Does the stopping time sequence given by the hitting time of $n$ reduce any continuous local martingale?

Question

I stumbled upon this, but it doesn't feel right. Is the following true?

Let $X$ be a cont. local martingale, $S_n= \inf \{ t | |X_t|\geq n \} $. Then $X^{S_n}$ is a martingale for each $n\in \mathbb{N}$.

The reasoning being that $$\mathbb{E}[\sup_{0\leq s \leq t} |X^{S_n}_t|]\leq n<\infty \text{ for each $t\geq 0$}.$$

So is this true? One thing I could see going wrong is having $S_n{\uparrow} \infty $ not hold. Any assistance would be appreciated!

Answer 1

Fix $n \in \mathbb{N}$. Let $Y_t = X_t^{S_n}$; then $Y_t$, being a stopped local martingale is a local martingale. It is, moreover, a bounded local martingale, which renders it into a true martingale. To see this, let $(\tau_k)$ be a localising sequence for $Y_t$ and note that for each $k$, $E(Y_{t \land \tau_k} \, | \, \mathcal{F}_s) = Y_{s \land \tau_k}$. Since $Y$ is bounded, we may use dominated convergence to get:

$$Y_s = \lim_{k\to \infty} Y_{s \land \tau_k} = \lim_{k\to \infty} E(Y_{t \land \tau_k} \, | \, \mathcal{F}_s) = E(Y_{t} \, | \, \mathcal{F}_s)$$

as desired.