Bounded convergence theorem

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Josia Pietsch 2023-07-15 00:57:02 +02:00
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3 changed files with 26 additions and 8 deletions

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@ -136,10 +136,10 @@ However, some subsets can be easily described, e.g.
Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
Then, by \autoref{cjensen}, $|Y| \le \bE[ |X| | \cG]$.
Hence $\bE[|Y|] \le \bE[|X|]$.
It follows that $\bP[|Y| > k] < \delta$
for $k$ suitably large,
since $\bE[|X|] \le \infty$.
Note that $\{Y > k\} \in \cG$.
By Markov's inequality,
it follows that $\bP[|Y| > k] < \delta$
for $k > \frac{\bE[|X|]}{\delta}$.
Note that $\{|Y| > k\} \in \cG$.
We have
\begin{IEEEeqnarray*}{rCl}
\bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
@ -200,7 +200,7 @@ However, some subsets can be easily described, e.g.
\end{IEEEeqnarray*}
for all $\delta > 0$ and suitable $k$.
Hence $\bP[|X_n| \ge k] < \delta$ by Markov's inequality.
Hence $\bP[|X_n| > k] < \delta$ by Markov's inequality.
Then by \autoref{lec19f4} part (a) it follows that
\[
\int_{|X_n| > k} |X_n| \dif \bP \le \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le 2 \epsilon.
@ -232,7 +232,7 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
\begin{theorem}
\label{ceismartingale}
Let $X \in L^p$ for some $p \ge 1$
and $\bigcup_n \cF_n \to \cF$.
and $\bigcup_n \cF_n \to \cF \supseteq \sigma(X)$.
Then $X_n \coloneqq \bE[X | \cF_n]$ defines a martingale which converges
to $X$ in $L^p$.
\end{theorem}

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@ -45,7 +45,8 @@
\int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
\end{IEEEeqnarray*}
as $\bP$ is regular,
i.e.~$\forall \epsilon > 0 . ~\exists k . ~\bP[|X|^p \in [-k,k] \ge 1-\epsilon$.
i.e.~$\forall \epsilon > 0 . ~\exists k . ~
\bP[|X|^p \in [-k,k]] \ge 1-\epsilon$.
Take some $\epsilon > 0$ and $M$ large enough such that
\[
@ -111,7 +112,7 @@ we need the following theorem, which we won't prove here:
\int_A X \dif \bP
&=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\
&=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\
&\overset{\text{for }n_k \ge m}{=}& \int_{k \to \infty} \bE[X_m \One_A].
&\overset{\text{for }n_k \ge m}{=}& \lim_{k \to \infty} \bE[X_m \One_A].
\end{IEEEeqnarray*}
Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
and by \autoref{ceismartingale},

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@ -215,6 +215,23 @@ from the lecture on stochastic.
\end{subproof}
\end{refproof}
\begin{theorem}[Bounded convergence theorem]
\label{thm:boundedconvergence}
Suppose that $X_n \xrightarrow{\bP} X$ and there exists
some $K$ such that $|X_n| \le K$ for all $n$.
Then $X_n \xrightarrow{L^1} X$.
\end{theorem}
\begin{proof}
Note that $|X| \le K$ a.s.~since
\[\bP[|X| \ge K + \epsilon] \le \bP[|X_n - X| > \epsilon]
\xrightarrow{n \to \infty} 0.\]
Hence
\begin{IEEEeqnarray*}{rCl}
\int |X_n - X| \dif \bP &\le& \int_{|X_n - X| \ge \epsilon} |X_n - X| \dif \bP + \epsilon\\
&\le &2K\bP[|X_n - X| \ge \epsilon] +\epsilon.
\end{IEEEeqnarray*}
\end{proof}
\subsection{Some Facts from Measure Theory}
\begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]