119 lines
4.1 KiB
TeX
119 lines
4.1 KiB
TeX
\begin{refproof}{ceismartingale}
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By the tower property (\autoref{cetower})
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it is clear that $(\bE[X | \cF_n])_n$
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is a martingale.
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First step:
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Assume that $X$ is bounded.
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Then, by \autoref{cejensen}, $|X_n| \le \bE[|X| | \cF_n]$,
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hence $\sup_{\substack{n \in \N \\ \omega \in \Omega}} | X_n(\omega)| < \infty$.
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Thus $(X_n)_n$ is a martingale in $L^{\infty} \subseteq L^2$.
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By the convergence theorem for martingales in $L^2$ % TODO REF
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there exists a random variable $Y$,
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such that $X_n \xrightarrow{L^2} Y$.
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Fix $m \in \N$ and $A \in \cF_m$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\int_A Y \dif \bP
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&=& \lim_{n \to \infty} \int_A X_n \dif \bP\\
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&=& \lim_{n \to \infty} \bE[X_n \One_A]\\
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&=& \lim_{n \to \infty} \bE[\bE[X | \cF_n] \One_A]\\
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&\overset{A \in \cF_n}{=}& \lim_{\substack{n \to \infty\\n \ge m}} \bE[X \One_A]\\
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\end{IEEEeqnarray*}
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Hence $\int_A Y \dif \bP = \int_A X \dif \bP$ for all $m \in \N, A \in \cF_m$.
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Since $\cF = \sigma\left( \bigcup \cF_n \right)$
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this holds for all $A \in \cF$.
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Hence $X = Y$ a.s., so $X_n \xrightarrow{L^2} X$.
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Since $(X_n)_n$ is uniformly bounded, this also means
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$X_n \xrightarrow{L^p} X$.
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Second step:
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Now let $X \in L^p$ be general and define
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\[
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X'(\omega) \coloneqq \begin{cases}
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X(\omega)& \text{ if } |X(\omega)| \le M,\\
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0&\text{ otherwise}
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\end{cases}
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\]
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for some $M > 0$.
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Then $X' \in L^\infty$ and
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\begin{IEEEeqnarray*}{rCl}
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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 \vocab{regular}, \todo{Definition?}
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i.e.~$\forall \epsilon > 0 \exists k . \bP[|X|^p \in [-k,k] \ge 1-\epsilon$.
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% Take some $\epsilon > 0$ and $M$ large enough such that
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% \[
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% \int |X - X'| \dif \bP < \epsilon.
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% \]
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% Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
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% Then $X_n' \xrightarrow{L^p} X'$ by the first step.
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% It is
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% \begin{IEEEeqnarray*}{rCl}
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% \|X_n - X_n'\|_{L^p}^p &=& \bE[\bE[X - X' | \cF_n]^{p}]\\
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% &\overset{\text{Jensen}}{\le}& \bE[\bE[(X- X')^p | \cF_n]\\
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% &=& \|X - X'\|_{L^p}^p\\
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% &<& \epsilon.
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% \end{IEEEeqnarray*}
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Hence
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\[
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\|X_n - X\|_{L^p} \le |X_n - X_n'|_{L^p} + |X_n' - X'|_{L^p} + | X - X'|_{L^p} \le 3 \epsilon.
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\]
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Thus $X_n \xrightarrow{L^p} X$.
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\end{refproof}
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For the proof of \autoref{martingaleisce},
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we need the following theorem, which we won't prove here:
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\begin{theorem}[Banach Alaoglu]
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\label{banachalaoglu}
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Let $X$ be a normed vector space and $X^\ast$ its
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continuous dual.
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Then the closed unit ball in $X^\ast$ is compact
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w.r.t.~the ${\text{weak}}^\ast$ topology.
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\end{theorem}
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\begin{fact}
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We have $L^p \cong (L^q)^\ast$ for $\frac{1}{p} + \frac{1}{q} = 1$
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via
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\begin{IEEEeqnarray*}{rCl}
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L^p &\longrightarrow & (L^q)^\ast \\
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f &\longmapsto & (g \mapsto \int g f \dif d\bP)
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\end{IEEEeqnarray*}
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We also have $(L^1)^\ast \cong L^\infty$,
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however $ (L^\infty)^\ast \not\cong L^1$.
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\end{fact}
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\begin{refproof}{martingaleisce}
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Since $(X_n)_n$ is bounded in $L^p$, by \autoref{banachalaoglu},
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there exists $X \in L^p$ and a subsequence
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$(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
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\[
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\int X_{n_k} Y \dif \bP \to \int XY \dif \bP
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\]
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(Note that this argument does not work for $p = 1$,
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because $(L^\infty)^\ast \not\cong L^1$).
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Let $A \in \cF_m$ for some fixed $m$ and write
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$Y = \One_A$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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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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\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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we get the convergence.
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\end{refproof}
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