Doob L1 typo

This commit is contained in:
Josia Pietsch 2023-07-16 01:15:14 +02:00
parent 5691d2c553
commit 3e816f515e
Signed by untrusted user who does not match committer: josia
GPG key ID: E70B571D66986A2D
3 changed files with 17 additions and 4 deletions

View file

@ -8,10 +8,21 @@ Exercise 4.3
10.2 10.2
Martingales converging a.s.~but not in $L^1$. \begin{example}[Martingale not converging in $L^1$]
Let $\Omega = [0,1]$, $\bP = \lambda\upharpoonright [0,1]$.
Define $X_n \coloneqq 2^n \cdot \One_{[0,2^n]}$,
and let $(\cF_n)_n$ be the canonical filtration.
Then $(X_n)_{n}$ is a Martingale
with $\bE[X_0] = 1$,
but $X_n \xrightarrow{a.s.} 0$.
\end{example}
Stopping times Stopping times
\begin{example}[{Martingale such that $\bE[X_T] \neq \bE[X_0]$}]
Consider the simple random walk and $T = \inf \{n : X_n \ge 1\}$.
Obviously $X_T = 1$.
\end{example}

View file

@ -193,7 +193,9 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
\end{definition} \end{definition}
\begin{corollary} \begin{corollary}
Suppose that $f: \R \to \R$ is a convex function such that $f(X_n) \in L^1(\bP)$. Suppose that $f: \R \to \R$ is a convex function such that $f(X_n) \in L^1(\bP)$.
Suppose that $(X_n)_n$ is a martingale\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}. Suppose that $(X_n)_n$ is a martingale%
\footnote{In this form it means, that there is some filtration,
that we don't explicitly specify}.
Then $(f(X_n))_n$ is a sub-martingale. Then $(f(X_n))_n$ is a sub-martingale.
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}

View file

@ -140,7 +140,7 @@ First, we need a very important inequality:
Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$ Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
denote the \vocab{running maximum}. denote the \vocab{running maximum}.
\begin{enumerate}[(1)] \begin{enumerate}[(1)]
\item Then \[ \forall \ell > 0 .~\bP[X_n^\ast \ge \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ell\}} |X_n| \dif \bP \le \frac{1}{\ell} \bE[|X_n|]. \] \item Then \[ \forall \ell > 0 .~\bP[X_n^\ast \ge \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ge \ell\}} |X_n| \dif \bP \le \frac{1}{\ell} \bE[|X_n|]. \]
(Doob's $L^1$ inequality). (Doob's $L^1$ inequality).
\item Fix $p > 1$. Then \[ \item Fix $p > 1$. Then \[
\bE[(X_n^\ast)^p] \le \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p]. \bE[(X_n^\ast)^p] \le \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p].