On the filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t\geq 0}, P)$, let $Z=(Z_t)_{t\geq 0}$ be a continuous-time local martingale (assume $Z_0=0$).
i.e. there exists a sequence of increasing $\mathbb{R}_+$-valued $(\mathcal{F}_t)_{t\geq 0}$-stopping times $(T_n)_{n\geq 1}$ with $T_n \uparrow \infty$ a.s. such that $Z^{T_n}$ is an $(\mathcal{F}_t)_{t\geq 0}$-martingale for all $n\geq 1$.
Question:
- Let $\mathcal{T}=(t_n)_{n\geq 0}\subseteq \mathbb{R}_+$ be any increasing sequence such that $t_n \uparrow \infty$. Is it true that $\tilde{Z}=(Z_{t})_{t \in \mathcal{T}}$ is a local martingale over $(\Omega, \mathcal{F}, (\mathcal{F}_{t})_{t \in \mathcal{T}}, P)$?
i.e. does there exist a sequence of increasing $\mathcal{T}$-valued $(\mathcal{F}_{t})_{t \in \mathcal{T}}$-stopping times $(S_n)_{n\geq 1}$ with $S_n \uparrow \infty$ a.s. such that $\tilde{Z}^{S_n}$ is an $(\mathcal{F}_{t})_{t \in \mathcal{T}}$-martingale for all $n\geq 1$?
- If the answer to 1 is false in general, then does there at least exist some discretization $\mathcal{T}=(t_n)_{n\geq 0}\subseteq \mathbb{R}_+$ such that the result is true?
If $Z$ is a martingale, then the answer is clearly yes. When $Z$ is only a local martingale, it seems natural to define $S_n := t_0$ when $T_n \leq t_0$ and $S_n := t_i$ when $T_n \in (t_{i-1}, t_i]$ for $i\geq 1$, for all $n\geq 1$. The $(S_n)_{n\geq 1}$ defined in this way would be an increasing $\mathcal{T}$-valued $(\mathcal{F}_{t})_{t \in \mathcal{T}}$-stopping times with $S_n \uparrow \infty$ a.s., but it does not seem clear why $\tilde{Z}^{S_n}$ needs to be a martingale.