Fixed point of transition kernel generates martingale

Question

Let $P^{h}, h \geqslant 0$ be a transition kernel for some homogenous Markov process $X_t$, $\mathbb{E}|X_t|<\infty$: $$ P_{X_{t+h},X_t}(A,B) = \int\limits_{A}P^h(x,B)P_{X_t}(dx) $$ where $P_{X_{t+h},X_{t}}(A,B) = \mathbb{P}(X_{t+h}\in A, X_{t} \in B)$ and $P_{X_t}(C) = \mathbb{P}(X_t \in C)$. Then $P^h$ can be considered as a linear operator acting on continuous bounded functions: $$ P^h f(x) = \int f(y)P^h(x,dy) $$ Let $f(x)$ be a fixed point of $P^h$ for any $h \geqslant 0$: $P^h f = f$. Now let $\mathcal{F}_{t} = \sigma(X_{s} \mid s \leqslant t)$ be a sigma-algebra generated by $\{X_s \mid s \leqslant t\}$. I want to show that $$ \mathbb{E}( f(X_{t+s}) \mid \mathcal{F}_t) = f(X_t) $$ for any $s \geqslant 0$. I can do it informally: $$ \mathbb{E}( f(X_{t+s}) \mid X_{t}=x ) = \int f(y) P^{s}(x,dy) = P^s f(x) = f(x). $$ Please help me to do it strictly.

Answer 1

You are using some non-usual formulas for the kernels of the Markov process. I think, it better to start with the following: $$ \mathsf P_x(X_{t+h}\in B|\mathscr F_t) = \mathsf P_{X_t}(X_h\in B)=P^h(X_t,B) $$ where the first identity is the definition of the Markov property, and the second one is the definition of the kernel. Clearly, just by writing integrals you have the very same expression for the expectations (at least for bounded measurable functions): $$ \mathsf E_x\left[f(X_{t+h})|\mathscr F_t\right] = \mathsf E_{X_t}[f(X_h)]=(P^hf)(X_t) \stackrel{harm}{=}f(X_t), $$ where $\stackrel{harm}{=}$ holds under the assumption $f$ is harmonic: $P^sf =f$.