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world:useful-bounds [2022/11/17 17:33] rdouc [Doob's inequalities] |
world:useful-bounds [2022/11/17 17:44] rdouc [Doob's inequalities] |
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\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}]\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. |
+ | * 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. |
$\eproof$ | $\eproof$ |