> [!definition] > > Let $\bracs{X_n}_0^\infty$ be an integrable, discrete-time [[Stochastic Process|stochastic process]] adapted to a [[Filtration|filtration]] $\bracs{\cf_n}_0^\infty$, then $\bracs{X_n}_0^\infty$ forms a $\bracs{\cf_n}_0^\infty$-**martingale** if for all $n \ge 0$, > $ > \ev\bracs{X_n|\cf_{n - 1}} = X_{n - 1} \quad (a.s) > $ > the [[Conditional Expectation|expectation]] of the future given the present is exactly the present. > > A stochastic process is a *super/sub*-martingale if the expectation of the future given the present is **less/greater** than the present. > [!definition] > > Let $(\Omega, \cf, \{\cf_t\}, \bp)$ be a filtered probability space, and $\bracs{X_t: t \ge 0}$ be a continuous-time stochastic process, progressively measurable with respect to $\{\cf_t\}$. $\{X_t\}$ is a $\{\cf_t\}$-martingale if > 1. For all $t \ge 0$, $X_t \in L^1$. > 2. The [[Conditional Expectation|CE]] $\ev[X_t|\cf_s] = X_s$ for all $0 \le s \le t < \infty$. > > If for all $0 \le s \le t < \infty$, $\ev[X_t|\cf_{s}] \le X_s$/$\ev[X_t|\cf_s] \ge X_s$ instead, then $\{X_t\}$ is a super/submartingale. > [!theorem] > > Let $\{X_t\}$ be a continuous time (super/sub) martingale with respect to $\{\cf_t\}$, and $\seq{t_n}$ be a strictly increasing sequence of times. Then $\seq{X_{t_n}}$ is a discrete time (super/sub) martingale. > [!theorem] > > Let $\bracs{X_t}$ be a continuous time $\real^d$-valued $\bracs{\cf_t}$-martingale and $\varphi: \real^d \to \real$ be a convex function. Then $\varphi(X_t)$ is a submartingale. > > If $\{X_t\}$ is a $\real$-valued submartingale and $\varphi$ is non-decreasing, then $\varphi(X_t)$ is also a submartingale. > > In particular, if $\bracs{X_t}$ is a martingale/non-negative submartingale, then $\abs{X_t}^p$ with $p \ge 1$ is also a submartingale. If $\bracs{X_t}$ is a submartingale then $(X_t - a)^+$ is also a submartingale. > [!theorem] > > Let $\bracs{X_t}$ be a continuous time $\bracs{\cf_t}$-martingale/non-negative submartingale[^1] with [[RCLL]] sample paths, and $\tau$ be a $\bracs{\cf_t}$-stopping time such that $\tau \le t$. Then $X_\tau$ is uniformly integrable. > > *Proof*. Let $n \ge 1$ and $Y^{(n)}_m = X_{m2^{-n}}$ and $\mathcal G_m = \cf_{m2^{-n}}$, then $Y^{(n)}$ is a discrete time $\bracs{\mathcal G_m}$-martingale/non-negative submartingale. Let $\tau$ be a stopping time, and assume without loss of generality that $\tau$ takes integer values and $\tau \le t$. Define $k(\tau) = \fl{2^n\tau}+1$, then $k(\tau) = m+1$ if and only if $\tau \in [m2^{-n}, (m+1)2^{-n})$. This event is in $\cf_{(m+1)2^{-n}}$ and $k(\tau)$ is a $\bracs{\mathcal G_m}$-stopping time. In addition, $Y_{k(\tau)}^{(n)} = X_{2^{-n}(\fl{2^n\tau}+1)} \to X_\tau$ as $n \to \infty$. > > Since $k(\tau) \le (\tau+1)2^{n}$, by [[Hunt's Theorem]], > $ > \abs{Y_{k(\tau)}^{(n)}} \le \ev\braks{\abs{Y_{(\tau+1)2^n}}|\mathcal G_{k(\tau)}} = \ev\braks{\abs{X_{\tau+1}}|\mathcal G_{k(\tau)}} > $ > Let $A \ge 0$, then > $ > \begin{align*} > \ev\braks{\abs{Y_{k(\tau)}^{(n)}}; \abs{Y_{k(\tau)}^{(n)}} > A} &\le \ev\braks{X_{t+1}; \abs{Y_{k(\tau)}^{(n)}} > A} \\ > &\le \ev\braks{X_{t+1}; \sup_{s \in [0, t+1]}\abs{X_s} > A} > \end{align*} > $ > By [[Fatou's Lemma]], > $ > \ev\braks{\abs{X_\tau}; \abs{X_\tau} > A} \le \limv{n}\ev\braks{\abs{Y_{k(\tau)}^{(n)}}; \abs{Y_{k(\tau)}^{(n)}} > A} \le \ev\braks{\abs{X_{t+1}}; \sup_{s \in [0, t+1]}\abs{X_s} > A} > $ > Breaking down the integral and using Doob's maximal inequality, > $ > \begin{align*} > &\ev\braks{\abs{X_{t+1}}; \sup_{t \in [0, t+1]}\abs{X_t} > A} \\ > &\le \ev\braks{\abs{X_{t+1}}; \abs{X_{t+1}} > \sqrt{A}} + \ev\braks{\abs{X_{t+1}}; \abs{X_{t+1}} \le \sqrt{A}, \sup_{s \in [0, t+1]}\abs{X_s} > A} \\ > &\le \ev\braks{\abs{X_{t+1}}; \abs{X_{t+1}} > \sqrt{A}} + \sqrt{A}\bp\bracs{\sup_{s \in [0, t+1]}\abs{X_s} > A} \\ > &\le \ev\braks{\abs{X_{t+1}}; \abs{X_{t+1}} > \sqrt{A}} + \frac{\sqrt{A}}{A}\ev\braks{\abs{X_{t+1}}} > \end{align*} > $ [^1]: Such that $|X|$ is a submartingale.