Example of $\mathbb{E}[X_t|\mathcal{F}_0]=X_0$ but $X_t$ is not a martingale?

Question

Edit: Some great examples in the comments and in the answer below by @NullUser: it would be great to see even more examples, if anyone can come up with more please. Happy to award a small bounty on this if incentive needed. Trying to collect as many different examples as possible.

I am looking for examples of processes that satisfy the condition that:

$$\mathbb{E}\left[X_t|\mathcal{F}_{t_0}\right]=X_{t_0}$$

But where $X_t$ is not a martingale, i.e. where for some $0<s\leq t$:

$$\mathbb{E}\left[X_t|\mathcal{F}_s\right]\neq X_s$$

I can't think of any such process at the moment, although I distinctly remember seeing examples of such processes in the past. As many interesting examples as possible would be very much appreciated here,

Thank you so much.

Answer 1

If we let $X_0 = \mu$ and $(X_t)$ be i.i.d. random variables with mean $\mu$ for $t > 0$ (it may make sense to do this in discrete time so we don't have to worry about questions like whether or not we can find uncountably many i.i.d. random variables on the same probability space), then we have $\mathbb{E}[X_t|\mathcal F_0]=\mu=X_0$, but also $\mathbb{E}[X_t|\mathcal F_s] = \mu \ne X_s$ unless we chose a trivial distribution.

Answer 2

If $\mathcal{F}_0$ is trivial, then the condition you are asking for is simply that $E[X_t] = X_0$ but $X_t$ is not a martingale.

Take any integrable process $Y_t$ with $Y_0 = 0$, and a uniform $\pm1$ valued $U$ independent of $Y_t$. Then the process $X_t:=UY_t$ has this property, and is usually not a martingale.

Indeed, $E[UY_t] = \frac12 E[Y_t] + \frac12 E[-Y_t] = 0$ for all $t$.