> [!theorem] > > Let $\bracs{X_n}_0^\infty$ be a discrete time [[Martingale|martingale]] with respect to $\bracs{\cf_n}_1^\infty$, $T \in \nat$ and $S_1 \le S_2 \le T$ be [[Stopping Time|stopping times]], then $\ev\braks{X_{T_2}|\cf_{T_1}} = X_{T_1}$. > [!theorem] > > Let $\bracs{X_t}$ be a continuous time $\bracs{\cf_t}$-martingale/non-negative submartingale with [[RCLL]] sample paths, and $\tau_1 \le \tau_2 \le T$ be $\bracs{\cf_t}$-stopping times, then $\ev\braks{X_{\tau_2}|\cf_{\tau_1}} = X_{\tau_1}$ (martingale) or $\ev\braks{X_{\tau_2}|\cf_{\tau_1}} \ge X_{\tau_1}$ (submartingale). > > *Proof*. Let $n \in \nat$, $Y_{m}^{(n)} = X_{m2^{-n}}$, $\mathcal G_{m} = \cf_{m2^{-n}}$. Let $k(\tau_j) = \fl{2^n\tau_j}+1$, then each $k(\tau_j)$ is a $\bracs{\mathcal G_m}$-stopping time with $Y_{k(\tau_j)}^{(n)} \to X_{\tau_j}$ as $n \to \infty$. In this case, $\bracs{Y_{k(\tau_j)}^{(n)}}$ is uniformly integrable for each $n \in \nat$. By [[Martingale Convergence Theorems|MCT (II)]], $Y_{k(\tau_j)}^{(n)} \to X_{\tau_j}$ in $L^1$. By the discrete Hunt's theorem, > $ > \ev\braks{Y_{k(\tau_2)}^{(n)}|\mathcal G_{k(\tau_1)}} = (\ge) Y_{k(\tau_1)}^{(n)} > $ > Since $\cf_{\tau_1} \subset \mathcal G_{k(\tau_1)}$, by towering > $ > \begin{align*} > \ev\braks{Y_{k(\tau_2)}^{(n)}|\mathcal F_{\tau_1}} &= \ev\braks{\ev\braks{Y_{k(\tau_2)}^{(n)}|\mathcal G_{k(\tau_1)}}|\cf_{\tau_1}} \\ > &= (\ge) \ev\braks{Y_{k(\tau_1)}^{(n)}|\cf_{\tau_1}} > \end{align*} > $ > since $L^1$ convergence implies the convergence of conditional expectations, taking the limit yields the desired result.