world:useful-bounds
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world:useful-bounds [2022/11/17 17:33] 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} | ||
| 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, | ||
| Line 116: | Line 121: | ||
| \end{align} | \end{align} | ||
| - | 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}] | + | \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] |
| $$ | $$ | ||
| - | We prove **(ii)**. Using \eqref{eq:fond} | + | 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}, | ||
| \begin{align*} | \begin{align*} | ||
| \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon}&\leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}]=\sum_{k=0}^n \PE[X_k \indiacc{\tau_\epsilon=k}]\\ | \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon}&\leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}]=\sum_{k=0}^n \PE[X_k \indiacc{\tau_\epsilon=k}]\\ | ||
| &\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.1668702800.txt.gz · Last modified: 2022/11/17 17:33 by rdouc
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