Author: Eiko

Time: 2025-01-02 08:41:33 - 2025-01-02 08:41:33 (UTC)

Kolmogorov’s Maximal Inequality

Let \(X_i\) be i.i.d and \(S_n\) be its partial sum. Then

\[\mathbb{P}\left(\max_{1\le i\le n} |S_i|\ge \varepsilon\right) \le \frac{\mathbb{E}S_n^2}{\varepsilon^2} \]

Proof.

Let \(\Lambda = \left\{ \max_{1\le i\le n} |S_i| \ge \varepsilon \right\}\) and \(\Lambda_k = \{ |S_{1,\dots,k-1}| < \varepsilon, |S_k|\ge \varepsilon\}\). It is clear that

\[ 1_\Lambda = \sum 1_{\Lambda_k} \]