Do we have $E[X_T|X_S]=X_{T\land S}$ for stopping times T,S?

Question

Let $X_t$ be a Right continuous Martingale. ($t\in (0,\infty)$). Let $T,S$ be two arbitrary stopping time. Do we have \begin{align*} E[X_T|\mathcal{F}_S] = X_{T\land S} \end{align*} By optional sampling theorem, we have $E[X_T|\mathcal{F}_{T\land S}]=X_{T\land S}$. And $\mathcal{F}_S\cap \mathcal{F}_T= \mathcal{F}_{S\land T}$. However, still cannot connect these with the statement above.

Does anyone have any idea or comments?

Thanks!

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Kind of figure out. Assume the U.I. condition (optional sampling also assumes this). So $X_{T\land t}$ is a R-Martingale, and the limit $X_T$ exists. Apply optional sampling to $X_{T\land \infty}=X_T$ we get \begin{align*} E[X_{T}|\mathcal{F}_S]=X_{T\land S} \end{align*} Guess this is the right answer.

Answer 1

The statement is not true for $X$ a Brownian motion, $T=\inf\{t\ge0\mid X_t=1\}$ and $S=0$.