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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.

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up vote 3 down vote accepted

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, $$ \lim\limits_{n\to\infty}\mathrm E(X_n)=\sum_{k=0}^Nk\pi(k)=Nw=N\frac{v}{u+v}. $$

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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].$

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I see! So ${\bf E}_{\pi}[X_2]$ = ${\bf E}_{\pi}[X_1] \Longleftrightarrow N\alpha_{x_1} = N\alpha_{x_0} \Longleftrightarrow \alpha_{x_1} = \alpha_{x_0}$ so the formula given by @did is the probability that will have $\alpha_{x_1} = \alpha_{x_0}$? – Moderat Sep 19 '12 at 23:14
@jmi4 Your comment baffles me. ${\bf E}_{\pi}[X_2]={\bf E}_{\pi}[X_1]$ is true, but appears nowhere in my post. $N\alpha_{x_1} = N\alpha_{x_0}$ is false when $x_0\neq x_1$, as simple calculations will show. – Byron Schmuland Sep 20 '12 at 0:17
@jmi4 You should try to follow my suggestion "You can work out...". – Byron Schmuland Sep 20 '12 at 12:59
Never mind. I understand now your hint: "when the initial state is chosen randomly using $\pi$." This corresponds to the work by did above where $\pi(A) = v/(u+v)$. I think my comment came from the fact that initially $p(x,y)$ is binomial $(N, \alpha_x)$. – Moderat Sep 20 '12 at 22:59
Since it's homework, I only give hints. Perhaps they are a little too mysterious. I will add a bit to my solution. – Byron Schmuland Sep 20 '12 at 23:02

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