0.1 The Maximal Inequality
If \((X_{n},\mathcal{F}_{n})\) is a submartingale, \(A=\{ \max_{0\leq m \leq n}(X_{m})\geq \lambda \},~\lambda>0\), then 1. \(\lambda \mathbb{P}(\max_{0 \leq m \leq n}(X_{m})\geq \lambda)\leq \mathbb{E}[X_{n}^+ \mathbb{1}_{A}]\leq \mathbb{E}[X_{n}^+].\) 2. \(\mathbb{E}[(\max_{0\leq m \leq n}(X_{m}^+))^p]\leq \left( \frac{p}{p-1} \right)^p \mathbb{E}[(X_{n}^+)^p]\) for \(p>1\). 3. The Minimal Inequality: For submartingale \((X_{n}, \mathcal{F}_{n})\), \(\lambda \mathbb{P}(\min_{0\leq m \leq n}(X_{m})\leq -\lambda)\leq \mathbb{E}X_{n}^+ - \mathbb{E}X_{0}\), \(\lambda>0\). 4. Doob-Kolmogorov Inequality for Martingales: Let \((X_{n}, \mathcal{F}_{n})\) be an \(L_{2}\) martingale: \(\mathbb{E}X^2<\infty\) for all \(n\). Then \[ \mathbb{P}\left(\max_{0 \leq m \leq n}(X_{m}^2)\geq \lambda^2\right) \leq \frac{\mathbb{E}X_{n}^2}{\lambda^2}. \]
\begin{proof} For the proof see [[files-notes-pstat213bc-lecture-notes-raya-pdf|PSTAT213BC Lecture Notes (Raya), page 102]] \end{proof}