This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
world:useful-bounds [2022/11/17 21:19] rdouc [Maximal Kolmogorov inequality] |
world:useful-bounds [2022/11/18 11:14] (current) rdouc [Doob's inequalities] |
||
---|---|---|---|
Line 83: | Line 83: | ||
$\bproof$ | $\bproof$ | ||
+ | <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 121: | 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. | + | 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}, | * We now turn to the proof of **(ii)**. Using \eqref{eq:fond}, | ||
\begin{align*} | \begin{align*} |