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On the Expected Number of Equilibria in a Multi-player Multi-strategy Evolutionary Game

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Abstract

In this paper, we analyze the mean number \(E(n,d)\) of internal equilibria in a general \(d\)-player \(n\)-strategy evolutionary game where the agents’ payoffs are normally distributed. First, we give a computationally implementable formula for the general case. Next, we characterize the asymptotic behavior of \(E(2,d)\), estimating its lower and upper bounds as \(d\) increases. Then we provide a closed formula for \(E(n,2)\). Two important consequences are obtained from this analysis. On the one hand, we show that in both cases, the probability of seeing the maximal possible number of equilibria tends to zero when \(d\) or \(n\), respectively, goes to infinity. On the other hand, we demonstrate that the expected number of stable equilibria is bounded within a certain interval. Finally, for larger \(n\) and \(d\), numerical results are provided and discussed.

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Acknowledgments

We would like to thank members of the WMS seminar at the CASA at Eindhoven University of Technology, especially Dr. R. van Hassel and Prof. A. Blokhuis for useful discussion. Part of this work was done under the support of the F.W.O. Belgium (TAH, postdoctoral fellowship 05_05 TR 7142).

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Authors and Affiliations

  1. Mathematics Institute, Zeeman Building, University of Warwick, Coventry, CV4 7AL, UK

    Manh Hong Duong

  2. School of Computing, Teesside University, Borough Road, Middlesbrough, TS1 3BA, UK

    The Anh Han

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  1. Manh Hong Duong
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Appendix

Appendix

1.1 Properties of \(a_k\)

In the following, we prove that \(a_{2d-4-k} = a_k\) for all \(0 \le k \le 2d-4\), i.e., (22). First, we transform \(a_k\) as follows

$$\begin{aligned} a_k&=\sum \limits _{\begin{array}{c} 0\le i\le d-1\\ 0\le j\le d-2\\ i+j=k \end{array}}\begin{pmatrix} d-1\\ i \end{pmatrix}^2\begin{pmatrix} d-2\\ j \end{pmatrix}^2-\sum \limits _{\begin{array}{c} 0\le i'\le d-2\\ 0\le j'\le d-2\\ i'+j'=k-1 \end{array}}\begin{pmatrix} d-2\\ i' \end{pmatrix}\begin{pmatrix} d-1\\ i'+1 \end{pmatrix}\begin{pmatrix} d-2\\ j' \end{pmatrix}\begin{pmatrix} d-1\\ j'+1 \end{pmatrix} \\&= \begin{pmatrix} d-2 \\ k \end{pmatrix}^2 + \sum \limits _{\begin{array}{c} 0\le i\le d-2\\ 0\le j\le d-2\\ i+j=k-1 \end{array}}\begin{pmatrix} d-1\\ i+1 \end{pmatrix}^2\begin{pmatrix} d-2\\ j \end{pmatrix}^2-\sum \limits _{\begin{array}{c} 0\le i'\le d-2\\ 0\le j'\le d-2\\ i'+j'=k-1 \end{array}}\begin{pmatrix} d-2\\ i' \end{pmatrix}\begin{pmatrix} d-1\\ i'+1 \end{pmatrix}\begin{pmatrix} d-2\\ j' \end{pmatrix}\begin{pmatrix} d-1\\ j'+1 \end{pmatrix}\\&= \begin{pmatrix} d-2 \\ k \end{pmatrix}^2 + \sum \limits _{\begin{array}{c} 0\le i\le d-2\\ 0\le j\le d-2\\ i+j=k-1 \end{array}}\begin{pmatrix} d-1\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j \end{pmatrix}\left( \begin{pmatrix} d-1\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j \end{pmatrix} - \begin{pmatrix} d-2\\ i \end{pmatrix}\begin{pmatrix} d-1\\ j+1 \end{pmatrix} \right) \\&= \begin{pmatrix} d-2 \\ k \end{pmatrix}^2 + \sum \limits _{\begin{array}{c} 0\le i \le j \le d-2\\ i+j=k-1 \end{array}}\left( \begin{pmatrix} d-1\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j \end{pmatrix} - \begin{pmatrix} d-2\\ i \end{pmatrix}\begin{pmatrix} d-1\\ j+1 \end{pmatrix} \right) ^2 \end{aligned}$$

Now we prove that \(a_{2d-4-k} = a_{k}\). Indeed, we have

$$\begin{aligned} a_{2d - 4 - k}&= \begin{pmatrix} d-2 \\ 2d - 4 - k \end{pmatrix}^2 + \sum \limits _{\begin{array}{c} 0\le i \le j \le d-2\\ i+j=2d-k-5 \end{array}}\left( \begin{pmatrix} d-2\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j \end{pmatrix} - \begin{pmatrix} d-2\\ i \end{pmatrix}\begin{pmatrix} d-1\\ j+1 \end{pmatrix} \right) ^2 \\&\quad \text {(we use here the transformations i = d-3 - i and j = d-3 - j)}\\&= \begin{pmatrix} d-2 \\ 2d - 4 - k \end{pmatrix}^2 + \sum \limits _{\begin{array}{c} -1\le j \le i \le d-3\\ i+j=k-1 \end{array}}\left( \begin{pmatrix} d-1\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j+1 \end{pmatrix} - \begin{pmatrix} d-2\\ i+1 \end{pmatrix}\begin{pmatrix} d-1\\ j+1 \end{pmatrix} \right) ^2 \\&= \begin{pmatrix} d-2 \\ 2d - 4 - k \end{pmatrix}^2 + \sum \limits _{\begin{array}{c} -1\le j \le i \le d-3\\ i+j=k-1 \end{array}}\left( \begin{pmatrix} d-1\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j \end{pmatrix} - \begin{pmatrix} d-2\\ i \end{pmatrix}\begin{pmatrix} d-1\\ j+1 \end{pmatrix} \right) ^2. \\ \end{aligned}$$

Since for \(j = -1\) and \(i = k\), we have

$$\begin{aligned} \left( \begin{pmatrix} d-1\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j \end{pmatrix} - \begin{pmatrix} d-2\\ i \end{pmatrix}\begin{pmatrix} d-1\\ j+1 \end{pmatrix} \right) ^2 = \begin{pmatrix} d-1 \\ k \end{pmatrix}^2, \end{aligned}$$

and for \(j = d-2\) and \(i = k-1 - (d-2)\), we have

$$\begin{aligned} \left( \begin{pmatrix} d-1\\ i+1 \end{pmatrix}\begin{pmatrix} d-2\\ j \end{pmatrix} - \begin{pmatrix} d-2\\ i \end{pmatrix}\begin{pmatrix} d-1\\ j+1 \end{pmatrix} \right) ^2&= \left( \begin{pmatrix} d-1\\ k-(d-2) \end{pmatrix}- \begin{pmatrix}d-2\\ k-1-(d-2) \end{pmatrix} \right) ^2 \\&= \begin{pmatrix} d-2\\ k-(d-2) \end{pmatrix}^2 \\&= \begin{pmatrix} d-2\\ 2d-4-k \end{pmatrix}^2 , \end{aligned}$$

it follows that \(a_{2d-4-k} = a_k\) for all \(0 \le k \le 2d-4\).

1.2 Detailed Computation of \(f(1)\)

We use the following identities involving the square of binomial coefficients.

$$\begin{aligned}&M_d(1)=\sum \limits _{k=0}^{d-1}\begin{pmatrix} d-1\\ k \end{pmatrix}^2=\begin{pmatrix} 2(d-1)\\ d-1 \end{pmatrix},\\&A_d(1)=(d-1)^2M_{d-1}(1)=(d-1)^2\begin{pmatrix} 2(d-2)\\ d-2 \end{pmatrix},\\&B_d(1)=\sum \limits _{k=1}^{d-1}k\begin{pmatrix} d-1\\ k \end{pmatrix}^2=\frac{d-1}{2}\begin{pmatrix} 2(d-1)\\ d-1 \end{pmatrix}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} f(1)&=\frac{1}{\pi }\frac{\sqrt{A_d(1)M_d(1)-B_d(1)^2}}{M_d(1)}\\&=\frac{1}{\pi }\frac{\sqrt{(d-1)^2\begin{pmatrix} 2(d-1)\\ d-1 \end{pmatrix}\left[ \begin{pmatrix} 2(d-2)\\ d-2 \end{pmatrix}-\frac{1}{4}\begin{pmatrix} 2(d-1)\\ d-1 \end{pmatrix}\right] }}{\begin{pmatrix} 2(d-1)\\ d-1 \end{pmatrix}}\\&=\frac{d-1}{\pi }\times \sqrt{\frac{\begin{pmatrix} 2(d-2)\\ d-2 \end{pmatrix}}{\begin{pmatrix} 2(d-1)\\ d-1 \end{pmatrix}}-\frac{1}{4}}\\&=\frac{d-1}{\pi }\times \sqrt{\frac{d-1}{2(2d-3)}-\frac{1}{4}}\\&=\frac{d-1}{2\pi \sqrt{2d-3}}, \end{aligned}$$

where we have used the identity \(\begin{pmatrix} 2(n+1)\\ n+1 \end{pmatrix}=\frac{n+1}{2(2n+1)}\begin{pmatrix}2n\\ n \end{pmatrix}.\)

1.3 Alternative Proof of the Fifth Property in Lemma 2

We show here an alternative proof of the fifth property without using the first one in Lemma 2.

Since \(M_d\left( \frac{1}{t}\right) =t^{2(1-d)M_d(t)}\), we have

$$\begin{aligned} M_d'\left( \frac{1}{t}\right)&=-t^{3-2d}\left[ 2(1-d)M_d(t)+tM_d'(t)\right] ,\\ M_d''\left( \frac{1}{t}\right)&=t^{4-2d}\left[ 2(3-2d)(1-d)M_d(t) +2(3-2d)tM_d'(t)+t^2M_d''(t)\right] . \end{aligned}$$

Therefore, from (26), we have

$$\begin{aligned} B_d\left( \frac{1}{t}\right) =\frac{1}{2}M_d'\left( \frac{1}{t}\right) =-t^{3-2d}\left[ (1-d)M_d(t)+\frac{1}{2}tM_d'(t)\right] , \end{aligned}$$

and

$$\begin{aligned} A_d\left( \frac{1}{t}\right)&=\frac{t}{4}\left[ M_d'\left( \frac{1}{t}\right) +\frac{1}{t}M_d''\left( \frac{1}{t}\right) \right] \\&=\frac{1}{4}t^{4-2d}\left[ 4(1-d)^2+(5-4d)tM_d'(t)+t^2M_d''(t)\right] . \end{aligned}$$

Hence,

$$\begin{aligned} A_d\left( \frac{1}{t}\right) M_d\left( \frac{1}{t}\right) -B_d\left( \frac{1}{t} \right) ^2&=\frac{1}{4}t^{6-2d}\left[ 4(1-d)^2M_d(t)^2+(5-4d)tM_d'(t)M_d(t)\right. \\&\qquad \left. +t^2M_d''(t)M_d(t)\right] -t^{6-4d}\left[ (1-d)M_d(t)+\frac{1}{2}tM_d'(t)\right] ^2\\&=\frac{1}{4}t^{6-2d}\left[ tM_d(t)M_d'(t)+t^2M_d(t)M_d''(t) -t^2M_d'(t)^2\right] \\&=\frac{1}{4}t^{8-2d}\left[ \frac{1}{t}M_d(t)M_d'(t) +M_d(t)M_d''(t)-M_d'(t)^2\right] . \end{aligned}$$

Therefore,

$$\begin{aligned} f\left( \frac{1}{t}\right)&=\frac{1}{\pi }\frac{\sqrt{A_d\left( \frac{1}{t} \right) M_d\left( \frac{1}{t}\right) -B_d\left( \frac{1}{t}\right) ^2}}{M_d \left( \frac{1}{t}\right) }\\&=\frac{1}{\pi }\frac{t^{4-2d}\sqrt{\frac{1}{4t}\left[ M_d(t)M_d'(t) +tM_d(t)M_d''(t)-tM_d'(t)^2\right] }}{t^{2-2d}M_d(t)}\\&=t^2\frac{1}{\pi }\frac{\sqrt{\frac{1}{4t}\left[ M_d(t)M_d'(t)+ tM_d(t)M_d''(t)-tM_d'(t)^2\right] }}{M_d(t)}\\&=t^2f(t). \end{aligned}$$

1.4 Proof of Lemma 3

Denoting \(\varSigma = 1+\sum _{k=1}^{n-1} t_k^2\), we have

$$\begin{aligned} \det L&= \frac{1}{\varSigma ^{2(n-1)}} \det \begin{pmatrix} \varSigma -t_1^2 &{} -t_1 t_2&{} \dots &{} -t_1 t_{n-1} \\ -t_2 t_1&{} \varSigma -t_2^2 &{} \dots &{} -t_2 t_{n-1} \\ \dots &{} \dots &{} \dots &{} \dots \\ -t_{n-1} t_1&{} -t_{n-1} t_2 &{} \dots &{}\varSigma -t_{n-1}^2 \end{pmatrix}\\&= \frac{1}{t_1 \dots t_{n-1}} \frac{1}{\varSigma ^{2(n-1)}} \det \begin{pmatrix} t_1(\varSigma -t_1^2) &{} -t_1 t_2^2&{} \dots &{} -t_1 t_{n-1}^2 \\ -t_2 t_1^2&{} t_2( \varSigma -t_2^2) &{} \dots &{} -t_2 t_{n-1}^2 \\ \dots &{} \dots &{} \dots &{} \dots \\ -t_{n-1} t_1^2&{} -t_{n-1} t_2^2 &{} \dots &{}t_{n-1}(\varSigma -t_{n-1}^2) \end{pmatrix}\\&= \frac{1}{t_1 \dots t_{n-1}} \frac{1}{\varSigma ^{2(n-1)}} \det \begin{pmatrix} t_1 &{} -t_1 t_2^2&{} \dots &{} -t_1 t_{n-1}^2 \\ t_2 &{} t_2( \varSigma -t_2^2) &{} \dots &{} -t_2 t_{n-1}^2 \\ \dots &{} \dots &{} \dots &{} \dots \\ t_{n-1} &{} -t_{n-1} t_2^2 &{} \dots &{}t_{n-1}(\varSigma -t_{n-1}^2) \end{pmatrix}\\&= \frac{1}{\varSigma ^{2(n-1)}}\det \begin{pmatrix} 1 &{} - t_2^2&{} \dots &{} - t_{n-1}^2 \\ 1 &{} \varSigma -t_2^2 &{} \dots &{} - t_{n-1}^2 \\ \dots &{} \dots &{} \dots &{} \dots \\ 1 &{} - t_2^2 &{} \dots &{}\varSigma -t_{n-1}^2 \end{pmatrix}\\&=\frac{1}{\varSigma ^{2(n-1)}} \det \begin{pmatrix} 1 &{} - t_2^2&{} \dots &{} - t_{n-1}^2 \\ 0 &{} \varSigma &{} \dots &{} 0 \\ \dots &{} \dots &{} \dots &{} \dots \\ 0 &{} 0 &{} \dots &{}\varSigma \end{pmatrix}\\&= \frac{1}{\varSigma ^n}. \end{aligned}$$

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Duong, M.H., Han, T.A. On the Expected Number of Equilibria in a Multi-player Multi-strategy Evolutionary Game. Dyn Games Appl 6, 324–346 (2016). https://doi.org/10.1007/s13235-015-0148-0

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