Let
- $E$ be at most countable and equipped with the discrete topology and $\mathcal E$ be the Borel $\sigma$-algebra on $E$
- $X=(X_n)_{n\in\mathbb N_0}$ be a discrete Markov chain with values in $(E,\mathcal E$) and distributions $(\operatorname P_x)_{x\in E}$
- $\tau_x^0:=0$ and $$\tau_x^k:=\inf\left\{n>\tau_x^{k-1}:X_n=x\right\}$$ for $x\in E$ and $k\in\mathbb N$
Let $$\varrho(x,y):=\operatorname P_x\left[\tau_y^1<\infty\right]\color{blue}{=\operatorname P_x\left[\exists n\in\mathbb N:X_n=y\right]}\;.$$
I want to prove, that $$\operatorname P_x\left[\tau_y^k<\infty\right]=\varrho(x,y)\varrho(y,y)^{k-1}\;\;\;\text{for all }k\in\mathbb N\tag 1$$ using the strong Markov property: $$\operatorname E_x\left[f\circ (X_{\tau+t})_{t\in \mathbb N_0}\mid\mathcal F_\tau\right]=\operatorname E_{X_\tau}\left[f\circ X\right]\tag 2$$ for all $x\in E$, $\sigma(X)$-stopping times $\tau$ and bounded, $\mathcal E^{\otimes\mathbb N_0}$-measurable $f:E^{\mathbb N_0}\to\mathbb R$.
I want to prove $(1)$ by induction over $k\in\mathbb N$. Since, $k=1$ is trivial, we only need to care about $k-1\to k$. Since $$\left\{\tau_y^{k-1}<\infty\right\}\cap\left\{\tau_y^k<\infty\right\}=\left\{\tau_y^k<\infty\right\}$$ and $$\left\{\tau_y^{k-1}<\infty\right\}\in\mathcal F_{\tau_y^{k-1}}\;,$$ we've got $$\operatorname P_x\left[\tau_y^k<\infty\right]=\operatorname E_x\left[1_{\left\{\tau_y^{k-1}<\infty\right\}}\color{red}{\operatorname P_x\left[\tau_y^k<\infty\mid\mathcal F_{\tau_y^{k-1}}\right]}\right]\;,\tag 4$$ by definition of the conditional expectation. Now, I think, that we somehow need to apply $(2)$ with $\tau=\tau_y^{k-1}$ to the $\color{red}{\text{red}}$ term in order to obtain $$\operatorname E_x\left[1_{\left\{\tau_y^{k-1}<\infty\right\}}\color{red}{\operatorname P_x\left[\tau_y^k<\infty\mid\mathcal F_{\tau_y^{k-1}}\right]}\right]=\operatorname E_x\left[1_{\left\{\tau_y^{k-1}<\infty\right\}}\varrho(y,y)\right]\;,$$ but I can't figure out how I need to choose $f$.
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