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 would be immensely appreciated.
Just to make it easy for further analysis, i have simplified in the below,
$$E_{b_1,...,b_t}(b_t) = E_{b_1,..,b_{t-1}}(P(b_{t-1}) (1+xb_{t-1}) + (1-P(b_{t-1})) (xb_{t-1})) = E_{b_1,...,b_{t-1}}(P(b_{t-1})+xb_{t-1})$$ $$ = \sum_{i = 1}^{t} x^{t-i} E_{b_1,...,b_{i-1}}(P(b_{i-1}))$$
$$E_{b_1,...,b_{i}}(P(b_{i})) = E(F(\kappa^{i+1} a + (1-\kappa) b_{i}(\kappa))) $$
$$ = E_{b_1,...,b_{i-1}} ( \ \ P(b_{i-1}) F(\kappa^{i+1} a + (1-\kappa) (1+\kappa b_{i-1})(\kappa)) + (1-P(b_{i-1})) F(\kappa^{i+1} a + (1-\kappa) \kappa b_{i-1}(\kappa)) )$$
$$ = \sum_{b_1,...,b_i} \prod_{j = 1}^i G(\kappa^{j+1} a + (1-\kappa) b_{j}(\kappa))$$
If $b_{j}(x) = 1+xb_{j-1}(x)$, $$G(\kappa^{j+1} a + (1-\kappa) b_{j}(\kappa)) = F(\kappa^{j+1} a + (1-\kappa) b_{j}(\kappa)) $$
If $b_{j}(x) = xb_{j-1}(x)$, $$G(\kappa^{j+1} a + (1-\kappa) b_{j}(\kappa)) = 1-F(\kappa^{j+1} a + (1-\kappa) b_{j}(\kappa)) $$
$$E_{b_1,...,b_t}(b_t) = \sum_{i = 1}^{t} x^{t-i} \sum_{valid \ b_1,...,b_{i-1}} \prod_{j = 1}^{i-1} G(\kappa^{j+1} a + (1-\kappa) b_{j}(\kappa))$$