 So, in the previous lecture, we introduced this class of strategies called mixed strategies and these were a generalization of our earlier strategies, the earlier strategies were then called pure strategies. So, the mixed strategy remember was a mixed strategy is a probability distribution on the set of pure strategies. Now, for we were talking in the context of a zero sum game. So, if we recall then if for the row player then we had a zero sum game aid which denoted a zero sum game and then for the row player mixed strategies for the row player this was a set Y which comprises of all vectors Y in now that have m rows. So, column vector of length m such that every component is greater than equal to 0 and if you sum over all the components you get 1. And similarly, so this was for the row player and similarly for the column player you had a set Z which was all the Z's in Rn which is the number of columns such that Z is greater than equal to 0 and the sum of Zj's is equal to 1. So, these are the mixed strategies for the column player. And then we said let us define similarly security strategies and security levels. So, the security strategy for the row player the row player it was a Y star in Y such that Y star transpose Az is less the maximum of Y star transpose Az is less than equal to the maximum of Y transpose Az for all Y in capital Y. And similarly for the column player it is a strategy Z star such that the minimum over Y in Y of Y transpose Az star is greater than equal to the minimum over Y in Y Y transpose Az for all Z in capital Z. So, now assuming that these security strategies exist this was these were called the security levels the mixed security levels. So, this was this is what we called Vm upper bar Vm upper bar of A and this is what we called Vm lower bar of A. And now assuming such the security strategies and security levels exist we got that we derived last time this series of inequalities. Now, one thing is actually quite straightforward here is that the mixed security levels actually exist provided I am I be a little more careful with my notation here all I need to do is I replace the max here by a soup and replace the min here by an int then the mixed security levels will exist. But what I will show today is that actually the way they are defined here they also the mixed security levels in this way also exist. So, first let us just consider what exactly is the issue I mean why is there why do we need to look at this carefully the why do we why is it not automatic that a security level exists. So, if you see what we are doing here the the Y star Y star basically is the next is the Y that minimizes this expression it minimizes the max over Z of Y transpose A Z. So, Y star solves basically the minimum over Y minimization of over Y of the maximization of over Z of Y transpose A Z. Now, if you look at this this thing here this is now a function of Y a question really is when does a function of Y like this have a minimum in a over a set capital Y. Capital Y is an infinite set. So, it is possible that this function does not attain its minimum over this particular set. I will give you an example for instance look at this suppose this function of Y here was like this this is why the function of Y behaves this way it reduces up until this point let us say some point here some point Y bar and then out here at this point it jumps and then increases this way. So, the value at Y bar is actually this the the the dotted value here. So, the function decreases till Y bar but at Y bar the value is not this this point but rather this point in that case question is does the minimum of of this function does this function have a minimum and suppose you want to minimize this function over this set Y and let us say Y is this no this all this set here is Y is capital Y. So, does this function have a minimum over capital Y the function does not have a minimum over capital Y but it has it has an infimum over capital Y. So, there is a least value for this function and that least value is this but that least value is not attained in the set capital Y. So, there is no point in the set capital Y whose function value is exactly equal to the the yellow value I have marked. So, this is a case where minimum is not attained or infimum is not attained. What are the other reasons for which can you imagine other situations where where you may not get a minimum in the set what other situations could arise. So, I am not sure which exact case you are referring to but here is one possible case another case where the where the infimum is there is finite but it is not attained. So, here is for example another case. So, suppose I take the set Y as an open set. So, my Y now capital Y is from this point till this point. So, let us say let us suppose this is small a and this is small b. So, if I consider Y to be an open set then you have the set Y. So, suppose Y is an open set where it is an interval from small a to small b. So, this being an open set means that small a and small b are not included and now suppose the function that I am minimizing is just this. In this case what is the least value of what is the infimum of this function over this set the infimum of this function is this. But is there a point that attains that particular value in this set? No, if b was in the set then it would have been but b is not in the set. So, again here so in this case the culprit for non-existence of non-attainment of the infimum is actually that the set has a problem. Whereas here in the earlier example the reason for the non-attainment was that the function had a had a jump in it. The function was broken whereas here the problem is that the set is open. What about there is also a third case let me draw that third case also here. So, this is what I am attempting to draw here is a set y which is let us say from 0 all the way till infinity and this function that keeps decreasing and eventually you know say let us say settles finally at some value like this. This is the value that it finally settles at. But what do you mean by settle that that is the limiting value of this function. So, as y goes to infinity that is the value of this function. So, now is there a y as a real number that attains this particular value there isn't. So, a classic example of this would be say a function which say take e to the minus x take the function e to the minus y give it and y in capital Y which is 0 to infinity. Now, what is the infimum of this function over capital Y it is 0. But is there a y that gives you e to the minus y equal to 0 there is no real y like that. So, the only way you can get 0 is actually if you put y equal to infinity but that is effective that is not even that is not a real number. So, another culprit for non-existence non-attainment for the infimum in the set is that the set sort of runs away as you keep the function value keeps decreasing and you have to go further and further and further and this to find a lower and lower value. But this process becomes endless and you eventually never get to a to you know at the least value. So, the point is there are it turns out that these are the only three possibilities. So, if you can plug these three possibilities then there will always be any that a function will always have will attain its minimum over a certain. So, these three so there is a theorem that basically comprises of this that basically tells you this and this is the sort of starting point of essentially any theory and optimization and theorem is due to Weierstrass. Let x be a subset of Rn and let f from x to R be continuous. If x is closed and bounded then f attains its infimum over x. So, consider a function f and consider a set x in which is a subset of Rn and let f be a function from x to R and f is assumed to be continuous. Now, if we if x is closed and bounded then f must attain its infimum over x means that there must be a point in x. So, that means what this means is there exists an x star which is in x such that f of x star is equal to the infimum over x x in x of f of x. So, what this is basically what this is telling you is that there is that whatever is the infimum value this infimum value firstly is finite and it is being attained by some point in the set itself. So, all we need to do now for claiming that so since we said that y star basically solves this problem all we have to argue is that y star actually belongs to capital Y because essentially this is we want to say that if y then it would mean that there is actually a minimum to this particular problem. So, in short we want to argue that if you look at this function here this function when minimized over y in capital Y gives you a y is its minimum value is attained by a y star in capital Y. So, for so means there is a y star that is there is a y star in capital Y such that if I take the max over z in capital Z of y star transpose A z that is equal to this here. So, all in order to do this what I need to basically argue is that this function here. So, the function that I am looking to mean that I am talking about now the function is a function of y and that is just this function that is the max over y transpose A z or max over sorry max over z of y transpose A z. Now, this if you remember what did we see last time about what did we say last time about the about this function we wrote this out we said we said maximum if you look at the maximum over A z of y transpose A z what was this actually equal to this was the maximum over j of y transpose A j. So, why was that because y transpose A is a was a column vector z is a set of is capital Z is the set of probability distributions and therefore, the maximum is going to be attained at when you put j is going to be attained at t with with by a probability distribution that full puts full mass on the largest component. So, this was this was actually equal to so max over z in capital Z of y transpose A z was basically was essentially was equal to the largest component of the vector y transpose A. So, now let us go back and look at this boxed expression here. So, the boxed expression then is actually is in fact equal to the max over j of y transpose A j. So, now what are these actually what exactly have we written out here this is basically this here is a collection of linear functions how many linear functions are these there are they are indexed by j. So, there is one for every j. So, there are n linear functions and you are taking the maximum of those linear functions each which is so each of these is actually a linear function. So, linear function each of these is a linear function of y and you take what we are what what is written out here is the maximum of n linear functions and so that is actually just the maximum of finite. So, now if you take so each of these linear functions is it a continuous function of y it is because it is after all just a linear function of y it is it is actually a polynomial inverse. So, it is a linear it is a continuous function of y and what you are doing is taking the maximum of finitely many of them. So, if you take the maximum of finitely many continuous functions what you get is a continuous function. So, therefore, this is a continuous function of y. So, therefore, this is a continuous function of y. So, the boxed expression is a continuous function of y. Now, what about so you are you are looking for the infimum of a continuous function now all right. Now, are you looking for an infimum of a continuous function over now a closed and bounded set or what kind of a set we are looking for we are minimizing this over a over a set capital Y what kind of a set is capital Y. So, the capital Y is a probability distribution is the set of all probability distribution. So, naturally it is a bounded set because every component is between 0 and 1 all right and it is also closed because if you take the limit of probability distributions it is also a probability distribution. So, capital Y is closed and bounded and the boxed function is continuous. So, what this means is you can directly apply Weistrass theorem. Weistrass theorem is basically asking you to check two things it is asking for it is saying that if your function is continuous and you are minimizing it over a closed and bounded set then there is a there is a then its minimum is attained okay. So, the function the boxed function is continuous that is what we just argued the minimizing it over a closed and bounded set capital Y. So, there is going to be a Weistar such that it that Weistar attains the attains this okay. So, in short from this basically tells us from So, Weistrass theorem. So, from Weistrass theorem there exists a security strategy we I just argued for the row player, but similarly one can argue for the column player also. So, there exists a security strategy for each alright. So, in short we are we are we are clean we do not security strategies as we just defined actually through the in fact exist alright. So, so we are so we are we therefore we so this the inequality that I just wrote out this whole inequality now is now proper this chain of inequalities is now proper okay that there is that v upper bar is greater than equal to v m upper bar which is greater than equal to v m lower bar which is greater than equal to v lower bar this is fine okay alright. So, now we come to the we can again now that security strategies exist and so on we can again talk of a saddle point define a saddle point. So, a saddle point in mixed strategies is a pair of Weistar comma Z star such that so this is the analogous set of inequalities holds. So, Weistar transpose A z is less than equal to is less than equal to Weistar transpose A z star and Weistar transpose A z star is it is in turn less than equal to Y transpose A z star which means that so a saddle point is a pair of strategies such that if the column player is playing z star the row player would want to respond with Y star and if the row player is playing Y star the column player would want to respond with z star okay. So, we now have we have now defined a saddle point in mixed strategies and as before we again have that there exists a saddle point Y star z star if and only if v m upper bar is equal to v m lower bar the proof follows exactly the way we did for pure strategies okay. So, we just defined a saddle point and we said that well there a saddle point exists if and only v m upper bar is equal to v m lower bar. Moreover, we will also have all the other properties if v m upper bar is equal to v m lower bar then saddle points will be comprised of security strategies and so on all of that and every pair of security strategies is a saddle point all of that will also hold okay. But remember we have not yet so this all we have we are saying is that there exists a saddle point if and only if this holds okay we just defined what a saddle point should be and we said that where we can find a saddle point if and only if this this equation okay. So, the crux of the matter now is that we have basically come to the exactly the same juncture that we had earlier except that we now are in a different space earlier we were talking of pure strategies we are now talking of we are talking of mixed strategies we have security levels again defined in term security levels and security strategies all defined in terms of mixed strategies okay. So, earlier what we show what we found was there are games okay there are matrices A for which v upper bar is not equal to v lower okay and we extended the space to mix strategies and then we got this chain of inequality. Now, what I will show today is that this here this is this in fact is an equality.