- Let $(M_{t})_{t\geq 0}$ a local martingale.
1) Do we have that $(M_t)_{t\in [0,T]}$ is a martingale for all $T>0$ ?
2) Do we have that for all sequence of stopping time $(\tau_n)$ s.t. $\tau_n\nearrow \infty $, that $(M_{\tau_n\wedge t})_{t\geq 0}$ is a Martingale ? Or just that there exist such a sequence $(\tau_n)$ ? (the problem it's I can't find any counter example).
- Is it possible that a process $(M_{t})_{t\in[0,T]}$ is a local martingale or such a process is either a martingale either not a martingale ?
Let $W$ be a Wiener process and and $\tau = \inf\{ t : W_t = −1 \}$. Define
$$ X_t = \begin{cases} W_{\min(t/(1-t),\tau)} &\text{for } 0 \le t < 1,\\ -1 &\text{for } 1 \le t < \infty. \end{cases}$$
In this thread, there is answered that this process is a local martingale with respect to the filtration $$\tilde{\mathcal{F}}_t := \begin{cases} \sigma(W_u; u \leq t/(1-t)) & t \in [0,1) \\ \sigma(W_u; u \geq 0), & t \geq 1. \end{cases}$$ and localizing sequence $$S_n = (\frac{n}{n+1})I(\tau \geq n) + (\frac{\tau}{\tau+1}+n)I(\tau<n)$$
1) For $T>1$ we have that $(X_t)_{t\in [0,T]}$ is a local martingale, but we have that $$\Bbb E [X_t] = \begin{cases} 0 &: t< 1\\ -1 &: t\in [1,T] \end{cases}$$ Therefore $(X_t)_{t\in [0,T]}$ can't be a martingale.
2) Setting $\tau_n := n$, $n\in \Bbb N$, is a sequence of stopping times with $\tau_n \nearrow \infty$, but $(X_{\tau_n \wedge t})_{t\geq 0}$ is not a martingale, because
$$\Bbb E [X_{\tau_n \wedge t}] = \Bbb E [X_{n \wedge t}] = \begin{cases} 0 &: t< 1\\ -1 &: t \geq 1\end{cases}$$