\lecture{17}{2023-06-15}{}

\subsection{Doob's Martingale Convergence Theorem}


\begin{definition}[Stochastic process]
    A \vocab{stochastic process} is a collection of random
    variables $(X_t)_{t \in T}$ for some index set $T$.
    In this lecture we will consider the case $T = \N$.
\end{definition}

\begin{goal}
    What about a ``gambling strategy''?
    
    Consider a stochastic process $(X_n)_{n \in \N}$.

    Note that the increments $X_{n+1} - X_n$ can be thought of as the win
    or loss per round of a game.
    Suppose that there is another stochastic process
    $(C_n)_{n \ge 1}$  such that $C_n$ is determined
    by the information gathered up until time $n$,
    i.e.~$C_n$ is measurable with respect to $\cF_{n-1}$.
    If such process $C_n$ exists, we say that $ C_n$ is
    \vocab[Stochastic process!previsible]{previsible}.
    Think of $C_n$ as our strategy of playing the game.
    Then $C_n(X_n - X_{n-1})$ defines the win in the $n$-th game,
    while
    \[
    Y_n \coloneqq  \sum_{j=1}^n C_j(X_j - X_{j-1})
    \] 
    defines the cumulative win process.
\end{goal}
\begin{lemma}
    If $(C_n)_{n \ge 1}$ is previsible and $(X_n)_{n \ge 0}$
    is a (sub/super-) martingale
    and there exists a constant $K_n$ such that $|C_n(\omega)| \le  K_n$.
    Then $(Y_n)_{n \ge 1}$ is also a (sub/super-) martingale.
\end{lemma}
\begin{proof}
    Exercise. \todo{Copy Exercise 10.4}
\end{proof}
\begin{remark}
    The assumption of $K_n$ being constant can be weakened to
    $C_n \in  L^p(\bP)$, $X_n \in  ^q(\bP)$ with $\frac{1}{p} + \frac{1}{q} = 1$.
\end{remark}

Suppose we have $(X_n)$ adapted, $X_n \in  L^1(\bP)$,
$(C_n)_{n \ge 1}$ previsible.
We play according to the following principle:
Pick two real numbers $a < b$.
Wait until $X_n \le  a$, then start playing.
Stop playing when $X_n \ge b$.
I.e.~define
\begin{itemize}
    \item $C_1 \coloneqq  0$,
    \item $C_n \coloneqq  \One_{\{C_{n-1} = 1\}} \cdot \One_{\{X_{n-1} \le b\}} + \One_{\{C_{n-1} = 0\} } \One_{\{X_{n-1}  < a\}}$.
\end{itemize}

\begin{definition}
Fix $N \in \N$ and let
\[U_n^X([a,b]) \coloneqq \# \{\text{Upcrossings of $[a,b]$ made by $n \mapsto X_n(\omega)$ by time $n$}\},\]
i.e.~$U_n([a,b])(\omega)$ is the largest $k \in \N_0$ such that we can find a
sequence
$0 \le  s_1 < t_1 < s_2 < t_2 < \ldots < s_k < t_k \le N$
such that $X_{s_j}(\omega) < a$ and $X_{t_j}(\omega) > b$ for all $1 \le  j \le  k$.
\end{definition}
Clearly $U_N^X([a,b]) \uparrow$ as $N$ increases.
It follows that the monotonic limit $U_\infty([a,b]) \coloneqq  \lim_{N \to \infty} U_N([a,b])$ exists pointwise.
\begin{lemma} % Lemma 1
    \label{lec17l1}
    \[
    \{\omega | \liminf_{N \to  \infty} Z_N(\omega) < a < b <
    \limsup_{N \to \infty} Z_N(\omega)\} \subseteq 
    \{\omega: U^{Z}_\infty([a,b])(\omega) = \infty\}
    \] 
    for every sequence of measurable functions $(Z_n)_{n \ge 1}$.
    
\end{lemma}
\begin{lemma} % 2
    \label{lec17l2}
    Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$.
    Then \[Y_N \ge  (b -a) U_N([a,b]) - (X_N - a)^{-}.\]
\end{lemma}
\begin{proof}
    Every upcrossing of $[a,b]$ increases the  value of $Y$ by $(b-a)$,
    while the last intverval of play  $(X_n -a)^{-}$ overemphasizes the loss.
\end{proof}

\begin{lemma} %3
    \label{lec17l3}
    Suppose $(X_n)_n$ is a supermartingale.
    Then in the above setup
    \[(b-a) \bE[U_N([a,b])] \le  \bE[(X_n - a)^-].\]
\end{lemma}
\begin{proof}
    This is obvious from \autoref{lec17l2}
    and the supermartingale property.
\end{proof}
\begin{corollary}
    Let $(X_n)_n$ be a
    \vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$},
    i.e.~$\sup_n \bE[|X_n|] < \infty$.
    Then $(b-a) \bE(U_\infty) \le  |a| + \sup_n \bE(|X_n|)$.
    In particular, $\bP[U_\infty = \infty] = 0$.
\end{corollary}
\begin{proof}
    By \autoref{lec17l3}
    we have that
    \[(b-a) \bE[U_N([a,b])] \le  \bE[ | X_N| ] + |a| \le  \sup_n \bE[|X_n|] + |a|.\]
    Since $U_N(\cdot) \ge  0$ and $U_N(\cdot ) \uparrow U_\infty(\cdot )$,
    by the monotone convergence theorem
    \[
    \bE(U_n([a,b])] \uparrow \bE[U_\infty([a,b])].
    \] 
\end{proof}

Assume now, that our process $(X_n)_{n \ge 1}$ is a supermartingale
bounded in $L^1(\bP)$.
Let
\[
\Lambda \coloneqq  \{\omega | X_n(\omega) \text{ does not converge to anything in $[-\infty,\infty]$}\}.
\] 
We have
\begin{IEEEeqnarray*}{rCl}
    \Lambda &=& \{\omega | \liminf_N X_N(\omega) < \limsup_N X_N(\omega)\}\\
            &=& \{\omega | \liminf_N X_N(\omega) < a < b \limsup_N X_N(\omega)\} \\
            &=& \bigcup_{a,b \in \Q} \underbrace{\{\omega | \liminf_N X_N(\omega) < a < b < \limsup_N X_N(\omega)\}}_{\Lambda_{a,b}} \\
\end{IEEEeqnarray*}

We have $\Lambda_{a,b} \subseteq  \{\omega : U_{\infty}([a,b])(\omega) = \infty\} $ by lemma 1. % TODO REF
By lemma 3 % TODO REF
we have $\bP(\Lambda_{a,b}) = 0$, hence $\bP(\Lambda) = 0$.
Hence there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s.}  X_\infty$.

\begin{claim}
    $\bP[X_\infty \in  \{\pm \infty\}] = 0$.
\end{claim}
\begin{subproof}
    It suffices to show that $\bE[|X_\infty|] < \infty$.
    We have.
    \begin{IEEEeqnarray*}{rCl}
        \bE[|X_\infty|] &=& \bE[\liminf_{n \to \infty} |X_n|]\\
                        &\overset{\text{Fatou}}{\le }& \liminf_n \bE[|X_n|]\\
                        &\le & \sup_n \bE[|X_n|]\\
                        &<& \infty.
    \end{IEEEeqnarray*}
\end{subproof}

We have thus shown
\begin{theorem}[Doob's martingale convergence theorem]
    \label{doobmartingaleconvergence}
    \label{doob}
    Any supermartingale bounded in $L^1$ converges almost surely to a
    random variable, which is almost surely finite.
    In particular, any non-negative supermartingale converges a.s.~to a finite random variable.
\end{theorem}
The second part follows from
\begin{claim}
    Any non-negative supermartingale is bounded in $L^1$.
\end{claim}
\begin{subproof}
    We need to show $\sup_n \bE(|X_n|) < \infty$.
    Since the supermartingale is non-negative, we have $\bE[|X_n|] = \bE[X_n]$ 
    and since it is a supermartingale $\bE[X_n]  \le \bE[X_0]$.
\end{subproof}

\todo{rearrange proof}
