s23-probability-theory/inputs/lecture_20.tex
2023-06-27 17:08:59 +02:00

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