Time distibution in Markov chain Let $E=\{A,B\}$ be a set and $X_{1,t}, X_{2,t}, X_{3,t}$ three independent Markov chains on the set $E$ with respective transition probability $P^{(1)}, P^{(2)}, P^{(3)}$ where $$P^{(i)}=\begin{bmatrix}p^{(i)} & 1-p^{(i)}\\
1-p^{(i)} & p^{(i)}
\end{bmatrix}$$
Let $P = P^{(1)}\otimes P^{(2)}\otimes P^{(3)}$ where $\otimes$ is the Kronecker product. 
The process $Y_t = X_{1,t}\times X_{2,t}\times X_{3,t}$ (product of the three Markov chain) is a Markov chain with transition probability $P$ and state space $E_Y = E\times E\times E = \{AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB\}$ and we can use the standard Markov theory to compute anything we want. But, here are my questions :
1 - As $ ''AAB '' $ and $ ''ABA '' $ are the same value equal to $A^2B$, then $E_Y$ can be reduce to $\{A^3, A^2B, AB^2, B^3\}$. Is $Y_t$ still a Markov chain if we combine the transition probability in other to set it in that reduced space?
2 - How can I determine the time distribution probability of begin in a particular state. In other word, how can I compute $$\mathbb{P}[Y_{t+1} \neq A^2B, Y_t = A^2B, Y_{t-1} = A^2B, \dots, Y_1 = A^2B | Y_0=A^2B]$$ for each $t\in \mathbb{N}$.
 A: *

*One can only aggregate states like $(A,B,A)$ and $(A,A,B)$ if the probability to reach the event "$(A,B,A)$ or $(A,A,B)$" is the same for all starting values $Y_{t-1}$ of the chain $Y_t$. See the details in the lumpability page on Wikipedia.

*Since the $X_{i,t}$ are independent, $$\Bbb P(Y_t=y'|Y_{t-1}=y)=\prod_{i=1}^3\Bbb P(X_{i,t}=y_i'|X_{i,t-1}=y_i)$$
A: I found solution for my question. So I'm proposition an answer.
1- $Y_t$ is not always a markov chain. And it even be a Markov chain but not homogenus.
2- $\mathbb{P}[Y_{t+1} \neq A^2B, Y_t =A^2B,\dots, Y_1=A^2B| Y_0=A^2B] = \nu P_\mathcal{B}(I-P_\mathcal{B})\mathbf{1}$ where $\nu$ is the initial probability of $X_t$ on the set $\mathcal{B}:=\{AAB,ABA,BAA\}$, $\mathbf{1} = \begin{bmatrix}1\\1\\1\end{bmatrix}$ and $P_\mathcal{B}$ is the transition probability on $\mathcal{B}$ in other that :
$$P = \begin{bmatrix} P_\mathcal{B} & P_{\mathcal{B}\mathcal{B}^c}\\
P_{\mathcal{B}^c\mathcal{B}} & P_{\mathcal{B}^c}\end{bmatrix}$$
$\mathcal{B}^c = E_Y - \mathcal{B}$
