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

$$\mathbb{P}(\underset{1\le i\le n}{max}|S_i|\ge \epsilon )\le \frac{\mathbb{E}{S}_n^2}{{\epsilon }^2}$$

Proof.

Let $\mathrm{\Lambda }=\{\underset{1\le i\le n}{max}|S_i|\ge \epsilon \}$ and ${\mathrm{\Lambda }}_k=\{|S_{1,\dots ,k-1}|<\epsilon ,|S_k|\ge \epsilon \}$. It is clear that

$$1_{\mathrm{\Lambda }}=\sum 1_{{\mathrm{\Lambda }}_k}$$