fixed doob's lp inequality

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Josia Pietsch 2023-07-17 20:56:30 +02:00
parent c27e6b5f34
commit 9884449627
Signed by untrusted user who does not match committer: jrpie
GPG key ID: E70B571D66986A2D

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@ -143,7 +143,8 @@ Now let $p \ge 1$ be not necessarily $2$.
First, we need a very important inequality:
\begin{theorem}[Doob's $L^p$ inequality]
\label{dooblp}
Suppose that $(X_n)_n$ is a sub-martingale.
Suppose that $(X_n)_n$ is a martingale
or a non-negative submartingale.
Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
denote the \vocab{running maximum}.
\begin{enumerate}[(1)]
@ -159,7 +160,7 @@ First, we need a very important inequality:
In order to prove \autoref{dooblp}, we first need
\begin{lemma}
\label{dooplplemma}
Let $p > 1$ and $X,Y$ non-negative random variable
Let $p > 1$ and $X,Y$ non-negative random variables
such that
\[
\forall \ell > 0 .~ \bP[Y \ge \ell] \le
@ -213,8 +214,10 @@ In order to prove \autoref{dooblp}, we first need
\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
\label{lec18eq2star}
\end{equation}
Since $(X_n)_n$ is a sub-martingale, $(|X_n|)_n$ is also a sub-martingale
(by \autoref{cjensen}).
We have that $(|X_n|)_n$ is a submartingale,
by \autoref{cor:convexmartingale}
in the case of $X_n$ being a martingale
and trivially if $X_n$ is non-negative.
Hence
\begin{IEEEeqnarray*}{rCl}
\bE[\One_{E_j}(|X_n| - |X_{j}|) | \cF_j]