 Welcome back to this NPTEL course on game 3. In the previous session, we introduced the notion of evolutionary stable strategy. We continue discussing about this. In fact, in the last session, we ended with an example, but of course we have not given a complete details. I will recall this example again. So it is a Hock-Dove example. In the Hock-Dove game, what we have is that it is a species having two behaviors. One is Hock behavior and other is Doe behavior. If two species of Hock behavior encounter, they fight for the resource. The resource has a work V and if they fight, one of them has to concede. In fact, one of them will lose the resource to the other one. So the cost of losing is C. Therefore, one species will get V, other will get minus C because we are in an averaging framework. So therefore, on an average, the Hocks get V minus C by 2. Now if Hock and Doe encounter, Hock gets the resource, Doe gets nothing because it is not fighting. And when two Does encounter, they split the resource. So therefore, the game is going to be 2 by 2 game, Hock-Dove, Hock-Dove. If H, H means it is V minus C by 2 and if Hock and Doe means it is Hock will get V, Doe will get 0. Of course, V minus C by 2, V minus C by 2, D gets nothing and H gets V and both Doe means is V by 2, V by 2. Now, what exactly is the evolutionary stable strategy? In fact, computing the Nash equilibrium itself is an interesting exercise, which you should try to do it. So evolutionary stable strategy. So evolutionary stable strategy satisfies the following thing. It satisfies pi x epsilon y plus 1 minus epsilon x is strictly greater than pi y epsilon y plus 1 minus epsilon x for all epsilon between 0 and epsilon bar of course, y in delta. Now equivalent definition is this is equivalent to saying that x is symmetric Nash equilibrium and pi xy is strictly greater than pi yy for all y in best response of x. These are the two conditions. So compute the Nash equilibrium of this game particularly when you need to look at the symmetric Nash equilibrium and then verify one of these two conditions and we will get the this thing. So I will leave a general thing as an exercise but in particular what I will do right now is take this matrix A to be minus 1 to 0, 1. We look at this specific situation and then calculate. So we look at the mixed strategy half-half and what is pi yx? This is going to be we calculate the value this is nothing but minus y1 by 2 plus y1 plus 1 minus y1 by 2 which if you calculate this is going to be half for all y in delta. If you take half-half as this thing and pi yx is going to be simply this. So basically we are using y transpose A half-half that is exacting this expression. Now if we calculate pi x, y minus pi y, y this if you really calculate using this formula or whatever it is is going to be 2 y1 minus half square which is strictly greater than or equals to 0. So here this equality holds only if y1 is equals to half that means y and x are one and the same. If y and x are the same then this particular thing is going to be equal to 0 otherwise this is strictly greater than 0. Now this implies the second condition of the theorem that we have proved in the previous session. So now the next thing that needs to check is half-half is symmetric Nash equilibrium. If half-half is a symmetric Nash equilibrium and this condition and from whatever we have done it previously in the previous session immediately infer we can infer that half-half is going to be a evolutionary stable strategy. So this is a simple exercise but all of you should try solving for general this thing is what is the symmetric Nash equilibrium you computed and then what is the ESS here in this game and what are the conditions on V and C which gives you this one. So this is just a computational exercise so I will skip the details. Now let us look at another game. We will look at now rock paper scissors. So the rock paper scissors game has this thing 0 1 minus 1 minus 1 0 1 1 minus 1 0. In fact we can easily verify that 1 by 3 1 by 3 1 by 3 is a symmetric Nash equilibrium. This is the in fact this is the unique symmetric unique Nash equilibrium here. So this is not a hard thing to prove so this is going to be a symmetric Nash equilibrium. Now is this so let us check what is going to be pi x E 1 if player 2 plays column 1 against 1 over 3 1 over 3 1 over 3 so therefore 1 over 3 into 0 1 over 3 into minus 1 1 over 3 into 1 so this is 0. In fact this is also same as pi E 1 E 1. So the condition that we had that if E 1 is in fact a best response to 1 over 3 1 over 3 1 over 3 because this has an equalizing property because this is a completely mixed equilibrium so every pure strategy will give you the same value but we require whenever such a thing happens E 1 is a best response to this thing we need we know that pi x E 1 should be bigger than strictly bigger than pi E 1 E 1 but that is not true here. Therefore 1 over 3 1 over 3 1 over 3 is not ESS. So this is a an example where there is no evolutionary stable strategy. Now there is some interesting consequences of this definition is let me recall the definition of a support of a mixed strategy is basically is nothing but all pure strategies E i such that x i is greater than 0. Support of a mixed strategy is nothing but the pure strategies which are played with positive probability okay. Let us say lambda x is basically the convex hull of S x. So lambda x is the convex hull of all the pure strategies which have a positive probability in the x. So this lambda x is convex hull of S x. Now we have the following interesting thing. Theorem if x belongs to delta is ESS then there is no other symmetric Nash equilibrium in lambda x. That means if x is a ESS then x is going to be the only ESS where all these pure strategies that are played in x are going to be played. So there is no other. This is in fact not a hard thing because x is a ESS that means x is a symmetric Nash equilibrium and hence all the pure strategies in S x are going to be best responses to x. Therefore this lambda x itself is going to be a best response because every pure strategy is a best response to x. Therefore any convex combination of those best responses is also in best responses. So therefore lambda x is going to lie inside this best response. Therefore because all of them so the condition pi x y is strictly greater than pi y y for every y in lambda x y cannot be symmetric Nash equilibrium. If y is symmetric Nash equilibrium this inequality cannot happen. So therefore if y is a symmetric Nash equilibrium pi y y should be bigger than or equals to pi x y for each x y but we have other way inequality. So therefore this thing. So this is an interesting fact about this ESS. If you have one ESS within that support that is going to be the only ESS. In some sense if you have two ESS of a game they will be completely designed. The supports will be in a sense designed. The next we can actually prove the following theorem. So let me the following are equivalent. So x is ESS. The second is there exists a neighborhood x such that f x y greater than f y y for all y in u such that y is not equals to x. So what exactly is this condition? If you have a x as a ESS this is equivalent to saying that there is a neighborhood u of x such that in that neighborhood f x y is strictly greater than f y y of course this is true for all y in u with y not equals to x. So in fact this is a nice exercise of convexity. So all these are pi's I have been putting a wrong notations here. All these are pi's I think in the previous we have put correctly this is pi. So the whole idea of the proof using this thing is that suppose this is x if you take any other thing you take a convex combination of this first any convex combination of this will be if you take some y here. So 1 minus epsilon x plus epsilon y will belongs to u for small epsilon. Because epsilon when epsilon is small this is going to be x this is contributing 0 that means it should come here. So the continuity arguments and all says that when epsilon is sufficiently small this must be in u. So use that fact in the previous theorem and from there you can conclude this. So this I will leave it as an exercise and we will go to prove the next part. So here we are now interested in introducing a replicator dynamics. So evolutionary stable strategy is basically a strategy of the incumbents incumbent and then if some mutants enter the population then whether this incumbent strategy survives or not ESS simply talks about that. But how the ESS is arrived at in the society or in that system it does not tell you the replicator dynamics gives you a way of dealing with this fact. So let us look at the replicator equation. So we consider a large population. So they are playing this pure strategy C1, E2, Em. These are the M pure strategies this large population the people in the population are playing one of these pure strategies. Let us assume the population state is Xt which is nothing but X1t, X2t, Xmt. So remember X1t is basically the fraction of population playing E1. X2 is the fraction of population playing E2, Xm is the fraction of population playing Em. So Xd is this thing the state of the population is this. Now in a sense Xit is nothing but Nit by Nt. At any point of time Nit is the number of people number of the population playing i the size of the population playing i by total size of the population this is the t here is the time we are now considering the system over a time. So t let us say at time t the average payoff to an individual adapting Ei in a random match. So let us say an individual in the population adapts to Ei and his randomly matched then his average payoff is nothing but pi Ei Xt. Pi Ei Xt is going to be his average payoff. So the population average is we have been seeing already is pi Xt Xt. So if you randomly pick people they are actually going to be according to Xt. So therefore pi Xt Xt gives you the population average whereas if a random individual adapting you if he is playing with this population he is going to get pi Ei Xt. So I denote this sigma Ei X to be this difference. So what now we would like to say that the relative rate of change so the relative rate of change Xi dot t by Xit this is measure of evolutionary success of Ei strategies. If a person has to change to Ei or not that particular factor is given by this Xi dot t by Xit. So in a sense this has to be sigma Ei X. So when this happens basically the success of Ei because Xi dot is the rate at which the population is changing the Xi dot t by Xit is giving you the relative rate of change of is a success of Ei strategies. So this is a kind of a Darwinian idea. So with this we get this equation. So this equation is called this if I write it multiply this one this is nothing but Xit sigma Ei X I will write it whatever is there here pi Ei Xt minus pi Xt Xt. So this is known as replicator equation. Now let us check one interesting fact is that what is the summation of Xi dot t where I runs from 1 to m this is nothing but if I sum over all these things pi of this that is nothing but pi Xi t of pi Ei Xt minus pi Xt Xt. If you really calculate thus this is going to be 0. So in fact Xi pi Ei Xt summation of that is simply pi Xt Xt this is pi Xt Xt Xi dot t Xi t summation of that is pi Xt Xt. So therefore this is going to be 0 therefore Xi t or summation is nothing but summation Xi 0. Initially if Xi at 0 Xi is a probability distribution that means summation Xi at 0 is 1 then summation Xi t continues to be 1. Therefore the replicator dynamics is a trajectory on the simplex. So therefore Xt belongs to delta for all t. So now the question is does there exist a solution to replicator equation. So if you look at it here this equation pi Ei Xt this is a linear in this variable. So therefore this is a nice function similarly this is a bilinear function. So therefore nice function Lipschitz continuity and everything holds. So therefore the right hand side of this differential equation this differential equation replicator equation is a Lipschitz continuous function. So therefore the replicator equation admits unique solution in delta whenever X at 0 belongs to delta. If you start in delta it remains in delta and in fact there is a unique solution. So now let us look at it. If X is an ESS what is going to happen to it? If X is an ESS if you look at this particular thing pi Ei Xt minus pi Xt Xt what will happen to this one? So let us look at it. So let us take the set delta 0 is all X in delta such that sigma Ei X is 0 for all i in sx. So I am only looking at those X in delta for which sigma Ei X is 0 for i in the support of X. If i is not in the support of X I do not care whether it is 0 or not. So this is nothing but the stationary points of replicator equation. So what is now interesting thing here is that now we can make the following statement which whose proof we will continue in the next session is the following thing. Let X belongs to delta if X is symmetric Nash equilibrium then it is a stationary point of replicator equation. So the converse is true when one of the following holds one X is in the interior of delta second X is a limit state of trajectory lying in the interior of delta 3 is X is Lyapunov stable state of replicator equation. So we will prove this thing theorem in the next session but first I would like to say this part if X is a symmetric Nash equilibrium then it is automatically a stationary point here. So if X is a symmetric Nash equilibrium for all of its support we know that they are in the best response and by the definition of the sigma Ei X this has to be 0. Therefore the symmetric Nash equilibrium are nothing are the stable or the stationary points. Now whether they are stable or not that requires further things and in fact ESS is connected with the stability which we will see in the next session. With this we will conclude this session and we will continue in the next session. Thank you.