Synopsis: This page will show that a particular mixed strategy that is composed of all possible acceptable costs, each to be played at a unique frequency is evolutionarily stable in the symmetrical war of attrition against any pure strategy (unique maximum cost) or other mix of pure strategies. We will term the stable mixed strategy "var". We will see that var is characterized by:
The approach on this page will be to
Please note that this is the most mathematical section of the Game Theory Website. It must be so because we will need to derive an equation that describes potentially an infinite number of behaviors (an infinite number of different maximum acceptable costs). In finding this equation and later in showing that 'var' is an ESS, we will make use of simple differential and integral calculus. I have tried to explain why these techniques are used and further, to explain how they are used so that any interested student, regardless of whether or not they are familiar with calculus, should be able to follow the arguments. As importantly, I hope to convince students of the benefits to any biologist gained by understanding basic calculus. |
| Note: In addition to the hypertext, Prof. Kevin Mitchell of Hobart and William Smith Colleges has provided an excellent overview of integration and probability density functions. This is available as a PDF document. |
On the last page we learned that in the symmetrical war of attrition, each unique cost x that an animal is prepared to pay (or time it is willing to display) is a pure strategy. Thus, there are potentially an infinite number of pure strategies each defined by a different cost x.
We also learned that no pure strategy is an ESS in the war of attrition. Given this, could there be a mixed ESS?
In looking for this mixed ESS, we must realize that any pure strategy is a candidate for inclusion in the mixed ESS. In fact, we expect that every possible pure strategy should belong to the mix (i.e., all possible maximum acceptable costs should support the mix). The reason for this is simple -- we learned earlier that under the right circumstances, any fix(cost) strategy can increase and/or mixes of these strategies can appear -- it's just that none of these are evolutionarily stable. So, we expect that any stable mix will contain all possible strategies as supporting strategies. Press here to review and to get a glimpse of the ESS we pursue!
Definitions: PURE STRATEGY is defined as some unique maximum acceptable cost between zero and infinity. SUPPORTING STRATEGIES: all pure strategies that are members of an equilibrial mix. See Bishop and Cannings (1978). A synonym for supporting strategy is component strategy. |
In characterizing a mix, we must know the likelihood that a given player might encounter each of these supporting strategies. While it is possible that these frequencies are the same for each supporting strategy, it would seem far more likely that many if not all supporting strategies would occur at their own unique frequencies. The only rules are that:
Thus, we can summarize the mix as:
| eq. 1: |
where a, b to n are supporting strategies and prob(cost(a)), etc. is either the frequency of the strategy in the population or the probability that a mixed strategist "adopts" that particular cost in a given contest.
Notice the last point -- as we learned earlier when we considered the "Hawks and Doves" game, there are two ways to produce an equilibrial mix. To this list, we'll add a third. A population that is evolutionarily stable could be:
As we start to look for a way to describe the mix, we seem to face a daunting task. We expect all possible costs to be members of this mix. Thus, there are an infinite number of supporting strategies each potentially at its own unique frequency.
So, we will not be able to use the simple technique to find the mix that we learned with Hawks and Doves. Instead of only needing a couple of linear equations to find two frequencies, we need a function that can give us the correct frequency for an infinite number of different supporting strategies! What follows is a general description of the methods used by Maynard Smith (1974) to find this function.
| Please read this section carefully; it sets the foundation, establishes terminology, and reviews the mathematics used throughout the rest of our treatment of the war of attrition. Exposition that is not crucial (i.e., can be taken on faith) is located on supplementary pages. Follow links to these pages when you are confident of the basics -- they're worth looking at, when you are ready. |
Here we go. We shall use the payoff that a specific supporting strategy expects to receive when competing against the 'mix' to find the function that gives us the equilibrial frequency of each strategy supporting the mix.
So, we start with a pure strategy that is a member of the mix. (See above to review why a pure strategy can be part of the equilibrial mix.)
Now, imagine that fix(x=m) is about to play a series of contests at random against other individuals (supporting strategies) from that mix. So, fix(x=m)'s opponent in any contest can be understood to be "mix" itself.
| Remember, it doesn't matter whether fix(x=m)'s opponent is a pure or mixed strategist: in either case we know the result is that only one strategy can be played by an opponent in a given game and the chance that a particular strategy (maximum cost) will be faced is given by the characteristics of the equilibrium (review). |
Let's find an equation for the payoff fix(x=m) receives against any other supporting strategy in the mix, E(fix(x=m), mix). Starting, in general terms:
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Lifetime Net Benefits to Focal Strategy in Wins Minus Lifetime Costs to Focal Strategy in Losses
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In finding these equations, let's make one other important assumption -- we will assume that the resource has a constant value in any given contest.
Constant Resource Values? You may think that it is obvious that a resource value should be constant in any contest. There certainly are many if not most situations where this is true. But, think for a moment and you'll realize that it is quite possible for a resource to become depleted during a contest. For example, individuals may be contesting a resource that one of them already is using or that naturally depletes in value over time independent of anything the contestants are doing. Or, while two individuals contest for a resource, it is possible that another individual, perhaps a member of a different species depletes it. So, while reasonable for most situations, the assumption that for a contest V= constant may not always be justified. |
Finding Expected Lifetime Net Benefits: Benefits are only obtained by the focal strategist when she wins -- i.e., when the focal strategist is willing to pay a higher cost than her opponent from the mix (x < m where m = the cost the focal strategist will pay):
| eq. #3a: Net Benefit to fix(x=m) in a win = (V - x) |
where V is the resource value and x is the cost the opponent from "mix" is willing to pay.
Unfortunately, equation #3a is not sufficient for our needs. The complexity of the war of attrition intervenes!
Recall that the mix is composed of an infinite number of component strategies. Fix(x=m) only faces one of these supporting strategies in any given contest. Thus, equation #3a only describes the net gain in one specific contest. You should realize that this particular contest will probably be quite rare given the many different strategists that fix(x) could face from the mix. Thus, one particular contest and its benefits will have little if any important lifetime effect on fix(x=m)'s fitness. Single contests cannot describe the net benefit that the focal supporting strategy expects to gain from a large number (a lifetime) of contests.
To get an accurate measurement of lifetime net gains, we need to take into account all types (costs) of contests that fix(x=m) will win and the probability of each:
(If you are having any trouble with this, please press here and read some more.)
Let's re-express eq. 3b using the notation of calculus (If you aren't familiar with calculus, don't fret because it will be fully explained!). We will use calculus because it will let us solve this complex problem (algebra just won't work here) and because it will ultimately give us an exact answer .
First, the definitions of a number of symbols (most we have seen before):
and to reiterate:
Equation
for Net Benefit to Focal Supporting Strategy Now, for those of you who haven't had calculus or who need a review, let's see what eq 3c means. First off, realize that it expresses the same ideas as does eq. 3b. With that assurance, let's start with the expression to the right of the integration sign (the integration sign is the S-like symbol with m above and 0 below it -- more about it below):
This expression calculates the lifetime net benefit in winning a contest of a given cost x. Recall that V = resource value and that x is the maximum cost that a particular opponent is willing to pay. Thus, as in eq. 3a, (V - x) is the net gain to fix(x=m) (see below for note about wins). For example, if in a given contest V = 1.0 and x = 0.001 fitness units, the net gain in winning this contest is 0.999 fitness units. Note that we could just as well write(V-x) as (V-m) and we will later on. Remember that it is not certain that fix(x=m) will play any particular supporting strategy in the mix. Instead, the probability of playing against a particular strategy x supporting the mix is p(x)dx where p(x) is the function that we want to discover to complete the description of the mix. The notation dx that follows p(x) simply means that we will multiply p(x) times an infinitesimally small value of cost. So, solving p(x)dx will give us the chance that our focal fix(x=m) strategist faces any particular value of x from the mix.Be careful not to assume this means some variable "d" times the cost x that"mix" adopted in this game. Also, don't make the common mistake of thinking that dx increases as x increases. It is a constant, tiny amount of cost. Finally, there is the integration sign:
Notice how the limits of the integration are crucial for defining what is a victory by fix(x=m) over mix. As long as the x from the opposing "mix" is less than m, then fix(x=m) wins and the expression calculates the added lifetime net benefit of this win. To summarize: for any contest where x < m, we
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Expected Lifetime Costs for Losses: Benefits were the hard part of the E(focal supporting strategist, mix) equation. Calculation of lifetime costs to focal strat fix(x=m) in contests it loses to the mix (i.e., a mix strategy opponent) is much easier.
As before, the logic is simple. Fix(x=m) loses whenever x, the cost the opponent from mix in any particular contest is willing to pay, is greater than m. All of these contests end with a cost = m. Therefore, for any one losing contest:
So, unlike the equation for net benefit, the costs in any loss are always the same. But, we're not done because as with net benefits, we need to take into account the proportion of the time fix(x=m) encounters an opponent that (in this case) it loses to:
Lifetime Costs of Losing to the Mix where m is the maximum cost that our focal supporting strategy will pay and the function Q(m) gives the lifetime proportion of times that fix(x=m) loses to another member of the mix. Now:
Once again, some explanation of this equation:
To recapitulate, to get lifetime expected cost of losses, we simply multiply this cost times the chance that fix(x=m) will lose. Notice that as with net benefits, the function p(x) is central.
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So, to get the expected lifetime payoff to fix(x=m) vs. the equilibrial mix, we simply substitute the two equations for net benefit and cost:
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Now we have the payoff equation (eq. #5 ) that contains the function p(x). How does one solve to find the function p(x)? It is not terribly difficult but then neither is it central to our story. At some point, if you are interested, you should take a look. But for the moment, we'll proceed directly to the next section where we'll introduce the result that Maynard Smith obtained for p(x) and we'll discuss it in considerable detail.
Press here to view an outline of how the solution of eq. 5 for p(x) is found. |
Recall that Maynard Smith's goal was to find a function, p(x), that would supply the frequencies of each supporting strategy (cost, x) for an equilibrium in the war of attrition. To get p(x) he solved eq. 5 and obtained the following result:
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where p(x) is the probability density function (dimensions of probability per unit cost), x is cost, V is resource value and e is the base of the system of natural logarithms (e about equals +2.713). We will also write this expression as 1/V*exp(-x/V) where exp(-x/V) is the same thing as writing e to the negative (x/V). Important note: Remember that exp(-x/V) is the equivalent of 1/exp(x/V). Negative exponents are the same thing as the inverse of the expression. So 2^ -2 = 1/2^2 = 1/4! |
Eq. #6 is an example of type of function called a probability density function.
Negative exponential distributions are an example of a very important group of functions called Poisson distributions. |
However, it does not give frequencies of different maximum acceptable costs. Instead, true to its name, it gives probability density: probability per unit x. To make this a bit more concrete, solutions to eq. #6 give probability (or frequency) per unit cost.
Another Note -- Probabilities and Frequencies: I was not pulling a fast one when I equated probabilities and frequencies. A quick review -- remember that a frequency is simply the proportion of the total made up by one particular class. For example, if 20 out of 1000 in a war of attrition will pay a cost of up to 0.08 fitness units, the frequency of individuals paying a maximum cost of 0.08 is 0.02 . By the same token, if we were to randomly pick an individual from this population, the chance of picking an individual who would pay a maximum of 0.08 would be 0.02 (2%). The main difference in common usage between the terms probability and frequency is that probabilities are usually theoretically expected proportions while frequencies are often actual measured values. However, probability values are often used synonymously with expected frequencies in theoretical distributions; that is what we will be doing for the rest of this section. |
How do we get simple probability (frequency)? We need to multiply p(x) by cost. Now the earlier equations that contained p(x) (e.g., eq. #5) should make a bit more sense. Notice that they contained the expression p(x)dx which means to:
A word about probabilities and ranges of cost. Since cost is a continuous variable, for any exact value of cost the frequency of contestants who play that exact value is exceptionally low (unless we are dealing with the exceptional case of p(x)d(x)=1.0 -- also see the grey box above) . Probability accumulates as a continuous variable changes. Thus, the greater the range of the costs that we consider, the greater the frequency of individuals between those costs (alternately, the greater the probability that a mixed strategist will quit between these two costs in a given contest).
As you probably (no pun) know, integration would be the best technique to apply to the problem of finding the frequency of individuals willing to pay or not pay a certain cost x. Recall that when we integrate, we invoke proven mathematical techniques that have the effect of adding together the results of solving for p(x)dx at each x(each tiny step). (Actually, the way I just described the process is a bit more like the way a computer would accomplish this operation, but in any case, it gives you the right idea about what integration accomplishes.) Thus:
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where p(x) is the probability density function (dimensions of probability per unit cost) and dx is a tiny increment of cost. What eq. #7 says to do is:
To gain a bit more understanding, let's see an example. Let's solve eq. #7 using the rules of integral calculus (if you've taken calculus this will be familiar, if not, just realize that we apply some rules to get the expected result)
Here's a step by step analysis:
Now since:
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Now you should have the basic idea about how we go from the probability density function (eq. #6) to probability. Since we're on the subject, let's see how we calculate the cumulative (total) probabilities (frequencies) of playing up to or beyond any particular cost (we alluded to these calculations earlier when we wrote expressions for net benefit to a supporting strategist (eq. 3c and 4)). Let's also see how to integrate eq. 6 to get an expression that tell us the chance that an individual plays up to a certain time.
First, let's find an expression for the total proportion of individuals in the mix who are expected to have quit between costs between zero and cost x=m (this of course is the same as giving the chance that a mixed strategist will quit by cost m). This is called the cumulative probability distribution of quitting times, P(m)
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Let's discuss this equation. Upon integrating eq. 8a we get a formula from which we can readily calculate P(m) for any particular cost (x=m): (press here to see the steps of the integration of eq. 8a) To reiterate: when we solve eq. 8b for any cost, the result will be the total proportion of a population of mixed strategists who would have quit as of cost m. Again, remember that this does not mean that they all quit at cost m. Instead, P(m) includes those quitting at cost m AND all that have quit before cost m. Here are plots of P(m) for three resource values (V) over a range of costs between x = 0 and x = 10:
Notice that in all cases the initial chance of having quit is (of course) zero. As contest costs accumulate, it becomes more likely that one will have quit since costs start to exceed the maximum different supporting strategies are willing to pay. (Note: we have talked about individuals who quit at cost = 0; assume that what really happens is that they quit after a small cost, 0 + dx, is paid). Another way to think about these plots is to imagine 1000 identical 'mix' strategists starting a display game. At time zero, all are playing so zero have quit. A short time later some have quit, as time goes on a greater and greater proportion have quit and so the overall chance that an individual who started the game will have quit gradually increases. The other thing to note is the effect of V on quitting. As V gets larger, individuals quit at lesser rate (fewer quit per increase in cost x). This should make sense -- a contestant should be less likely to give up over a valuable resource. In fact, the rate of quitting is proportional to 1/V; more about this below.
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Hopefully this is all starting to make a lot of sense. Now let's look at the converse of the cumulative probability of having quit as cost x=m (alternately -- the total frequency of quitters as of cost x=m). The converse would be the cumulative frequency who have not quit as of cost m (a.k.a. "probability of not having quit", or the probability of enduring to a certain cost); we call this Q(m) and we saw it earlier with the equation for net cost to any supporting strategy vs. "mix":
OK, if P(m) is the cumulative chance that an individual will have quit as some cost, then 1- P(m) will be the chance that they are still playing. We'll call this Q(m): the probability of enduring up (not having quit) to a certain cost. Here is a graph for Q(m) when V=1:
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We can of course find Q(m) by integrating p(x) from m to infinity. Press here if you want to see this integration. Now, as with P(m), if we solve eq. #9 for a series of values of costs we can get a plot of the cumulative chance of enduring (not quit) as of any cost m. Review the plot for P(m) and then try to imagine how this graph should look. After you have thought about this, press here to see the plot of Q(m) vs. cost. |
Notice that eqs. 8 and 9 both give us cumulative probabilities. This means that both give frequencies/probabilities starting at zero up to some cost x=m (thus, if that cost x is infinity, then the cumulative chance of having quit by that cost is 1.0 and the cumulative chance of not having quit is 0).
But what if we simply want to know the chance that an individual will quit over some specific cost range -- for example, between cost x1 = 0.50000 and cost x2 = 0.50001. This is especially useful in understanding how a computer solves the war of attrition such as in the war of attrition simulation that accompanies this page.
eq. #10: Calculation of deltaP(x) notice that this is the same expression as:
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So, we have now gone over the equations that can give us various probabilities or frequency distributions in the war of attrition. All of these are the "children" of eq. 6, the probability density function that Maynard Smith derived to describe the mixed ESS. We will use these functions in the discussions that come below or on related pages (for instance, we will use eq. 10 on a related page that considers how a computer would solve the war of attrition).
In the next section, we will talk about what eq. 6 really means: what does it say about mixed strategies in the war of attrition. After we have a full description of this mix, we will turn ourselves to our final task -- proving that the mix is an ESS.
Questions About Chances of Continuing 1. Name the probability distributions that we saw earlier that give (i) chances of continuing to a certain cost or (ii) quitting as of a certain cost. Answer 2. If eq. 11 gives the chance of continuing for a unit of cost, write an expression that gives the chance of quitting per unit cost. Answer
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We are now at a point where we can understand the characteristics of the mixed equilibrium. As mentioned previously, this equilibrium could consist of either:
Thus, eq. #6 has a key role in describing the equilibrium.
In this section we will focus on the characteristics of the equilibrium. How should members of a population at this equilibrium act?
Important Convention For convenience we are going to think about our population in terms of the second possibility just discussed -- we will regard the equilibrial population as consisting entirely of mixed strategists, all of whom are capable of playing any maximal cost with a probability ultimately described by eq. 6. |
Since other mixes are possible we'll give this particular mix a name 'var' for variable cost strategist.
A Note About Strategy Names
Some of this is reiteration of what was just said but please glance over it so that you are familiar with the strategy names and definitions we will use from here on out. The names and symbols we will use for the strategies are a bit different than those used by Maynard Smith and Bishop and Cannings. They are meant to be more descriptive and therefore easier for someone to remember; hopefully this use will not result in any confusion. to those familiar with these author's work. I do this with some reluctance but have found that my students seem to have an easier time this way as compared to using symbols such as I and J or the generic term "mix". So:
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What are the characteristics of our mixed strategy "var"?
1. Like other strategies, 'Var' is highly secretive! There can NO INFORMATION TRANSFER from var to its opponent THAT MIGHT SIGNAL WHEN 'VAR' WILL QUIT.
2. Var strategists may potentially play any cost -- from no cost to (theoretically) an infinite cost. We discussed the reasons for this in the first section of this page (review).
3. 'Var' strategists have a constant rate of continuing over each unit of cost. The chance of continuing is proportional to 1/V; this quantity is also known as the rate constant (press here if you want to read a bit more about rate constants). The chance of continuing per unit cost:
eq. 11: (note that this equation is the same as eq. 9 when Q(m) is solved for x= m = 1) |
Thus, with regard to the chance of var's continuing to display:
If you don't spend a lot of time dealing with exponents, these last two statements might confuse you. It is very important that you keep in mind the fact that x/V is part of a negative exponent. Thus:
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4. Now, since the behavior of a 'var' strategist is determined by a certain chance of quitting with each unit of cost, and since var never tips its hand, you should realize that an opponent will never know exactly when a 'var' strategist will quit -- anymore than you, me or anyone can always correctly guess when a "fair" coin will turn up "heads". Thus, knowing when something will happen is quite different from knowing the chance of some event. This is the essence of the problem var's opponents face!
5. Another result of a constant chance of continuing per unit cost (i.e., a constant chance of quitting per cost) is that the chance of accepting greater costs (i.e., of playing from the start through to cost x) decreases exponentially (for any value of V less than infinity, i.e., for any exp(-1/V) < 1.0). The effect of this is that there is virtually no chance that a var strategist will be willing to pay a cost that is very large compared to V.
6. To summarize, the opponent:
Link to an Illustration of "Var - Like" Behavior The last statement is perhaps the most crucial in understanding the behavior of 'var' strategists. Central to it are the ideas of constant probability of continuing the game and independence of decisions from one moment (cost) to the next. You will also explore this in great detail when you run the simulations. For the moment, however, take the time to read an example illustrating how a strategy like 'var' works. |
Questions About the Mixed Strategy Var 1. Compare what a contestant sees when it confronts a population consisting entirely of 'var' strategists as compared to a population that is an equilibrial mix of pure supporting fix(x) strategies. Would the contestant see any difference in these two situations? Answer 2. How would you express the idea of constant rate of quitting with respect to a population of pure strategists who together produce an equilibrium? Answer 3. Why is it crucial that no information as to var's intention to continue or quit a contest be passed on to its opponent? Answer 4. How do you estimate the probability that a var strategist will win a contest of cost x? Answer 5. How do you estimate the probability that a var strategist will lose a contest of cost x? Answer 6. How do you estimate the probability that a var strategist loses by paying a cost between x and x+dx? Answer All of the remaining questions call for solutions to equations derived from eq. 6, the probability density function that describes var. You will need a calculator or spreadsheet with natural logs. Alternatively, you can use the number 2.72 whenever you need e. 7. Should the chance of encountering a member of the "stable mix" with a quitting cost between 0.60 and 0.61 be greater or less than encountering an individual with a quitting cost between 0.60 and 0.62? Explain. Answer 8a. What is the chance of encountering a member of the stable mix with a quitting time between a cost of 0.60 and 0.61 if V=1? V=0.5? Compare these answers with the next question. Answer 8b. What is the the chance of encountering a member of the mix who quits between a cost of 1.0 and 1.01 if V=1? V=0.5? Compare these answers with the last answers. Why the difference? -- the size of the cost interval is the same Answer. |
We now know the general characteristics of the mixed strategy we call 'var' -- the range of its maximum display costs, the probability of playing each of these costs, and the relationship of these probabilities to the resource. And we know that the equation that eq. #6, which describes var's behavior sprung from the assumption that:
E(any fix, var) = E(any mix, var) = E(var, var) = constant |
Finally, we know that Bishop and Cannings (1978) have showed that this assumption must correct for any ESS in the symmetrical war of attrition (see Bishop-Cannings theorem).
However, simply showing that the 'var' strategy has some behavior consistent with being an ESS is not the same thing as showing that it is an ESS. Recall the two general rules for finding ESSs we learned about earlier . 'Var' is an ESS (cannot be invaded if sufficiently common) if:
Now, in the case of 'var' we are only interested in rule #2 since we already know that part a of rule #2 is true. In fact,'Var' is derived from part a! And of course rule #2 is not consistent with rule #1. But just because 'var' is derived from rule #2(a) does not mean that it must be consistent with rule #2(b). And if 'var' vs. any fix(x) is not consistent with part B, then var is not an ESS (see box below).
If 'Var' Were Not an ESS, What Would It Be? If 'var ' vs. any fix(x) is only consistent with rule 2 part A, it is equilibrial. This is because if E(var,fix(x)) > E(fix(x),fix(x)) is false, then the only interpretation that is also consistent with rule 2A is that E(var,fix(x)) = E(fix(x),fix(x)). So, the common interactions would have the same fitness consequences on each party (no advantage to either) and the rare interactions would also give no advantage to either strategy. Note that the payoffs in common vs. rare interactions would not have to equal each other, the only equality needed is that common are equal for both as are rare. The result is that selection could not change the strategy frequencies and we would say that the population was equilibrial. (The only way that frequencies can change are by mutation, immigration or emigration.) |
So, to show that 'var' is an ESS all we need to do is to show that rule #2 part b holds:
Rule 2, part b: E(mix,fix(x)) > E(fix(x),fix(x)) |
What will follow is a mathematical proof that rule 2b is in fact true and therefore that 'var' is an ESS in the war of attrition. Once again, there will be a bit of calculus to enhance the argument but anyone should at least be able to follow the outline of the proof. As before the calculus is all explained, furthermore, much of it is very similar to what we have seen earlier. And, to make the concepts clearer, a number of graphs will be presented.
Once again,'var' is an ESS if:
Rule 2, part b: E(var,fix(x)) > E(fix(x),fix(x)) |
is true.
So, we will need to find expressions for E(var,fix(x)) and E(fix(x),fix(x)) and determine whether or not the difference between the two is always a positive number -- i.e.,
| eq. 12: E(mix,fix(x)) - E(fix(x),fix(x)) > 0 |
Now, recall eq. #2 from earlier. The payoff to a given strategy in a certain type of contest is always:
eq. #2: E(focal strat., opponent) = Lifetime Net Benefits to Focal Strategy in Wins |
So, let's find the net benefit and cost equations for E(mix,fix(x)) and E(fix(x),fix(x)) and then substitute them into eq. 2 before finally solving to see if we have an ESS. We'll use the same general symbols and operations that we used in finding E(fix(x), mix (i.e., 'var')) earlier.
Part One: Calculation of Net Benefits
The benefits needed to calculate these payoffs are easy to find and so they represent a good place for us to start. First, recall that we assume that the value of the resource is constant in any given contest; further we assume that it has the same value to both contestants. As usual, we will symbolize it as V. Here are the net benefits for each type of interaction.
Net Benefits to Var in Contests vs. Fix(x): Remember that var does not enter a contest possessing a particular maximum cost that it is willing to pay. Instead, at each instant it has a constant probability of quitting proportional to 1/V. Thus, it is unpredictable as to exactly when it will quit.
Now remember that in wars of attrition, winners, like losers, pay costs. These costs lower the net (realized) value of the resource to the winner (press here to review our assumptions about costs):. We'll call the maximum cost the fix(x) strategist is willing to pay m). So, against a given fix(x=m) strategist, 'var' wins whenever it is willing to pay more (i.e., whenever it continues to play after fix(x=m) quits). Thus, when 'var' wins, it will always win V-m. But it is not certain that 'var' will play to a higher (winning) cost than fix(x=m) since var uses a probability function to determine when to quit. So, 'var' expects to get:
| eq. 3b: net Benefit = (V - m) * (Chance of winning) |
Recall from earlier that the chance that 'var' has not quit as of paying any cost x= m is Q(m):
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Recall that this equation finds the chance that var has not quit as of cost m by adding up all of the probabilities of 'var' quitting at costs greater m. |
So, after substituting eq. 13 into the net benefit equation (#3b), the benefit to 'var' is:
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Notes about the equation: Notice that (V-m) is placed outside of the integration sign. That is because in the case of 'var' against a given fix(x=m), 'var' can never expect to win anything except V-m. So, (V-m) is a constant for a contest that can last up to any given cost m. And 'var' only wins when it has not quit as of m. And, of course, the purpose of the integration is simply to find the chance that var will still be playing as of cost x=m. Solving eq. 14a:
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Net Benefits in Fix(x) vs. Fix(x) Contests: In this contest we have two identical fix(x) strategists facing each other. Thus, they play to exactly the same cost x=m. Since we assume no other asymmetries, then it is best to assume that two identical individuals will each win 50% of the time -- they will in effect split the net benefits. Thus:
| eq. 15: B for fix(x) vs.fix(x) =0.5 * (V - m) =0.5 * (V - x) |
Part Two: Calculation of the Cost of Losing
Calculation of Cost to Var Strategists in Losses
to Fix(x): The calculations for lifetime loss costs to 'var' are a bit
more complicated than those for net benefit. The reason is that 'var'
can lose to a given fix(x=m) many ways!
Here's an example.
Let's express this idea mathematically:
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Let's be sure we understand what eq. 16a means:
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We can solve eq. 16a by inserting eq. 6 for p(x) and integrating:
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(press here to see the integration) If you understand calculus and/or if you are sure that you understand how costs are calculated, you can move on to the next section. If not, please visit the following link which will take you to a discrete calculation of net benefits and costs. |
Calculation of Cost to Fix(x) Strategists When vs. Fix(x): Once again, this is a very easy calculation. The contestants are identical -- both are willing to pay cost x=m. As we said in our consideration of benefits, we simply assume that each individual wins 0.5 (50%) times. So, half the time they lost and pay cost x=m:
eq. #17: Cost paid by a fix(x) in losing to a fix(x): = 0.5 * x = 0.5 * m |
Section A: E(fix(x=m), fix(x=m)): Let's start with fix(x) contests that end in ties (since they're easy). Now, since
eq. #2: Payoff(to Strat., when vs. a Strat.) = (Benefit from win) - (Cost from loss) if we simply substitute equations for benefit in winning (eq. 15) and cost in losing (eq. 17) we obtain: |
Section B: E(var, fix(x=m)): This time we substitute eqs. 14 and 16 into eq. #2 :
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and if we integrate this equation we obtain the following result: (You have seen the steps to this integration previously
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At this point you can either continue on to the final proof that 'var' is an ESS or you might find this a good place to take a side trip that explores the differences between the 'var' and fix(x) strategies by presenting graphs of benefits, costs and payoffs for each strategy. Press
here to go to a graphical presentation of |
Recall from above that to prove that 'var' is evolutionarily stable that we need to show that rule 2b is correct. Here we go:
Finding an Equation for the Difference in Payoffs Starting with rule 2b: E(var, fix(x)) > E(fix(x), fix(x)) and rearranging, we get: E(var, fix(x)) - E(fix(x), fix(x)) > 0 Now since: E(fix(x) ,fix(x)) = 0.5 V - m (review) and since E(var, fix(x)) = 2 * V* exp(-x/V) - V (eq. 19b) then: 2 * V* exp(-x/V) - V - 0.5*V - m > 0 which simplifies to: |
Now the big question -- is eq. 20 always positive as it must be if 'var' is an ESS?
We could start out by simply graphing it. If we do so for V=1 we will see that there is no place where E(var,fix(x)) < = E(fix(x), fix(x)):
(Looks like the "swoosh" doesn't it!).
Thus, it would appear that 'var' is stable. But not so fast -- this is for only one value of V. Is it possible that there are values of V where 'var' is not evolutionarily stable? After all, V does affect 'var's behavior.
As with finding the frequency of each maximum acceptable cost (when we looked for p(x)), solving for every possible V might appear to be a difficult problem (and approached that way, it is!). However, once again a bit of elementary calculus can come to our aid and comfort.
Mathematical Proof: To show that no point on eq. 20 is less than or equal to zero, we need to find the minimum value of eq. 20. This occurs where the slope of the graph is zero (the flat part of the graph above; on that graph it happens at a value somewhere near cost = 0.7).
Thus, 'var' is an ESS!
Graphical Illustration of the Proof: If you are not fully confident that you understand the proof, you will probably be reassured if you look at the graphs below of eq. 20 for different values of V. Remember, we have said that the minimum difference in fitness will always = 0.193*V and will always occur at cost = 0.693*V:
Notice that as V gets larger the minimum difference between the two payoffs increases. (If you are "Thomas from Missouri" and want me to show you the low V graphs in more detail, press here.)
So there you have it. For any cost paid by the winner, m, E(var,fix(x)) > E(fix(x), fix(x)). So since this is true and since E(var, var) = E(fix(x), var), then var is evolutionarily stable against any fix(x)!
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1. Write an expression for the lifetime cost to a var strategist of quitting at a cost of exactly x. Answer 2. Write an expression for the lifetime cost to a var strategist for losing contests where the winner was willing to pay m? Answer 3. What is E(var, fix(0)) in the case of a tie? Answer |
Things to Remember About the 'Var' Strategy
Perhaps the most striking thing about the var strategy is that its opponent never can know when it will quit. We have seen that the overall pattern of quitting is described by an exponential decay type of Poisson distribution with a rate constant equal to 1/V. Thus, an opponent can "learn" in general terms what its var opponent would do. It could "know" that it was most likely to quit early in a contest and that the chance of quitting per unit of contest display cost is exp(-1/V). From this, it is possible to calculate (or learn from experience) the expected outcome of contests of various costs.
However, even if it knew these things, it could never know whether or not 'var' really would quit with the next increment of cost. Thus, no amount of experience with 'var' strategists will allow an opponent any edge over it.
The other thing to reiterate about var is that there is a logic to its quitting. It is tied to the resource value -- the greater that value, the less likely that var will quit at any particular cost and as a consequence it is potentially willing to accept a higher cost contest. Also, since 'var' always quits most frequently early in contests, the chance that it will pay large costs relative to a resource value are low.
"Are You Feeling Lucky, Punk?" In the classic Clint Eastwood thriller, Dirty Harry, the Eastwood character asks a naer-do-well to predict the future and guess whether or not there bullets left in Eastwood's gun. So what do you think? Are you feeling lucky 1. The chance of getting killed in a scheduled commercial airline crash is roughly on the order of one in several million. It is about the same chance the earth has of being hit by a large meteor, small asteroid, or comet. Discuss whether or not someone who flies commercial airlines daily (e.g., a flight attendant or pilot) for years is more likely on her or his next flight to be in a fatal accident. Likewise, the earth has not been hit by a really big one for about 65 million years. Are we more likely to be hit now than we were say 60 million years ago (5 million after the last one). Are you more likely to win on your next lottery entry (tax on stupidity) if you haven't won in the past and less likely if you have won? What does all of this have to do with the war of attrition? Discussion |
There are a number of famous examples of animals that appear to be playing simple waiting games. We will not go into them here because they are well presented both in the literature and in just about every animal behavior text book. Perhaps the classic is the dung fly, Scatophaga stercoraria, studied heavy by Parker and Parker and Thompson (refs). The interested reader is urged to consult these papers or any number of behavioral ecology texts. We will finish this page, however, with the following question (which was addressed by Parker and Thompson):
? Suppose that someone demonstrated that animal waiting times corresponded to those predicted by eq. 9 Does that constitute sufficient proof that a mixed ESS described by eq. #9 exists? Explain. |
Problems dealing with the calculation of P(m)
1. What is the cumulative chance of quitting between a cost of 0 and infinity if V=1? V=5? V=0.5?
It makes no difference what the value of V is in this case. Any number to the infinite power is infinite and the inverse of infinity is essentially zero. Therefore P(m) = 1.0 in all cases:
P(m)=1 - (1 / e^(infinity))=1 - (1 / infinity) = 1 - 0 = 1
2. What is the cumulative chance of quitting between a cost of 0 and 0.6 if V=1? V=0.5?
For V=1: P(m = 0.6) = 1 - (1 / e^(0.6/0.5)) = 1 - (1 / e^(1.2) =
1 - (0.30) = 0.7
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Questions About Chances of Continuing
1. Name the probability distributions that we saw earlier that give (i) chances of continuing up to a certain cost or (ii) quitting as of a certain cost.
Ans: Q(m) and P(m), respectively
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2. If eq. 11 gives the chance of continuing for a unit of cost, write an expression that gives the chance of quitting per unit cost.
Ans: = 1 - exp(-1/V) -- recall that eq. 8 (chance
of quitting) is nothing more than 1 - eq. 9 (i.e., 1 - Q(m)). Now since eq. 9 is essentially
the same as eq.
11, then (1 - eq. 11) = (1 - exp(-1/V)) and this gives us the chance
of quitting.
So, for example, if V = 1, chance of quitting
per unit cost is 0.632.
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Questions About the Mixed Strategy Var
1. Compare what a contestant sees when it confronts a population consisting entirely of 'var' strategists as compared to a population that is an equilibrial mix of pure supporting fix(x) strategies. Would the contestant see any difference in these two situations?
Answer: No, they are equivalent. In both cases, the contestant has no idea which maximum cost it is facing (provided that encounters with different fix(x) supporting strategies are random in the mixed population and that in neither case the maximum cost is tipped before being reached).
2. How would you express the idea of constant rate of quitting with respect to a population of pure strategists who together produce an equilibrium?
Answer: One way would be to say that in any contest with members of this population, there is a constant chance per increment of cost that one' s opponent will quit. This corresponds to the idea that one's chance of opposing a given type of supporting strategist (maximum x) would be equal to its frequency in the population (as determined by integrating eq. #6). Supporting strategies with low maximum x values would be more common so you would be more likely to face them.
3. Why is it crucial that no information as to var's intention to continue or quit a contest be passed on to its opponent?
If the opponent has some reason to know var's intentions, there will be strong selective pressure for it to act in a way that thwarts var and serves its own best interests. For instance, if it is certain that var will not quit before reaching the opponents max cost, it will pay the opponent to quit immediately and cut its losses. Likewise, if var is certain to quit on the next move or over the next bit of cost, it will pay the opponent to wait var out and gain the resource (as compared to var who in this case gains nothing).
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4. How do you estimate the probability that a var strategist will win a contest of cost x?
This is equal to Q(m) since Q(m) gives the chance that var has not quit as of cost x=m.
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5. How do you estimate the probability that a var strategist will lose a contest of cost x?
This is equal to P(m) since P(m) gives the cumulative chance that var has already quit as of some cost x=m.
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6. How do you estimate the probability that a var strategist loses by paying a cost between x and x+dx?
This is equal to delta P(m) since delta P(m) gives the chance that var has endured to cost x=m without quitting but will quit before paying cost x+dx (i.e., m+dm) where dx or dm is some additional cost.
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Calculation of the Chance of Var Paying a Specific Cost
7. Should the chance of a var quitting between 0.60 and 0.61 be greater or less than the chance of quitting between 0.60 and 0.62? Explain.
It should be less for the smaller range of costs -- i.e., less in 0.60 to 0.61 than in 0.60 to 0.62. In this case, all we have done is make a cost interval larger by 0.01. So, there are more quitting times in this larger interval and therefore a greater total probability that an individual var will quit within this interval.
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8a. What is the chance of quitting within the specific cost interval of 0.60 and 0.61 if V=1? V=0.5?
for V=1: deltaP(m)=exp(-0.60 ) - exp(-0.61) = 0.00546
for V=0.5: deltaP(m)=exp(-0.60 / 0.5) - exp(-0.61 / 0.5) = 0.00596
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8b. What is the cumulative chance of quitting within the specific cost interval of 1.0 and 1.01 if V=1? V=0.5? Compare these answers with those you go in the last problem -- why is there a difference in probability even though delta m is the same (0.01) in both cases?
for V=1: deltaP(m)=exp(-1.0) - exp(-1.01) = 0.00366
for V=0.5: deltaP(m)=exp(-1.0 / 0.5) - exp(-1.01 / 0.5) = 0.00268
Notice that the chance of QUITTING WITHIN A SPECIFIC COST INTERVAL (delta P(m)) OF A CONSTANT RANGE (0.01) DECREASES AS THE AVERAGE COST OF THE INTERVAL INCREASES. This is not because the chance of quitting per 0.01 increment in cost has changed. Indeed, it is always proportional to 1/V, regardless of the interval.
So why the difference? The difference reflects the lower chance that an individual will actually have played to the higher cost. Thus, the chance of actually having played to x = 0.60 is P(0.6) = 0.549 but the chance of playing all the way to x = 1.00 is P(1.00) = 0.368. If you apply a constant chance of remaining over the next 0.01x to each of these numbers (if V = 1.0, it is 0.99) you will see that fewer actually quit in the second interval (because there are fewer there to quit!). There will be more about this in the text.
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Problems dealing with the calculation of costs to var in losses
1. Write an expression for the lifetime cost to a var strategist of quitting at a cost of exactly x.
Answer: This is given by p(x)dx and it is a very small number.
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2. Write an expression for the lifetime cost to a var strategist for losing contests where the winner was willing to pay m?
Var loses any contest that costs less than m. There are lots of ways this can happen -- each losing cost has a unique probability of occurrence based on 'var's probability density function. Thus:
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3. What is E(var, fix(0)) in the case of a tie?
Following our usual rule, each side wins 50% of the time. Since there is a 100% chance that var will play at time 0 and the cost = 0, then E(var,fix(o))=0.5*{(V-m)-m}= 0.5 * (V - 0) - 0 = 0.5V.
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Note about the term "Learn": I use the term learn loosely -- it could mean "learn" in the usual sense of learning and memory or it may be that we are simply talking about making an appropriate evolutionary response -- selection for responses that work against a fixed wait time. In either case, an appropriate response arises to a particular fixed strategy.
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"Are You Feeling Lucky, Punk?"
In the classic Clint Eastwood thriller, Dirty Harry, the Eastwood character asks a naer-do-well to predict the future and guess whether or not there bullets left in Eastwood's gun. So what do you think? Are you feeling lucky?
1. The chance of getting killed in a scheduled commercial airline crash is roughly on the order of one in several million. About the same chance the earth has of being hit by a large meteor, small asteroid, or comet. Discuss whether or not someone who flies commercial airlines daily (e.g., a flight attendant or pilot) for years is more likely on her or his next flight to be in a fatal accident. Likewise, the earth has not been hit by a really big one for about 65 million years. Are we more likely to be hit now than we were say 60 million years ago (5 million after the last one). Are you more likely to win on your next lottery entry (tax on stupidity) if you haven't won in the past and less likely if you have won? What does all of this have to do with the war of attrition?
All of these chances are independent. In these cases, there is a more or less constant probability per flight of a disaster (this might be the worst example of the three since clearly a poor pilot, bad weather, poor maintenance or whatever could change your odds) -- what happens on other flights does not affect the next one you get on. The same with asteroids and lottery tickets. As with 'var', a constant probability means that it can happen any time or maybe even not at all. The main difference between these examples and the war of attrition is that in the 'war' we are concerned with the distribution of quitting costs while in the other examples the emphasis is on the constant probability of some event.
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Copyright © 1999 by Kenneth N. Prestwich About Fair Use of these materials Last modified 3- 14 - 99 |
Specifically, this
is a definite integral. It says to add up all values of (V-x)*p(x)dx
between costs of x = 0 (the number underneath the integration sign)
up to x = m (above the integration sign). (Note -- it is a definite
integral because these limits are given -- when limits are not given (indefinite
integral) we integrate over all possible numbers. However, since costs can
only be positive or equal zero, we need to use this definite integral!).
