A continuous process s.t. $\mathbb E[X_{t+1}\mid \mathcal F_t]=X_t$ is a martingale? Can someone provide a example of $L^1$ continuous stochastic process s.t. $\mathbb E[X_{t+1}\mid \mathcal F_t]=X_t$ for all $t$, but $(X_t)$ is not a martingale ? I have the impression that it's a martingale, but my teacher says that it's not.
 A: A "stupid" example would be to consider a periodic function such as $X_t=sin(2\pi t)$. Obviously $\mathbb{E}[X_{t+1}\vert \mathcal{F}_t]=X_{t+1}=X_t$, but for (almost) any $\varepsilon >0$ different from $1$ you won't get $\mathbb{E}[X_{t+\varepsilon}\vert \mathcal{F}_t]=X_t$.
If you want to add some randomness in your process, consider any continuous non-constant function $f:[0,1]\rightarrow \mathbb{R}$ such that $f(0)=f(1)=0$, and a sequence of i.i.d. random variables $(Y_k)_{k\geq 0}$ such that $\mathbb{P}(Y_k=2)=\frac{1}{3}$ and $\mathbb{P}(Y_k=\frac{1}{2})=\frac{2}{3}$. Then set $X_t=Y_0f(t)$ if $t\in [0,1]$ and $X_t=Y_{\lfloor t\rfloor}X_{t-1}$ if $t>1$. By construction $\mathbb{E}[X_{t+1}\vert \mathcal{F}_t]=\left(2X_t\frac{1}{3}+\frac{X_t}{2}\frac{2}{3}\right)=X_t$, but for some $t$ and $\varepsilon <\lceil t\rceil -t$ you will have $\mathbb{E}[X_{t+\varepsilon}\vert \mathcal{F}_t]=X_{t+\varepsilon}\neq X_t$ (since we chose $f$ to be non-constant). And thus such process is continuous (you can even make it smooth) and integrable but not a martingale.
Edit : someone answered while I was writing... I should be faster next time ^^
A: Let $(Y_k)$ be a sequence of i.i.d. RVs with $\mathbb{P}(Y_k = 1) = \mathbb{P}(Y_k = -1) = 1/2$. Then define $X = (X_t)_{t\geq 0}$ by
$$ X_t = \left| \sin(\pi t) \right| \sum_{k=1}^{\lfloor t \rfloor} Y_k $$
and let $\mathcal{F}_t = \sigma(X_s : s \leq t)$ be the natural filtration. Then $X$ has continuous sample paths and satisfies
$$ X_{t+1} = X_t + \left| \sin(\pi t) \right| Y_{\lfloor t\rfloor + 1} $$
and hence $\mathbb{E}[X_{t+1}\,|\,\mathcal{F}_t] = X_t$. On the other hand, for $n \in \mathbb{Z}_{\geq 0}$ and $n < s < t < n+\frac{1}{2}$,
$$ \mathbb{E}[X_t \mid \mathcal{F}_s] = \left| \frac{\sin(\pi t)}{\sin(\pi s)} \right| X_s $$
is different from $X_s$ since $\left| \frac{\sin(\pi t)}{\sin(\pi s)} \right| > 1$. So $X$ cannot be a martingale.
