world:useful-bounds
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world:useful-bounds [2022/11/17 17:30] rdouc [Doob's inequalities] |
world:useful-bounds [2022/11/17 17:44] rdouc [Doob's inequalities] |
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| === Proof === | === Proof === | ||
| Define $\tau_\epsilon=\inf \set{k\in \nset}{X_k \geq \epsilon}$ with the convention that $\inf \emptyset =\infty$. Then, | Define $\tau_\epsilon=\inf \set{k\in \nset}{X_k \geq \epsilon}$ with the convention that $\inf \emptyset =\infty$. Then, | ||
| - | \begin{align*} | + | \begin{align} \label{eq:fond} |
| \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon}= \epsilon \PP\lr{\tau_\epsilon \leq n}&\leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}] | \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon}= \epsilon \PP\lr{\tau_\epsilon \leq n}&\leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}] | ||
| - | \end{align*} | + | \end{align} |
| - | We prove **(ii)**. Define $\tau_\epsilon=\inf \set{k\in \nset}{X_k \geq \epsilon}$ with the convention that $\inf \emptyset =\infty$. Then, | + | * We prove **(i)**. From \eqref{eq:fond}, we have |
| + | $$ | ||
| + | \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon} \leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}] \leq \PE[X_{\tau_\epsilon \wedge n}]\leq \PE[X_{\tau_\epsilon \wedge 0}]=\PE[X_0] | ||
| + | $$ | ||
| + | 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. | ||
| + | * 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}= \epsilon \PP\lr{\tau_\epsilon \leq n}&\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.txt · Last modified: 2022/11/18 11:14 by rdouc
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