Computing the mean of a stationary distribution for Wright-Fisher model The Question
Considering a population of $N$ organisms, there are two possible gene types, $A$ or $a$. An $A$ that is drawn ends up becoming an $a$ in the next generation with probability $u$ and remains an $A$ with probability $(1 -u)$. An $a$ that is drawn becomes an $A$ in the next generation with probability $v$, and stays an $a$ with probability $(1- v)$. 
I am given that the probability of producing $A$ on a single draw given that the current population has $x$ individuals with the $A$ allele is $$\alpha_x = {x \over N}(1 - u) + {N - x \over N} v$$ This means that the transition probability for $X_n$ is now $$\tag{$*$} p(x,y) = {N \choose y}(\alpha_x)^y (1 - \alpha_x)^{N - y}$$
For the transition probability in $(*)$, 


*

*Show that if $0 < u,v < 1$ then $\displaystyle \lim_{n \to \infty} p^n(x,y) = \pi(y)$ where $\pi$ is the unique stationary distribution. 


My Work: I showed that the chain is irreducible and aperiodic (using simple work that I won't include here for space considerations), thus it has a stationary distribution $\pi$, thus by the Convergence Theorem $p^n(x,y)$ converges to $\pi(y)$ in the limit. 


*

*Compute the mean of $\pi$, $${\bf E}[\pi] = \sum_{y = 0}^N y \, \pi(y) = \lim_{n \to \infty} {\bf E}[X_n]$$


My Work: 
I am given a hint that tells me to first compute ${\bf E}[X_1]$. I went to office hours, and my teacher told me that I should then use conditional expectation to compute ${\bf E}[X_2]$ at which point I might see a recurrent formula that I can take a limit of. 
Computing ${\bf E}[X_1]$, I have that 
\begin{align*}
{\bf E}[X_1] &= \sum_{x=0}^N x{\bf P}(X_1 = x \mid X_0 = x_0) \\
&= \sum_{x = 0}^N x \, p(x_0, x) \\
&= \sum_{x = 0}^N x \, {N \choose x}(\alpha_{x_0})^x(1 - \alpha_{x_0})^{N - x}
\end{align*}
At this point, I believe I have a Binomial random variable, so ${\bf E}[X_1] = N\alpha_{x_0}$

Having done this work, I'm not really sure where to go from here. $N$ is fixed, and I don't see how this will turn into a quantity of which I can take a limit. 
 A: After a large number of generations, each organism has been drawn a large number of times. The states of a given organism after each time when it is drawn perform a Markov chain on the state space $\{A,a\}$ with transition probabilities $1-u$, $u$, $1-v$ and $v$ for $A\to A$, $A\to a$, $a\to a$ and $a\to A$ respectively. Hence the distribution of this organism converges to the stationary distribution $\mu$ of this two-states Markov chain, which is given by $\mu(A)=w$ and $\mu(a)=1-w$, with $w=v/(u+v)$.
Now, the states of the $N$ different organisms become independent after a large number of generations since they forget their initial state and their dynamics are independent. Hence the number $X_n$ of $A$'s in the overall population converges in distribution to the binomial distribution $(N,w)$, and, in particular,
$$
\lim\limits_{n\to\infty}\mathrm E(X_n)=\sum_{k=0}^Nk\pi(k)=Nw=N\frac{v}{u+v}.
$$
A: You have correctly calculated ${\bf E}[X_1] = N\alpha_{x}$ when the initial state is $x$. In other notation, we write ${\bf E}_{x}[X_1] = N\alpha_{x}$. From this, you can work out an expression for the mean of $X_1$  when the initial state is chosen randomly using $\pi$. That is, 
$${\bf E}_{\pi}[X_1]=\sum_{x=0}^N \pi(x)\, {\bf E}_x[X_1].\tag1$$
Expand the right hand side of (1), and solve the equation $ \bar\pi= {\bf E}_{\pi}[X_0] = {\bf E}_{\pi}[X_1].$   
A: I have encountered this question as well, I was asked to find the stationary distribution. Here is something I discovered:
Following the idea of trying to find the limit expectation of $$X = \lim_{n \rightarrow \infty}X_n$$
since the Markov Chain is irreducible and aperiodic, we can be sure that there will be a stationary distribution and if $E(X)$ exists, we will have:
$$\lim_{n \rightarrow \infty}E(X_{n+1})=\lim_{n \rightarrow \infty}E(X_n) = E(X)$$
Consider a filtration $F_n$ that contains all the informtaion of $X_i : \{1\leq i \leq n \}$. Since $\{X_n\}$ is a Markov Chain, 
$$E(X_{n+1}|F_n) = E(X_{n+1}|X_n)$$
since the r.v. $X_{n+1}|X_n \sim $ Bin($N, \alpha_{X_{n}}$)
$$E(X_{n+1}|X_n) = N\alpha_{X_{n}} = X_n(1-u) + (N-X_n)v$$
therefore,
$$E(X_{n+1}) = E(E(X_{n+1}|X_n))=E(X_n)(1-u)+(N-E(X_n))v$$
take $n \rightarrow \infty$, we will have
$$E(X) = E(X)(1-u)+(N-E(X))v$$
$$E(X) = N\frac{v}{u+v}$$
Here we have the same answer everybody arrived at, and we can conclude that there will be a stationary distribution and we know its limit expectation. However, if we naively think the stationary distribution is simply Bin($N, \frac{v}{u+v}$), we are surprisingly wrong. Here is my way of proving that the stationary distribution does not necessarily follow a binomial distribution:
Suppose the stationary distribution is $X \sim $ Bin($N, \frac{v}{u+v}$), then we need to prove that
$$P(X = i) = \sum_{j = 0}^{N}{N \choose i}\alpha_j^i(1-\alpha_j)^{N-i}P(X = j)$$
i.e. we need to prove
$$(\frac{v}{u+v})^i(\frac{u}{u+v})^{N-i}=\sum_{j = 0}^{N}\alpha_j^i(1-\alpha_j)^{N-i}{N \choose j}(\frac{v}{u+v})^j(\frac{u}{u+v})^{N-j}$$
Look at the right hand side of the equation, it is not hard to see that
$$\sum_{j = 0}^{N}\alpha_j^i(1-\alpha_j)^{N-i}{N \choose j}(\frac{v}{u+v})^j(\frac{u}{u+v})^{N-j} = E(\alpha_J^i(1-\alpha_J)^{N-i})$$
where $J \sim$ Bin($N$,$\frac{v}{u+v}$). Now it boils down to prove
$$(\frac{v}{u+v})^i(\frac{u}{u+v})^{N-i} = E(\alpha_J^i(1-\alpha_J)^{N-i})$$
it is easy to see that $E(\alpha_J) = \alpha_{E(J)} = \frac{v}{u+v}$ and $E(1-\alpha_J) = 1 - \alpha_{E(J)}=\frac{u}{u+v}$, hence it further boils down to prove
$$\alpha_{E(J)}^i(1-\alpha_{E(J)})^{N-i} = E(\alpha_J^i(1-\alpha_J)^{N-i})$$
for all $i \in \{0,1,...,N\}$.
This is simply unacheivable most of the time since according to Jensen's inequality, for any convex function $G(x)$ and r.v. $X$ with expectation $E(X)$,
$$E(G(X)) \geq G(E(X))$$
i.e.
$$\alpha_{E(J)}^i(1-\alpha_{E(J)})^{N-i} \leq E(\alpha_J^i(1-\alpha_J)^{N-i})$$
whenever
$$g(x) = \alpha_x^i(1-\alpha_x)^{N-i}$$ is a convex function and 
$$\alpha_{E(J)}^i(1-\alpha_{E(J)})^{N-i} \geq E(\alpha_J^i(1-\alpha_J)^{N-i})$$
whenever
$$g(x) = \alpha_x^i(1-\alpha_x)^{N-i}$$ is a concave function. It is easy to see that $g(x)$ will always be concave or convex depend on the value of $N$ and $i$. Therefore, the equality will not hold. However,
the equality does hold when $u=v$, since $J \sim$ Bin($N$,$\frac{1}{2}$) which is a symmetric distribution. While $u \neq v$, the equality fails which means it contradicts our initial claim on $X$ being binomial. 
Hence we see that the stationary distribution of $X$ is not binomial unless $u = v$ which is a very surprising result given the form of the stationary expectation.
I do not think there will be an explicit gerneral form of the stationary distribution. But I did some coding, and here is something I have: ($N = 20$, $u = 0.1$, $v = 0.2$)

