# Fixed point of transition kernel generates martingale

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.

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Which book are you using to study Markov processes? – Ilya Jan 18 '13 at 7:33
@Ilya I don't use some particular book, I try to obtain given theorems myself, but when I have no ideas I look Shiryaev's "Probability". – Nimza Jan 18 '13 at 13:50

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: $P^sf =f$.
Thank you for the answer! I have a question: why $\mathscr{F}_{t} = \sigma(X_{s} \mid s\leqslant t)$ is equal to $\sigma(X_t)$? I think that's because $X_t$ is a Markov process, but how to show this? – Nimza Jan 18 '13 at 13:47
My second question: why $\mathsf E_{X_t} f(X_{t+h}) = (\mathsf P^h f)(X_t)$? LHS is $\mathscr F_{X_t}$-measurable, then there exists Borel function $\varphi(x)$ such that $\mathsf E_{X_t} f(X_{t+h}) = \varphi(X_t)$. By definition $\mathsf E(f(X_{t+h}) \mid X_t=x) := \varphi(x)$. But why $\varphi(x) = P^h f (x)$? – Nimza Jan 18 '13 at 13:57
@Nimza: the answer to the first question is no. But conditioning on these $\sigma$-algebras for $X_u$ with $u\geq t$ is equivalent. When $u<t$ the conditioning is not equivalent. – Ilya Jan 18 '13 at 16:45
@Nimza: for the second question, this follows from the fact that $\mathsf P_{X_t}(X_h\in B) = P^h(X_t,B)$ by the definition of the transition kernel $P^h$ as I as said above. Would you agree on this? Then be integrating $f$ w.r.t. this measure you get the formula for the expectation. – Ilya Jan 18 '13 at 16:51