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.