I have trouble with showing the Markov and homogeneity property of a homogenous Markov chain if we allow somewhat arbitrary sets left and right of the condition bar. To put it formally:
Let $(X_n)_{n \geq 0}$ be a homogenous, time discrete Markov chain. Let $\tau_y := \min\{n \geq 1: X_n = y\}$ be the (random) return time to $y$ for $y \in E$ with state space $E$. Define $\mathbb{P}^x(\tau_y \leq k) := \mathbb{P}(\tau_y \leq k|X_0 = x)$ for $x \in E$ to be the probabilty that the chain visits $y$ if started in $x$ (or in the case of $x = y$ visits itself again after starting).
My scripture now states, that for some fixed $k \in \mathbb{N}$
$$\exists \alpha \in [0,1]: \forall x \in E: \mathbb{P}^x(\tau_y \leq k) \geq \alpha \implies \forall n \in \mathbb{N}, x \in E: \mathbb{P}^x(\tau_y > nk) \leq (1-\alpha)^n$$
We then use induction over $n$ to show the proposition. Trivial base case and induction hypothesis (IH) aside, the induction step goes like this:
$$\begin{array}{ll} \mathbb{P}^x(\tau_y > (n+1)k) &= \mathbb{P}^x(\tau_y > (n+1)k|\tau_y > nk) \cdot \mathbb{P}^x(\tau_y > nk) \\ &= \mathbb{P}^x(\tau_y > k) \cdot \mathbb{P}^x(\tau_y > nk) \\ &\stackrel{(IH)}{\leq} (1-\alpha) \cdot (1-\alpha)^n = (1-\alpha)^{n+1} \end{array}$$
The step that I refuse to believe without evidence is $\mathbb{P}^x(\tau_y > (n+1)k|\tau_y > nk) = \mathbb{P}^x(\tau_y > k)$. In my scripture, this is put off as a use of the Markov property and homogeneity, but I can't really show why that must be true.
I have already shown both the Markov and homogeneity property for expressions of the form $\mathbb{P}(X_n = i_n,\ldots,X_{l+1} = i_{l+1}|X_l = i_l,\ldots,X_0=i_0)$ and from there it is easy to generalize to events on the left side that can be expressed as a countable union of finitely many $``X_i=\ldots"$-terms on the left side, but the above probability $\mathbb{P}^x(\tau_y > (n+1)k|\tau_y > nk)$ having a possibly infinite amount of terms on both the left and right side and there being a non-$X_i$ expression on the right leaves me questioning.
I revoke my statement on having found a proof in I this book by A. Klenke, I mixed it up with a different problem. Maybe someone different can point me to a proposition relevant to the question.
Any help is appreciated. Thank you.
Fix $n\ge 0$. On the event $\{\tau_y>nk\}$ you have $\tau_y=nk+\tilde\tau_y$, where $\tilde\tau_y$ is the first time that $(X_{m+nk})_{m\ge 1}$ returns to $y$; that is, $\tilde\tau_y:=\inf\{m\ge 1: X_{m+nk}=y\}$. By the Markov property applied at time $nk$, the events $\{\tau_y>nk\}$ and $\{\tilde\tau_y>k\}$ are conditionally independent, given the value of $X_{nk}$, and the latter has conditional probability equal to $$ \Bbb P(\tilde\tau_y>k\mid X_{nk}=z)=\Bbb P^z(\tau_y>k)< 1-\alpha. $$ Amplification: $\mathcal F_j$ is the history of $X$ up to time $j$. Because $\tau_y$ is a stopping time, the event $\{\tau_y>nk\}$ is $\mathcal F_{nk}$-measurable. $$ \eqalign{ \Bbb P(\tau_y>(n+1)k) &=\Bbb P(\tau_y>nk, \tilde\tau_y>k)\cr &=\Bbb E\left[ 1_{\{\tau_y>nk\}} \Bbb P(\tilde\tau_y>k\mid \mathcal F_{nk})\right]\cr &=\Bbb E\left[ 1_{\{\tau_y>nk\}} \Bbb P^{X_{nk}}( \tau_y>k)\right]\cr &\le\Bbb E\left[ 1_{\{\tau_y>nk\}} (1-\alpha)\right]\cr &=(1-\alpha)\Bbb P( \tau_y>nk).\cr } $$