When do we have $E[X_{n+1}\mid X_n] = E[X_{n+1}\mid\mathscr{F}_n]$?

Question

Under what conditions does it hold that

$$E[X_{n+1}\mid X_n] = E[X_{n+1}\mid\mathscr{F}_n]$$

if we are given a stochastic process $X = (X_n)_{n \geq 0}$ on a filtered probability space $(\Omega, \mathscr{F}, (\mathscr{F}_n)_{n \in \mathbb N}, \mathbb{P})$ where $\mathscr{F}_n = \mathscr{F}_n^X := \sigma(X_0, X_1, \ldots, X_n)$

?

I was under the impression that the equality held true only for Markov processes, but I guess there may be other conditions.

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Markov property is:

$$E[f(X_{t})\mid X_s] = E[f(X_{t})\mid\mathscr{F}_s]$$

$\forall 0 \le s \le t$ and $\forall f: \mathbb R \to \mathbb R$ bounded and measurable.

So, if $X_0, X_1, \ldots, X_n, \ldots$ is a Markov process, then we have

$$E[X_{n+1}\mid X_n] = E[X_{n+1}\mid\mathscr{F}_n]$$

but what are other sufficient conditions?

Answer 1

This is by no means a full answer to your question, but I've at least found an example of a non-markovian process that satisfies that equation. (We've run out of space in the comments).

Take the process $X$ where $X_0, X_1$ are iid positive random variables and $X_2$ is a normal random variable with variance $X_0 + X_1$ and mean zero.

Then $E[X_2|X_1] = 0$ and $E[X_2|\mathcal{F_1}]= 0$.