> [!theorem] > > Let $(\Omega, \cf, \bp, \bracs{\cf_t})$ be a [[Filtration|filtered]] [[Probability|probability]] space and $\bracs{X_t}$ be a $\bracs{\cf_t}$-(sub)[[Martingale|martingale]] with [[RCLL]] sample paths. > > Let $\eps > 0$ and $T > 0$, then > $ > \bp\bracs{\sup_{t \in [0, T]}X_t > \eps} \le \frac{1}{\eps}\ev\braks{X_T: \sup_{t \in [0, t]}X_t > \eps} > $ > *Proof*. Let $D_n = \bracs{m2^{-n}T: 0 \le m \le 2^n}$, $Y_m^{(n)} = X_{m2^{-n}t}$ Then $\{Y_m^{(n)}\}$ is a (sub)martingale with respect to $\bracs{\cf_q: q \in D_n}$. By RCLL sample paths, > $ > \bp\bracs{\max_{q \in D_n}X_{q} > \eps} \upto \bp\bracs{\sup_{t \in [0, t]}X_t > \eps} \quad n \to \infty > $ > by the discrete Doob's maximal inequality, > $ > \bp\bracs{\sup_{t \in [0, t]}X_t > \eps} \le \limv{n}\frac{1}{\eps}\ev\braks{X_T: \max_{q \in D_n}X_q > \eps} = \frac{1}{\eps}\ev\braks{X_T: \sup_{t \in [0, t]}X_t > \eps} > $ > [!theorem] > > If in addition to the previous assumptions, $X$ is a martingale or a non-negative submartingale (such that $\abs{X}$ is a submartingale), and uniformly integrable, then for every $\eps > 0$, > $ > \bp\bracs{\sup_{t \ge 0}\abs{X_t} > \eps} \le \frac{1}{\eps}\ev\bracs{\abs{X_\infty}; \sup_{t \ge 0}\abs{X_t} > \eps} > $ > *Proof*. Let $\eps > 0$, then > $ > \begin{align*} > \bp\bracs{\sup_{t \ge 0}\abs{X_t} \ge \eps} &= \limv{T}\bp\bracs{\sup_{t \in [0, T]}\abs{X_t} > \eps} \\ > &= \limv{T}\frac{1}{\eps}\ev\braks{\abs{X_T}: \sup_{t \in [0, T]}\abs{X_t} > \eps} \\ > &\le \limv{T}\frac{1}{\eps}\ev\braks{\abs{X_\infty}: \sup_{t \in [0, T]}\abs{X_t} > \eps} \\ > &\le \frac{1}{\eps}\ev\braks{\abs{X_\infty}: \sup_{t \ge 0}\abs{X_t} > \eps} > \end{align*} > $