# Question on $\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]]$

I have a question on these lecture notes: http://page.math.tu-berlin.de/~scheutzow/WT3main.pdf Page 46-47 Lemma 4.15

In this proof, we use, amongst other things that:

$$\mathbb E_{x}[1_{B_{n-1}}(X_{t_{n-1}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]\lvert \mathcal{F}_{t_{n-2}}]=\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]](*)$$

which comes from the Markov Property which states for any and bounded measurable function $$f$$ we have:

$$\mathbb E_{x}[f((X_{t+h})_{t\geq 0})\lvert \mathcal{F}_{h}]=\mathbb E_{X_{h}}[f((X_{t})_{t\geq 0})]$$

But my issue in $$(*)$$ would be that in our case the function $$f$$ is indeed: $$f(X_{t_{n-1}})=1_{B_{n-1}}(X_{t_{n-1}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]$$ and thus

$$\mathbb E_{x}[f(X_{t_{n-1}})\lvert \mathcal{F}_{t_{n-2}}]=\mathbb E_{X_{t_{n-2}}}[f(X_{t_{n-1}-t_{n-2}})]=\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}-t_{n-2}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]]$$

Note the difference in this computation compared to $$(*)$$. My question is rather why would

$$\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]$$ remain the same when evaluated under $$\mathcal{F}_{t_{n-2}}$$ even though it is a function of $$f(X_{t_{n-1}})$$. Is this a mistake or am I simply missing something?

The notation is a bit confusing (although, it is typical to these settings). Here is the same proof presented differently. First, for $$g\in \text{b}\mathcal{E}$$, a transition semigroup $$(P_t)_{t\ge 0}$$, and $$s, the Markov property is $$\mathsf{E}_x[g(X_{t})\mid \mathcal{F}_{s}]=\mathsf{E}_{X_s}[g(X_{t-s})]=P_{t-s}g(X_s) \quad(\mathsf{P}_x\text{-a.s.}),$$ which is a function of $$X_s$$. Applying this property recursively, for $$f_1,\ldots,f_n\in \text{b}\mathcal{E}$$, one has

\begin{align} &\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-1}(X_{t_{n-1}})f_n(X_{t_n})] \\ &\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-1}(X_{t_{n-1}})\mathsf{E}_x[f_n(X_{t_n})\mid \mathcal{F}_{t_{n-1}}]] \\ &\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots (f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-1}})] \\ &\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-2}(X_{t_{n-2}})\mathsf{E}_x[(f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-1}})\mid \mathcal{F}_{t_{n-2}}]] \\ &\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots (f_{n-2}P_{t_{n-1}-t_{n-2}}f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-2}})] \\ &\qquad =\ldots \\ &\qquad=\mathsf{E}_x[(f_1P_{t_2-t_1} f_{2}\ldots P_{t_{n-1}-t_{n-2}}f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_1})] \\ &\qquad= P_{t_1}[f_1P_{t_2-t_1}f_2 \ldots P_{t_{n}-t_{n-1}}f_n](x). \end{align}

For example, in the fourth line we apply the Markov property to $$g(x)=(f_{n-1}P_{t_{n}-t_{n-1}}f_n)(x) \in \text{b}\mathcal{E},$$ evaluated at $$X_{t_{n-1}}.$$