We consider a system whose state is described by a finite-state, homogeneous, discrete-time Markove chain $X_k$, $k\in \Bbb N$. $X_0$ is given. Suppose the state space of $X_k$ is $S_X=\{e_1,\cdots ,e_N\}$, where $e_i$ are unit vectors in $\Bbb R^N$ with unity as the $i$-th element and zeros elsewhere.
Let $\mathcal F_k$ be the $\sigma$-algebra generated by $X_0,X_1,\cdots,X_k$ then $\{\mathcal F_k\}$ is a filtration. By the Markov property, we have $$ P(X_{k+1}=e_j|\mathcal F_k)=P(X_{k+1}=e_j|X_k). $$
Write $a_{ji}=P(X_{k+1}=e_j|X_k=e_i)$, $A=(a_{ji})$. Then $$E[X_{k+1}|\mathcal F_k]=E[X_{k+1}|X_k]=AX_k.$$
Define $V_{k+1}=X_{k+1}-AX_k$. So that $X_{k+1}=AX_k+V_{k+1}$.
The state process $X$ is not observed directly. We suppose there is a fuction $c(.,.)$ with finite range and we observed the values $Y_{k+1}=c(X_k,w_{k+1})$. The $w_k$ are a sequence of i.i.d. random variables, with $V_k$, $w_k$ being mutually independent. Suppose the range of $c(.,.)$ consists of $M$ points. Then we can identify the range of $c(.,.)$ with the set of unit vectors $S_Y=\{f_1,\cdots,f_M\}$, where $f_i\in \Bbb R^M$ are unit vectors with unity as the $i$-th element and zeros elsewhere. Let $\mathcal G_k$ be the sigma algebra generated by $X_0,\cdots, X_k,Y_1,\cdots,Y_k$. Then we have $$ P(Y_{k+1}=f_j|\mathcal G_k)=P(Y_{k+1}=f_j|X_k). $$
Write $c_{ji}=P(Y_{k+1}=f_j|X_k=e_i)$, $C=(c_{ji})\in\Bbb R^{M\times N}.$
Then $E[Y_{k+1}|X_k]=CX_k.$
Write $Y_k=(Y_k^1,\cdots,Y_k^M)^T\in\Bbb R^M$. Note $\sum_{i=1}^M Y_k^i=1$. Wirte $c_{k+1}^i=E[Y_{k+1}^i|\mathcal G_k]=\sum_{j=1}^N c_{ij}<e_j,X_k>$, and $c_{k+1}=(c_{k+1}^1,\cdots,c_{k+1}^M)^T.$ Then $c_{k+1}=E[Y_{k+1}|X_k]=CX_k$.
Assume that $c_l^i>0$ for $1\le i\le M$, $l\in \Bbb N$.
Define $\lambda_l=\prod_{i=1}^M\left(\frac{1}{Mc_l^i}\right)^{Y_l^i}$ and $\Lambda_k=\prod_{l=1}^k \lambda_l$.
We define a new probablilty measure $\bar{P}$ on $(\Omega,\cup_{l=1}^\infty \mathcal G_k)$ by putting the restriction of the Radon-Nikodym derivative $\frac{d\bar P}{dP}$ to the sigma algebra $\mathcal G_k$ equal to $\Lambda_k$. Thus
$$ \frac{d\bar P}{dP}|_{\mathcal G_k}=\Lambda_k.$$
I want to ask two questions. First, why $\bar P(Y_{k+1}=1)=\frac{1}{M}$, and why $E(\lambda_{k+1}X_{k+1}|\mathcal G_k)=AX_k$.
Appreciate any suggestions.