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world:useful-bounds [2022/11/17 21:19]
rdouc [Maximal Kolmogorov inequality]
world:useful-bounds [2022/11/17 21:23]
rdouc [Doob's inequalities]
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 $\bproof$ $\bproof$
 +<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, 
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 \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon} \leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}] \leq \PE[X_{\tau_\epsilon \wedge n}]\leq \PE[X_{\tau_\epsilon \wedge 0}]=\PE[X_0] \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon} \leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}] \leq \PE[X_{\tau_\epsilon \wedge n}]\leq \PE[X_{\tau_\epsilon \wedge 0}]=\PE[X_0]
 $$ $$
-where we used in the second inequality that $(X_n)$ is non-negative and in the third inequality that $(X_{\tau_\epsilon \wedge n})$ is a supermartingale. ​ +where we used in the second inequality that $(X_n)$ is non-negative and in the third inequality that $(X_{\tau_\epsilon \wedge n})$ is a supermartingale. ​The proof then follows by letting $n$ goes to infinity. ​
   * We now turn to the proof of **(ii)**. Using \eqref{eq:​fond},​   * We now turn to the proof of **(ii)**. Using \eqref{eq:​fond},​
 \begin{align*} \begin{align*}
world/useful-bounds.txt · Last modified: 2022/11/18 11:14 by rdouc