 Welcome back to this NPTEL course on game theory. In the previous session we have introduced replicator dynamics and we also stated a theorem. We will now complete the proof of the theorem. So let me recall what the statement of the theorem is. The statement is the following thing. Let X be a mixed strategy and if X is symmetric Nash equilibrium then it is a stationary point of the replicator dynamics. The converse is true any one of the following conditions. So one X is in the interior of delta X is an interior point that means X i is greater than 0 for all i. The second is X is limit state of trajectory lying in the interior of delta. So the X is basically the limit point of a trajectory which lies entirely inside the interior of delta. The third point is X is Lyapunov stable state of replicator equation. So this is the statement that we ended with in the previous lecture. Now we will try to prove this one. The replicator dynamics is given by the following fact X i dot is nothing but X i into F E i comma X minus F X comma X where F is the payoff function where F X Y A is the corresponding symmetric matrix that we are considering. So let us try to prove this fact. So let us assume that X is let X be symmetric Nash equilibrium then clearly F E i X is same as F X X for all E i in the support of X. If support of X means X i is greater than 0 then E i we say that that is in the support of X. So if X is in E i then X is E i is the best response to this thing. So therefore F E i X is always same as F X X. So therefore for all i with X i greater than 0 what we have is that F E i X minus F X X is 0. This is the right hand side here. This implies X i into F E i X minus F X X is 0. So that means the right hand side is 0, right hand side of the replicator equation is 0. This is 0 for every i with X i greater than 0. Now it is also true for all i with X i equals to 0 that F E i X minus F X X into X i is 0 because X i is 0 automatically this does not matter whether it is 0 or not this is going to be 0. Therefore X is a stationary point of replicator equation. So that is the first step. Now the next thing we need to look at the converse part. So we need to show that under the one of these three conditions the X is going to be a symmetric Nash group. So let us look at it. Suppose if X is an interior point of this delta then what happens is that now X is also stationary implies X i F E i X minus F X X is 0. Now because X is interior point X i is strictly greater than 0 which implies F E i X minus F X X is 0. What this says is that E i is the best response to X. This immediately tells me that X is best response to X. Each E i is a best response to X and hence convex combination of them is also a best response to this thing and therefore X becomes best response. Therefore X is symmetric Nash equilibrium. So that is the first point the second point second thing. Now assume stationary point X is a limit state that means it is a limit of a trajectory that is X is nothing but limit T going to infinity of some X t where X t is a trajectory of the replicator equation where X t satisfies replicator equation and it lies in the interior of delta for all time greater than equals to 0. Note that X i t is nothing but X i t naught e power integral t naught to t F E i X s minus F X s d s. So this is basically the variation of parameters formula for the differential equations. So I will not go into the details of the differential equations. So this can be seen a simple consequence of differential equation. So this is true for all i. What we will do is that let if possible let X be not a symmetric Nash equilibrium. Then there is a pure strategy E j which is in the support of X satisfying F E j X is strictly greater than F X X that means the pure strategy E j will give more payoff than X itself F E j X is strictly greater than F X X. This implies that we define 2 delta to be the F E j X minus F X X 2 delta is I define this is strictly greater than here. I am just taking the delta to be half of this one which is strictly greater than 0. So this immediately implies that there exists a neighborhood would you such that F E j Y minus F Y Y is greater than equals to delta for all Y in U Y naught equals to X. Basically what I am saying is that because this is a continuous function this is strictly greater than 0 which is 2 delta. So therefore there will be a neighborhood where this inequality holds with strictly greater than this values are always greater than equals to delta. This is true in fact for all i in fact when Y is equals to X this is in fact 2 delta. So even then it is true automatically. Since X is equals to limit T going to infinity of this trajectory that is the definition of it. So what we have is that there is T naught such that X t is in U for all t greater than equals to T naught because X t is converging to X naught and X naught is inside U, U in the neighborhood is around X. So therefore in the large for large T X t will be inside U and that is it. Therefore what we have now is that F E j X t minus F X t X t this is greater than equals to delta for all t because than equals to T naught. Now we use this fact into this equation because this is strictly greater than equals to that. So if I put that here what we have got here is that X j t is greater than equals to X j 0 into E power delta t minus T naught. So because this term is greater than equals to delta so therefore this is greater than equals to delta so delta t minus T naught therefore X j t greater than equals to X j t naught E power delta into t minus T naught that is exactly this. This is true for all t because than equals to T naught. So once we have this one what it now says is it is a contradiction because X j t greater than equals to this one as T going to infinity this is going to be converging to infinity. So this contradicts because as T goes to infinity then X t converges to this X naught some this point X. So each for each j X j t converges to X j but what it is saying here is that X j t is blowing up because E power delta t minus T naught as T converges to infinity this blows up. So therefore X j t is blowing up which is a contradiction to our fact. So therefore X is symmetric Nash equilibrium. So this proves the second bulleted point. Now finally remains to prove the third one. So this is again is just right forward thing so let us take X be Lyapunov stable stationary point of replicator equation. Now it is again exactly the same argument if X is not symmetric Nash equilibrium then we follow exactly the previous line then there will be pure strategy E j and T naught such that the whatever condition that we have proved earlier. So we can actually get the same thing in fact we can essentially the replicate these arguments here because X is Lyapunov stable state therefore it is also a limit point and then you can exactly follow the same lines and then again the same contradiction comes and this happens therefore X has to be symmetric Nash equilibrium. So this proof automatically follows from the previous argument itself. So thus we have now proved this theorem. So let me say that what exactly this theorem is giving is that every symmetric Nash equilibrium is a stationary point of replicator equation. Now the stationary point of a replicator equation need not be equilibrium symmetric Nash equilibrium and it is symmetric Nash equilibrium under the following three conditions. One is if X is a interior point of delta and a stationary point then X is a symmetric Nash equilibrium. The second point says that if X is the limit state of a trajectory and which the trajectory must lie inside interior of a delta then it is a symmetric Nash equilibrium and the final thing is if it is a Lyapunov stable state then also this becomes. So this theorem is in that sense is characterizes what could be the stationary points are in particular the relation between the stationary points of the replicator equation and the symmetric Nash equilibrium. Now finally we will prove the following important theorem is that if X is ESS, if X is an evolutionary stable strategy then it is asymptotically stable state of replicator equation. What it says is that evolutionary stable strategy if you take it, it is going to be asymptotically stable state. That means around any neighborhood or around X you take it and if the trajectory starts inside that neighborhood it remains close to that trajectory. So this is an important concept from differential equation this is asymptotically stable state and we will now prove that our ESS is going to be asymptotically stable. The proof is basically follows by constructing a Lyapunov function of course here is some prerequisites that I am assuming it this prerequisites are from differential equations even though we do not use any of the technical details we just simply apply show that something is Lyapunov function and then apply to some known results in differential equations. Let us go to the proof of this fact. Let X be ESS. So for the sake of notational simplifications I will introduce notations sigma xy to be fxy minus fyy. So this is a notation that I would like to use it. So now since X is at ESS so from the one of the previous theorems which we have proved in the earlier class what we can show is that there exists a neighborhood u of X such that sigma xy is greater than 0 for all y in u minus X. So I will just take this conclusion which is a consequence of one of the previous theorem that we have proved in the class so that we will take this one. Now the thing is we are going to use the Lyapunov's direct method. Consider O to be set of all y in delta such that the support of X is contained in support of y. So in fact O is going to be a relative neighborhood of X that means this O will be intersection of a neighborhood intersection delta. So this is the, now we define Lyapunov function V from O to R by V of y is summation i in the support of X xi log xi by yi. So this is an entropy function actually. So I am defining for all pure strategies which are in the support of X the corresponding xi into log of xi by i and then some overall such i's. So this I define it as Vy. Now in fact this verification is that V is continuous. So this is a straightforward fact and then V of X is going to be 0 because xi by xi that is 1 so therefore you can easily see V of X is 0. These are the immediate facts that we can see it. Next for y in O different from X what we have is that V of y is going to be minus of summation i in Sx xi log yi by xi. So I am taking log yi by xi that is why the minus sign came. So this is greater than minus summation i in Sx of xi into yi by xi minus 1. So this is following from the fact that log R is less than R minus 1 for all R not equals to 1. Use this fact and then this inequality comes and this is nothing but 1 minus summation i in Sx of yi. So this is going to be 0 therefore V is a non-negative function. So the next thing we need to show that the time derivative of V of yt along a trajectory the derivative of V is strictly negative. This is the next condition to verify to apply the Lyapunov's direct method. So let us look at it d by dt of V of yt. So this is just direct verification summation i dV by dy of yt into y dot t. So this is nothing but of course if I calculate these values minus xi by yi t, yi t sigma ei yt. So I am just substituting the function V and also y dot t this is the solution of the replicatory equation. So therefore this comes. So if we really simplify this fact this is going to be minus sigma x yt. So therefore this is strictly negative. Therefore along this trajectory this is strictly negative. So now the Lyapunov's direct method implies that this because this we have constructed a Lyapunov function and this is the time derivative is strictly negative. And of course here we are using the fact that x is ESS. So keep that in mind y is this strictly negative that comes from the fact that this x is a ESS. So this is a small step that you need to fill in. So and then the Lyapunov direct method can be applied. Once you apply that then we say we get that this x is a asymptotic stable state. This follows now from the Lyapunov's direct method. So this is the main very important result in this evolutionary game theory which says that every evolutionary stable strategy is a asymptotic stable point of the replicator dynamics. It is good. Now we will just see some simple example, some example we will give to say that the converse need not be true. So the question is, suppose is every asymptotically stable point, is it going to be a ESS? The answer is no. So for that you take the matrix A to be the following thing. It is a 3 by 3 symmetric game. Take this one. So here in this thing no pure strategies, none of them is symmetric equilibrium. So in fact f of E1y is going to be y1 plus phi y2 f of E2y is nothing but y2 plus phi y3 f of E3y. We can calculate these values phi y1 plus 4 y3. Just calculate these values. This is a straight forward calculation. And in fact using these calculates, you can actually do the following verifications again is that f of E1y is same as f of E2y if and only if you can verify these things easily. This is f E2y is same as f E3y if and only if 6 y1 is 1. Similarly f E3y is same as f E1y if and only if 9 y2 is 4. Now using these facts, the game has one symmetric Nash equilibrium which is given by 1 by 6, 4 by 9, 7 by 18. So use these inequalities and then show that this is a symmetric Nash equilibrium. This is in fact 6 y1 is equals to 1. So 1 by 6 is coming. So basically if this is a symmetric Nash equilibrium, then interior symmetric Nash equilibrium, then f E1y should be same as f U2y. All these things should be same because every pure strategies is a best response. So you follow that to say that this is symmetric Nash equilibrium. So we can also verify that X is not ESS. So we can easily verify that this X is not going to be ESS here. So this is a multiple way you can see it. So for example you can say that f X E3 is 4 X3 which is 28 by 18 whereas f E3 E3 is going to be 4 which is strictly greater than 28 by 18. So this tells that this X is not going to be ESS. In fact one of the condition of the previous theorems while it here. Nonetheless we can show that X is asymptotically stable point of replicator equation. In this case the replicator equation becomes the following thing. Let me write it here y1 dot is equals to minus y1 y1 square plus y2 square plus 4 y3 square plus 5 y1 y2 plus 5 y2 y3 plus 5 y3 y1 minus y1 minus 5 y2 that is the equation corresponding to y1 y2 dot is going to be minus y2 of y1 square plus y2 square plus 4 y3 square plus 5 y1 y2 plus 5 y2 y3 plus 5 y3 y1 minus y2 minus 5 y3. This is the equation corresponding to y2 then y3 dot is going to be minus y3 into y1 square plus y2 square plus 4 y3 square plus 5 y1 y2 plus 5 y2 y3 plus 5 y3 y1 minus 5 y1 minus 4 y3. So this is going to be the system of equation which is the replicator equation in this thing. So in fact to how you can see that this is asymptotically stable point is the following thing. You take the RHS of the replicator equation is a map from R32 R3. So it is if gradient matrix if you calculate that is going to be minus 7 by 12 2 by 9 minus 37 by 36 minus 2 minus 32 by 27 minus 14 by 27 7 by 36 minus 77 by 54 minus 91 by 108. If you calculate the right hand side you take this as a function of y1 y2 y3 and then its gradient at a point you will have this value at our ESS point EX if you calculate this is going to be this thing. And then look at its eigenvalue look at its eigenvalues calculate the eigenvalues of this and then because we know that if the eigenvalues the real part of the eigenvalue is negative then the corresponding point is going to be a asymptotically stable point that comes from the differential equation and then verify based on that X is asymptotically stable point from the theory of differential equations. So with this we will conclude the evolutionary stable strategy in the next class we will look at another topic. Thank you.