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world:useful-bounds [2022/11/17 17:41]
rdouc [Doob's inequalities]
world:useful-bounds [2022/11/18 11:14] (current)
rdouc [Doob's inequalities]
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 ==== 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}
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 \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, 
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   * We prove **(i)**. From \eqref{eq:​fond},​ we have    * We prove **(i)**. From \eqref{eq:​fond},​ we have 
 $$ $$
-\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}]=\PE\lrb{\PE[X_{\tau_\epsilon \wedge n}|\mcf_0]}\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*}
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 &\leq \sum_{k=0}^n \PE[\PE[X_n|\mcf_k] \indiacc{\tau_\epsilon=k}]=\sum_{k=0}^n \PE[X_n \indiacc{\tau_\epsilon=k}]=\PE[X_n \indiacc{\tau_\epsilon\leq n}] \leq  \PE[X_n] &\leq \sum_{k=0}^n \PE[\PE[X_n|\mcf_k] \indiacc{\tau_\epsilon=k}]=\sum_{k=0}^n \PE[X_n \indiacc{\tau_\epsilon=k}]=\PE[X_n \indiacc{\tau_\epsilon\leq n}] \leq  \PE[X_n]
 \end{align*} \end{align*}
-which completes ​the proof. ​+where we used in the last inequality that $(X_n)$ is non-negative. The proof is completed
 $\eproof$ $\eproof$
world/useful-bounds.1668703307.txt.gz · Last modified: 2022/11/17 17:41 by rdouc · Currently locked by: 114.119.142.125,10.179.80.219