A question about the translation property Markov kernel Given that ${X_n}$ is a Markov chain, and a Markov kernel with translation property $p(y+x,E+x)=p(y,E)$.
Question: How to show $Y_n=X_n-X_{n-1}$ are i.i.d?
I'm trying to use Markov Property and Monotone Class Theorem, but haven't got any helpful result. In fact, I get in trouble when apply the translation property on integral.
 A: Fix $n \in \mathbb{N}$ and measurable sets $B_1,\ldots,B_n$. Using the tower property, we get
$$\begin{align*} \mathbb{P}(Y_1 \in B_1,\ldots,Y_n \in B_n) &= \mathbb{E} \left[ \mathbb{E} \left( \prod_{j=1}^n 1_{B_j}(Y_j) \mid \mathcal{F}_{n-1} \right) \right] \\ &= \mathbb{E} \left[ \prod_{j=1}^{n-1} 1_{B_j}(Y_j) \mathbb{E}(1_{B_n}(Y_n) \mid \mathcal{F}_{n-1}) \right].\tag{1} \end{align*}$$
By the Markov property and the translational invariance,
$$\begin{align*} \mathbb{E}(1_{B_n}(Y_n) \mid \mathcal{F}_{n-1}) &= \mathbb{E}(1_{B_n}(X_n-X_{n-1}) \mid X_{n-1}) \\ &= \mathbb{E}^{X_{n-1}}(1_{B_n}(X_1-X_0)) \\ &= p(x,x+B_n) \bigg|_{x=X_{n-1}} = p(0,B_n). \tag{2} \end{align*}$$
Taking expectation at both sides, we find in particular
$$\mathbb{P}(Y_n \in B) = p(0,B_n) = \mathbb{E}(1_{B_n}(Y_n) \mid \mathcal{F}_{n-1}). \tag{3}$$
Plugging this into $(1)$ yields
$$\mathbb{P}(Y_1 \in B_1,\ldots,Y_n \in B_n) = \mathbb{P}(Y_n \in B_n) \mathbb{P}(Y_1 \in B_1,\ldots,Y_{n-1} \in B_{n-1}).$$
By iterating the argumentation, the claim follows.
