\lecture{20}{2023-06-27}{}
\begin{refproof}{ceismartingale}
    By the \yaref{cetower}
    it is clear that $(\bE[X | \cF_n])_n$
    is a martingale.

    First step:
    Assume that $X$ is bounded.
    Then, by \yaref{cjensen}, $|X_n| \le  \bE[|X| | \cF_n]$,
    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 $L^2$
    (\yaref{martingaleconvergencel2})
    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 $\sigma(X) = \bigcup \cF_n$
    this holds for all $A \in  \sigma(X)$.
    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 regular,
    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{\yaref{cjensen}}{\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 \yaref{martingaleisce},
we need the following theorem, which we won't prove here:
\begin{theorem}[Banach Alaoglu]
    \yalabel{Banach Alaoglu}{Banach Alaoglu}{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\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 \yaref{banachalaoglu},
    there exists $X \in L^p$ and a subsequence
    $(X_{n_k})_k$ such that for all $Y \in L^q$,
    where as always $\frac{1}{p} + \frac{1}{q} = 1$,
    \[
    \int X_{n_k} Y \dif \bP \to  \int XY \dif \bP
    \]
    (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 choose $Y = \One_A$.
    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}{=}& \bE[X_m \One_A].
    \end{IEEEeqnarray*}
    Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
    and by \yaref{ceismartingale},
    we get the convergence.
\end{refproof}

\begin{example}+[\vocab{Branching Process}; Exercise 10.1, 12.4]
    Let $(Y_{n,k})_{n \in \N_0, k \in \N}$ be i.i.d.~with values in $\N_0$
    such that $0 < \bE[Y_{n,k}] = m < \infty$.
    Define
    \[
    S_0 \coloneqq  1, S_n \coloneqq \sum_{k=1}^{S_{n-1}} Y_{n-1,k}
    \]
    and let $M_n \coloneqq  \frac{S_n}{m^n}$.
    $S_n$ models the size of a population.

    \begin{claim}
        $M_n$ is a martingale.
    \end{claim}
    \begin{subproof}
        We have
        \begin{IEEEeqnarray*}{rCl}
            \bE[M_{n+1} - M_n | \cF_n]
            &=& \frac{1}{m^n} \left( \frac{1}{m}\sum_{k=1}^{S_{n}}  \bE[X_{n,k}] - S_n\right)\\
            &=& \frac{1}{m^n}(S_n - S_n).
        \end{IEEEeqnarray*}
    \end{subproof}

    \begin{claim}
        $(M_n)_{n \in \N}$ is bounded in $L^2$ iff $m > 1$.
    \end{claim}
    \todo{TODO}
    \begin{claim}
        If $m > 1$ and $M_n \to M_\infty$,
        then
        \[
        \Var(M_\infty) = \sigma^2(m(m-1))^{-1}.
        \]
    \end{claim}
    \todo{TODO}
\end{example}

\subsection{Stopping Times}

\begin{definition}[Stopping time]
    \label{def:stopping-time}
    A random variable $T: \Omega \to \N_0 \cup  \{\infty\}$ on a filtered probability space $(\Omega, \cF, \{\cF_n\}_n, \bP)$ is called a \vocab{stopping time},
    if
    \[
    \{T \le  n\} \in  \cF_n
    \]
    for all $n \in \N$.
    Equivalently, $\{T = n\}  \in \cF_n$ for all $n \in \N$.
\end{definition}

\begin{example}
    A constant random variable $T = c$ is a stopping time.
\end{example}

\begin{example}[Hitting times]
    For an adapted process $(X_n)_n$
    with values in $\R$ and $A \in  \cB(\R)$, the \vocab{hitting time}
    \[
    T \coloneqq \inf \{n \in \N : X_n \in A\}
    \]
    is a stopping time,
    as
    \[
    \{T \le  n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
    \]

    However, the last exit time
    \[
    T \coloneqq  \sup \{n \in \N : X_n \in  A\}
    \]
    is not a stopping time.
\end{example}


\begin{example}
    Consider the simple random walk, i.e.
    $X_n$ i.i.d.~with $\bP[X_n = 1] = \bP[X_n = -1] = \frac{1}{2}$.
    Set $S_n \coloneqq \sum_{i=1}^{n}  X_n$.
    Then
    \[
    T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le  B\}
    \]
    is a stopping time.
\end{example}

\begin{fact}
    If $T_1, T_2$ are stopping times with respect to the same filtration,
    then
    \begin{itemize}
        \item $T_1 + T_2$,
        \item $\min \{T_1, T_2\}$ and
        \item $\max \{T_1, T_2\}$
    \end{itemize}
    are stopping times.
\end{fact}
\begin{warning}
    Note that $T_1 - T_2$ is not a stopping time.
\end{warning}

\begin{remark}
    There are two ways to look at the interaction between a stopping time $T$
    and a stochastic process $(X_n)_n$:
    \begin{itemize}
        \item The behaviour of $ X_n$ until $T$, i.e.
            \[
            X^T \coloneqq  \left(X_{T \wedge n}\right)_{n \in \N}
            \]
            is called the \vocab{stopped process}.
        \item  The value of $(X_n)_n)$ at time $T$,
            i.e.~looking at $X_T$.
    \end{itemize}
\end{remark}
\begin{example}
    If we look at a process
    \[ S_n = \sum_{i=1}^{n}  X_i \]
    for some $(X_n)_n$,
    then
    \[ S^T = (\sum_{i=1}^{T \wedge n} X_i)_n \]
    and
    \[ S_T = \sum_{i=1}^{T}  X_i. \]
\end{example}

\begin{theorem}
    If $(X_n)_n$ is a supermartingale and $T$ is a stopping time,
    then $X^T$ is also a supermartingale,
    and we have $\bE[X_{T \wedge n}] \le  \bE[X_0]$ for all $n$.
    If $(X_n)_n$ is a martingale, then so is $X^T$
    and $\bE[X_{T \wedge n}] = \bE[X_0]$.
\end{theorem}
\begin{proof}
    First, we need to show that $X^T$ is adapted.
    This is clear since
    \begin{IEEEeqnarray*}{rCl}
        X^T_n &=& X_T \One_{T < n} + X_n \One_{T \ge n}\\
              &=& \sum_{k=1}^{n-1}  X_k \One_{T = k} + X_n \One_{T \ge n}.
    \end{IEEEeqnarray*}

    It is also clear that $X^T_n$ is integrable since
    \[
    \bE[|X^T_n|] \le  \sum_{k=1}^{n} \bE[|X_k|] < \infty.
    \]

    We have
    \begin{IEEEeqnarray*}{rCl}
        &&\bE[X^T_n - X^T_{n-1} | \cF_{n-1}]\\
        &=& \bE\left[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1}  X_k \One_{\{ T = k\} }
        - X_{n-1}(\One_{T \ge  n}  + \One_{\{T = n-1\}})\right.\\
        &&\left.+ \sum_{k=1}^{n-2}  X_k \One_{\{T = k\} }  \middle| \cF_{n-1}\right]\\
        &=&  \bE[(X_n - X_{n-1}) \One_{\{ T \ge  n\} } | \cF_{n-1}]\\
        &=& \One_{\{ T \ge  n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})
        \begin{cases}
            \le 0\\
            = 0 \text{ if $(X_n)_n$ is a martingale}.
        \end{cases}
    \end{IEEEeqnarray*}
\end{proof}

\begin{remark}
    \label{roptionalstoppingi}
    We now want a similar statement for $X_T$.
    In the case that $T \le M$ is bounded,
    we get from the above that
    \[
    \bE[X_T] \overset{n \ge M}{=} \bE[X^T_n]  \begin{cases}
        \le \bE[X_0] & \text{ supermartingale},\\
        = \bE[X_0] & \text{ martingale}.
    \end{cases}
    \]
    However if $T$ is not bounded, this does not hold in general.
\end{remark}
\begin{example}
    Let $(S_n)_n$ be the simple random walk
    and take $T \coloneqq  \inf \{n : S_n = 1\}$.
    Then $\bP[T < \infty] = 1$, but
    \[
    1 = \bE[S_T] \neq  \bE[S_0] = 0.
    \]
\end{example}

\begin{theorem}[Optional Stopping]
    \yalabel{Optional Stopping Theorem}{Optional Stopping}{optionalstopping}
    Let $(X_n)_n$ be a supermartingale
    and let $T$ be a stopping time
    taking values in $\N$.

    If one of the following holds
    \begin{enumerate}[(i)]
        \item $T \le M$ is bounded,
        \item $(X_n)_n$ is uniformly bounded
            and $T < \infty$  a.s.,
        \item $\bE[T] < \infty$
            and $|X_n(\omega) - X_{n-1}(\omega)| \le K$
            for all $n \in \N, \omega \in \Omega$ and
            some $K > 0$,
    \end{enumerate}
    then $\bE[X_T] \le  \bE[X_0]$.

    If $(X_n)_n$ even is a martingale, then
    under the same conditions
    $\bE[X_T] = \bE[X_0]$.
\end{theorem}
\begin{proof}
    (i) was already done in \yaref{roptionalstoppingi}.

    (ii): Since $(X_n)_n$ is bounded, we get that
    \begin{IEEEeqnarray*}{rCl}
        \bE[|X_T - X_0|] &\overset{\text{dominated convergence}}{=}& \lim_{n \to \infty} \bE[|X_{T \wedge n} - X_0|]\\
                         &\overset{\text{part (i)}}{\le}& 0.
    \end{IEEEeqnarray*}

    (iii): It is
    \begin{IEEEeqnarray*}{rCl}
        |X_{T \wedge n}- X_0| &\le& | \sum_{k=1}^{T \wedge n}  X_k - X_{k-1}|\\
                              &\le & (T \wedge n) \cdot  K\\
                              &\le & T \cdot K < \infty.
    \end{IEEEeqnarray*}

    Hence, we can apply dominated convergence and obtain
    \begin{IEEEeqnarray*}{rCl}
        \bE[X_T - X_0] &=& \lim_{n \to \infty} \bE[X_{T \wedge n} - X_0].
    \end{IEEEeqnarray*}
    Thus, we can apply (ii).

    The statement about martingales follows from
    applying this to $(X_n)_n$ and $(-X_n)_n$,
    which are both supermartingales.
\end{proof}
\begin{remark}+
    Let $(X_n)_n$ be a supermartingale and $T$ a stopping time.
    If $(X_n)_n$ itself is not bounded,
    but  $T$ ensures boundedness,
    i.e. $T < \infty$ a.s.~and $(X_{T \wedge n})_n$
    is uniformly bounded,
    the \yaref{optionalstopping} can still be applied, as
    \[
        \bE[X_T] = \bE[X_{T \wedge T}]
        \overset{\yaref{optionalstopping}}{\le} \bE[X_{T \wedge 0}]
            = \bE[X_0].
    \]
\end{remark}