Mean Independence of self-contained random variable

Question

Let \begin{align} Y_t = X_t - \mathbb{E}[X_t | \mathcal{F}_t], \text{ where } \mathcal{F}_t = \{Y_{\tau}, \tau \leq t \} \end{align} Does this imply mean-independence of $X_t$ and $\mathcal{F}_t$, i.e. \begin{align} \mathbb{E}[X_t] = \mathbb{E}[X_t | \mathcal{F}_t]? \end{align}

I did not come up with a suitable approach any help would be greatly appreciated. I am not even sure if everything is well-defined here.

I came across this question due to this paper. On page 26, we need the equivalence of equation (58) and $\omega_t + \alpha_t \cdot (g^2(\omega_t) + M^4(t))$ for the iteration algorithm to be in the form of equation (8) on page 7. This implies mean-independence and is used in several other occasions as well.

Answer 1

Let us first show that $X_t=E(X_t|\mathcal F_t)$ instead of $EX_t=E(X_t|\mathcal F_t)$. Note that $Y_t$ is $\mathcal F_t$ measurable . Hence $X_t=Y_t+E(X_t|\mathcal F_t)$ is also $\mathcal F_t$ measurable. This gives $X_t=E(X_t|\mathcal F_t)$.

Now we get $Y_t=0$ so $\mathcal F_t$ is a trivial sigma field. (Every set in it has probability $0$ or $1$). This implies that any random variable, in particular $X_t$, is independent of $\mathcal F_t$ so we end up with $X_t=E(X_t|\mathcal F_t )=EX_t$.