# About sigma-algebras, filtrations, and the tower property of conditional expectation

Let $$(X_t)$$ be a sample-continuous stochastic process and $$f$$ a real measurable bounded function.

Let $$(\mathcal F_t)$$ be the filtration generated by $$(X_t)$$ and $$(\mathcal G_t)$$ the filtration generated by $$(f(X_t))$$.

Let $$0. By the tower property of conditional expectation, we have

$$E[f(X_t)| \mathcal G_s] = E[E[f(X_t)| \mathcal F_s] | \mathcal G_s]=E[E[f(X_t)| \mathcal G_s] | \mathcal F_s].$$

Now suppose that $$(X_t)$$ is Markov w.r.t its own filtration, i.e, $$E[f(X_t)| \mathcal F_s]=E[f(X_t)| X_s]$$. This means that $$E[f(X_t)| \mathcal G_s]=E[E[f(X_t)| X_s] | \mathcal G_s]=E[E[f(X_t)| \mathcal G_s] | X_s]$$

First question: does this mean that $$\mathcal G_s \subseteq \sigma(X_s)$$ ? (this would be some sort of converse to the tower property)

Second question: does this mean that $$E[E[f(X_t)| X_s] | \mathcal G_s]=E[E[f(X_t)| X_s] | \sigma(f(X_s))]$$?

The $$\sigma$$-algebra $${\cal G}_s$$ is generated by the sets $$\{f(X_t)\in B\}$$ where $$t\le s$$ and $$B\in {\cal B}(\mathbb R)\,.$$ Because every such set can be written as $$\{X_t\in f^{-1}(B)\}$$ it is in $$\sigma(X_t)\,.$$ Since this holds for all $$t\le s$$ this shows $${\cal G}_s\subseteq{\cal F}_s\,.$$ It is not necessarily true that $${\cal G}_s\subseteq\sigma(X_s)$$ which is smaller than $${\cal F}_s\,.$$ The tower property of conditional expectations is not needed to examine such relationships.
The answer to your last question is no because $${\cal G}_s$$ is generated not only by $$f(X_s)$$ but by all $$f(X_t)$$ with $$t\le s\,.$$