> [!theorem] > > Let $(\Omega, \cf, \bracs{\cf_n}_0^\infty, \bp)$ be a [[Filtration|filtered]] [[Probability|probability]] space and $\bracs{X_n}_0^\infty$ be a $\bracs{\cf_n}_0^\infty$-adapted integrable [[Stochastic Process|stochastic process]], then the following are equivalent: > 1. $\bracs{X_n}_0^\infty$ is a [[Martingale|supermartingale]]. > 2. For any bounded [[Stopping Time|stopping time]] $T$, and any stopping time $S$, $\ev\bracs{X_T|\cf_S} \le X_{\min(S, T)}$ [[Almost Everywhere|a.s.]] > 3. For any stopping time $T$, the process $\bracs{X_{\min(T, n)}}_{0}^\infty$ is a supermartingale. > 4. For any bounded stopping times $S$ and $T$ with $S \le T$ a.s., $\ev(X_T) \le \ev(X_S)$. > > ### First Three > > Suppose that (1) holds. Let $S$ be a stopping time and $T$ be a bounded stopping time. Let $n \in \nat$ such that $\bp\bracs{T \le n} = 1$, then > $ > X_T = X_{\min(S, T)} + \sum_{k = 0}^n(X_{k + 1} - X_k)\cdot\one_{\bracs{S \le k < T}} > $ > Let $A \in \cf_S$ and $k \in \nat$, then $A \cap \bracs{S \le k} \in \cf_k$ and $\bracs{T > k} \in \cf_k$, giving > $ > \begin{align*} > \ev\braks{(X_{k + 1} - X_k) \cdot \one_{\bracs{S \le k < T} \cap A}} &= \int \ev\bracs{(X_{k + 1} - X_k) \cdot \one_{\bracs{S \le k < T} \cap A}|\cf_k} \\ > &= \int_{\bracs{S \le k < T} \cap A} \ev\bracs{X_{k + 1} - X_k|\cf_k} \\ > &\le 0 > \end{align*} > $ > so for any $A \in \cf_S$, > $ > \int_A \ev\bracs{X_T|\cf_S} = \int_A X_T \le \int_A X_{\min(S, T)} > $ > Since this applies to every set in $\cf_S$, and $X_{\min(S, T)}$ is $\cf_S$-measurable, > $ > \ev\bracs{X_T|\cf_S} \le X_{\min(S, T)} > $ > almost surely. > > Suppose that (2) holds. Let $T$ be a stopping time, $n \in \nat$ and $S = n - 1$, then $\cf_S = \cf_{n - 1}$ and $\min{(T, n)}$ is a bounded stopping time. By assumption, > $ > \begin{align*} > \ev\bracs{X_{\min(T, n)}|\cf_{n - 1}} &= \ev\bracs{X_{\min(T, n)}|\cf_{S}} \\ > &\le X_{\min(\min(T, n), S)} \\ > &= X_{\min(T, n - 1)} > \end{align*} > $ > so $\bracs{X_{\min(T, n)}}_{0}^\infty$ is a supermartingale. > > Suppose that (3) holds. Let $T = n$, then $X_{\min(T, n)} = X_n$ and $X_{\min(T, n - 1)} = X_{n - 1}$. By assumption that $\bracs{X_{\min(T, n)}}_{0}^\infty$ is a supermartingale, > $ > \ev\bracs{X_n|\cf_{n - 1}}\le X_{n - 1} > $ > we have $\bracs{X_n}_0^\infty$ being a supermartingale as well. > > ### Optional Stopping > > Suppose that (2) holds. Let $S$ and $T$ be bounded stopping times with $S \le T$, then > $ > \begin{align*} > \ev\bracs{X_T|\cf_S} &\le X_{\min(S, T)} = X_{S} \\ > \int \ev\bracs{X_T|\cf_S} &\le \int X_S \\ > \ev(X_T)&\le \ev(X_S) > \end{align*} > $ > > Suppose that (4) holds. Let $n \ge 1$ and $A \in \cf_{n - 1}$. Choose $T = n$ and $S = (n - 1)\one_{A} + n\one_{A^c}$, then $S$ is also a stopping time. Since $S \le T$, > $ > \ev(X_n) = \ev(X_T) \le \ev(X_S) > $ > where > $ > \int X_n \le \int X_S = \int_A X_{n - 1} + \int_{A^c}X_n > $ > so > $ > \int_{A}X_n \le \int_A X_{n - 1} > $ > for all $A \in \cf_{n - 1}$. Therefore $\ev\bracs{X_n|\cf_{n - 1}} \le X_{n - 1}$ almost surely.