All right, so this was the
definition we saw at the end last time.
I’ve tried to write it a bit larger, it just repeats the
second of those definitions. This is the definition that
connects the notion of evolutionary stability,
for now in pure strategies, with Nash Equilibrium.
Basically it says this, to check whether a strategy is
evolutionarily stable in pure strategies,
first check whether (Ŝ,Ŝ) is a symmetric
Nash Equilibrium. And, if it is,
if it’s a strict Nash Equilibrium we’re done.
And if it’s not strict, that means there’s at least
another strategy that would tie with Ŝ against Ŝ,
then compare how Ŝ does against this mutation with
how the mutation does against itself.
And if Ŝ does better than the mutation
than the mutation does against itself then we’re okay.
One virtue of this definition is it’s very easy to check,
so let’s try an example to see that and also to get us back
into gear and reminding ourselves what we’re doing a
bit. So in this example–sort of
trivial game but still–the game looks like this.
And suppose we’re asked the question what is evolutionarily
stable in this game? So no prizes for finding the
symmetric Nash Equilibrium in this game.
Shout it out. What’s the symmetric Nash
Equilibrium in this game? (A,A).
So (A,A) is a symmetric Nash Equilibrium.
That’s easy to check. So really the only candidate
for an evolutionarily stable strategy here is A.
So the second thing you would check is: is (A,A) a strict Nash
Equilibrium? So what does strict Nash
Equilibrium mean? It means that if you deviate
you do strictly worse. So is (A,A) strict Nash?
Well is it strict Nash or not? It’s not.
So if you deviate to B, you notice very quickly that
U(A,A) is equal to U(B,A). Is that right?
It’s just tie, both of these get 1.
So it’s not a strict Nash Equilibrium so we have to check
our third rule. What’s our third rule?
So we need to check how does A do against the possible
deviation which is here–which is B here, how does that compare
with B against itself? So U(A,B) the payoff to A
against B is 1 and the payoff to B against itself is 0:
1 is indeed bigger than 0, so we’re okay and A is in fact
evolutionarily stable. So: just a very,
very simple example to illustrate how quick it is to
check this idea. I want to spend all of today
going over more interesting examples, so having the payoff
of having invested in this last time.
To start off with, let’s get rid of this rather
trivial example. I want to think about evolution
as it’s often applied in the social sciences.
So one thing you might talk about in the social sciences is
the evolution of a social convention.
Sometimes you’ll read a lot in sociology or political science
about the evolution of institutions or social
conventions and things like this–maybe also in
anthropology–and I want to see a trivial example of this to see
how this might work and to see if we can learn anything.
The trivial example I’m going to think about is driving on the
left or the right; on the left or right side of
the road. So this is a very simple social
convention. I think we all can agree this
is a social convention and let’s have a look at the payoffs in
this little game. So you could imagine people
drive on the left or drive on the right, and if you drive on
the left and the other people are driving on the right,
you don’t do great and nor do they;
and if you drive on the right and they’re driving on the left
you don’t do great, and nor do they.
If you both drive on the right, you do fine.
But if you both drive on the left you do a little better,
because you look a little bit more sophisticated.
So this is our little game. We can see that this,
this could be an evolutionary game.
So you could imagine this emerging–you can imagine a
government coming in and imposing a law saying everyone
has to drive on the left or everyone has to drive on the
right, but you could also imagine at
the beginning of roads in different societies in different
parts of the world, people just started doing
something and then they settled down to one convention or
another. You can see how evolutionary
stability’s going to play a role here.
Well perhaps we should just work it through and see what
happens. So what are the potential
evolutionarily stable things here?
What are the potentially evolutionarily stable things?
Let’s get some mikes up. What’s liable to be
evolutionarily stable in this setting?
Anyone? Student: Left,
left and right, right are both candidates.
Professor Ben Polak: Good, so our obvious candidates
here are left, left and right,
right. These are the two candidates.
More formally, left is a candidate and right
is a candidate, but you’re right to say that
left, left and right, right are both Nash Equilibria
in this game. What’s more,
not only are they Nash Equilibria but what kind of Nash
Equilibria are they? They’re strict Nash Equilibria.
The fact they’re strict, both are strict,
so indeed left is evolutionarily stable and right
is evolutionarily stable. Let’s just talk through the
intuition for that and make it kind of clear.
So suppose you’re in a society in which everybody drives on the
left, so this is England and suppose a mutation occurs and
the mutation you could think of as an American tourist.
An American tourist is dropped into English society,
hasn’t read the guidebook carefully,
starts driving on the right, and what happens to the
tourist? They die out pretty rapidly.
Conversely, if you drop an unobservant Brit into America
and they drive on the right, they’re going to get squashed
pretty quickly, so it’s kind of clear why
everyone driving on the left or why everyone driving on the
right is each of these are perfectly good evolutionarily
stable social conventions. But despite that simplicity,
there’s kind of a useful lesson here.
So the first lesson here is you can have multiple evolutionarily
stable settings, you could have multiple social
conventions that are evolutionarily stable.
We shouldn’t be surprised particularly if we think there’s
some evolutionary type force, some sort of random dynamic
going on that’s generating social conventions other than
stable, we should not be surprised to
see different social conventions in different parts of the world.
In fact we do, we do see parts of the world
like England and Japan, and Australia where they drive
on the left and we see parts of the world like France and
America where they drive on the right,
so we can have multiple evolutionarily stable
conventions. There’s another lesson here,
which is you could imagine a society settling down to a
social convention down here. You could imagine ending up at
a social convention of right, right.
What do we know about the social convention of everyone
driving on the right? We know it’s worse than the
social convention of everyone driving on the left,
at least in my version. So what do we regard,
what can we see here? We can see that they’re not
necessarily efficient. These need not be equally good.
So it’s hard to resist saying that American society’s driving
habits, if we think about the alternatives to evolution,
are a good example of unintelligent design.
So when you’re talking about this in a less formal way,
in your anthropology or political science,
or sociology classes you want to have in the back of your
mind, what we mean by evolutionary stability and that
it doesn’t necessarily mean that we’re going to arrive at
efficiency. In some sense this is exactly
analogous to what we discussed when we looked at games like
this with rational players playing if you think about
coordination games. These are essentially
coordination games. That was a fairly
straightforward example. To leave it up there let me,
there’s another board here somewhere, here we go,
let’s look at yet another example, slightly more
interesting. So today I’m going to spend
most of today just looking at examples and talking about them.
So here’s another example of a game we might imagine and once
again it’s a two-by-two game, a nice and simple game,
and here the payoffs are as follows.
So down the diagonal we have 0,0, 0,0 and off the diagonal we
have 2,1 and 1,2. So what is this game
essentially? It’s very, very similar to a
game we’ve seen already in class, anybody?
This is essentially Battle of the Sexes.
I’ve taken the Battle of the Sexes Game, The Dating Game,
I’ve just twiddled it around to make it symmetric,
so this is a symmetric version of Battle of the Sexes or our
dating game. You can think of this game also
in the context of driving around.
So you could imagine that one version of this game,
you sometimes hear this game referred to as chicken.
What’s the game chicken when cars are involved?
Anybody? What’s the game chicken when
cars are involved? It’s probably a good thing that
people don’t know this. No one knows this?
Okay, all right somebody knows it.
Can I get the mic way back there?
Yeah shout it out. Right, so you can imagine,
here we are on a road that perhaps isn’t big enough to have
driving and driving on the right.
These two cars face each other, they drive towards each other,
both of them going in the middle of the road,
and the loser is the person who swerves first.
If you think of A as being the aggressive strategy of not
swerving and B as being the less aggressive,
more benevolent strategy, if you like,
of swerving–so the best thing for you is for you to be
aggressive and the other person to swerve,
and then at least they remain alive.
Conversely, if you’re benevolent and they swerve,
you remain alive but they win and unfortunately now,
if you’re both aggressive you get nothing, and the way we’ve
written this game, if you’re both benevolent you
get nothing. You can imagine making some
more negatives here. So this is a game that seems
kind of important in nature, not just in teenage male
behavior but in animal behavior, since we’re talking about
aggression and non-aggression. So what’s evolutionarily stable
in this game? Well remember our starting
point is what? Our starting point is to look
for symmetric Nash Equilibria. So are there any symmetric Nash
Equilibria in this game and if so what are they?
There are some Nash Equilibria in pure strategies,
there are some Nash Equilibria in this game,
for example, (A,B) is a Nash Equilibrium and
(B,A) is a Nash Equilibrium, but unfortunately they’re not
symmetric, and so far we’re focusing on
games that are symmetric, this is random matching.
There’s no asymmetry in the rows–in the row and column
player, although in the handout–not the handout,
in the reading packet I made for you–they do also look at
some asymmetric versions of games,
but for now we’re just looking at symmetry.
So neither (A,B) nor (B,A) will serve our purpose because
they’re not symmetric Nash Equilibria.
In fact, there is no symmetric pure strategy Nash Equilibrium
in this game. So it can’t be that if this was
a species, if this was an animal that came in,
that had two possible strategies, aggression or
passivity, it can’t be the case in this particular game that you
end up with 100% aggression out there–100% aggressive genes out
there–or 100% unaggressive genes out there.
In either case, if you had 100% aggressive
genes out there, then it would be doing very,
very badly and you get an invasion of passive genes and if
you had 100% passive genes out there,
you’d get an invasion of aggressive genes.
You can’t have a pure ESS, a pure evolutionarily stable
gene mix out there. So what does that suggest?
It suggests we should start looking at mixed strategies.
There is a mixed strategy, there is a symmetric mixed
strategy Nash Equilibrium in the game and we could go through and
work it out. We all know now–probably
you’ve been laboring through the homework assignments–so you all
know how to find a mixed strategy Nash Equilibrium in
this game, you have to set the mix to make
the other player indifferent between her two strategies.
But this is a game we’ve seen already, it’s essentially Battle
of the Sexes, so you probably remember from a
week ago what that equilibrium mix is.
Can anyone remember what the equilibrium mix in Battle of the
Sexes? Its (2/3,1/3),
turns out that (2/3,1/3) is a Nash Equilibrium here.
So if you go back to your notes a week ago you’ll find something
very much like that was a Nash Equilibrium in the original
version of Battle of the Sexes. A week ago it would have been
(2/3,1/3), (1/3,2/3) because things weren’t symmetric.
Now I’ve made things symmetric, it’s just (2/3,1/3) for both
players. You can check it out.
So what’s this telling us? It’s telling us that there’s at
least an equilibrium in this game in which 2/3 of the genes
are aggressive and 1/3 of the genes are unaggressive.
But does that mean anything? What could that mean?
In terms of biology what could that mean?
Well, so far, we’ve been looking at
evolutionarily stable pure strategies and the
evolutionarily stable pure strategies correspond to
evolutionarily stable situations in nature which are what are
called monomorphic. Monomorphic means one shape or
one type out there, but you can also have
situations in nature where there are actually stable mixed types
and they’re called polymorphic. It’s probably not hyphenated,
it’s probably just one word. So you can have a monomorphic
population, that’s what we’ve focused on so far,
but you could also have a mixed population.
Now for this to be a mixed population we better change the
definition accordingly. We’ll come back and talk about
what it means in a second a bit more, but first we better make
sure we have an okay definition for it.
So what I’m going to do is I’m going to drag this definition
down and do something you’re not meant to do usually in teaching,
I’m going to use the eraser to correct my definition.
So here I have my pure strategy definition and let me just
change it into a definition that will allow for these polymorphic
populations. So I’m going to change this
into a P, I’m going to change this pure into mixed,
and everywhere you see an Ŝ I’m going to put a P,,
and over here too and everywhere you see an S’ I’m
going to put a P.” The reason I’m doing this this
way is I want to emphasize that there’s nothing new here,
I’m just writing down the same definitions we had before,
except I’m now allowing for the idea of populations being mixed,
and I’m also, just to note in passing,
I’m also allowing for the possibility that a mutation
might be mixed. Did I catch them all?
So I’ve just gone through the definition you have and I’ve
switched everything from pure to mixed. So in our example does the mix
2/3,1/3 satisfy the definition above?
Let’s go through carefully. So (2/3,1/3) is a Nash
Equilibrium so we’ve satisfied part A, it’s a symmetric Nash
Equilibrium, so we’re okay there.
Is this equilibrium a strict equilibrium?
Is this mixed population 2/3 aggressive and 1/3 unaggressive,
or this mixed strategy, is it a strict equilibrium?
How do deviations do against it? Anybody?
Go ahead. Student: Equally well.
Professor Ben Polak: Equally well,
thank you. So it can’t be a strict Nash
Equilibrium, because if we deviated to A we’d do as well as
we were doing in the mix, or if we deviated to B,
we’d do as well as we’re doing in the mix.
Another way of saying it is, an A mutation does exactly as
well against this mix as the mix does against itself,
and a B mutation does exactly as well.
In fact, that’s how we constructed the equilibrium in
the first place. We chose a P that made you
indifferent between A and B, so in a mixed Nash Equilibrium
it can’t be strict since it is mixed.
In a mixed Nash Equilibrium, a genuinely mixed Nash
Equilibrium by definition, you’re indifferent between the
strategies in the mix. So to show that this is in fact
evolutionarily stable we’d have to show rule B,
so we need to show–we need to check–let’s give ourselves some
room here–we need to check how the payoff of this strategy,
let’s call it P, how P does against all possible
deviations and compare that with how those deviations do against
themselves. We have to make this comparison
– how does this mix do against all other possible mixes versus
how those mixes do against themselves?
We’d have to do this, unfortunately,
we’d have to check this for all possible–and now we have to be
careful–all possible mixed mutations P”.
So that would take us a while, it’s actually possible to do.
If you do enough math, it isn’t so hard to do.
So rather than prove that to you, let me give you a heuristic
argument why this is the case. I’m going to try and convince
you without actually proving it that this is indeed the case.
So here we are with this population, exactly 2/3 of the
population is aggressive and 1/3 of the population is passive.
Suppose there is a mutation, P”, that is more aggressive
than P, it’s a relatively aggressively mutation.
For example, this mutation may be 100%
aggressive, or at least it may be very, very highly
aggressively, maybe 90% aggressive or
something like that. Now I want to argue that that
aggressive mutation is going to die out and I’m going to argue
it by thinking about this rule. So I want to argue that the
reason this very aggressive mutation dies out is because the
aggressive mutation does very badly against itself.
Is that right? If you have a very aggressive
mutant, the very aggressive mutants do very,
very badly against themselves, they get 0.
And that’s going to cause them to die out.
What about the other extreme? What about a deviation that’s
very passive? So a very nice mutation,
a very passive type, for example,
it could a 100% B or you know 99% B or 98% B,
how will that do? Well it turns out,
in this game again, it doesn’t do very well against
itself and in addition, the original mix,
the mixed P that is more aggressive than this very
passive mutation does very well against the mutation.
So the mix that’s in there, the mix that’s in the
population already is relatively aggressive compared to this very
passive mutation and so the incumbent,
the relatively aggressive incumbents are doing very,
very well on average against the mutation,
and hence once again, this equality holds.
So just heuristically, without proving it,
more aggressive mutations are going to lose out here because
they do very badly against themselves,
and more passive mutations are going to do badly because they
make life easy for P which is more aggressive.
So it wasn’t a proof but it was a heuristic argument and it
turns out indeed to be the case. So in this particular game,
a game you could imagine in nature, a game involving
aggression and passivity within this species,
it turns out that in this example the only equilibrium is
a mixed equilibrium with 2/3 aggressive and 1/3 unaggressive.
And this raises the question: what does it mean?
What does it mean to have a mix in nature?
So it could mean two different things.
It could mean that the gene itself is randomizing.
It could mean that the strategy played by the particular ant,
squirrel, lion, or spider is actually to
randomize, right, that’s possible.
But there’s another thing it could mean that’s probably a
little bit more important. What’s the other thing it could
mean? It could mean that in the
stable mix, the evolutionarily stable population,
for this particular spider say, it could be that there are
actually two types surviving stably in these proportions.
If you go back to what we said about mixed strategies a week
ago, we said one of the possible interpretations of mixed
strategies is not that people are necessarily randomizing,
but that you see a mix of different strategies in society.
Again in nature, one of the impossible
interpretations here, the polymorphic population
interpretation, is that, rather than just have
all of the species look and act alike, it could be there’s a
stable mix of behaviors and/or appearances in this species.
So let me try and convince you that that’s not an uninteresting
idea. So again, with apologies I’m
not a biologist, I’ve spent a bit of time on the
web this weekend trying to come up with good examples for you
and the example I really wanted to come up with,
I couldn’t find on the web, which makes me think maybe it’s
apocryphal, but I’ll tell you the story anyway.
It’s not entirely apocryphal, it may just be that my version
of it’s apocryphal. So this particular example I
have in mind, is to do with elephant seals,
and I think even if it isn’t true of elephant seals,
it’s definitely true of certain types of fish,
except that elephant seals make a better story.
So imagine that these elephant seals–it turns out that there
are two possible mating strategies for male elephant
seals. By the way, do you all know
what elephant seals are? They’re these big–people are
looking blankly at me. You all have some rough image
in your mind of an elephant seal?
You’ve all seen enough nature shows at night?
Yes, no, yes? Okay, so there are two male
mating strategies for the male elephant seal.
One successful male mating strategy is to be the head,
the dominant, or a dominant elephant male,
male elephant seal, and have as it were,
a harem of many female elephant seals with which the male mates
with. For the males in the room don’t
get too happy, these are elephant seals,
they’re not you guys. So one possible successful
strategy is to be a successful bull elephant seal and have
many, many, many potentially wives, so to be a polygamist.
Presumably to do that well a good idea, a thing that would go
well with that strategy is to be huge.
So you could imagine the successful male elephant seal
being an enormous animal. It looks like a sort of a
linebacker in football and basically fights off all other
big elephant seals that show up. But it turns out,
I think I’m right in saying if I did my research correctly,
this is true among northern elephant seals but not true
among southern elephant seals, so the Arctic not the
Antarctic, but someone’s going to correct me.
Once it’s on the web I’m going to get floods of emails saying
that I’ve got this wrong. But never mind.
So it turns out that this is not quite evolutionarily stable.
Why is this not evolutionarily stable?
What’s the alternative male strategy that can successfully
invade the large bull harem keeper elephant seal?
Any guesses? Anyone looking for a successful
career as an elephant seal, as a male elephant seal?
Say that again? Student: [Inaudible]
Professor Ben Polak: Good, so good thank you.
Did people catch that? Good, an alternative strategy
is to be a male elephant seal who looks remarkably like a
female elephant seal. Instead of looking like a
linebacker, they look like a wide receiver.
I’m offending somebody in the football team.
You get the idea, right? What do they do?
They sneak in among these large numbers of male elephant seals
and they just mate with a few of them.
So they look like a female seal, and they can hide among
the female elephant seals in the harem,
and they mate with a few of them and provided this is
successful enough it’ll be evolutionarily stable for the
female elephant seal to want to mate with that too.
Now I forget if actually this is exactly right,
but it’s certainly–I did enough research over the weekend
to know it’s right at least in some species,
and the nicest part of this story is that at least some
biologists have a nice technical name for this strategy that was
well described by our friend at the back,
and the name for this strategy is SLF, and since we’re on film
I’m going to tell you what the S and the L are,
but you’ll have to guess the rest.
So this is sneaky, this is little,
and you can guess what that is. So this turns out to be
actually quite a common occurrence.
It’s been observed in a number of different species,
perhaps not with the full added color I just gave to it.
So having convinced you that polymorphic populations can be
interesting, let’s go back to a case,
a more subtle case, of aggression and
non-aggression, because that seems to be one of
the most important things we can think of in animal behavior.
So let’s go back and look at a harder example of this,
where we started. So as these examples get
harder, they also get more interesting.
So that’s why I want to get a little bit harder. So the chicken game,
the Battle of the Sexes game, is not a particularly
interesting version of aggression and non-aggression.
Let’s look at a more general version of aggression versus
non-aggression, and let’s look at a game that’s
been studied a lot by biologists and a little bit by economists
called hawk-dove. And again, just to stress,
we’re talking about within species competition here,
so I don’t mean hawks versus doves.
I mean thinking of hawk as being an aggressive strategy and
dove as being a dovish, a passive strategy.
So here’s the game and now we’re going to look at more
general payoffs than we did before.
So this is the hawk strategy, this is the dove strategy–hawk
and dove–and the payoffs are as follows.
V + C–sorry, start again. (V-C)/2 and (V-C)/2 and here we
get V/2 and V/2, and here we get V and 0 and
here we get 0 and V. So this is a generalization,
a more interesting version of the game we saw already.
Let’s just talk about it a little bit.
So the idea here is there’s some potential battle that can
occur among these two animals and the prize in the battle is
V. So V is the victor’s spoils and
we’re going to assume that V is positive.
And unfortunately, if the animals fight–so if the
hawk meets another hawk and they fight one another–then there
are costs of fighting. So the costs of fighting are C,
and again, we’ll assume that they’re positive.
So this is the cost of fighting. This more general format is
going to allow us to do two things.
We’re going to look and ask what is going to be
evolutionarily stable, including mixtures now.
And we’re also going to be allowed to ask,
able to ask, what happens,
what will happen to the evolutionarily stable mix as we
change the prize or as we change the cost of fighting?
Seems a more interesting, a richer game.
Okay, so let’s start off by asking could we have an
evolutionarily stable population of doves?
So is D an evolutionarily stable strategy?
I’ll start using the term ESS now.
So ESS means the evolutionarily stable strategy.
Is D in an ESS. So in this game,
could it be the case that we end up with a population of
doves? Seems a nice thing to imagine,
but is it going to occur in this game in nature?
What do people think? How do we go about checking
that? What’s the first step?
First step is to ask, is (D, D) a Nash Equilibrium?
If it’s evolutionarily stable, in particular,
(D,D) would have to be a Nash Equilibrium.
That’s going to make it pretty easy to check,
so is (D, D) a Nash Equilibrium in this game?
It’s not, but why not? Because if you had a mutation,
keep on tempted to say deviation, but you want to think
of it as mutation. If I had a mutation of hawks,
the hawk mutation against the doves is getting V,
whereas, dove against dove is only getting V/2,
so it’s not Nash. So we can’t have a
evolutionarily stable population of doves, and the reason is
there will be a hawk mutation, an aggressive type will get in
there and grow, much like we had last week when
we dropped Rahul into the classroom in Prisoner’s Dilemma
and he grew, or his type grew.
So second question: is hawk an evolutionarily
stable strategy? So how do we check this?
What we have to look at once again–and ask the
question–this is the first question to ask is:
is (H,H) a Nash Equilibrium? So is it a Nash Equilibrium?
Well I claim it depends. I claim it’s a Nash Equilibrium
provided (V-C)/2 is at least as large as 0.
Is that right? It’s a Nash Equilibrium–it’s a
symmetric Nash Equilibrium –provided hawk against hawk
does better, or does at least as well as, dove against hawk.
So the answer is yes, if (V-C)/2 is at least as big
as 0. So now we have to think fairly
carefully, because there’s two cases.
So case one is the easy case which is when V is strictly
bigger than C. If V is strictly bigger than C,
then (V-C)/2 is strictly positive, is that right?
In which case, what kind of a Nash Equilibrium
is this? It’s strict right?
So if V is bigger than C then (H, H) is a strict Nash
Equilibrium. The second case is if V is
equal to C, then (V-C)/2 is actually equal to 0,
which is the same as saying that the payoff of H against H
is equal to the payoff of dove against hawk.
That correct? So in that case what do we have
to check? Well I’ve deleted it now,
it’ll have to come from your notes, what do I have to check
in the case in which there’s a tie like that? What do I have to check?
I have to check–in this case I need to check how hawk does
against dove, because dove will be the
mutation. I need to compare that with the
payoff of dove against dove. So how does hawk do against
dove? What’s the payoff of hawk
against dove? Anybody?
Payoff of hawk against dove. It shouldn’t be that hard.
It’s on the board: hawk against dove.
Shout it out. V, thank you.
So this is V and how about the payoff of dove against dove?
V/2, so which is bigger V or V/2?
V is bigger because it’s positive, so this is bigger so
we’re okay. So what have we shown?
We’ve shown, let’s just draw it over here,
we’ve shown, that if V is at least as big as
C, then H is an evolutionarily stable strategy.
So in this game, in this setting in nature,
if the size of the prize to winning the fight is bigger than
the cost that would occur if there is a fight,
then it can occur that all the animals in this species are
going to fight in an evolutionarily stable setting.
Let me say it again, if it turns out in this setting
in nature that the prize to winning the fight is bigger,
or at least as big as, the cost of fighting,
then it will turn out that it be evolutionarily stable for all
the animals to fight. The only surviving genes will
be the aggressive genes. What does that mean?
So what typically do we think of as the payoffs to fight and
the costs of fighting? Just to put this in a
biological context. The fight could be about what?
It could be–let’s go back to where we started from–it could
be males fighting for the right to mate with females.
This could be pretty important for genetic fitness.
It could be females fighting over the right to mate with
males. It could also be fighting over,
for example, food or shelter.
So if the prize is large and the cost of fighting is small,
you’re going to see fights in nature.
But we’re not done yet, why are we not done?
Because we’ve only considered the case when V is bigger than
C. So we also need to consider the
case when C is bigger than V. This is the case where the cost
of fighting are high relative to the prize in the particular
setting we’re looking at. So again let’s go back to the
example, suppose the cost of fighting could be that the
animal could lose a leg or even its life,
and the prize is just today’s meal, and perhaps there are
other meals out there. Then we expect something
different to occur. However, we’ve already
concluded that even in this setting it cannot be the case
only to have doves in the population.
We’ve shown that even in the case where the costs of fighting
are high relative to the prizes, it cannot be evolutionarily
stable only to have dove genes around;
passive genes around. So in this case,
it must be the case that if anything is evolutionarily
stable it’s going to be what? It’s going to be a mix.
So in this case, we know that H is not ESS and
we know that D is not ESS. So what about a mix? What about some mix P?
We could actually put P the in here.
We can imagine looking for a mix P, 1- P that will be stable.
Now how we do go about finding a possible mixed population that
has some chance or some hope of being evolutionarily stable?
So here we are we’re biologists. We’re about to set up an
experiment. We’re about to either
experiment, or about to go out and do some field work out
there. And you want to set things up.
And we’re asking the question: what’s the mix we expect to
see? What’s the first exercise we
should do here? Well if it has any hope to be
evolutionarily stable, what does it have to be?
It has to be a symmetric Nash Equilibrium.
So the first step is, step one find a symmetric mixed
Nash Equilibrium in which people will be playing P (,1- P).
It’s symmetric, so both sides will be playing
this. So this is good review for the
exam on Wednesday. How do I go about finding a
mixed equilibrium here? Shouldn’t be too many blank
faces. This is likely to come up on
the exam on Wednesday. Let’s get some cold calling
going on here. How do I find the mix strategy?
Just find anybody. How do we find–how do I find a
mixed strategy equilibrium? Student: Just use the
other player’s payoff. Professor Ben Polak: I
use the other player’s payoffs and what do I do with the other
person’s payoffs? Student: You set them
equal. Professor Ben Polak: Set
them equal, okay. So here it’s a symmetric game.
It’s really there’s only one population out there.
So all I need, I need the payoff of hawk
against P or (P, 1-P), I need this to be equal
to the payoff of dove against this P.
So the payoff of hawk is going to be what?
Just reading up from up there–let’s use our pointer–so
hawk P of the time will meet another hawk and get this
payoff. So they’ll get–so P of the
time they’ll get a payoff of (V-C)/2 and 1- P of the time
they’ll meet a dove and get a payoff of V.
And dove against this same mix (P, 1- P): P of the time they’ll
meet a hawk and get nothing and 1-P of the time they’ll meet
another dove and get V/2. Everyone happy with the way I
did that? That should be pretty familiar
territory to everybody by now, is that right?
So I’m going to set these two things equal to each other since
they must be equal, if this is in fact a mixed
strategy equilibrium. And then I’m going to play
around with the algebra. So as to save time I did it at
home, so trust me on this–this is implication with the word
trust on top of it–trust me that I got the algebra right or
check me at home. This is going to turn out to
imply that P equals V/C. So it turns out that there is
in fact a mixed Nash Equilibirum.
There’s a mixed Nash Equilibrium which is of the
following form, V/C and 1-V/C played by both
players. Is this a strict Nash
Equilibrium? I’ve found the Nash
Equilibrium, is it strict? Everyone should be shouting it
out. Is it strict?
It can’t be strict because it’s mixed right?
By definition it’s not. It can’t be strict because we
know that deviating to H, or, for that matter,
deviating to D yields the same payoff.
So it can’t be a strict Nash Equilibrium.
So we need to check something. So we need to check not strict;
so we need to check whether U of P against P” is bigger than
U of P” against itself, and we need to check this for
all possible mutations P”. Again, that would take a little
bit of time to do in class, so just trust me on it and once
again, let me give you the heuristic
argument I gave to you before. It’s essentially the same
argument. So the heuristic argument I
gave to you before was: imagine a P’,
a mutation, that is more aggressive than our candidate
equilibrium. If it’s more aggressive then
it’s going to do very, very badly against itself
because C is bigger than V in this case.
So it’s actually going to get negative payoffs against itself.
Since it gets negative payoffs against itself,
it turns out that will cause it to die out.
Conversely, imagine a mutation that’s relatively soft,
that’s relatively dovish, this mutation is very good for
the incumbents because the incumbents essentially beat up
on it or score very highly on it.
So once again, the more dovish mutation will
die out. So again, that isn’t a proof
but trust the argument. We need to show this but it
does in fact turn out to be the case.
So what have we shown here? It didn’t prove the last bit,
but what we’ve argued is that, in the case in which the cost
of fighting in nature are bigger than the prizes of winning the
fight, it is not the case that we end
up with a 100% doves. So we don’t end up with no
fights, for example: no fights is not what we would
expect to observe in nature. And we don’t end up with a 100%
fights: 100% fights is not what you expect to see in nature.
What we end up with is a mixture of hawks and doves,
such that V/C is the proportion of hawks.
So the fights that occur are essentially V/C squared.
We can actually observe those fights in nature.
What lessons can we draw from this?
So biology lessons. So we used a lot of what we’ve
learned in the last day or so to figure out what the ESS was
there. We kind of did the nerdy part.
Now let’s try and draw some lessons from it. So the first thing we know
is–I’ve hidden what we care about here–so we know that if V
is smaller than C, then the evolutionarily stable
mixed population has V/C hawks. So let’s just see how much of
this makes sense. So as V goes up,
as the prizes go up–if you took the same species and put
them into a setting in which the prizes tended to be larger–what
would we expect to see? Would we expect to see the
proportion of hawks up or down? Up right: as V goes up,
we see more hawks. What else do we see?
Not so surprisingly as C goes up–we look at settings where
the cost of fighting is higher–we tend to see more
doves. Now this is in ESS,
so more hawks in the evolutionarily stable mix.
And more doves in the evolutionarily stable mix.
It’s possible, of course that the species in
question can recognize these two different situations and be
coded differently, to behave differently in these
two different situations but that’s beyond the class for now.
Perhaps a more interesting observation is about the
payoffs. Let’s look at the actual
genetic fitness of the species overall.
So in this mix what is the payoff?
Well how are we going to figure out what is the payoff?
So the payoff in this mix, we can actually construct by
looking at the payoff to dove. It doesn’t really matter
whether you look at the payoff to dove or the payoff to hawk.
Let’s look at the payoff to dove.
So the payoff was what? It was, if you were dove,
then 1-V/C of the time, you met another dove and in
that instance you get a payoff of V/2 and it must be the payoff
to being a dove is the same since they’re mixing.
This is the payoff. So what’s happening to this
payoff as we increase the cost of fighting?
What happens as C rises? So just to note out,
what happens as C goes up? So you might think naively,
you might think that if you’re in a setting,
be it a social evolutionary setting or a biology,
a nature evolutionary setting, you might think that as the
cost of fighting goes up for you guys in society,
or for the ants, antelopes, or lions we’re
talking about, you might think the payoffs in
society go down. Costs of fighting go up,
more limbs get lost and so on, sounds like that’s going to be
bad for the overall genetic fitness of the species.
But in fact we don’t find that. What happens as C goes up?
The payoff goes up. As C goes up,
the payoff goes up. As we take C bigger,
this begets smaller, which means this is bigger.
Everyone see that? Just look at that term
(1-V/C)(V/2), it’s actually increasing in C.
So how does that work? As the costs of fighting go up,
it’s true that if you do fight, you’re more likely to lose a
finger, or a limb, or a claw,
or whatever those things are called, or a foot or whatever it
is you’re likely to lose. But the number of fights that
actually occur in this evolutionarily stable mix goes
down and it goes down sufficiently much to compensate
you for that. Kind of a remarkable thing.
So these animals that actually are going to lose a lot through
fighting are actually going to do rather well overall because
of that mix effect. I feel like it’s one of those
strategic effects. Now of course that raises a
question, which is what would happen if a particular part of
this species evolved that had lower costs of fighting?
It could regrow a leg. Sounds like that would do
pretty well, and that would be bad news for the species as a
whole. Third thing we can observe here
is what’s sometimes called identification. So what does identification
mean here? It means that by observing the
data in nature, by going out and filming these
animals behaving for hours and hours,
or changing their setting in a lab and seeing how they
interact, or changing their setting in the field and seeing
how they interact, we can actually observe
something, namely the proportion of fights.
Perhaps we can do better now and actually look at their
genetics directly since science has evolved, and we can actually
back out the V and the C. By looking at the proportion of
hawk genes out there or hawkish behavior out there,
we can actually identify what must be the ratio of V to C.
We can tell what the ratio V/C is from looking at data.
We started off with a little matrix, I’ve just written in V’s
and C’s, I didn’t put any numbers in there.
We can’t tell what V is, we can’t tell what C is,
but we can tell what the ratio is by looking at real world
data. So if you spend enough hours in
front of the TV watching nature shows you could back this out.
Not literally, you need to actually do some
serious work. So this is a useful input of
theory into empirical science. You want theory to be able to
set up the science so you can back out the unknowns in the
theory: and that’s called identification,
not just in biology but in Economics as well.
Now there’s one other thing you’d like theory to be other
than identifiable. You’d like theory to be
testable. You’d like theory to make
predictions that were kind of outside of the sample you
started with. Is what I’m saying familiar to
everyone in the room. This is a very familiar idea
I’m hoping to everybody, slightly philosophy but a very
familiar idea. You have a new theory.
It’s one thing for that theory to explain the existing facts,
but you’d like it to predict new facts.
Why? Because it’s a little easy–it
might be a little bit too easy to reverse engineer a model or
theory to fit existing facts, but if it has to deal with new
facts, that’s kind of exciting, that’s a real test.
Does that make sense? So you might ask about this
theory, you might say, well it’s just a whole bunch of
‘just so stories’. Does anyone know what I mean by
a ‘just so story’? It’s children’s stories,
written by Kipling, and people sometimes accuse a
lot of evolutionary theory as being ‘just so stories’,
because you know what the fact is already, and you reverse the
game, you come up with a story afterwards.
That doesn’t sound like good science.
So you’d like this theory, this theory that matches Game
Theory with evolution, to predict something that we
hadn’t seen before. And then we can go out and look
for it, and see if it’s there, and that’s exactly what we now
have. So our last example is a
slightly more complicated game again. The slightly more complicated
game has three strategies and the strategies are called,
I’ll tell you what the strategies are called in a
second actually, I’ll give you the payoffs first
of all. So once again this is a game
about different forms of aggression and we’ll look at
other interpretations in a second.
Once again, V is going to be the prize for winning,
0 is going to be the prize for losing, and 1 is if it’s a tie.
I’m just short sighted enough that I can’t read my own
writing. So I hope I’ve got this right,
there we go. So this is the game,
and we’re going to assume that the prize V is somewhere between
1 and 2. So V you can think of as
winning, 0 is losing, and 1 is if it’s a tie.
Does anyone recognize what this game essentially is?
It’s essentially rock, paper, scissors.
And it turns out that when biologists play rock,
paper, scissors they give it a different name,
they call it “scratch, bite, and trample.”
Scratch, bite, and trample is essentially the
tactics of the Australian football team.
So scratch, bite, and trample are the three
strategies, and it’s a little bit like rock,
paper, and scissors. How do we change it?
First, we added 1 to all the payoffs to make sure there are
no negatives in there and second we added a little bit more than
1 to winning. If we added 1 to
everything–sorry, a little bit less than 1 to
winning. So if we added 1 to everything
then V would have been 2, but we kept V somewhere between
1 and 2. So this is certainly a game,
you can imagine in nature, there’s three possible
strategies for this species and the payoff matrix happens to
look like this. So where’s my prediction going
to come from? Well since this is rock,
paper, scissors we know that there’s really only one hope for
an evolutionarily stable strategy.
Since it’s essentially rock, paper, scissors what would,
if there is an evolutionarily stable strategy or
evolutionarily stable mix, what must it be?
(1/3,1/3,1/3). So the only hope for an ESS is
(1/3,1/3,1/3)–let’s put that in here.
So (1/3,1/3,1/3) and you can check at home that that indeed
is a mixed-strategy equilibrium. And the question is:
is this evolutionarily stable? So we know it’s a Nash
Equilibrium that I’ve given you, and we know it’s not a strict
Nash Equilibrium. Everyone okay with that?
It can’t be a strict Nash Equilibrium because it’s mixed.
So if this is an ESS it must be the case–we’d have to check
that–let’s call this P again like we’ve been doing–we have
to check that the payoff from P against any other P’ would have
to be bigger than the payoff from P’ against itself.
We need that to be the case. So let P’ be scratch.
So let’s compare these things. So U of P against scratch is
what? Well you’re playing against
scratch, you are 1/3 scratch, 1/3 bite, 1/3 trample.
So 1/3 of the time you get 1,1/3 of the time you get
nothing, and 1/3 of the time you get V.
Is that right? So your payoff is (1+V)/3.
How would we do if we’re scratch against scratch?
The payoff of scratch against scratch is what?
No prizes for this, what’s the payoff of scratch
against scratch? 1.
Which is bigger, (1+V)/3 or 1? Well look, V is less than 2 so
(1+V)/3 is less than 1. So 1 is bigger.
So in this game the only hope for an evolutionarily stable mix
was a (1/3,1/3,1/3) and it isn’t stable.
So here’s an example. In this example there is no
evolutionarily stable strategy. There’s no evolutionarily
stable mix. Then the obvious question is,
what does that mean in nature? Can we find a setting that
looks like rock, paper, scissors in nature in
which nothing is ES? If nothing is ES what’s going
to happen? We’re going to see a kind of
cycling around. You’re going to see a lot of
the scratch strategy, followed by a lot of the
trample strategy, followed by a lot of the bite
strategy and so on. So it turns out that’s exactly
what you see when you look at these, the example I’ve left you
on the web. There’s an article in nature in
the mid 90s that looked at a certain type of lizard.
And these lizards come in three colors.
I forget what the colors are, I know I wrote it down.
One is orange; one is yellow;
and one is blue. And these lizards have three
strategies. The orange lizard is like our
big elephant bull: it likes to keep a harem of
many–or a large territory with many female lizards in it to
mate with. But that can be invaded by our
SLF strategy which turns out to be the yellow lizard.
The yellow lizard can invade and just mate with a few of
these female lizards. But when there are too many of
these sneaky yellow lizards, then it turns out that they can
be invaded by a blue lizard: and the blue lizard has much
smaller territories, it’s almost monogamous.
So what happens in nature is you get a cycle,
orange invaded by yellow, invaded by blue:
harem keeper invaded by sneaky, invaded by monogamous invaded
by harem keeper again. Indeed, the population does
cycle around exactly as predicted by the model,
so here’s an example of evolutionary theory,
via Game Theory, making a prediction that we can
actually go off and test and find.
This, for biologists, was like finding a black hole,
it’s a really cool thing. We’ll leave evolution here:
mid-term on Wednesday, we’ll come back to do something
totally different next week. See you on Wednesday.