Review of Why Pure Strategies Are
Not Evolutionarily Stable in the War of Attrition

Assume that you have a pure strategy A which involves accepting a maximum cost that is less than half of the resource value, V. Assume that the initial population is entirely A strategists. Before we start, let's remember the rules for evolutionary stability:

Rule 1 (common interactions): E(var, var) > E(fix(x),var) OR Rule 2: IF (common interactions): (part a): E(var, var) = E(fix(x),var) THEN (rare interactions): (part b): E(var,fix(x)) > E(fix(x),fix(x))

Now assume that "less than A" which has a maximum acceptable cost less than that of A tries to invade.

• "Less than A" always loses to A
• therefore its total payoff is a (loss) equal to its maximum cost
  E("less than A", A) =( - "less than A cost"). Since it is a rare invader, it has no contests with itself.
• By contrast, in A's most common interaction against other A strategists, A comes out ahead since it splits these contests with its opponent. Thus,
  E(A,A) = 0.5V - costA = positive number (since A < 0.5V, see above)
• Now, since cost E(A,A) > E(less than A, B) A is stable against the invader "less than A".

Now, let's assume that a new invader, B arrives. B' s max cost is greater than A's but is still less than 0.5 V.

• B wins in each B vs. A contest and obtains E(B,A) = V- costA.
• By contrast, when A plays another A (most common) it gains E(A,A) = 0.5 V-A.
• Since this is less than E(B,A) = V - A, A is not stable against B and B successfully invades.

OK, assume that B has taken over and is very common. Now the big question:

? What if another strategy (we'll call it C) that waits just a bit longer than B shows up?

The answer, of course, is that we will have a repeat of the situation when B invaded A! Thus, C will now successfully invade B and so B is not a pure ESS.

If you continue to follow this logic, you may come to the conclusion that a strategy that is willing to pay an infinite cost would be a pure ESS. Not so fast:

Situation 3:These Queues are Getting Too Long!

Imagine that our population continues to be invaded by individuals that are willing to wait longer to win. Now, costs are increasing with longer waits:

m = Display Costs = k * t

but the value of the resource is still the same. Thus, the net gain for winning is becoming less and less the longer one waits to win.

Imagine that we finally get to a waiting time that is so long that it is greater than 1/2 the value of the resource, i.e.:

m(Long) > 0.5*V

m(Long) is still a winning value with respect to taking the resource when against any strategist playing a lower maximum cost. But that ability to win doesn't mean that it is stable.

Let's say a new mutant appears that does not wait or display at all , we can call this one maxCost(x=0). Here's a payoff matrix for contests between maxCost(x=0) and m(Long):

	m(Long)	maxCost(x=0)
m(Long)	< 0 (negative)	V
maxCost(x=0)	0	0.5 * V

So if m(Long) represents a strategy where display times are more costly than 0.5 V, will it be stable against invasion by individuals who simply do not display?

• Once again, m(Long) is common, maxCost(x=0) is rare.
• Looking down the first column which shows the most common interactions for each strategy, we can see that maxCost(x=0) can invade a population of m(Long) strategists since E(maxCost(x=0),m(Long)) > E(m(Long).m(Long)).
• This will be true for any strategy where the maximum cost is greater than 0.5 V
• One other point about No Display. Notice that if 0.5 V>m(Display), then the matrix looks like this:

	m(Long)	maxCost(x=0)
m(Long)	> 0 (positive)	V
maxCost(x=0)	0	0.5 * V

and a population of maxCost(x=0) can be invaded! Thus we are back to the situation where increasingly higher maximum cost strategists can invade.

Take Home: We seem to have a dog chasing its tail! Every strategy has other strategies that can invade it. It is impossible to image a large, mature population playing this game that would contain all of these strategies at some frequency, each coexisting with the others. There would be cases where each strategy would lose and other cases where it would win. The payoffs would be different with each type of opponent. If you understand this example, you now have a 'bare-bones" understanding of the mixed ESS in the war of attrition.

A moment's thought will lead you to realize that the result will be a mix of different strategies.