 back to the NPTEL course on game theory. In this session we discuss about evolutionary stability. So, here we consider a non-zero sum games and in particular we look at a symmetric game. So, we consider a symmetric game with a payoff matrix A. So, let me introduce some notation. So, the set of all pure strategies are we are writing it E1, E2, E, M. There are M pure strategies because it is a symmetric game. This is M by M game and mixed strategies are nothing but delta which are given by X in R, M such that X1 plus to X, M is 1 and all of them are non-negative. This is the set of all mixed strategies and the payoff function mixed payoff function pi X, Y is nothing but X transpose Ay. This is nothing but of course if I write it Aij Xiyj of course ij both of them from 1 to M. So, we are considering a symmetric game and these are the notations that mean. So, what is this evolutionary game? So, we are interested in evolutionary games. So, let us consider a large population. There is a large population and then all the people have to choose one of these strategic choices available to them. So, let us assume that they are all playing the pure strategies. So, each person is associated with some pure strategy. So, now you randomly pick some a pair of people. So, if then if you randomly pick a pair of people then they will play this game with their strategies. So, now what is the population state here? A population state is basically nothing but the fraction of people who are playing the let us say pure strategy 1, fraction of people playing pure strategy 2 likewise. So, that is going to be the state of the population. So, the state of the population essentially tells you the fraction of people who are playing the pure strategy even fraction of people who are playing E2 likewise. Now, suppose if some set of people who are somehow forced to play something else fraction of people. Now, when those fraction of people are playing let us say some other strategy. That means they are programmed to play something else. Now, when you look at this population now again randomly pick. So, now the some of them are playing the original incumbent strategy and some people who are particularly from the this fraction of people they will be playing something else overall what exactly is the average payoff a person gets. Now, if this average payoff that a person gets how is this different from the average that a person who is modified to use something else we would like to compare with them. When you compare these 2 things what we are going to get is this notion of evolutionary stable strategy. So, let me introduce you this notation completely. So, the large population and they are playing this symmetric game. So, incumbent strategies this is x they said this is x 1, x 2, x m. So, x 1 tells you the fraction of people who are playing the pure strategy 1 this is fraction of people who are playing x 2, e 2 this is the fraction of people who are playing em. Now, so epsilon fraction of individuals they we call them as a mutants. So, let us say they play. So, in the new population the new state 1 minus epsilon plays x and epsilon are playing y. So, what is the average fitness here? So, the x individuals the average fitness they will get is basically pi x and epsilon y plus 1 minus epsilon x because he is playing from this x a person who is playing and then the fraction of people are programmed to play y and the remaining people are playing x. So, therefore, if you pick an individual there that distribution is going to be epsilon y plus 1 minus epsilon x. So, pi x epsilon y this is basically. Now, what about the y individuals they will be getting pi y epsilon plus 1 minus epsilon x this is going to be the sink. Now, what the definition is that x is said to be evolutionary stable strategy denoted by E SS if for each y not equals to x there exists some number epsilon bar which is something between 0 and 1 such that this pi x epsilon y plus 1 minus epsilon x this should be bigger than pi y epsilon x plus 1 minus epsilon x for all epsilon less than epsilon bar. So, in fact, 0 less than epsilon. So, whenever this a fraction is the mutant population is playing y the average fitness the x individuals is getting that is this pi x epsilon y plus 1 minus epsilon x if their average fitness is higher than the average fitness that the mutant individuals get then we call that particular incumbent strategy as evolutionary stable. So, what are some interesting consequences and before that this I would like to say that this epsilon bar is called invasion barrier. So, if epsilon bar if the mutant population becomes larger then this x may not satisfy this condition. So, this condition is required to satisfy only for a small epsilon bar and this epsilon bar depends on y it need not be uniform if epsilon bar is independent of y then this is called uniform invasion barrier. So, let us now rewrite this. So, what is the condition that we have pi x epsilon y plus 1 minus epsilon x is strictly greater than pi y epsilon y plus 1 minus epsilon x. This is true for all 0 less than epsilon less than epsilon bar and of course now let me rewrite this one because pi is linear in any of the variable when the other variable is fixed if I do that one this is nothing but epsilon pi y epsilon pi x y plus 1 minus epsilon times pi x x. Now, this one thus the right hand side is going to be equal to epsilon pi y pi y y plus 1 minus epsilon pi y x. So, therefore this term is greater than close to this. So, I can rewrite this this implies epsilon pi x y minus pi y y plus 1 minus epsilon pi x x minus pi y x this is greater than 0. So, let me rewrite the next slide epsilon times pi x y minus pi y y plus 1 minus epsilon pi x x minus pi y x this is greater than 0. This is true for all epsilon between 0 and epsilon bar of course for all y of course epsilon bar is depends on a y. Now, let epsilon go x to 0 if I let epsilon goes to 0 this term simply goes to 0 and this term converges to pi x x minus pi y x greater than or equals to 0. So, this implies pi x x is greater than equals to pi y x for each y in delta. Now, remember our game is a symmetric game. Therefore, pi x x greater than equals to pi y x is the payoff function that player 1 is getting and same the other player is also going to get. So, therefore, this when player both the players are when player 1 is fixing x player 2 when player 2 fixes x for player 1 deviating from x to y is not profitable. The unilateral deviation is not profitable for player and being a symmetric game the same thing happens with other player this immediately implies x comma x is a Nash equilibrium. In fact, this is a symmetric Nash equilibrium. So, what is interesting thing here is that the proposition is that every ESS is a symmetric Nash equilibrium. If x is a evolutionary stable strategy, then it is always a symmetric Nash equilibrium. In fact, in the following proposition following theorem, we in fact, provide one characterization for the ESS. So, let me make the statement theorem the following are equivalent the one x is ESS x is symmetric Nash equilibrium. And if y is a best response of x then pi x y is strictly greater than pi y y this is true for all of course as I said for all y which is best response. What is the best response of x here? It means pi x comma x is same as pi y comma x whenever y is best response to x pi x x is same as pi y x then pi x y has to be bigger than strictly bigger than pi y y. So, what is here few points which I would like to say I am simply calling x as a symmetric Nash equilibrium. So, remember this is a symmetric game both players are going to play same strategy in a symmetric equilibrium. Therefore, x what I mean by x is symmetric Nash equilibrium is actually x comma x as a symmetric Nash equilibrium. Now, let us try to prove. So, we already have shown the following inequality if x is ESS implies epsilon times pi x y minus pi y y plus 1 minus epsilon pi x comma x minus pi y comma x this is greater than 0 this comes from the definition of ESS. Now, if we observe it from here already we mentioned that letting epsilon goes to 0 we show that x is a symmetric Nash equilibrium. So, therefore, this part is already done. Now, to show the next part this part note that if this condition pi xx is equals to pi y x happens this becomes 0. Therefore, this term disappears what we will get is that epsilon times pi x y minus pi y y greater than 0 now epsilon can be taken away. So, this is exactly the condition that is required. Therefore, 1 implies 2 is clear. Now, what is now required is to prove the other way. So, now we will prove the other way. So, now 2 implies 1. So, what are the condition that we have x is symmetric Nash equilibrium then we have pi x y is strictly bigger than pi y y whenever pi x comma x is same as pi y comma x. So, this is now let us try to prove that y belongs to delta and of course, y not equals to x that we have to take. Now, look at the pi of x comma epsilon y plus 1 minus epsilon x minus pi of y comma epsilon y plus 1 minus epsilon x. So, this is nothing but so pi x y epsilon and pi y y. So, therefore, that is going to be epsilon into pi x y minus pi y y. Yes, we already have seen this inequality then that and then 1 minus epsilon pi xx minus pi y x. Now, first thing is that we know that x is a symmetric Nash equilibrium. Therefore, this is always greater than equals to 0. So, this is always greater than equals to 0. Then what we really need to show is that we need to show this is greater than 0 that is what we need to show. So, therefore, this is always positive if y is a best response this is going to be 0. So, let us assume if y if y belongs to best response of x then this is a this is anyhow greater than equals to 0, but this can this is also greater than 0 by the hypothesis. Therefore, this is also strictly greater than 0 this is greater than equals to 0 this strictly greater than 0. Therefore, this whole is strictly greater than 0. Now, let us take y not best response. If y is not a best response then first thing is that this is going to be strictly greater than 0 and we have no idea what happens to this. So, this can be anything negative positive whatever it is, but no important thing is that as epsilon becomes smaller and smaller this butler term the term in the red box goes to 0 whereas this term is strictly greater than 0. Therefore, for sufficiently small epsilon this entire sum will be strictly greater than 0. Therefore, for any y there is always some epsilon bar such that for all epsilon less than epsilon bar this term is going to be greater than 0 that proves to implies 1. So, therefore, so this immediately proves that this proposition is true. So, whenever x is an ESS x is also symmetric Nash equilibrium further this condition holds and this condition is this if and only if. So, now we have proved this. Now, we will see some interesting examples. One important example here in evolutionary game theory is Hawke-Dove. So, in a Hawke-Dove game what happens is that this is a population of words the two behaviors there are two behaviors one is hawk behavior and other is dove behavior. So, when two hawks encounter that means two species having this hawk behavior encounter they fight for a resource when they both of them fight for a resource one of them will lose the resource to the other only one of them will get it. Whereas, when hawk and dove encounter the dove automatically gives the resource to hawk and dove does not get any injury or anything. Similarly, when two dose encounter they just simply split the resource. So, let us say if the resource is worth v and cost of fighting is c. So, what happens if when two hawks come both of them are fighting. So, therefore, the resource is v and one of them will concede. Therefore, the c the cost. So, therefore, the other guy gets v and other is this c. So, in a sense v minus c is basically the resource on an average we get v minus c by 2. And if hawk and dove encounter that is going to be the hawk will get v and dove will get nothing and both are dose then they simply split the resource. Now, this is the game that we will continue discussing this in the next lecture and we will see the evolutionary stable strategy of this particular game and see how the behavior what exactly is the evolutionary stable. We will stop here.