By construction, the candidate stationary distribution given by Eqs. (
19) and (
20) is a probability distribution since it satisfies the condition in Eq. (
6) and for any state
\(\textbf{a}\),
\(\textrm{u}_{\textbf{a}}\) is between 0 and 1. Again, since
\(\beta \) is finite the process will have a unique stationary distribution. Again, to show that the candidate stationary distribution is the unique stationary distribution, we need to check if Eq. (
5) holds. That is, the condition in Eq. (
62) must hold for all states
\(\textbf{a}\). We re-introduce some notations that we will use in this proof:
$$\begin{aligned} \bar{a}^j&:= \{\textrm{D},\textrm{C}\} \setminus \{a_j\} \end{aligned}$$
(71)
$$\begin{aligned} \alpha (a)&:= {\left\{ \begin{array}{ll} 1 \quad \text {if} \quad a = \textrm{C}\\ 0 \quad \text {if} \quad a = \textrm{D}\end{array}\right. } \end{aligned}$$
(72)
$$\begin{aligned} \mathcal {C}(\textbf{a})&= \sum _{j=1}^N \alpha (a_j) . \end{aligned}$$
(73)
For this process, since there are only two actions, the first term in the right-hand side of Eq. (
62) can be simplified as
$$\begin{aligned} \textrm{u}_{\textbf{a}} \textrm{T}_{\textbf{a},\textbf{a}}&= \textrm{u}_\textbf{a}- \textrm{u}_{\textbf{a}} \sum _{k=1}^{N} \textrm{T}_{(a_k, \textbf{a}_{-k}),(\bar{a}_{k}, \textbf{a}_{-k})} \end{aligned}$$
(74)
$$\begin{aligned}&= \textrm{u}_{\textbf{a}} - \frac{\textrm{u}_{\textbf{a}}}{N} \sum _{k=1}^N \frac{1}{1 + \displaystyle e^{\textrm{sign}(\bar{a}_{k}) \beta f(N_k)}} . \end{aligned}$$
(75)
where the function
\(\textrm{sign}(.)\) is defined as in Eq. (
16) and
f(
j) is the difference in payoffs between playing
\(\textrm{D}\) and
\(\textrm{C}\) when there are
j co-players playing
\(\textrm{C}\). The term
\(N_k\) represents the number of co-players of
k that play
\(\textrm{C}\) in state
\(\textbf{a}\). That is,
$$\begin{aligned} N_k := \sum _{j \ne k} \alpha (a_j) . \end{aligned}$$
(76)
The second term in the right-hand side of Eq. (
62) can be simplified as
$$\begin{aligned} \sum _{\textbf{b}\ne \textbf{a}} \textrm{T}_{\textbf{b}, \textbf{a}} \textrm{u}_{\textbf{b}}&= \sum _{k=1}^N \textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})} \textrm{u}_{(\bar{a}_k, \textbf{a}_{-k})} \end{aligned}$$
(77)
$$\begin{aligned}&= \frac{1}{N \Gamma } \sum _{k=1}^N \textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})} \displaystyle \prod _{j=1}^{\mathcal {C}((\bar{a}_k, \textbf{a}_{-k}))} e^{-\beta f(j-1)} \end{aligned}$$
(78)
$$\begin{aligned}&= \frac{1}{N \Gamma } \sum _{k=1}^N \textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})} \displaystyle \left( \prod _{j=1}^{N_k} e^{-\beta f(j-1)} \right) \cdot e^{-\beta \alpha (\bar{a}_k)f(-\alpha (a_k)+ N_k}) . \end{aligned}$$
(79)
From Eq. (
78) to Eq. (
79), we took out one term from the product that is present in our candidate distribution. This term accounts for the
\(k^{th}\) players action in the neighboring state
\((\bar{a}_k, \textbf{a}_{-k})\) of
\(\textbf{a}\). For simplicity, we represent
\(\textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})}\) with just
\(\textbf{T}\) in the next steps. We continue the simplification of Eq. (
79) in the next steps by introducing terms that cancel each other,
$$\begin{aligned} \sum _{\textbf{b}\ne \textbf{a}} \textrm{T}_{\textbf{b}, \textbf{a}} \textrm{u}_{\textbf{b}}&= \frac{1}{N \Gamma } \sum _{k=1}^N \textbf{T}\cdot \displaystyle \left( \prod _{j=1}^{N_k} e^{-\beta f(j-1)} \right) \cdot \frac{e^{-\beta \alpha (\bar{a}_k)f(-\alpha (a_k)+ N_k)}}{e^{-\beta \alpha (a_k)f(-\alpha (\bar{a}_k) +N_k)}} \cdot e^{-\beta \alpha (a_k)f(-\alpha (\bar{a}_k) + N_k)} . \end{aligned}$$
(80)
The newly introduced term in Eq. (
80) can be taken inside the product. Note that this term is 1 if the
\(k^{th}\) player plays
\(\textrm{D}\) in the state
\(\textbf{a}\). When this term is taken inside the product bracket, products of exponent
\(e^{-\beta f(j-1)}\) can be performed for
j ranging from 1 to the number of cooperators in state
\(\textbf{a}\),
\(\mathcal {C}(\textbf{a})\). This product is then the candidate stationary distribution probability
\(\textrm{u}_\textbf{a}\). That is,
$$\begin{aligned} \sum _{\textbf{b}\ne \textbf{a}} \textrm{T}_{\textbf{b}, \textbf{a}} \textrm{u}_{\textbf{b}}&= \frac{1}{N \Gamma } \sum _{k=1}^N \textbf{T}\cdot \displaystyle \left( \prod _{j=1}^{N_k} e^{-\beta f(j-1)} \cdot e^{-\beta \alpha (a_k)f(-\alpha (\bar{a}_k) + N_k)} \right) \cdot \frac{e^{-\beta \alpha (\bar{a}_k)f(-\alpha (a_k)+ N_k)}}{e^{-\beta \alpha (a_k)f(-\alpha (\bar{a}_k) +N_k)}} \end{aligned}$$
(81)
$$\begin{aligned}&= \frac{1}{N} \sum _{k=1}^N \textbf{T}\cdot \left( \frac{1}{\Gamma } \prod _{j=1}^{\mathcal {C}(\textbf{a})} e^{-\beta f(j-1)}\right) \cdot \frac{e^{-\beta \alpha (\bar{a}_k)f(-\alpha (a_k)+ N_k)}}{e^{-\beta \alpha (a_k)f(-\alpha (\bar{a}_k) +N_k)}} \end{aligned}$$
(82)
$$\begin{aligned}&= \frac{1}{N} \sum _{k=1}^N \textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})} \cdot \textrm{u}_\textbf{a}\cdot \frac{e^{-\beta \alpha (\bar{a}_k)f(-\alpha (a_k)+ N_k)}}{e^{-\beta \alpha (a_k)f(-\alpha (\bar{a}_k) +N_k)}} . \end{aligned}$$
(83)
The fraction inside the sum in Eq. (
83) can be simplified using the
\(\textrm{sign}(.)\) function (in
16) leading to further simplification of Eq. (
83):
$$\begin{aligned} \sum _{\textbf{b}\ne \textbf{a}} \textrm{T}_{\textbf{b}, \textbf{a}} \textrm{u}_{\textbf{b}}&= \frac{1}{N} \sum _{k=1}^N \textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})} \cdot \textrm{u}_\textbf{a}\cdot e^{sign(a_k) \beta f(N_k)} . \end{aligned}$$
(84)
In Eq. (
84), we can replace the element of the transition matrix
\(\textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})}\) with
$$\begin{aligned} \textrm{T}_{(\bar{a}_k, \textbf{a}_{-k}),(a_k, \textbf{a}_{-k})} = \frac{1}{1 + \displaystyle e^{\textrm{sign}(a_k) \beta f(N_k)}} . \end{aligned}$$
(85)
Using the expression for the transition matrix element from Eq. (
85) into Eq. (
84) and by using Eq. (
75), we can simplify further:
$$\begin{aligned} \sum _{\textbf{b}\ne \textbf{a}} \textrm{T}_{\textbf{b}, \textbf{a}} \textrm{u}_{\textbf{b}}&= \frac{\textrm{u}_\textbf{a}}{N} \sum _{k=1}^N \frac{1}{1 + \displaystyle e^{\textrm{sign}(a_k) \beta f(N_k)}} \cdot e^{sign(a_k) \beta f(N_k)} \end{aligned}$$
(86)
$$\begin{aligned}&= \frac{\textrm{u}_\textbf{a}}{N} \sum _{k=1}^N \frac{1}{1 + \displaystyle e^{\textrm{sign}(\bar{a}_k) \beta f(N_k)}} \end{aligned}$$
(87)
$$\begin{aligned}&= \textrm{u}_\textbf{a}- \textrm{u}_\textbf{a}\textrm{T}_{\textbf{a},\textbf{a}} . \end{aligned}$$
(88)
The final step in the previous simplification shows that Eq. (
62) holds for any
\(\textbf{a}\in \{\textrm{C},\textrm{D}\}^N\). Therefore, the candidate distribution we propose in Eq. (
19) is the unique stationary distribution of the symmetric
N-player game with two strategies.
\(\square \)