E2X? E2X is the second max of this thing. So in fact it can be written in the following way minimum J is equals to 1 to M of max I not equals to J E X S I or E I we can put it. So let me write this is nothing but E I X S I or E X S I. So you take pick any J and look at the maximum of all the I's such that I not equals to J. So for example to illustrate this one let us take J is equals to 1. So what is this one? Max of I not equals to 1 E X S I that means exclude E X S 1 and take the maximum of the remaining N minus 1. Now suppose look at this one. Suppose the one corresponds to the maximum of that E1X let us say then here this will be the second maximum because E X S 1 is removed and maximum among them. Now in all the other things E X S 1 will appear because when you take J to be different from 1 then 2, 3 and the 1 is always there so those appear here. So therefore the when I take the minimum of them this is nothing but the second max among the excess functions. Now like that E 3 X will be the minimum of J K. Now we take 3 indices and then you take J different from K and then take the max I not equals to J not equals to K of the E X S I. Now because you take 2 at a time you are removing 2 of them and then looking at the maximum. Now if both J and K are exactly the first max and second max then that will give you the minimum among this one and then like that. Now this way you can write all the things. Now the excess functions now clearly can be written in this form. Now most important thing to note here is that these excess functions E 1 X E 2 X E 3 X these are all continuous functions. So all of these are continuous functions. So therefore the maxima and minima the reason is set of all imputations if you really go back and see it because they are efficient and other things they are positive and all they are that is a set of imputations is a compact set of all the allocation vectors because every person is satisfying individual rationality they are above something and then the sum of them is less than equals to V n. Therefore the imputations are always the set of imputations is always a compact set. Now these are all continuous functions. So therefore every function all these E 1, E 2, E X they enjoy this existence of maxima as well as minimum that we have used it earlier also. So therefore okay now E 1 X for example let us say look at the E 1 X let us say it attains minimum value. So E 1 X attains its minimum value because E 1 X over set of all allocations X is a continuous function over a compact set. So therefore it will have a minimum value. Now that means E 1 X at this minimum wherever this minimum is achieved that is going to be the point where the excess function can have the minimum value. In fact whenever such if the minimizer is unique then this minimizer is going to be nuclear less. So there is nothing else to be done when you have such a minimizer here but the problem is that this minimizer need not be unique there may be a set of minimizers. So once you know that there is some set of minimizers what you need to go here is that look at the next level E 2 X okay and then E 2 X is again a continuous function. Now look at the set of all minimizers that E 1 X is minimized and then on that particular things then look at the second function E 2 that also will have a minimizer and go on. So once for example if U 2 X has a unique minimum then automatically that allocation is now fixed. Of course remember when I go for minimizing E 2 I only need to look at the set of minimizers of E 1 and therefore E 1 is already minimized that means E 1 X will be same for any 2 allocations in the set of minimizers and then look at E 2 if the minimizer of E 2 over that is unique then that gives you the nuclear less if the set of minimizers are not unique then I go to E 3 and then go on one by one. Now in this fashion if you really go it there is a some set where all these things are minimized therefore this idea gives existence. Now the problem here is that how about uniqueness. So in fact the uniqueness can actually be proved but it is slightly tricky we will not go into the details but the idea of this uniqueness is that you have this set of minimizers as a convex sets and then the intersection of all these convex somehow that they become really unique. So I will not go into the details but the details can be for example found in the original paper of Schmidler. Schmidler is the one who developed this notion of a nuclear less. So original paper of Schmidler or in fact we can look at for example Schmiedler's book of game theory. In fact this is the book the material from this book is used heavily in this course. So now of course we will not discuss about the uniqueness but existence is proved. Now note that before we go further I would like to make a point here that the cooperative games we have multiple notions we are defining it so and particularly the nucleus and Shapley value are unique but then if you look at it the computing nucleus is harder than computing the Shapley value because Shapley value is given by a formula and computing nucleus is a little harder. Now there is one very interesting question that people ask here is that are there some learning methodologies or some non-cooperative framework to these cooperative solutions. So this is a very interesting problem in fact there is a lot of work happening but we will not touch upon this but in fact a good point to look at it is to start with this Nair Harri's book or even Maier-Sens book on game theory. So with this we will stop in the next session we will consider another problem from commutative games.