This question is cross-posted from <https://math.stackexchange.com/questions/4711799/asymptotic-behavior-of-a-markov-process-on-the-set-of-0-1-polynomials>


I am trying to study the asymptotic behavior of a stochastic process defined on the space of single variable polynomials whose coefficients are either $0$ or $1$.


Letting $\mathbb{B}=\{0,1\}$, I will denote by $\mathbb{B}[x]$ this set of polynomials, and I will denote by $\mathbb{B}[x]\_{t}\subset\mathbb{B}[x]$ the subset of such polynomials whose degree is less or equal to $t$. Moreover, given a polynomial $b\_t(x)\in \mathbb{B}[x]\_{t}$ notice that $xb\_t(x)$ and $1+xb\_t(x)$ are elements of $\mathbb{B}[x]\_{t+1}$: these are the two possible "shifted" version of the polynomial.


Fix two parameters $\kappa,a\in (0,1)$. Define a process $(A\_t)\_{t=0}^\infty\subset\Delta(\mathbb{B}[x])$ as follows


* $A\_0=\delta\_0\in\Delta(\mathbb{B}\_{0}[x])$. In words $A\_0$ is a Dirac on the null polynomial.
* $A\_t\in\Delta(\mathbb{B}\_{t-1}[x])$ is defined as follows:
\begin{align\*}
A\_{t+1}(b\_{t})=\begin{cases}
A\_{t}(b\_{t-1})P(b\_{t-1})\quad &if\quad b\_{t}=1+xb\_{t-1}\\
A\_{t}(b\_{t-1})(1-P(b\_{t-1})) \quad &if\quad b\_{t}=xb\_{t-1}\\
0\quad & else
\end{cases}
\end{align\*}
where, letting $b\_{t-1}(\kappa)$ be the evaluation of $b\_{t-1}$ at $\kappa$, $P(b\_{t-1})=F(\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa))$ where $F:[0,1]\to[0,1]$ is a decreasing function such that $F(x)=1$ for $x$ close to 0, $F(1/2)=1/2$ and $F(x)=1$ for $x$ close to 1.


Hence, transition probabilities are as follows. $b\_{t-1}\in\mathbb{B}[x]\_{t-1}$ transitions to $1+xb\_{t-1}\in\mathbb{B}[x]\_{t}$ with probability $P(b\_{t-1})=F(\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa))$ or to $xb\_{t-1}\in\mathbb{B}[x]\_{t}$, with probability $1-P(b\_{t-1})$.


This should be a Markov process, where intuitively, the state $b\_{t-1}$ is split in the two "shifted" states $1+xb\_{t-1}$ and $xb\_{t-1}$. Since each distribution is supported in a new set, there is no hope that there is an ergodic distributions. In simulations,though, it appears that, asymptotically $A\_t$ concentrates on those polynomials in $\mathbb{B}[x]\_{t-1}$ such that the value $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ is closer to $1/2$.


This is intuitive, given the definition of $P$. While for polynomials such that the value $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ is very close to $0$, surely or with high probability the mass $A\_t(b\_{t-1})$ is transfered completely on $1+xb\_{t-1}$ (and viceversa for those who have $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ close to $1$) the polynomials with $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$ close to $1/2$ are those where the mass is split equally. There is a sort of "balance" on this states.


But how can I show this formally?


Any help or reference would be immensely appreciated.



Asymptotic behavior of a Markov process on the set of $\{0,1\}$-polynomials
Just a comment really, but too long to fit in a comment box! I think you can simplify the formulation considerably. Let $X\_t$ be your $\kappa^{t}a+b\_{t-1}(\kappa)(1-\kappa)$. Then you can describe the process $X\_t$ as a Markov chain in the following way: $X\_1=\kappa a$, and for $t\geq 1$,  

\begin{equation\*}
X\_{t+1}=\begin{cases} 
\kappa X\_t &\text{with probability $1-F(X\_t)$};\\
\kappa X\_t + (1-\kappa) &\text{with probability $F(X\_t)$}.
\end{cases}
\end{equation\*}
Now $(X\_t)\_{t\geq 1}$ is a nice Markov process taking values in $[0,1]$, and you can indeed look for a stationary distribution to which $X\_t$ converges in distribution as $t\to\infty$. Because of your assumption on $F$, such a stationary distribution will be supported on some sub-interval $[u,v]$ with $0<u<1/2<v<1$ (but it's not the case that it's concentrated on the point $1/2$).


