This shows you the differences between two versions of the page.
Both sides previous revision Previous revision | |||
world:useful-bounds [2022/11/17 21:23] rdouc [Doob's inequalities] |
world:useful-bounds [2022/11/18 11:14] rdouc [Doob's inequalities] |
||
---|---|---|---|
Line 123: | Line 123: | ||
* 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. The proof then follows by letting $n$ goes to infinity. | 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. |