 Welcome back to NPTEL course on game theory, so we continue our discussion about mixed strategies. So let us recall once again there are 2 players we are talking about mixed strategy, the strategy spaces X and Y, mixed strategy of player 1 is probability distribution on X, mixed strategy for player 2 is probability distribution on Y. So this space we denoted by PX, this will denote by Py. Now as I said last time if X is a continuum then describing this PX requires the knowledge of measure theory or probability theory, so we avoid going into those technicalities. So we concentrate on the case where X and Y are finite sets. So let us say X and Y are finite sets. So this is where the game becomes, this game can be expressed by matrices, matrix, so look at it. So X is let us assume 1, 2, M, Y is 1, 2, maybe K then what the payoff function is basically pi from X cross Y to R, so we have the following thing, pi 11, pi 12, pi 1K, pi 21, pi 22, pi M1, pi M2, pi Mk, now think about this as a M by K matrix, so let me denote this by A. So the player 1 is choosing the rows and player 2 this is about the player 1 and player 2 choice are basically the columns, the player 1 chooses one of the row and player 2 chooses one of the column that is exactly the X is basically the set of rows of this matrix and Y is the set of columns of this matrix and the payoff function is simply the entries in this matrix. This entry tells you that if the player 1 chooses the row 1 and player 2 chooses the column 1 then this is the amount that player 1 gets. So these are called matrix games, so they are given by a single matrix. So there are 2 players, row player and column player, so the row player is nothing but our player 1, column player is nothing but player 2, so you have a matrix, player 1 is choosing the rows, player 2 is choosing the columns and the corresponding entry gives you the payoff of player 1 or the row player and the same thing is what paid by the player 2. Now matching pennies is a matrix game because there are only 2 choices, so in this case the matrix will be 1 minus 1 minus 1 1, so if the player 1 chooses row 1, player 2 chooses column 1 then player 1 gets 1, player 2 gets minus 1, similarly this thing and similarly the rock papers is just game is another matrix game. So here let us recall this is rps, rps, this is the diagonal entries will be 0, if when rock and paper means the rock can be hidden by paper, so therefore this is minus 1, this is 1 paper and rock, so this is minus 1 and paper and scissors, this is plus 1 and scissors and rock means this is minus 1, scissors and paper means this is 1, this is the matrix game where the matrix is given by 3 by 3 matrix which is given like this, so this is a rock paper scissors game, so like that there are so many matrix games. So now we concentrate on matrix games, so these are given by a single matrix, we have already seen that in these matrix games for example as an example is matching pennies there is no saddle point equilibrium and that is where we started discussing the mixed strategies. There is a mixed strategies, player 1 is choosing rows, so he can choose any of the rows with certain probabilities, so he chooses any row with certain probabilities. So let us say row 1, let us say he chooses this with probability x1, row 2 with probability x2, so row m with probability xm and we want all this sum to 1 and of course all of them are non-negative, this is a mixed strategies and what is a mixed strategy for column player? Basically he choose column 1 with probability y1, column 2 with probability y2 like this column k he will choose with probability yk and we want summation yj to be 1, of course j runs from 1 to k and all these things are non-negative, so this is a mixed strategy. So we will say the mixed strategies of player 1, let me denote it by delta m and player 2 delta k, so the delta m of course we can easily see that this is a subset of rm, this is a subset of rk. Now let us look at some interesting properties, delta m is convex and compact, of course same way is convex and compact. So why is this convex? We will only prove for this case and the other things similar, so xz let us take 2 in delta m and we also take some lambda in 0 comma 1. So what is x? x is nothing but x1, x2, xm and we know that the summation xi is 1 and of course all of them are non-negative. Similarly z is z1, z2, zm and of course zi they are all non-negative and summation zi is 1. Now what is lambda x plus 1 minus lambda z, this easily you can calculate, so this calculation is straight forward and now we can easily verify that the summation of all these things lambda x1 plus 1 minus lambda z1 plus lambda x2 plus 1 minus lambda z2 plus so and so lambda xm plus 1 minus lambda zm, this one can actually verify this, it becomes 1 and leave this as a simple exercise, we can show it easily. So this says that this delta m is convex. Now also recall this delta m it is a simple exercise again once you know little bit of real numbers the delta m is closed as well as bounded. The boundedness is trivial because the entries of delta m all of them are less than or equals to 1. Any vector in delta m if you take it it is a vector in m dimensional rm and each entry in this vector is a non-negative number less than or equals to 1 and all of them sum to 1. So therefore all the entries are less than or equals to 1 and then that you can use it to show that the boundedness, what exactly the boundedness that is a little bit of a topological thing. So let us not go into those details and closeness again comes from multiple ways one can see it. So for example this delta m is basically all those vectors such that the sum of the components is 1 so one can also see that as a hyperspace. So hyperplane a linear relation satisfying and then that is the solution space of that I will leave it as simple exercises. So the delta m is convex similarly delta k is also convex of course compact. So that is next what is the suppose player 1 chooses a mixed strategy x in delta m and player 2 chooses y in delta k how much player 1 will get this we have not yet defined. So we need to define this one we only know this the payoff function only when they use their choices if they choose rho 1 or rho 2 or anything but now we are we are saying that the they can choose the those rows with certain probabilities. So now let us look at it so we need to define let me call it as a pi x comma y is this is what we need to get an expression for this. So let us do that one what is pi x y pi x y rho 1 is chosen with probability x 1 or let me put as rho i with x i column j is chosen with probability x not x y j therefore I and column j are chosen with probability x i into y j why is this product x i y j because as I as we discussed earlier in this scheme the most important thing is that the when a player is choosing his strategy the other players are completely unaware of it. So they are choosing completely independently simultaneously. So therefore if the rho i is chosen with a probability x i and column j is chosen with a probability y j rho i and column j together are chosen with probability x i into y j the independence is really playing a whole here. Therefore player 1 will get a i j the i j entry of the payoff matrix with probability x i y j therefore player 1 receives a i j with probability x i into y j this is as now sum over all j 1 to k and sum over all i 1 to m. So let us look at understand this one. So he receives a i j with probability x i into y j that is basically the his expected payoff. This is expected utility that player 1 gets under this mixed strategy x, y another mixed strategy pay. So the player 1 has chosen x player 2 has chosen y and then what player 1 will get is that summation i is equals to 1 to m summation j is equals to 1 to k a i j into x i y j. So let us understand once again why is this term because player 1 let us go back to this player 1 is getting a i j when player 1 chooses rho i and column j by player 2. Now the rho i is chosen with a probability x i and column j is chosen with a probability y j therefore the rho i and column j together are chosen with probability x i into y j and hence he will get a i j with probability x i into y j. And now this should be sum over all rows and columns that is why rho i over rho i and over columns j. Now this is known as the expected profit. This is let me denote it by pi x y. So let us try to understand this little more. So let us write it pi x comma y is nothing but summation i is equals to 1 to m j is equals to 1 to k a i j x i y j. So in fact the people can see it as the following thing. This is a matrix A this is m by k then y 1 y 2 y k this is the column vector I am writing it and I will also write here x 1 x 2 x m as a rho vector this is m by 1. So this is 1 by m this is k by 1. So now this is a like a matrix multiplication this we can also write it as the vector x A y or we also can put it as x transpose A y x transpose A y. So remember when I say y we generally write every vector as a column vector. So that is why y and x transpose whenever we are writing we are writing it as a row then I put it as a x transpose these are the various ways people do is. So if we understand this notation very clearly we avoid this transpose but you can write it as x A y or whatever way we like. So this is basically the matrix notation that we have used. This is known as expected payoff that player 1 receives from player 2 when player 1 follows the mixed strategy x and player 2 follows the mixed strategy y. Now the matrix game A now that has become a game delta m delta k and pi. So now this is compact and convex. So this is also compact and convex and what about this what kind of function this is. So let us check that what is this pi pi x y is nothing but I write I generally prefer writing inner products x A y. So we can easily see we can verify pi is linear in x when y is fixed. Similarly it is linear in y when x is fixed. So in fact the most important what is the linear in x means what here it is a if for example because this is the inner product if you have you might have already seen in some courses that if inner product x plus x prime A y is same as x A y plus x prime A y. So this is the linearity of course if I put alpha here beta here then this will be alpha this will be beta that is the linearity then similarly with respect to y. Now if you observe this is much stronger than the convexity or concaveity. So in fact what we can say here is that pi is convex in y for sure and it is concave in x. This we can I will leave it as an exercise just follow the definition of convexity and concavity and show that pi is convex in y and concave in x. In fact the linearity actually says that the pi is both concave as well as convex. So we can see. Now because it is if you recall the Minmax theorem can be applied. So what does the Minmax theorem say if the strategy sets x and y are compact and convex and the payoff function is concave in the maximizing variable convex in the minimizing variable then there is a saddle point equilibrium. Now all that conditions hold true here and hence from Minmax theorem saddle point equilibrium exists. So the Minmax theorem for this matrix game Minmax theorem for matrix games is basically due to von Neumann I think around 1918 has proved this result around this time and this is basically the first result formal result which actually paid way for the subject game theory. So now once we understand this matrix games and this notion of equilibrium here and we know that this saddle point equilibrium always exists from the Minmax theorem. Now what remains for us now is to prove this Minmax theorem. So there are multiple ways of proving this result. So we will prove some way which does not use much technical details because any of this existence of equilibrium they require a fixed point. We will try to avoid the fixed point theorems here. Of course in the matrix game case this can be proved using the separation theorems in the convex in the convexity we use something called separation theorems. So that requires us to introduce more notation. So instead of that we use basic calculus and optimization. So we will try to prove the Minmax theorem under the convex assumption. So now we will try to see the proof of it and we will continue the in the next session.