This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
world:useful-bounds [2022/11/17 17:34] rdouc [Doob's inequalities] |
world:useful-bounds [2022/11/17 17:41] rdouc [Doob's inequalities] |
||
---|---|---|---|
Line 116: | Line 116: | ||
\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}]\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}]\leq \PE[X_{\tau_\epsilon \wedge 0}]=\PE[X_0] | ||
$$ | $$ | ||
- | where we used that $(X_{\tau_\epsilon \wedge n})$ is an integrable super-martingale. | + | 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 prove **(ii)**. Using \eqref{eq:fond}, | + | * 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}]\\ |