\lecture{16}{2023-06-13}{}

% \subsection{Conditional expectation}

\begin{refproof}{ceroleofindependence}
    Let $\cH$ be independent of $\sigma(\sigma(X), \cG)$.
    Then for all $H \in \cH$, we have that $\One_H$
    and any random variable measurable with respect to either $\sigma(X)$
    or $\cG$ must be independent.

    It suffices to consider the case of $X \ge 0$.
    Let $G \in \cG$ and $H \in \cH$.
    By assumption, $X \One_G$ and $\One_H$ are independent.
    Let $Z \coloneqq  \bE[X | \cG]$.
    Then
    \begin{IEEEeqnarray*}{rCl}
        \underbrace{\bE[X;G \cap H]}_{\coloneqq \int_{G \cap H} X \dif \bP} &=& \bE[(X \One_G) \One_H]\\
                                                                                                  &=& \bE[X \One_G] \bE[\One_H]\\
                                                                                                  &=& \bE[Z \One_G] \bP(H)\\
                                                                                                  &=& \bE[Z; G \cap H]
    \end{IEEEeqnarray*}

    The identity above means, that the measures $A \mapsto \bE[X; A]$
    and  $A \mapsto \bE[Z; A]$
    agree on the  $\sigma$-algebra $\sigma(\cG, \cH)$ for events
    of the form $G \cap H$.
    Since sets of this form generate $\sigma(\cG, \cH)$,
    these two measures must agree on  $\sigma(\cG, \cH)$.
    The claim of the theorem follows by the uniqueness of conditional expectation.

    To deduce the second statement, choose $\cG = \{\emptyset, \Omega\}$.
\end{refproof}


\subsection{The Radon Nikodym Theorem}

First, let us recall some basic facts:
\begin{fact}
    Let $(\Omega, \cF, \mu)$ be a \vocab[Measure space!$\sigma$-finite]{$\sigma$-finite
    measure space},
    i.e.~$\Omega$ can be decomposed into countably many subsets of finite measure.
    Let $f: \Omega \to [0, \infty)$ be measurable.
    Define $\nu(A) \coloneqq \int_A f \dif \mu$.
    Then $\nu$ is also a $\sigma$-finite measure on $(\Omega, \cF)$.\todo{Application of mct}
    Moreover, $\nu$ is finite iff $f$ is integrable.
\end{fact}
Note that in this setting, if $\mu(A) = 0$ it follows that $\nu(A) = 0$.

The Radon Nikodym theorem is the converse of that:
\begin{theorem}[Radon-Nikodym]
    \label{radonnikodym}

    Let $\mu$ and $\nu$ be two $\sigma$-finite measures
    on $(\Omega, \cF)$.
    Suppose
    \[
    \forall A \in  \cF . ~ \mu(A) = 0 \implies  \nu(A) = 0.
    \]
    Then
    \begin{enumerate}[(1)]
        \item there exists $Z: \Omega \to [0, \infty)$ measurable,
            such that
            \[\forall  A \in  \cF . ~ \nu(A) = \int_A Z \dif \mu.\]
        \item Such a $Z$ is unique up to equality a.e.~(w.r.t.  $\mu$).
        \item $Z$ is integrable w.r.t.~$ \mu$ iff $\nu$ is a finite measure.
    \end{enumerate}

    Such a $Z$ is called the \vocab{Radon-Nikodym derivative}.
\end{theorem}
\begin{definition}
    Whenever the property $\forall  A \in  \cF, \mu(A) = 0 \implies \nu(A) = 0$
    holds for two measures $\mu$ and $\nu$,
    we say that $\nu$ is \vocab{absolutely continuous}
    w.r.t.~$\mu$.
    This is written as $\nu \ll \mu$.
\end{definition}

With \autoref{radonnikodym} we get a very short proof of the existence
of conditional expectation:
\begin{proof}[Second proof of \autoref{conditionalexpectation}]
    Let $(\Omega, \cF, \bP)$ as always, $X \in L^1(\bP)$ and $\cG \subseteq  \cF$.
    It suffices to consider the case of $X \ge 0$.
    For all $G \in \cG$, define $\nu(G) \coloneqq  \int_G X \dif \bP$.
    Obviously, $\nu \ll \bP$ on $\cG$.
    Then apply \autoref{radonnikodym}.
\end{proof}


\begin{refproof}{radonnikodym}
    We will only sketch the proof.
    A full proof can be found in the official notes.

    \paragraph{Step 1: Uniqueness} \notes
    \paragraph{Step 2: Reduction to the finite measure case}
        \notes
    \paragraph{Step 3: Getting hold of $Z$}
    Assume now that $\mu$ and $\nu$ are two finite measures.
    Let
    \[
        \cC \coloneqq  \left\{f: \Omega \to [0,\infty] \middle| %
        \forall A \in \cF.~\int_A f \dif \mu \le  \nu(A)\right\}.
    \]

    We have $\cC \neq  \emptyset$ since $0 \in \cC$.
    The goal is to find a maximal function $Z$ in $\cC$.
    Obviously its integral will also be maximal.

    \begin{enumerate}[(a)]
        \item If $f,g \in \cC$, than $f \lor g$ (the pointwise maximum)
            s also in $\cC$.
        \item Suppose $\{f_n\}_{n \ge 1}$ is an increasing sequence in $\cC$.
            Let $f$ be the pointwise limit.
            Then $f \in \cC$.
        \item For all $f \in \cC$, we have
            \[
            \int_\Omega f \dif \mu \le \nu(\Omega) < \infty.
            \]
    \end{enumerate}

    Define $\alpha \coloneqq  \sup \{ \int f \dif \mu : f \in \cC\} \le \nu(\Omega) < \infty$.
    Let $f_n \in \cC, n\in \N$ be a sequence
    with $\int f_n \dif \mu \to \alpha$.
    Define $g_n \coloneqq \max \{f_1,\ldots,f_n\} \in \cC$.
    Applying (b), we get that the pointwise limit, $Z$,
    is an element of $\cC$.

    \paragraph{Step 4: Showing that our choice of $Z$ works}
    Define $\lambda(A) \coloneqq  \nu(A) - \int_A Z \dif \mu \ge 0$.
    $\lambda$ is a measure.
    \begin{claim}
        $\lambda = 0$.
    \end{claim}
    \begin{subproof}
        Call $G \in \cF$ \emph{good} if the following hold:
        \begin{enumerate}[(i)]
            \item $\lambda(G) - \frac{1}{k}\mu(G) > 0$.
            \item $\forall  B \subseteq G, B \in \cF. ~ \lambda(B) - \frac{1}{k}\mu(B) \ge 0$.
        \end{enumerate}
        Suppose we know that for all $A \in  \cF, k \in \N$
        we have
        $\lambda(A) \le  \frac{1}{k} \mu(A)$.
        Then $\lambda(A) = 0$ since $\mu$ is finite.

        Assume the claim does not hold.
        Then there must be some $k \in \N$, $A \in \cF$
        such that $\lambda(A) - \frac{1}{k} \mu(A) > 0$.
        Fix this $A$ and $k$.
        Then $A$ satisfies condition (i) of being good,
        but it need not satisfy (ii).

        The tricky part is to make $A$ smaller such that it also
        satisfies (ii).\notes
    \end{subproof}
\end{refproof}



\section{Martingales}

\subsection{Definition}

We have already worked with martingales, but we will define them
rigorously now.

\begin{definition}[Filtration]
    A \vocab{filtration} is a sequence $(\cF_n)$ of $\sigma$-algebras
    such that $\cF_n \subseteq \cF_{n+1}$ for all $n \ge 1$.
\end{definition}
Intuitively, we can think of a $\cF_n$ as the set of information
we have gathered up to time $n$.
Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.

\begin{definition}
    Let $(\cF_n)$ be a filtration and
    $X_1,\ldots,X_n$ be random variables such that $X_i \in L^1(\bP)$.
    Then we say that $(X_n)_{n \ge 1}$ is an $(\cF_n)_n$-\vocab{martingale}
    if the following hold:
    \begin{itemize}
        \item $X_n$ is $\cF_n$-measurable for all $n$.

           ($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ).
        \item $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{=} X_n$
            for all $n$.
    \end{itemize}

    $(X_n)_n$ is called a \vocab{sub-martingale},
    if it is adapted to $\cF_n$ but
    \[
    \bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n.
    \]
    It is called a \vocab{super-martingale}
    if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$.
\end{definition}
\begin{corollary}
    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}.
    Then $(f(X_n))_n$ is a sub-martingale.
    Likewise, if $f$ is concave, then $((f(X_n))_n$ is a super-martingale.
\end{corollary}
\begin{proof}
    Apply \autoref{cjensen}.
\end{proof}

\begin{corollary}
    If $(X_n)_n$ is a martingale,
    then $\bE[X_n] = \bE[X_0]$.
\end{corollary}


\begin{example}
    The simple random walk:

            Let $\xi_1, \xi_2, ..$ iid,
            $\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$,
            $X_n \coloneqq  \xi_1 + \ldots + \xi_n$
            and $\cF_n \coloneqq  \sigma(\xi_1, \ldots, \xi_n) = \sigma(X_1, \ldots, X_n)$.
            Then $X_n$ is $\cF_n$-measurable.
            Showing that $(X_n)_n$ is a martingale
             is left as an exercise.
\end{example}
\begin{example}
    See exercise sheet 9.
    \todo{Copy}
\end{example}
