 Welcome back to the NPTEL course on game theory. In the previous lecture we have proved the min-max theorem for finite games. Now let us explore some properties of the saddle points. So we want to start with the following proposition. Suppose x star, y star and x hat, y hat are two saddle points. The two saddle point it will be a of the matrix game A. Then x star transpose A, y star is same as x hat transpose A, y hat. So how do we prove this one? So this is a very important point. Why is that whatever if there are two saddle point it will be both give the same payoff to the player one and same payoff to player two. So how do you prove it? In fact the proof is not hard, the proof follows from the following thing. This follows from the fact x star transpose A, y star is same as minimum y in delta 2 max x in delta 1 of x transpose A, y which is also same as max x in delta 1 min y in delta 2 of x transpose A, y. Once I observe this fact this proposition becomes immediate. Why is this true? Let us look at the definition of x star, y star, x star, y star. So just for the sake of simplicity let me start using this language which we introduced earlier pi x, y is nothing but x transpose A, y. So with this notation pi x star, y star has the following property. Pi x, y is less than equals to pi x star, y star, pi x, y star which is less than equals to pi x star, y for all x in delta 1, y in delta 2. So now look at this fact. This immediately tells me that max x in delta 1 of pi x, y star is less than equals to pi x star, y star. Now if I look at this fact, this tells me that this pi x star, y star is also less than or equals to minimum over y of delta 2 of pi x star, y. So let me write down in the next line again. What we have is max x in delta 1 of pi x, y star which is same as pi x star, y star which is same as minimum y in delta 2 of pi x star, y. This is what we have it. Now look at this one. This immediately tells me this is basically maximum of this quantity at y star. So therefore min y in delta 2 max x in delta 1 of pi x, y is certainly less than or equals to max x in delta 1 pi x, y star. This is true. Similarly for this thing I can make another, this thing is that min y in delta 2 pi x star, y is certainly less than or equals to max x in delta 1, min y in delta 2 of pi x, y. Putting all these two things what we have is that min y in delta 2 max x in delta 1 of pi x, y is less than or equals to pi x star, y star which is less than or equals to max x in delta 1, min x in delta 2 of pi x, y. Now the interesting thing is that for any function pi x, y the max min is always smaller than this. But max x in delta 1, min y in delta 2 of pi x, y is always less than or equals to min y in delta 2 max x in delta 1 of pi x, y. Recall this is defined as v minus of a, this is defined as v plus a. What we can always prove without any problem which is, we have done it earlier and I leave it as an exercise now is that v minus a is always less than or equals to v plus a. But in the previous thing what we have proved is that but we proved v plus a is less than or equals to pi x star, y star which is less than or equals to v minus, this immediately tells me that v minus a is same as v plus a with same as pi x star, y star. That means if x star and y star are two saddle point equal or is a pair of saddle point equilibrium, pair of strategies of player 1 and player 2 which is a saddle point equilibrium then the value corresponding to that x star and y star is same as the lower value which is equal to upper value. And this immediately this proves that any two saddle point equilibrium, any two saddle point will give same value. What I mean is that pi x star, y star is same as pi x hat, y hat for any two sp, x star, y star and x hat, y hat. So this proves the theorem. So this is an interesting property which we will come back to this again and again later that any two saddle point equilibrium gives you the same value. Now there is another interesting property. If x is hat, y hat and x star, y star are two saddle point equilibria then x hat, y star is also a saddle point equilibrium. So this is again another interesting property. We have a two saddle point equilibrium and x hat, the optimal strategy for the player 1 in the first saddle point equilibrium and the optimal strategy for the second player in the second strategy if you take them then this thing. This is known as the exchangeability which is a very important property which says that the in the zero sum games the saddle point equilibrium have exchangeable property. So the proof is again not very hard. In fact if you really the proof is not hard, the proof follows essentially from the fact that if you really go back here we have this from here because these two are equal what we have is the following thing. Pi x star, y star is nothing but the max x in delta 1 pi x, y star and pi x hat, y star you need to show that will satisfy the same inequality here. So we have we already have the following pi x star, y star is same as of course mean this thing is the max y in delta 2 of pi x star y. So x star is basically the outer minimizer of this one. So therefore one fact that we can say this is nothing but mean x in delta 1 max y in delta 2 of pi x star y. So this is certainly less than or equals to max y in delta 2 of pi x hat y and pi max pi x hat y this is nothing but pi x hat for example y hat. So what it says that the therefore this inequality is true therefore what we have is that pi x star, y star is same as this max y in delta 2 pi x star y which is same as max y in delta 2 of pi x hat y and in fact this is same as this mean x in delta 1 max y in delta 2 of pi x y. So therefore both x star and x hat are the outer minimizers here. Therefore x hat y star is a saddle point equilibrium. A little more details required here some more details not hard would leave them as exercise this thing. So this is another very interesting property of this saddle points where we know that the saddle points have this exchangeability property. The another way to put this whatever we have been saying is that if we really look at it for any y in delta 2 let us look at set of all x star in delta 1 such that pi x star y is nothing but max over x in delta 1 of pi x y. For a fix if a player 1 fixes his strategy to y what are all the x star which maximizes this pi x y let me call them this set this is nothing but the best response of player 1 when player 2 fixes strategy pi this is known as best response of player 1 when player 2 plays y. Similarly best response of second player when player 1 fixes x this is nothing but all y star in delta 2 such that pi x y star is nothing but minimum over y in delta 2 of pi x y. So this is the best response of player 2 when player 1 plays x. Now would like an interesting property that x star y star is saddle point equilibrium if and only if x star belongs to best response of player 1 when player 2 fixes to y star and y star is in best response of second player when player 1 fixes to x star. So this in fact the previous argument show that this is true I leave this as a exercise. But before going further would like to say that this best response of player 1 when player 2 fixes y is a convex and compact set another way is another thing best response of second player for when player 1 fixes x is also a convex and compact set. So this follows from this definition. So this is basically if you look at it x star x star are nothing but the maximizers of pi x y and remember pi x y is a linear function x variable. So you are maximizing a linear function over a convex and compact set therefore the set of minimizers is convex as well as compact and hence this Br1 y is convex and compact. So that is exactly what is written here. So therefore these best responses are convex and compact set. So this best responses will become more and more handy when we go for non-zero some games. But right now I would like to point out that a point x star y star a pair of mixed strategies x star and y star is a saddle point equilibrium if and only if x star is a best response to player 1 when player 1 fixes player 2 plays y star y star is a best response to player 2 when player 1 plays x star. Now another point I would like to say that even though we are proving these things for a matrix games these results go through even when we go for a general games like of course we need to make sure that the payoff function satisfies the concave convex property which we have assumed in the min-max theorem in a general setup. If you go back to previous lectures you will see that we have assumed the payoff function is concave in the maximizing variable and convex in the minimizing variable and under that assumptions these results will hold true. And proving them for in that general setup is not hard and I will leave that as a homework. The next we see the linear programming connection. So what would like to say is the following thing. Now we are restricting to matrix games so the payoff function for us is pi x y is x transpose a y this is the setup that we are in now and would like to say that this finding this saddle point equilibrium has a connection with linear programming. So the problem here right now we want to address is that we know that the matrix games admit a saddle point equilibrium or not only matrix game even if you assume the general conditions convex concave convex functions we know that the equilibrium strategy saddle point equilibrium exists. But how do we compute it? So the computation is a altogether different issue and here in the finite matrix games we show that the computing the saddle point equilibrium is equivalent to solving certain linear program. So we will look at it. First note that for x in delta 1 we know that this min y in delta 2 the second player's min strategy is this x transpose a y is same as minimum of 1 less than equals to j less than equals to n x transpose a e j. So what is e j here? E j is a mix is a strategy for second player where e j is nothing but all of them are 0s one in the jth position and rest of them are 0s. So e j is a in a sense is a pure strategy or the jth column that player 2 is going to play. So e j is the pure strategy and similarly later we use this notation E i in delta 1 this is basically the pure strategy for player 1 in a sense this is the ith row or I can also say that these are all 0s one will be in the ith position and rest of them are 0. So E i and E j I am using this E i below here and j above here to distinguish them from player 1 and player 2. Because this is a vector in R m this is a vector in R n so that is also another reason why we are writing this. So what I would like to say here is that this minimum y in delta 2 of x transpose ay is same as minimum 1 less than equals to j less than equals to m n x transpose a e j. So y is this true y is nothing but y 1 e 1 plus y 2 e 2 plus y n e n. So any vector y is basically a convex or rather it is a convex combination of this e 1 e 2 e n these are the pure vectors of pure strategies of player 2. So this is there and remember x transpose a y is nothing but summation j is equals to 1 to n x transpose a e j that means x transpose ay is nothing but the convex combination of these terms. Therefore any minimizer of this cannot be if you take this as a minimum because it is a linear function it is a convex combination of this thing the value x transpose a y the minimum of this when you take it this has to be the minimum among these things it cannot be less than this thing. So therefore this will be true. So this is a very important fact. So therefore minimum y in delta 2 x transpose a y is same as minimum of x transpose a e j 1 less than equals to j less than equals to n. So this is a very important observation. Therefore finding optimal strategy for player 1 amounts to the following optimization problem. Maximize over x in delta 1 of course minimum 1 less than equals to j less than equals to n x transpose a e j. Subject to of course x satisfies these conditions x i greater than equals to 0 and summation x i i runs from 1 to m is 1. This is this thing. So now let us look at this thing this is the maximum we need and somehow we need to get a linear programming for from here. So if you look at it this one that is always less than equals to x transpose a e j for n e j 1 2. So therefore I am look this particular quantity if I take as a t if I take that to be t to be the minimum 1 less than equals to j less than equals to n x transpose a e j then we know that t is certainly less than equals to x transpose a e j for all j running from 1 to n. Now this quantity is nothing but the smallest of these two these things. So I would say that there is this t the number t is the minimum of this one and then I want to maximize t. So therefore the problem I can say that now the following thing I want to maximize t such that this condition holds such that t is less than equals to x transpose a e j j is equals to 1 to n and then of course x i greater than equals to 0 summation x i is 1. Now if you look at it we are maximizing this number t which is such that t less than equals to x transpose a e j this is n linear inequalities and then this is the non-negativity constraint and this is again the convexity constraint. So all of them are linear now this is a linear programming problem. So the player 1 simply has to find the t and x which maximizes this butler linear programming and then the player 1's problem becomes simply a linear problem. So now let us look at the player 2's. So let us say player 2 fixes y means he is looking at the following thing. He knows that the player if he plays y the player 1 is going to play x which maximizes x in delta 1 of this x transpose a y. Now as I we said previously this is same as max 1 less than equals to i less than equals to n e i transpose a y. Now his player 1's problem is mean y in delta 2 max x in delta 1 of x transpose a y which I can write now as mean y in delta 2 of max 1 less than equals to i less than equals to n e i transpose the same arguments like in the previous step we say that this is if I put it as a s this s is the maximum of that and then I want to minimize this s subject to s is the maximum of this. If s is maximum of that then s has to be greater than equals to each of that. So therefore I can now rewrite this problem as the minimize s subject to s is bigger than or equals to e i prime a y and y j greater than equals to 0 summation y j is 1 of course j is equals to 1 to n. Now this becomes another linear program. So the problem of solving the optimal strategies are saddle point equilibrium. Now amounts to solving two linear programs. So one is for the player 1 the other is for the player 2. So these are there are two linear programs. So now here is an interesting exercise show that these two linear programming problems are dual to each other. We will not go into these details the linear programming and other things. So at least we expect that people have studied some optimization and linear programming. So therefore if you have not seen the linear programming you must go through the linear programming and see that these two are dual to each other. This is a very important and interesting exercise here which I hope everyone will do this. So now what we have done in this lecture. So we explored certain properties of this saddle point equilibrium. The most important is that the saddle points have exchangeability property. That is a very very important thing. Why this is important we will see when we go for non-zero sum games. And the second and most important thing is that computing this saddle point equilibrium can be reduced to a linear program a pair of linear programs. So with this we will end this session and in the next session we start seeing some other interesting properties of zero sum games.