Theorem (Kolmogorov's martingale inequality)   Let $X(t)$ , for $0 \leq t \leq T$ , be a submartingale with continuous sample paths. Then for any constant $\alpha > 0$ ,$$ \PP \Bigl( \max_{0 \leq t \leq T} X(t) \geq \alpha \Bigr) \leq \frac{\E[ X(T)^+ ]}{\alpha}\,.$$

Proof. Let $\{ t_i \}_{i=1}^n$ be a partition of the interval $[0,T]$ . Let$$ B = \Bigl\{ \max_{1 \leq i \leq n} X(t_i) \geq \alpha \Bigr\}$$ and split $B$ into disjoint parts $B_i$ , defined by$$ B_i = \Bigl\{ X(t_j) < \alpha \text{ for all } j < i \text{ but } X(t_i) \geq \alpha \Bigr\}\,.$$ Also let $\{ \sF_t \}$ be the filtration under which $X(t)$ is a submartingale.

Then

Since the sample paths are continuous by hypothesis, the event$$ A = \Bigl\{ \max_{0 \leq t \leq T} X(t) \geq \alpha \Bigr\}$$ can be expressed as an countably infinite intersection of events of the form $B$ with finer and finer partitions $\{ t_i \}$ of the time interval $[0,T]$ . By taking limits, it follows $\PP(A)$ has the same bound as the probabilities $\PP(B)$ .

Corollary   Let $X(t)$ , for $0 \leq t \leq T$ , be a square-integrable martingale possessing continuous sample paths, whose unconditional mean is $m = \E[X(0)]$ . For any constant $\alpha > 0$ ,$$ \PP \Bigl( \max_{0 \leq t \leq T} \abs{X(t)-m} \geq \alpha \Bigr) \leq \frac{\Var[X(T)]}{\alpha^2}\,.$$