Let $(X_n)_{n\ge0}$ be a Markov chain taking values in a measurable space $(E,\mathcal{E})$ and with transition kernels $(P_n)_{n\ge 1}$.
Question: Is $P_n$ unique, for every $n\ge 1$?
My try: Fix $n\ge 1$ and a probability kernel $\tilde{P}_{n+1}$ s.t. for every bounded and measurable $f:E\to\mathbb{R}$ we have $E(f(X_{n+1})|\mathcal{F}_n)=\tilde{P}_{n+1}(X_n)$. Taking $f:=\mathbb{1}_A$ for $A\in\mathcal{E}$ arbitrary thus gives $P_{n+1}(X_n,A)=\tilde{P}_{n+1}(X_n,A)$ so on the image of $X_n$ we have $P_{n+1}=\tilde{P}_{n+1}$. How do I find $f$ such that I can extend this to $x\in E$ not being in the image of $X_n$? Or is the result actually not true after all?
Best Answer
If I understand correctly, then outside the image of $X_n$, we cannot say anything about $P_{n+1}$, and so uniqueness may not hold there.
We can see this with a discrete example. Suppose $E = \{e_1, e_2\}$ and $(X_n)_{n\ge 0}$ is a time-homogeneous Markov chain in which $e_1$ always transitions to $e_2$, and $e_2$ always stays put. Then for $n>0$, we just always have $X_n = e_2$. So we can change $P_n$ for $n>1$ to anything where $e_2$ still always transitions to $e_2$ (but $e_1$ does not always transition to $e_1$), and this doesn't actually affect the distribution of $X_n$ in any way.