Problem with definition of ESS

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Jeffrey C. Ely

Dec 3, 1993, 5:09:07 AM12/3/93
to

It seems to me that there is a problem with the standard definition of ESS.

DEFINITION:
A strategy s is an ESS if for each mutant strategy m, there exists
an e such that:

eU(m,m) + (1-e)U(m,s) < eU(s.m) + (1-e)U(s,s)

In this definition, e is meant to be interpreted as the probability of
being matched with a mutant. Presumably this probability is just the
fraction of the population made up of mutants.

Thus, a strategy is an ESS if for every mutant, we can find a fraction
small enough that if the mutant fraction of the population is at least
that small, the mutants will do worse than the population and therefore
be driven out.

But there is no guarantee that there will be a uniform upper bound which will
work for every possible mutant. That is, the mutant fraction of the
population may have to be taken arbitrarily small in order to take care of
all possible mutant strategies. But if the population is finite, there
*is* a lower bound to the actual fraction of the population which can be
made up of mutants. Specifically, in a population of size N, the
probability of meeting a mutant is at least 1/N-1 (I am assuming there are at
least 2 mutants in the population. If there were only 1, the mutant could
never meet another).

To illustrate the point, consider the following symmetric 2 player game:

Each player chooses a real number in the interval [0,1]. The payoffs are
symmetric, so if player i plays s, and player j plays t, the payoff to player
i will be U(s,t) and the payoff to player j will be U(t,s).

For each t in [0,1) U(t,t) = 2
U(1,s) = 1 for all s in [0,1]
U(s,1) is continuous and strictly increasing for s in [0,1]
For s <> t and s,t<1 U(s,t) = 0

In this game, 1 is an ESS because it is a strict Nash Equilibrium. However,
for any finite population of size N (however large), I can find a mutant m
which can profitably enter, since the probability of meeting a
mutant will be at least 1/N-1. I simply choose m to be close enough to 1.
Such mutants, when paired together, will earn a payoff of 2, better than the
payoff of 1 which the incumbents earn when they are paired together, and if m
is chosen close enough to 1, the penalty to the mutants when they meet an
incumbent can be made very small so that on the whole, the mutants do better
than the population.

My conclusion is that there is an implicit order-of-quantifiers type issue in
defining ESS. One can either first fix a mutant and then see if there is a
population large enough to kill it, or alternatively, one can fix a population
size and then see if there is a mutant which can enter given the size of the
population. It seems to me that the latter is the more reasonable order since
the population, however large, must be finite.

I have in mind a refinement of ESS which captures in a natural way the
restriction that the population must be finite (although it may be
arbitrarily large). If there is any interest, I will be glad to post it
or to discuss it via e-mail. (Note that the example given above suggests
that if we impose this restriction, some inefficient ESS's can be ruled
out. I have some results along these lines)

Jeff

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Jeff Ely* "He said the Sun's not yellow...
Department of Economics, UC Berkeley It's Chicken!"
je...@econ.berkeley.edu
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*No opinions, unemployed