Some questions about the properties of a non-irreducible markov chain

Question

I was inspired to come up with the following questions from project euler problem 267 (don't worry, I'm not asking for a solution, I just was trying to extend the idea contained in this problem further):

Given a random variable $X_n$, define a markov chain with the following probabilities:

$$P(X_n=X_{n-1}+2fX_{n-1}) = 0.5\\ P(X_n=X_{n-1}-fX_{n-1}) = 0.5$$ for some constant $f$.

This is obviously a discrete-space discrete-time process for $n\lt\infty$ and only 2 states are accessible from any given state.

My questions are:

1) Given $n \rightarrow \infty$, is it possible for any state $X_n=r\in\mathbb{R}_{>0}$ to be reached?

2) Given any state $X_n=r\in\mathbb{R}_{>0}$, is this state recurrent? If so, what is the periodicity?

3) Given a stopping time $N=n\in\mathbb{Z}_{>0} \ \ s.t. \ \ X_n=S$, what is the mean time to stopping?

Sorry for the basic questions, I am about as far away from a real mathematician as it gets, but I like doing this stuff. If anyone could answer a few, or point me in a good direction, that'd be great.

Answer 1

1. The answer is no. To see it, note that $X_{i+1}$ equals either $(1+f)X_i$ or $(1-2f)X_i$, so any number $X$ we can reach can be written as $X_0 \cdot (1+f)^a (1-2f)^b$, with $a$ and $b$ positive integers. The cardinality of the set of such $X$ is then equal to $|\mathbb{N}|<|\mathbb{R}|$.

2. If, and only if $f=0$, then $X_0$ is recurrent. The "if" part is trivial. The "only if" part is trickier. Hint: consider $Y_i=\log (X_i)$, and assume e.g. $f>0$. Then $(1-2f)<(1-f)<\frac1{1+f}$, so there is a certain number of steps $k$ such that with probability $p>0$ within $k$ steps $Y_i$ falls below $\log(S)$ without ever having become equal to it. Can you prove that if and when that happens, there is a probability strictly greater than $0$ that $Y_i$ will never "climb back up" to $\log(S)$ or above (meaning the chain will never stop)?

3. From 1, we know that "almost all" reals are simply inaccessible. So, you are asking: assuming an $S$ is accessible, i.e. it can be written as $S=X_0 (1+f)^a (1-2f)^b$, what's the mean stopping time to $S$? The answer is $\infty$, unless $f=0$ and $X=X_0$, using the same argument as in $2$.