\lecture{18}{2023-06-20}{}

Recall our key lemma for supermartingales from last time:
\[
    (b-a) \bE[U_N([a,b])] \le  \bE[(X_n - a)^-].
\] 
% TODO Ref

What happens for submartingales?
If $(X_n)_{n \in \N}$ is a submartingale, then $(-X_n)_{n \in \N}$ is a supermartingale.
Hence the same holds for submartingales, i.e.
\begin{lemma}
    A (sub-/super-) martingale bounded in $L^1$ converges
    a.s.~to a finite limit, which is a.s.~finite.
\end{lemma}

\begin{question}
    What about $L^p$ convergence of martingales?
\end{question}

\begin{example}[Branching process]
    Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with
    $\bP[Z_n = 1] = p \in (0,1)$.

    Fix $u > 0$. Let $X_0 = x > 0$.
    Define $X_{n+1} \coloneqq  u^{Z_{n+1}} X_{n}$.


    \paragraph{Exercise}
    Given $u \ge 0$, find $p = p(u)$
    such that  $(X_n)_n$ is a martingale w.r.t.~the canonical filtration.

    % TODO


    By \autoref{doobmartingaleconvergence}, there is an
    a.s.~limit $X_\infty$.
    By the SLLN, we have
    \[
    \frac{1}{n} \sum_{k=1}^{n}  Z_k \xrightarrow{a.s.} \bE[Z_1] = zp - 1.
    \] 
    Hence
    \[
        \left(\frac{X_n}{x}\right)^{\frac{1}{n}} = u^{\frac{1}{n} \sum_{k=1}^n Z_k}
        \xrightarrow{a.s.} u^{zp -1}.
    \] 
    Since $(X_n)_{n \ge 0}$ is a martingale, we must have $\bE[u^{Z_1}] = 1$.
    Hence $2p - 1 < 0$, because $u > 1$.

    Hence, if $\epsilon > 0$ is small, there exists
    $N_0(\epsilon)$ (possibly random)
    such that for all $n > N_0(\epsilon)$
    \[
        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le  u^{2p - 1}(1 + \epsilon) \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
    \]
    Hence it can not converge in $L^1$.
    % TODO Confusion

\end{example}

$L^2$ is nice, since it is a Hilbert space. So we will first
consider $L^2$.

\begin{fact}[Martingale increments are orthogonal in $L^2$ ]
    \label{martingaleincrementsorthogonal}
    Let $(X_n)_n$ be a martingale
    and let $Y_n \coloneqq  X_n - X_{n-1}$ 
    denote the \vocab{martingale increments}.
    Then for all $m \neq n$ we have that
    \[
    \langle Y_m | Y_n\rangle_{L^2} = \bE[Y_n Y_m] = 0.
    \]
\end{fact}
\begin{proof}
    Since $\bE[X_n | \cF_{n-1}] = X_{n-1}$ a.s.,
    by induction $\bE[X_n | \cF_{k}] = X_k$ a.s.~for all $k \le  n$.
    Play with conditional expectation.
    % TODO Exercise
\end{proof}

\begin{fact}[Parallelogram identity]
    % TODO
\end{fact}

\begin{theorem}
    Suppose that $(X_n)_n$ is a martingale bounded in
    $L^2$, i.e.~$\sup_n \bE[X_n^2] < \infty$.
    Then there is a random variable $X_\infty$ such that
    \[
    X_n \xrightarrow{L^2}  X_\infty.
    \] 
\end{theorem}
\begin{proof}
    Let $Y_n \coloneqq  X_n - X_{n-1}$ and write
    \[
    X_n = \sum_{j=1}^{n} Y_j.
    \] 
    We have
    \[
    \bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
    \] 
    by \autoref{martingaleincrementsorthogonal}
    (this is known as the \vocab{parallelogram identity}). % TODO Move
    In particular,
    \[
    \sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
    \] 

    Since $(X_n)_n$ is bounded in  $L^2$,
    there exists $X_\infty$ such that $X_n \xrightarrow{\text{a.s.}} X_\infty$
    by \autoref{doob}.

    It remains to show $X_n \xrightarrow{L^2} X_\infty$.
    For any $r \in \N$, consider
    \[\bE[(X_{n+r} - X_n)^2] = \sum_{j=n+1}^{n+r} \bE[Y_j^2] \xrightarrow{n \to \infty} 0\]
    as a tail of a convergent series.

    Hence $(X_n)_n$ is Cauchy, thus it converges in $L^2$.
    Since $\bE[(X_\infty - X_n)^2]$ converges to the increasing
    limit
    \[
    \sum_{j \ge n + 1}  \xrightarrow{n\to \infty} 0
    \] 
    we get $\bE[(X_\infty - X_n)^2] \xrightarrow{n\to \infty} 0$.
\end{proof}

Now let $p \ge 1$ be not necessarily $2$.
First, we need a very important inequality:
\begin{theorem}[Doob's $L^p$ inequality]
    \label{dooblp}
    Suppose that $(X_n)_n$ is a sub-martingale.
    Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
    denote the \vocab{running maximum}.
    \begin{enumerate}[(1)]
        \item Then \[ \forall  \ell > 0 .~\bP[X_n^\ast \ge  \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ell\}} |X_n| \dif \bP \le  \frac{1}{\ell} \bE[|X_n|].  \]
                (Doob's $L^1$ inequality).
        \item Fix $p > 1$. Then \[
        \bE[(X_n^\ast)^p] \le  \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p].
        \]
        (Doob's $L^p$ inequality).
    \end{enumerate}
\end{theorem}

We first need
\begin{lemma}
    \label{dooplplemma}
    Let $p > 1$ and $X,Y$ non-negative random variable
    such that
    \[
    \forall \ell > 0 .~ \bP[Y \ge \ell] \le \frac{1}{\ell} \int_{\{Y \ge \ell\} } x \dif \bP
    \]
    Then
    \[
    \bE[Y^p] \le  \left( \frac{p}{p-1} \right)^p \bE[X^p].
    \]
\end{lemma}
\begin{proof}
    First, assume $Y \in L^p$.

    Then
    \begin{IEEEeqnarray}{rCl}
        \|Y\|_{L^p}^p &=& \bE[Y^p]\\
                      &=& \int Y(\omega)^p \dif \bP(\omega)\\
                      &=&k \int_{\Omega} \left( \int_0^{Y(\omega)} p \ell^{p-1} \dif \ell \right) \dif \bP(\omega)\\
                      &\overset{\text{Fubini}}{=}& \int_0^\infty  \int_\Omega \underbrace{\One_{Y \ge \ell}\dif \bP\dif}_{\bP[Y \ge \ell]} \ell. \label{l18star}\\
    \end{IEEEeqnarray}

    We have
    \begin{IEEEeqnarray*}{rCl}
        \eqref{l18star} &\le & \int_0^\infty \frac{1}{\ell} \int_{\{Y(\omega) \ge \ell\}} \ell^p \dif \ell\\
                        &\overset{\text{Fubini}}{=}& \int_\Omega  X(\omega) \int_{0}^{Y(\omega)} p \ell^{p-2} \dif \ell\bP(\dif \omega)\\
                        &=& \frac{p}{p-1} \int X(\omega) Y (\omega)^{p-1} \bP(\dif \omega)\\
                        &\overset{\text{Hölder}}{\le}& \frac{p}{p-1} \|X\|_{L^p} \|Y\|_{p}^{p-1},
    \end{IEEEeqnarray*}
    where the assumption was used to apply Hölder.

    Suppose now $Y  \not\in L^p$.
    Then look at $Y_M = Y \wedge M$.
    Apply the case of $Y \in L^p$ and use the monotone convergence theorem.
\end{proof}

\begin{refproof}{dooblp}
    Let $E \coloneqq \{X_n^\ast \ge \ell\} = E_1 \sqcup \ldots \sqcup E_n$
    where
    \[
    E_j = \{|X_1| \le \ell, |X_2| \le \ell, \ldots, |X_{j-1}| \le \ell, |X_j| \ge \ell\}.
    \] 
    Then $\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP (\ast\ast)$.
    Since $(X_n)_n$ is a sub-martingale, $(|X_n|)_n$ is also a sub-martingale
    (by \autoref{jensen}).
    Hence
    \begin{IEEEeqnarray*}{rCl}
        \bE[\One_{E_j}(|X_n| - |X_{j}|) | \cF_j]
        &=& \One_{E_j} \bE[(|X_n| - |X_{j}|)|\cF_j]\\
        &\overset{\text{a.s.}}{\ge }& 0.
    \end{IEEEeqnarray*}
    By the law of total expectation, \autoref{totalexpectation},
    it follows that
    \[
    \bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0 (\ast\ast\ast).
    \] 
    
    Now
    \begin{IEEEeqnarray*}{rCl}
        \bP(E) &=& \sum_{j=1}^n \bP(E_j)\\
               &\overset{(\ast\ast) (\ast\ast\ast)}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
               &=& \frac{1}{\ell} \int_E |X_n| \dif \bP
    \end{IEEEeqnarray*}

    This proves the first part.

    For the second part, we apply the first part and 
    \autoref{dooplplemma} (choose $Y \coloneqq  X_n^\ast$).
\end{refproof}