Bounded convergence theorem
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@ -136,10 +136,10 @@ However, some subsets can be easily described, e.g.
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Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
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Then, by \autoref{cjensen}, $|Y| \le \bE[ |X| | \cG]$.
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Hence $\bE[|Y|] \le \bE[|X|]$.
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It follows that $\bP[|Y| > k] < \delta$
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for $k$ suitably large,
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since $\bE[|X|] \le \infty$.
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Note that $\{Y > k\} \in \cG$.
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By Markov's inequality,
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it follows that $\bP[|Y| > k] < \delta$
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for $k > \frac{\bE[|X|]}{\delta}$.
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Note that $\{|Y| > k\} \in \cG$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
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@ -200,7 +200,7 @@ However, some subsets can be easily described, e.g.
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\end{IEEEeqnarray*}
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for all $\delta > 0$ and suitable $k$.
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Hence $\bP[|X_n| \ge k] < \delta$ by Markov's inequality.
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Hence $\bP[|X_n| > k] < \delta$ by Markov's inequality.
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Then by \autoref{lec19f4} part (a) it follows that
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\[
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\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.
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@ -232,7 +232,7 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
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\begin{theorem}
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\label{ceismartingale}
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Let $X \in L^p$ for some $p \ge 1$
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and $\bigcup_n \cF_n \to \cF$.
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and $\bigcup_n \cF_n \to \cF \supseteq \sigma(X)$.
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Then $X_n \coloneqq \bE[X | \cF_n]$ defines a martingale which converges
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to $X$ in $L^p$.
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\end{theorem}
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@ -45,7 +45,8 @@
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\int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
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\end{IEEEeqnarray*}
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as $\bP$ is regular,
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i.e.~$\forall \epsilon > 0 . ~\exists k . ~\bP[|X|^p \in [-k,k] \ge 1-\epsilon$.
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i.e.~$\forall \epsilon > 0 . ~\exists k . ~
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\bP[|X|^p \in [-k,k]] \ge 1-\epsilon$.
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Take some $\epsilon > 0$ and $M$ large enough such that
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\[
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@ -111,7 +112,7 @@ we need the following theorem, which we won't prove here:
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\int_A X \dif \bP
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&=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\
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&=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\
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&\overset{\text{for }n_k \ge m}{=}& \int_{k \to \infty} \bE[X_m \One_A].
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&\overset{\text{for }n_k \ge m}{=}& \lim_{k \to \infty} \bE[X_m \One_A].
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\end{IEEEeqnarray*}
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Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
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and by \autoref{ceismartingale},
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@ -215,6 +215,23 @@ from the lecture on stochastic.
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\end{subproof}
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\end{refproof}
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\begin{theorem}[Bounded convergence theorem]
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\label{thm:boundedconvergence}
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Suppose that $X_n \xrightarrow{\bP} X$ and there exists
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some $K$ such that $|X_n| \le K$ for all $n$.
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Then $X_n \xrightarrow{L^1} X$.
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\end{theorem}
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\begin{proof}
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Note that $|X| \le K$ a.s.~since
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\[\bP[|X| \ge K + \epsilon] \le \bP[|X_n - X| > \epsilon]
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\xrightarrow{n \to \infty} 0.\]
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Hence
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\begin{IEEEeqnarray*}{rCl}
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\int |X_n - X| \dif \bP &\le& \int_{|X_n - X| \ge \epsilon} |X_n - X| \dif \bP + \epsilon\\
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&\le &2K\bP[|X_n - X| \ge \epsilon] +\epsilon.
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\end{IEEEeqnarray*}
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\end{proof}
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\subsection{Some Facts from Measure Theory}
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\begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]
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