Let
- $(\Omega,\mathcal A)$ be a measurable space and $\mathbb F=(\mathcal F_n)_{n\in\mathbb N_0}$ be a filtration on $(\Omega,\mathcal A)$
- $E$ be an at most countable set equipped with the discrete topology
- $X=(X_n)_{n\in\mathbb N_0}$ be a discrete Markov chain on $(\Omega,\mathcal A)$ with respect to $\mathbb F$ with values in $(E,\mathcal E)$
Let $x\in E$ be recurrent, $$\tau_x^k:=\inf\left\{n>\tau_x^{k-1}:X_n=x\right\}\;\;\;\text{with }\tau_x^0:=0\;,$$ $t_0:=x$ and $$t_k:=\tau_x^k-\tau_x^{k-1}\;\;\;$$ for $k\in\mathbb N$. How can we prove, that $(t_k)_{k\in\mathbb N_0}$ is independent and identically distributed with respect to $\operatorname P_x$?
Let $k\in\mathbb N_0$. Then, $\tau:=\tau_x^{k-1}$ is a $\mathbb F$-stopping time. Let $$\tilde X:=\left(X_{\tau+n}\right)_{n\in\mathbb N_0}\;.$$ The strong Markov property yields for all $n\in\mathbb N_0$ \begin{equation} \begin{split} \operatorname P_x\left[t_k=n\right]&=\operatorname P_x\left[\tilde X_1\ne x,\ldots,\tilde X_{n-1}\ne x\text{ and }\tilde X_n=x\right]\\ &=\operatorname P_{X_\tau}\left[X_1\ne x,\ldots,X_{n-1}\ne x\text{ and }X_n=x\right]\\ &=\operatorname P_x\left[X_1\ne x,\ldots,X_{n-1}\ne x\text{ and }X_n=x\right] \end{split}\tag 1 \end{equation} $\operatorname P_x$-almost surely, since $\operatorname P_x\left[\tau<\infty\right]=1$ by the recurrence of $x$. Since the right-hand side of $(1)$ doesn't depend on $k$, the $t_k$ are indeed identically distributed.
How can we show, that they are independent, too?