Doob's martingale inequality

Remarks

Let $\tau$ be the first $k \le n$ with $X_k \ge \lambda$, and split the event $A = \{\max_k X_k \ge \lambda\}$ into the disjoint events $\{\tau = k\}$. Since $\{\tau = k\} \in \mathcal{F}_k$, the submartingale property gives $E[X_n \mathbf{1}\{\tau = k\}] \ge E[X_k \mathbf{1}\{\tau = k\}] \ge \lambda P(\tau = k)$. Summing over $k$ gives the first inequality, and $X_n \mathbf{1}_A \le X_n^+$ gives the second. If $(M_k)$ is a martingale, then $\abs{M_k}$ is a submartingale, so $P(\max_{k \le n} \abs{M_k} \ge \lambda) \le E[\abs{M_n}]/\lambda$. For $M_k$ a sum of independent centred square-integrable variables, applying this to the submartingale $M_k^2$ gives Kolmogorov's inequality. The inequality is stronger than Markov's inequality for $X_n$ alone: it controls the whole path up to time $n$ by the last value. For a non-negative submartingale and $p > 1$, it implies Doob's $L^p$ inequality $E[\max_{k \le n} X_k^p] \le \bigl(\tfrac{p}{p-1}\bigr)^p E[X_n^p]$.

Going deeper

For a student

The inequality is the basic maximal inequality of martingale theory. It gives Kolmogorov's inequality, which is a step in the strong law of large numbers and the three-series theorem. Its $L^p$ form is used to prove $L^p$ convergence of martingales and to define the norm on spaces of continuous martingales. In continuous time it holds for right-continuous submartingales, with the maximum replaced by the supremum over $[0, t]$.