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 $0<s<t$. 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))] $?


1 Answer 1


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 $t\le s\,.$

A related question to yours is this one. It has no answer but at least some reference.

  • $\begingroup$ Someone wants to comment instead of just downvoting the answer? $\endgroup$
    – Kurt G.
    Feb 15, 2022 at 10:15
  • $\begingroup$ Your solution is correct. I also gave OP a few answers before to no avail (either no upvote or downvote as well). $\endgroup$
    – William M.
    Feb 18, 2022 at 19:02

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