Nash Equilibrium and evolutionary stable strategy. Welcome. We will talk now about very... two
concepts that are very important in this context. The references are a couple
of chapters from Easley and Kleinberg and one chapter from Novak. So, we are getting
used to the representation of payoffs in a theoretical
game using a payoff matrix which may be in one of these two forms, is a very good
exercise for you to be able to write one form from the other form. Here, we're representing
two strategies denoted by the letter S and the letter T and the payoffs
are represented in a general form with values a, b, c, and d.
Okay, so let's see, we are going to use this representation of payoffs to introduce some
definitions, OK? So here is the first definition of an equilibrium,
specifically a Nash equilibrium, named after John Nash. Two players
play a strategy that is a Nash Equilibrium when it is not possible for them to deviate,
to move away from this strategy and increase their payoff. In other words, in
a sense, they are locked in. There is nothing else they can play other
than this one strategy with the ultimate goal of increasing their payoff. That is the
main thing on these games, right? So, remember, the game requires high end players and strategies
that the players are going to play and there are expected returns or payoffs.
That's what the players are going after, maximizing their payoff.
So, this idea of equilibrium comes from thinking that once you play this strategy there is
absolutely nothing else that could be played that would increase payoffs
and, in that sense, players are locked in, it's like being frozen, right, that's that's
a deadlock or something like that so that idea
of being frozen is what brings about this notion of an equilibrium.
Another way of phrasing Nash equilibrium is by referring to not one strategy but referring
to a pair of strategies. Say, for example, S and T, we say that pair
is a Nash equilibrium if S is the best response to T and T is the best
response to S. In other words, one strategy is the best response to another
strategy while another strategy is the best response to one, this one strategy. Okay?
So, that's the idea of being in some type of locked in deadlock situation.
So, if players choose strategies that are best responses to each other, then
no player has an incentive to deviate to an alternate...
alternative strategy and the game is in an equilibrium state because there is no force
pushing the game towards a different outcome, OK? So this... this kind of wraps
up this these ideas of being at equilibrium. Okay. We have been using
the word equilibrium as an analogy for steady-state with differential equation models
equilibrium is defined as associated by the state in which there is no change. So, with
the differential equation model, an equilibrium is this one point at which the
derivative equals zero meaning that there is no change, okay?
Here, the idea of equilibrium is summarized in, with these words, the game is in an equilibrium
state because there is no other force that would push the
game towards a different outcome. Now, we would say that, using this general
form of the payoff matrix, the strategy S is a Nash Equilibrium if a is greater than
or equal to c. So, here is the payoff of a, so a is the payoff
obrained by a player that is playing the strategy S against another player who is also playing
the same strategy S and each of these players would get the payoff of this level of this
value of a. now if that value is bigger, if this value
is... here, if this value a is bigger than little c,
right, if a is bigger than c, then the payoff from playing S is larger than the payoff from
playing T against... against S, right? So, in that case
we would say that that strategy S is a Nash Equilibrium.
If you have that is it is actually a strict inequality is a strictly bigger than see them
you would call as I Straightened St., Nashik
If you have that it is it is actually a strict inequality; if a is strictly bigger than c,
then you would call as a strict strict Nash Equilibrium.
We can also say the pair of strategies S,S forms a Nash Equilibrium if a is greater than
or equal to c. OK? So, we can refer as one strategy being an equilibrium or we can
refer to a pair of strategies being an equilibrium just as we mentioned when we are defining
Nash Equilibrium before. Now, for these matrix or this matrix we say that
T, your strategy, is a Nash Equilibrium according
to the values. Now, we're going to compare the values.
Now we are comparing the values of d and b, right, so we're going to say that if d is
greater than or equal to b, then d is a Nash Equilibrium or we can say that
the pair (T,T) is a Nash equilibrium if d is greater than or equal to b.
We can also use the strict Nash, the definition of a strict Nash Equilibrium which applies
if you have a strict inequality. Now, consider a population of players that
are playing strategy S, okay, and assume that there is a mutant of type
T which is introduced into this large population
of S players. So, let's assume that there is an infinitesimal density,
that's the idea, that this mutant, this mutant is introduced at a really tiny
tiny small... it's an itty bitty small level of mutants with with the strategy T being
introduced into the population as an infinitesimal amount which we denote here
by epsilon. OK, this epsilon is a number between zero and one but it is probably
closer to zero than it is to one. That's the idea of being infinitesimal.
So, we would say that things that are T, there is
frequency epsilon of T players, the invaders, then the rest, right, is going to be one minus
epsilon so clearly epsilon plus one minus epsilon equals one. So, we're going
to say that there are epsilon of these players, are T players, and one minus epsilon are S
players. So what we do now that we have this, we multiply
the matrix with the values of payoff, okay, by the vector with densities. Remember
here, the first quantity is density of S, the second quantity is the density of
players in T. When we carry out this matrix vector multiplication this is the first row
times this column, that gives you a minus a times quantity one minus epsilon
plus b epsilon and the second entry, you get c times quantity one minus epsilon plus d
epsilon. So this is now what we call the fitness vector,
this is v F1 and F2. OK, this is from our previous calculations.
Now, the fitness of the S strategy is... so, we're looking at
again, this represents the densities of players with S and T which means that this would represent
F1 or F sub S, the fitness of strategy S, right? So, the
fitness of strategy S is greater meaning this quantity would be greater than
the fitness of strategy T if the first value of fitness is greater than the second value
of fitness. So, if a is greater than c, the fitness of
S exceeds the fitness of T. If a is exactly equal to c and, so this goes
together, a equals c and b is strictly bigger than d,
then, the fitness of S is greater than the fitness of T.
So, this, then, this condition brings us to a definition of another quantity that's called
evolutionary stable strategy. So we would say the strategy S is evolutionary
stable if either one of two conditions are met; one
either a is strictly bigger than c or a is equal to c and b is equal to d. If that happens,
we say that S is an evolutionary stable strategy. The definition
of this this is also called ESS, evolutionary stable strategy. This definition
guarantees that selection will oppose the invasion of T into S. Okay. So, let's consider
this scenario. We have, this is the example that we have been dissecting, using it for
several different instances. So, let's go back to that example of the body size in
the population of beetles. There there were two strategies; one they were
phenotypes small and large. So, if a few large beetles are introduced into a population of
small beetles then the large beetles will do extremely well. Remember that,
here, body size was the main driving factor in the competitions for food. The food
competitions, these were pairwise competitions. Two beetles meeting each other, the bigger
guy kind of had a better chance of keeping most of the food or something, right?
So, what happens with this, let's examine that
in the context of invasion, right, in the context of evolutionary stability.
So, since they rarely meet each other, then they get most of the food in almost every
competition they experience. Right? So, we're we're talking about
the large beetles, right, when large meets large, then there
is a significantly smaller payoff as what they get out of those nutrients of food, right,
significantly smaller when compared to when large meets small, right?
So, since large doesn't meet large that often, so they have... they have an evolutionary
advantage. The population of small beetle... small beetles cannot drive out the large size
beetles. So the strategy small is not evolutionary stable.
Okay? Strategy small is not evolutionary stable. From the strategy payoff matrix what we have
here we have that five, this entry, is less than this entry, okay, which violates
the definition of evolutionary stable strategy. So, this does not
satisfy the definition of evolutionary stable strategy.
Now, consider the following payoff matrix, right? In this case
so, what is different? The values here, small against small is five, large against large
is three. Let's see here, large against large is still
three, Small against small is still 5. So, these values are the same values but now I'm
sorting the strategies here on a different order. I list large first, small second, large
first, small second. I went to do that just to show you something that is consistent
with the way which we listed the definitions. If we now consider the payoff
matrix in this one; same values but we've rearranged the order of the strategies were
consistent, right, this strategy goes first here, this strategy goes second and
then I chose the columns. We keep that we are consistent with the order. This
strategy goes first, this strategy goes second, okay?
So clearly here, three is bigger than one and that implies that the definition of evolutionary
stable strategy applies in this sense. So, strategy large is evolutionary
stable. Let's digest this. In other words, in a population of large beetles, a
few small beetles will do... will do very badly, ok?
So, in a population of large beetles, now we think of the majority of the population
having the phenotype of large body size, if we introduce an infinitesimal...
an infinitesimal amount of beetles with small body size, then
those small beetles are going to do terrible. They will have a clear disadvantage in food
competitions, right, and they're going to be
losing almost every competition for food because of their body size. The population of large
beetles resists the invasion, resists the invasion of small beetles
because the large phenotype is considered an evolutionary stable strategy. So,
given a matrix like this one with values a, b, c, and d,
we say that a pair S,S is a Nash... is a Nash equilibrium when S is the best response to
a choice of S by the other player. We can also equivalently state that S,S is
a Nash equilibrium if a is strictly bigger than... I mean if
a is greater than or equal to c. So you are comparing the payoffs
on here in the first column, right? That's what we have been working with. We can also
say that that this is an evolutionary stable strategy
if a is strictly bigger than C or if a is equal to c and b
this these two things can be the same, and you have that b is strictly bigger than d.
So, if a strategy is evolutionary stable, then S is a strategy that is evolutionary
stable, then the pair of strategies S,S is a
Nash equilibrium, and this concludes our discussion on the concepts of Nash equilibrium and evolutionary
stable strategies. Thank you!