Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$
- $M$ be a local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$
- $\tau$ be an $\mathcal F$-stopping time
Is $M^\tau$ a local $\mathcal F$-martingale?
With the given assumptions, this claim can be found in the proof of Lemma 15.1 in Foundations of Modern Probability by Olav Kallenberg.
I know how we're able to prove the claim using the Optional Sampling Theorem, if $\tau$ takes only countable many values or $M$ is almost surely right-continuous, but I have no idea how I can prove the claim without one of these additional assumptions.
More concretely, he states that if $(\sigma_n)_{n\in\mathbb N}$ is an $\mathcal F$-localizing sequence for $M$, then $(M^\tau)^{\sigma_n}=(M^{\sigma_n})^\tau$ is an $\mathcal F$-martingale. So, maybe the question reduces to the question if a stopped martingale is a martingale, but, again, I'm only able to proof this with the Optional Sampling Theorem and one of its additional assumptions.
Best Answer
To prove a stopped martingale is a martingale, please use the Th.7.29, p.135 in Kallenberg's book. Since $N=M^{\sigma_n}$ is a martingale and both $N,-N$ are submartingales also, then for $t\ge s$, we have \begin{align} \mathsf{E}[(N^\tau)_t|\mathcal{F}_s]&=\mathsf{E}[N_{\tau\wedge t }| \mathcal{F}_s] \stackrel{(16)}=N_{(\tau\wedge t)\wedge s}\\ &=N_{\tau\wedge s}=(N^\tau)_s. \end{align} This means $(M^\tau)^{\sigma_n}$ is a martingale.