I encountered a specific kind of Markov chain with two parameters $\alpha,\beta\in(0,1)$. It works as follows: the variables $X_0,X_1,\dots$ live in the real interval $[0,1]$. When we have $X_n$, the value of $X_{n+1}$ is either $$\frac{\alpha X_n}{\alpha X_n + \beta (1-X_n)}$$ with probability $\alpha X_n+\beta (1-X_n)$, or $$\frac{(1-\alpha)X_n}{(1-\alpha) X_n + (1-\beta) (1-X_n)}$$ with probability $(1-\alpha)X_n+(1-\beta) (1-X_n)$.
From Python experiments, it seems that most of the times, $X_0<<1$ implies that $X_n<<1$ for all $n$, and $X_0\approx 1$ implies that $X_n\approx0$ for all $n$. However, there should be a small probability to 'jump to the other side'.
I would like to know the following:
- If $X_0$ is small (for example $X_0=0.01$), what is the probability (in terms of $X_1$) that for a large $n$, the value of $X_n$ is large (for example $X_n>0.99$)?
We start with $X_0=x$. Now one can prove with induction that there are $n+1$ possible values for $X_n$, which we shall label $X(n,0),X(n,1),\dots,X(n,n)$. We have $$X(n,a)=\frac{\alpha^a(1-\alpha)^{n-a} x}{\alpha^a(1-\alpha)^{n-a} x+\beta^a(1-\beta)^{n-a} (1-x)}.$$ Define $p(n,a)$ as the probability that $X_n=X(n,a)$. Then also by induction, we get $$p(n,a)=\binom{n}{a}\left(\alpha^a(1-\alpha)^{n-a} x+\beta^a(1-\beta)^{n-a} (1-x)\right).$$ Now the expected value of $X_n$ equals $$\sum_{a=0}^n p(n,a)\cdot X(n,a)=\sum_{a=0}^n \binom{n}{a}\alpha^a(1-\alpha)^{n-a}x=x.$$ Now let $n$ be large. From the De Moivre-Laplace Theorem, we know that the distributions of $\binom{n}{a}\alpha^a(1-\alpha)^{n-a}$ and $\binom{n}{a}\beta^a(1-\beta)^{n-a}$ are centered around $n\alpha$ and $n\beta$, respectively. So the only places where $p(n,a)$ is significant are around $n\alpha$ and $n\beta$. For $a\approx n\alpha$, we have $X(n,a)\approx 1$, and for $a\approx n\beta$, we have $X(n,a)\approx0$.
We conclude that the distribution of $X_n$ consists mainly of two peaks: one very close to $0$, and one very close to $1$. Since the expected value of $X_n$ is $x$, it follows that the probability that $X_n\approx 1$ is approximately $x$. So when we start with $X_0=0.01$, the probability that $X_n\approx 1$ after a large, fixed number of steps is $0.01$.