This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
world:useful-bounds [2022/11/17 17:44] rdouc [Doob's inequalities] |
world:useful-bounds [2022/11/17 21:21] rdouc [Maximal Kolmogorov inequality] |
||
---|---|---|---|
Line 74: | Line 74: | ||
==== Maximal Kolmogorov inequality ==== | ==== Maximal Kolmogorov inequality ==== | ||
+ | <WRAP center round tip 80%> | ||
Let \((M_k)_{k\in\nset}\) be a square integrable \((\mcf_k)_{k\in\nset}\)-martingale. Then, | Let \((M_k)_{k\in\nset}\) be a square integrable \((\mcf_k)_{k\in\nset}\)-martingale. Then, | ||
\begin{equation} | \begin{equation} | ||
Line 79: | Line 80: | ||
\end{equation} | \end{equation} | ||
- | $\bproof$ | + | </WRAP> |
+ | $\bproof$ | ||
+ | <note tip> | ||
+ | We provide a complete proof here but actually, we can also apply Doob's inequality below to the non-negative submartingale $M_n^2$ | ||
+ | </note> | ||
Let \(\sigma=\inf\set{k\geq 1}{|M_k| \geq \alpha}\) with the convention that $\inf \emptyset=\infty$. | Let \(\sigma=\inf\set{k\geq 1}{|M_k| \geq \alpha}\) with the convention that $\inf \emptyset=\infty$. | ||
Then, | Then, |