 Welcome back to this NPTEL course on game theory. Now, we consider the matrix games and we prove the phoneme and minmax theorem. So, the proof requires some facts from convexity, we will first derive those things and go on proving the minmax theorem. So, what we require is the following proposition, let C be a compact convex set of an Euclidean space and we assume that 0 is not in C, C does not contain the origin. Then there exists a vector z in Rn where Rn is this Euclidean space such that z dot x is greater than 0 for all x in C. In fact, I should put it. Now, let us go through the proof of this fact. Origin is not in C. So, let us say 0, C is something like this, this is origin. So, therefore, there will be a point which is the nearest. So, let me call this as a z, this point is the z. So, such a point always exists, the nearest point exists. So, this can be taken as an exercise. Once we know that this nearest point exists what it gives us is the following thing. What we have is that the distance from origin to z that is mod z square is always less than equals to, let us put a norm here. Norm z square is always less than equals to norm x square for all x in C. In fact, z is unique point here. The convex T in fact tells you that. Anyway, let us not worry about that but let us prove this next part. So, as we said this is z and this is origin 0. Take any point x here inside the C and then take the any point on this thing. So, take alpha in 0, 1 open interval 0, 1 and look at 1 minus alpha z plus alpha x. Now, this clearly belongs to C. Therefore, what we have is that norm z square is less than or equals to norm 1 minus alpha z plus alpha x square. Now, what is this value? This value let us calculate. This is nothing but 1 minus alpha whole square norm z square plus alpha square norm x square plus 2 alpha times 1 minus alpha inner product of z and x. Now, we substitute this into this equation star. If we do that one, what we have is this norm z square which is less than or equals to 1 minus alpha square norm z square plus alpha square norm x square plus 2 alpha 1 minus alpha z dot x. Now, taking this norm z square to this side, what we have is 0 less than or equals to 1 minus alpha whole square minus 1 that becomes alpha square minus 2 alpha. So, that can be written as alpha into alpha minus 2 norm z square plus alpha square norm x square plus 2 alpha 1 minus alpha z dot x. Now, alpha is in open interval 0 1 that is by definition we have considered this. So, therefore, 1 alpha can get cancelled dividing this by alpha what we have is 0 less than or equals to alpha minus 2 norm z square plus alpha square norm x square plus 2 into 1 minus alpha z dot x. Now, what we do is that let alpha goes to 0. Letting alpha goes to 0, then what we will get is that 0 less than or equals to minus 2 norm z square plus this becomes 0, this remains as 2 z dot x. And now 2 gets cancelled here, this implies norm z square is less than or equals to z dot x. This is for all x in C. This proves this proposition here I have we have only z dot x greater than 0, but here I only have said norm z square less than or equals to z dot x. So, therefore, there is one more step to do it what we have is what we proved norm z square is less than or equals to z dot x for all x in C. Now, note that z is not equals to 0. Since z is not equals to 0, this implies norm z is not equals to 0. Therefore, what we have is that 0 is strictly less than z dot x for all x in C. This proves the fact what we intended to prove here. So, this is one step that we require, then we need to prove another proposition let a be any matrix order n by n. Then alternatives that I am saying there exists x, okay here I have to there is a small mistake here, it is not n by n, it is m by n. There exists x in R m which of course x not equals to 0 and all the coordinates of x are non negative such that x prime a is greater than equals to 0, okay. That is one alternative and the other alternative is that there exists y in R n once again y not equals to 0, y greater than equals to 0 such that a y less than or equals to 0, okay. So, what this proposition is saying that given any matrix of order m by n there will be either a point x in R m which is non negative all the entries of x all the coordinates of x are non negative and it is strictly different from 0 such that x transpose a is greater than equals to 0 or there will be a y in R n again non negative coordinates and different from 0 such that a y is less than equals to 0, one of this always holds. So, let us prove this fact. So, let E 1, E 2, E n be the standard unit vectors in R n. What I mean here is that E i consists of all 0s but in the ith position it will be 1 and rest of them are 0s only in the ith position it is 1 and rest of the positions are. So, let us say these are all vectors in R n, okay. Let the rows of a be denoted by a 1, a 2, a m they call a is an m by n matrix. So, therefore there are m rows and each row is a vector in R n. So, let us look at them and I write them as a 1, a 2, a m. So, we take c to be the convex hull of minus E 1, minus E 2, minus E n then take a 1, a 2, a m. Look at all the negatives of a 1, a 2, e n and a 1, a 2, a m these are finitely many vectors and look at their convex hull. So, it is clearly c is compared of course convex, okay. Now, there are 2 cases. Now, 2 cases one is 0 belongs to c or 0 does not belongs to c. So, either c will contain 0 or not. So, let us consider the case where 0 is in c. So, case 0 belongs to c. Let us look at this case. So, if 0 is in c therefore there exist non-negative real numbers alpha 1, alpha 2, alpha m, beta 1, beta 2, beta n such that minus beta 1, E 1, minus beta 2, E 2, minus beta n, E n plus alpha 1, a 1 plus alpha 2, a 2 plus so and so alpha m, a m is equals to 0. So, such a non-negative real numbers I am assuming them to be non-negative. This comes because c is the convex hull of this. So, convex hull means they are the convex combinations of these vectors. So, that is basically the reason why such a thing happens and most importantly this beta 1 plus beta 2 plus beta n plus alpha 1 plus alpha m all of them equals to 1, okay. So, would like to climb not all alpha 1, alpha 2, alpha m are 0. All this alpha 1, alpha 2, alpha m these cannot be 0, all of them. Why is this? If all of them turn out to be 0 then what happens is that this linear combination here, these terms become 0. What it says is that minus beta 1, E 1, minus beta 2, E 2 so and so, minus beta n, E n is 0, but even E 2, E n are the standard unit vectors. They form a basis of R n. So, beta 1, beta 2, beta n have to be 0, but that will be contradicted by this fact. Therefore, all of this cannot be 0, alpha 1, alpha 2, alpha m these cannot be all of them together cannot be 0. So, that is the first thing. Therefore, what we have now is the following thing. Therefore, we have non-negative real numbers alpha 1, alpha 2, alpha m not all 0 and they satisfy the following thing alpha 1, A 1 plus alpha 2, A 2 plus alpha m, A m this is same as I will put it as a vector y z here where z is nothing but beta 1, beta 2, beta n which is in R n and remember z is non-negative. Now, what is this quantity? Therefore, this is nothing but alpha 1, alpha 2, alpha m if I take this as a vector x, what we have is x transpose A which is nothing but our z which is greater than equals to 0. Of course, sometimes I am putting transpose this is I have written it as a row vector, but in general we always take this as a transpose all these vectors we want them as a column vectors, you must understand from the context whether I am taking the this thing. So, but to be on a safer side we always take them as a column vectors. So, therefore x transpose A which is same as z which will be greater than equals to 0, this particular quantity is nothing but x transpose A. So, this proves the first part, first alternative let us go back and see this one, the first alternative is now proved under the case 0 is in C. Now, let us look at the other. So, now we look at the case 0 does not belongs to C. If 0 is not C then from the previous proposition there exists z such that z dot x is greater than 0 for all x in C. So, this is there. Now, remember what is C the definition of C if you look at it, it contains all these negatives of this unit vectors. So, by definition of C minus EI belongs to C, this implies z dot minus EI is greater than 0, this implies the z I is a negative number, z is are going to be negative number. Now, also note that AIs belongs to C by the definition of C again, this therefore AI dot z this should be greater than 0 for all I 1, 2 up to M, this implies that A z is greater than 0 and z has negative entries, this immediately implies that A y is negative with y is equals to minus z which have non-negative entries. So, this proves the second alternative. So, this proves second alternative, thus we have proved the basic ingredients required for the proof of this theorem. So, now we need to prove the min max theorem. The statement we have already made it earlier, what the statement says is that there exists x star, y star, the mixed strategies pair delta 1 cross delta 2 such that pi x star, y star, where pi x, y star for the player 1, x star maximizes when the player 2 fixes to y star and then for the player 2, y star minimizes when player 1 plays x star. So, this is the statement. In other words, the saddle point equilibrium exists in mixed strategies. So, we need to prove this fact. We will use this two facts to prove this one. So, we have two things, two cases. There exists x greater than equals to 0, of course, this is in Rm and of course, x is not equals to 0 such that x transpose A is greater than equals to 0. This is one alternative. The second alternative is there exists y greater than equals to 0, which is in Rn, again y not equals to 0 such that Ay is less than equals to 0. So, one of these two we have. So, let us look at the consequences of these two things. Take x bar to be x by summation xi, i is equals to 1 to m and y bar to be y by summation yj j is equals to 1 to m. Basically, we are normalizing this vector x and y. So, you can say that x bar belongs to delta 1, y bar belongs to delta 2. Now, they become a probability vectors. Now, what we have is that either x bar transpose A, this is greater than equals to 0 or Ay bar is less than equals to 0. One of these two alternatives happen. Now, the first case. So, recall first case is x bar A greater than equals to 0, x bar transpose A greater than equals to 0. This implies x bar transpose Ay is greater than equals to 0 for all y in delta n. This implies, this is true for all y. So, therefore the minimum y in delta n x bar transpose Ay is greater than equals to 0. This implies max x in delta 1 min y in delta n of x transpose Ay is greater than equals to 0. This is nothing but the lower value of the game. This we denoted by v minus of A is greater than equals to 0. So, this is the first thing. In the second case, the second case what we have is that let us look at it again here, Ay bar less than equals to 0. This immediately tells me that x transpose Ay bar is less than equals to 0 for all x in delta 1. This implies max x in delta 1 of this x transpose Ay bar is less than or equals to 0. So, this implies minimum y in delta 2 max x in delta 1 of x transpose Ay is less than equals to 0. And this is nothing but the upper value of the game which we denote by v plus A which is less than equals to 0. So, what we have? Thus we have either v minus A is greater than equals to 0 or v plus A is less than equals to 0. So, what we have is that the value, the lower value of the game either it is non-negative or the upper value of the game is non-positive. So, now what we can do here is that now because we started with a matrix A, take another matrix B where B is nothing but this A ij is minus a constant C. Basically A minus C C C C C you are reducing all the entries by a constant C. Now, here is an interesting exercise is that v minus of B is nothing but v minus of A minus C and v plus of B is nothing but v plus of A minus C. This is what happens. Now what it says that we know v minus A is less greater than equals to 0 or v plus A less than equals to 0. Now using that here what we get is that thus we must have v minus A greater than equals to C or v plus A less than or equals to C and now C is arbitrary. Moreover by definition v minus A is less than equals to v plus A. Now see the arbitrariness of C and this fact immediately tells us that v minus A has to be same as v plus A. Therefore, the two values, the lower value and upper value are same. Therefore, v minus A is equals to v plus A. Therefore, the game has a value. Now one further point to conclude here is existence of saddle point equilibrium. How do we show that the saddle point equilibrium exists here? So here I will leave it as a small exercise is that take the outer minimizer in y variable and outer maximizer in the x variable. So what I mean is that take x star in delta 1, y star in delta 2 such that following thing happens y in delta 2 of x star transpose A y is same as max x in delta 1 mean y in delta 2 of x transpose A y. Similarly, max x in delta 1 of x transpose A y star is same as mean y in delta 2 max x in delta 1 of x transpose A y. Choose x star and y star satisfying this such a thing exists because delta 1 delta 2 are compacts and this function the mixed extension mixed payoff function is a bilinear function which we have seen earlier. Of course, in earlier we know we identified this as pi x comma y this mixed extension mixed payoff is a bilinear function and this compactness everything ensures that such an x star and y star exist and the exercise here is show x star y star is saddle point equilibrium. In the proof you need to use the fact that these two quantities are one and the same. So once you prove this one the theorem the proof is complete. This proves the min max theorem for bi matrix games. So this proof only requires the convexity ideas. We conclude this session here we will come back to another topic in the next session.