Let $(X_n)_{n\in \mathbb{N}_0}$ be a Markov Chain on a discrete state space $S$ and transition semigroup $P = (p(x,y))_{x,y\in S}$. Let $C\subset S$ and let $k\in \mathbb{N}$. I'm supposed to check that the following equality holds, or correct it so that it does hold
$$P_x(X_1\notin C,...,X_k\notin C ) = P_x(X_1\notin C)\ \prod\limits_{j=2}^k E_x[P_{X_j}(\hat{X}_1\notin C)1_{\{X_j\notin C\}}] $$
with $\hat{X}\sim X$. So far my computations tell me that something is wrong here. For instance, for $k=2$.
$$ P_x(X_1\notin C, X_2\notin C ) = \sum_{y\notin C}\sum_{z\notin C}P_x(X_1=y,X_2=z)\\ = \sum_{y\notin C}\sum_{z\notin C}p(x,y)p(y,z) \\ = \sum_{y\notin C} p(x,y) P_y(X_1\notin C) \\ = E_x[P_{X_1}(\hat{X}_1\notin C)1_{\{X_1\notin C\}}]\\ = P_x(X_1\notin C)\ E_x[P_{X_1}(\hat{X}_1\notin C)|X_1\notin C] $$ It is pretty close to what was expected, but not the same. For $k=3$, what I get is even further from what is expected:
$$ P_x(X_1\notin C, X_2\notin C,X_3\notin C ) = \sum_{y\notin C}\sum_{z\notin C}P_x(X_1=y,X_2=z)\\ = \sum_{y\notin C}P_x(X_1=y, X_2\notin C,X_3\notin C )\\ = \sum_{y\notin C}P_x(X_1=y, X_2\notin C,X_3\notin C )\\ = \sum_{y\notin C}P_x(X_2\notin C,X_3\notin C| X_1=y)p(x,y) \\ = \sum_{y\notin C}P_y(X_1\notin C,X_2\notin C)p(x,y) \\ = \sum_{y\notin C}\sum_{z\notin C} p(y,z) P_z(X_1\notin C)p(x,y) $$
The last inequality holds using the computations for $k=2$.
In summary, I don't know how to get a result similar to the claimed equality. Any help would be appreciated.