3. The Stationary Distribution
Now the dominant eigenvalue is m, and we derive a recurrence relation for the corresponding left eigenvector
, the stationary distribution, where we set the right-most element equal to 1. It is straightforward to demonstrate that the final three elements of the eigenvector 
are
, 
and 1.
Observe that
where, throughout, 
is the 
identity matrix, 
is a k element column vector and 
is a k element row vector.
We then have
.
Also
and so
Now consider
We have
so 
is the required eigenvector.
We now have a recurrence relation for the 
which is
This is valid for 
using 
and
.
Suppose we write 
for the sum of the elements of 
so that 
is the stationary distribution of the Markov chain. We have immediately that
with
,
. This is sequence A001040 [5] specified as
, where our 
where the sequence is initiated with 
and
.
We can extract individual elements of the stationary distributions. Suppose that 
is the i’th element of
. Then we have that
for
, 
, with initial values 
and
, 
and 
and for
, we have 
and
. The sequence for 
is A058307, and 
is A058279 in [5].
Table 1 gives some values of
.
4. The Asymptotic Eigenvector
Having derived recurrence relations for the elements of the eigenvectors we now consider the limit as
. We begin with a simple Lemma.
Lemma
Suppose we have a recurrence relation of the form 
where the 
is not dependent on the
, and
. Suppose we have two sequences, 
and 
satisfying the recurrence relationship but initiated by different values i.e. by 
and 
respectively. Then 
as 
where 
is a constant, which depends on the initial values.
Proof
Since
We have
Now the denominator increases at a rate greater than
as one can see easily by considering the liniting case
, so the above expression tends to zero as 
and so 
as
.
Comment Much weaker conditions are necessary than those stated above for 
as
.
We can apply the lemma immediately to the elements of the stationary distribution, expressed in the integer form. The ratios for 0 and 1 elements for 
are 1, 
, 1.307692, 1.308824, 1.308789, 1308789 illustrating the speed with which convergence takes place. We have no expression for the asymptotic value but for m = 200 the ratio is approximately 1.3087893731. The ratios for 1 and 2 elements for 
are 1, 0.5, 0.529412, 0.528090, 0.528131, 0.0.528130 and for m = 200 approximately 0.5281297672.
In the absence of a simple way of evaluating the limiting ratios discussed above analytically we adopt a different method to derive the asymptotic stationary distribution, again expressed in integers. Suppose this is given by
, and define
. We have that
. Thus
, and
.
In a similar way we can obtain
, and
, and so on. It is clear that the signs alternate. For ease we introduce the following notation; we write 
when 
so that the sequences 
and 
for 
consist of positive integers. Similarly we write 
when 
so the sequences
and 
consist of positive integers. Thus we have
, 
, 
, 
, 
and so on, while
, 
, 
and so on.
The theorem below gives recurrence relations for
, 
, 
and
, in terms of A058279 and A058307 [5].
Theorem
For 
we have 
where
with 
([5] A058279)
Table 1. Eigenvectors for m = 3(1)8.
and 
where 
with 
and 
([5] A058307). Further 
and
.
Proof
We have that 
and 
, and we have already shown that
, 
, 
, 
, 
which satisfy the formula given in the statement of the theorem.
We prove that if the formula holds for 
for 
then it holds for 
for
, and thence for 
for
. Thence by induction.
Note first that
Suppose then that the formula for 
holds for
. Then since 
we have 
and substituting for the expressions given in the statement of the theorem we have
Now clearly we also have 
so we have
Table 2 gives the first 15 elements for the eigenvector for
. Some idea of the speed of convergence can be gained by observing that these values agree with the elements of the eigenvector for 
except in the final 2 decimal places.
5. Conclusion
We have derived the stationary distribution of the frequencies of the available strategies in a population in which mutations occur infrequently, for Parke’s model when the reward is 2+ and for integer valued strategies. These relate to certain known integer sequences. This work provides a base for further investigations for other values of the reward, and more complex invasion processes.
6. Discussion
Parker’s model, which is also known as the Scotch Auction, is often used in the conflict theory literature as an example of a simple model in which there is no ESS (evolutionarily stable strategy). The implication of this is that there is no population assembly which is resistant to invasion. Of course if such a contests actually occurs it is important to ask what will happen in the population. This is the question which is addressed in [3], and which generates the class of matrices considered here. The stationary distribution then corresponds to the frequency with which one would observe a population to be playing a specific strategy, except if one happened on a population in transition.
The class of cases discusses above arises from Parker’s model when we consider a fixed reward value
, and when the value of m, the range of possible strategies, is allowed to vary. It would be of interest to examine
Table 2. First 15 eigenvector elements for m = 200.
other possible values of V as m varies. For example, for
, the “Markov matrices” have 1’s for 
, 0’s for 
and diagonal elements to make the row sums m. It is hoped to treat these models in a subsequent paper.
We observe from the numerical values that the most frequent strategy value played is 1, that the distribution is uni-modal and that the strategies 
are played over 90% of the time; asymptotically approximately 0.90667 which agrees to five decimal places to the value for
, while the mean value is asymptotically approximately 1.1207 which agrees to five decimal places to the value for
. These latter figures confirm the rapidity of the convergence.