Construct a continuous local martingale $(M(t))_{t \in [0,1]}$ with $M(0)=0$, $M(t)$ is not always equal to $0$, and $M(t)$ is constant with positive probability.
Is the above construction possible?
I am thinking that the $M(t)$ can divide as constant part and non-constant part, then I can show the positive part prob higher than 0.
However, I don't know how to find such.
Please provide me a example.
Let $(B_t)_{t \geq 0}$ be a martingale with continuous sample paths (e.g. a Brownian motion), and let $X$ be a random variable which is independent from $(B_t)_{t \geq 0}$ and which satisfies $$\mathbb{P}(X= 1) = \frac{1}{2} \qquad \mathbb{P}(X=0) = \frac{1}{2}.$$
If we consider
$$M_t := X \cdot B_t$$
with the filtration
$$\mathcal{F}_t := \sigma(X, B_s; s \leq t),$$
then $(M_t)_{t \geq 0}$ has all desired properties. Clearly, $(M_t)_{t \geq 0}$ has continuous sample paths, it satisfies $$\mathbb{P}(\forall t \geq 0: \, \, M_t=0)= \frac{1}{2},$$
and $(M_t,\mathcal{F}_t)_{t \geq 0}$ is a martingale since
$$\mathbb{E}(M_t \mid \mathcal{F}_s) = X \cdot \mathbb{E}(B_t \mid \mathcal{F}_s) = X \cdot B_s = M_s$$
for all $s \leq t$.