lecture 18

inputs/lecture_17.tex

@@ -48,7 +48,7 @@ Stop playing when $X_n \ge b$.
 I.e.~define
 \begin{itemize}
     \item $C_1 \coloneqq  0$,
-    \item $C_n \coloneqq  \One_{\{C_{n-1} = 1\}} \cdot \One_{\{X_{n-1} \le b\}} + \One_{\{C_{n-1} = 0\} } \One_{\{X_{n-1}\}  < a}$.
+    \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}
@@ -139,6 +139,7 @@ Hence there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s.
 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.
@@ -151,16 +152,6 @@ The second part follows from
     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}
-
-
-
-\begin{example}[Branching process]
-    % TODO
-
-\end{example}
-
-

inputs/lecture_18.tex

@@ -0,0 +1,214 @@
+\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 $n > N_0(\epsilon)$
+    \[
+        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le  u^{2p - 1}(1 + \epsilon) \implies x [\underbrace_{< 1}{u^{2p - 1} (1+\epsilon)}]^n \xrightarrow{n \to \infty} 0.
+    \] 
+    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}

probability_theory.tex

@@ -41,6 +41,7 @@
 \input{inputs/lecture_15.tex}
 \input{inputs/lecture_16.tex}
 \input{inputs/lecture_17.tex}
+\input{inputs/lecture_18.tex}
 
 \cleardoublepage