I have the following from a book:
Assume that $$ P_x(\tau_C \circ \theta_{(k-1)N} > N|F_{(k-1)N}) = P_{X_{(k-1)N}}(\tau_C > N). $$ Integrating over $\{ \tau_C > (k-1)N\}$ using the definition of conditional probability we have $$ P_x(\tau_C > kN) = E_x\left(\mathbf{1}\{\tau_C \circ \theta_{(k-1)N} > N\} \cdot\mathbf{1}\{\tau_C > (k-1)N\}\right) $$
I'm a bit unsure how from that equality he gets the second equality we see. I can see that the LHS of the first equality multiplied by $\mathbf{1}\{\tau_C > (k-1)N\}$ and then taking the expectation wrt. x yields the RHS of the second equality, but how does the RHS of the first equality being "integrated over" as claimed produce the LHS of the second equality?
Here, $\tau_C$ is the first hitting time of some set $C$.
Thanks.
The second equality holds for every random process, be it Markov or not, since, for every $k\geqslant1$, the events $\{ \tau_C \gt kN\}$ and $\{\tau_C \circ \theta_{(k-1)N} \gt N\}\cap\{\tau_C \gt (k-1)N\}$ coincide.
Proof:
Introducing the canonical process $(X_i)_{i\geqslant0}$, $\{ \tau_C \gt n\}=\{X_i\notin C\ \text{for every}\ 1\leqslant i\leqslant n\}$ for every nonnegative $n$ and $\{\tau_C \circ \theta_m \gt n\}=\{X_i\notin C\ \text{for every}\ m+1\leqslant i\leqslant m+n\}$ for every nonnegative $n$ and $m$. Use the first identity for $n=kN$ and $n=(k-1)N$ and the second identity for $n=N$ and $m=(k-1)N$.