fixed some typos in lecture 18

inputs/lecture_18.tex

@@ -19,8 +19,8 @@ Hence the same holds for submartingales, i.e.
     What about $L^p$ convergence of martingales?
 \end{question}
 
-\begin{example}[\vocab{Branching process}]
+\begin{example}[A martingale not converging in $L^1$ ]
-    Fix $u > 0$ and let $p = \frac{1}{1+u}$.
+    Fix $u > 1$ and let $p = \frac{1}{1+u}$.
     Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with
     $\bP[Z_n = 1] = p$.
 
@@ -38,28 +38,27 @@ Hence the same holds for submartingales, i.e.
 
     By \autoref{doobmartingaleconvergence},
     there exists an a.s.~limit $X_\infty$.
-    By the SLLN, we have
+    By the SLLN, we have almost surely
     \[
     \frac{1}{n} \sum_{k=1}^{n}  Z_k \xrightarrow{a.s.} \bE[Z_1] = 2p - 1.
     \]
     Hence
     \[
         \left(\frac{X_n}{x}\right)^{\frac{1}{n}} = u^{\frac{1}{n} \sum_{k=1}^n Z_k}
-        \xrightarrow{a.s.} u^{zp -1}.
+        \xrightarrow{\text{a.s.}} u^{2p -1}.
     \]
-    Since $(X_n)_{n \ge 0}$ is a martingale, we must have $\bE[u^{Z_1}] = 1$.
+    Since $(X_n)_{n \ge 0}$ is a martingale, we have $\bE[u^{Z_1}] = 1$.
     Hence $2p - 1 < 0$, because $u > 1$.
-
+    Choose $\epsilon > 0$ small enough such that $u^{2p - 1}(1 + \epsilon) < 1$.
-    Hence, if $\epsilon > 0$ is small, there exists
+    Then there exists $N_0(\epsilon)$ (possibly random)
-    $N_0(\epsilon)$ (possibly random)
+    such that for all $n > N_0(\epsilon)$ almost
-    such that for all $n > N_0(\epsilon)$
     \[
-        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le  u^{2p - 1}(1 + \epsilon) %
+        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \overset{\text{a.s.}}{\le}
-        \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
+        u^{2p - 1}(1 + \epsilon) %
+        \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow[n \to \infty]{\bP} 0.
     \]
-    Thus it can not converge in $L^1$.
+    However, $X_n$ cannot converge to $0$ in $L^1$,
-    % TODO Make this less confusing
+    as $\bE[X_n] = \bE[X_0] = x > 0$.
-
 \end{example}
 
 $L^2$ is nice, since it is a Hilbert space. So we will first
@@ -67,7 +66,7 @@ consider $L^2$.
 
 \begin{fact}[Martingale increments are orthogonal in $L^2$ ]
     \label{martingaleincrementsorthogonal}
-    Let $(X_n)_n$ be a martingale
+    Let $(X_n)_n$ be a martingale with $X_n \in L^2$ for all $n$
     and let $Y_n \coloneqq  X_n - X_{n-1}$
     denote the \vocab{martingale increments}.
     Then for all $m \neq n$ we have that
@@ -76,10 +75,18 @@ consider $L^2$.
     \]
 \end{fact}
 \begin{proof}
+    As $\bE[Y_n^2] = \bE[X_n^2] - 2\bE[X_nX_{n-1}] + \bE[X_{n-1}^2] < \infty$,
+    we have $Y_n \in L^2$.
     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.
+    In particular $\bE[Y_n | \cF_k] = 0$ for $k < n$.
-    \todo{Exercise}
+    Suppose that $m < n$.
+    Then
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[Y_n Y_m] &=& \bE[\bE[Y_n Y_m | \cF_m]]\\
+                     &=& \bE[Y_m \bE[Y_n | \cF_m]]\\
+                     &=& 0
+    \end{IEEEeqnarray*}
 \end{proof}
 
 \begin{fact}[\vocab{Parallelogram identity}]
@@ -108,8 +115,7 @@ consider $L^2$.
     \[
     \bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
     \]
-    by \autoref{martingaleincrementsorthogonal}
+    by \autoref{martingaleincrementsorthogonal}.
-    % (this is known as the \vocab{parallelogram identity}). % TODO how exactly is this used here?
     In particular,
     \[
     \sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
@@ -128,7 +134,7 @@ consider $L^2$.
     Since $\bE[(X_\infty - X_n)^2]$ converges to the increasing
     limit
     \[
-    \sum_{j \ge n + 1}  \xrightarrow{n\to \infty} 0
+        \sum_{j \ge n + 1} \bE[Y_j^2] \xrightarrow{n\to \infty} 0
     \]
     we get $\bE[(X_\infty - X_n)^2] \xrightarrow{n\to \infty} 0$.
 \end{proof}
@@ -156,7 +162,8 @@ In order to prove \autoref{dooblp}, we first need
     Let $p > 1$ and $X,Y$ non-negative random variable
     such that
     \[
-    \forall \ell > 0 .~ \bP[Y \ge \ell] \le \frac{1}{\ell} \int_{\{Y \ge \ell\} } x \dif \bP
+    \forall \ell > 0 .~ \bP[Y \ge \ell] \le
+      \frac{1}{\ell} \int_{\{Y \ge \ell\} } X \dif \bP
     \]
     Then
     \[
@@ -168,17 +175,24 @@ In order to prove \autoref{dooblp}, we first need
 
     Then
     \begin{IEEEeqnarray}{rCl}
-        \|Y\|_{L^p}^p &=& \bE[Y^p]\\
+        \|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)\\
+         &=&\int_{\Omega} \left( \int_0^{Y(\omega)} p \ell^{p-1} \dif \ell
-                      &\overset{\text{Fubini}}{=}& \int_0^\infty  \int_\Omega \underbrace{\One_{Y \ge \ell}\dif \bP\dif}_{\bP[Y \ge \ell]} \ell. \label{l18star}\\
+          \right) \dif \bP(\omega)\\
+         &\overset{\text{Fubini}}{=}&
+           \int_0^\infty p \ell^{p-1}\underbrace{\int_\Omega \One_{Y \ge \ell}\dif \bP}_%
+             {\bP[Y \ge \ell]} \dif\ell. \label{l18star}
     \end{IEEEeqnarray}
 
-    We have
+    By the assumption it follows that
     \begin{IEEEeqnarray*}{rCl}
-        \eqref{l18star} &\le & \int_0^\infty \frac{1}{\ell} \int_{\{Y(\omega) \ge \ell\}} \ell^p \dif \ell\\
+        \eqref{l18star}
-                        &\overset{\text{Fubini}}{=}& \int_\Omega  X(\omega) \int_{0}^{Y(\omega)} p \ell^{p-2} \dif \ell\bP(\dif \omega)\\
+         &\le& \int_0^\infty p \ell^{p-2}
-                        &=& \frac{p}{p-1} \int X(\omega) Y (\omega)^{p-1} \bP(\dif \omega)\\
+         \int_{\{Y(\omega) \ge \ell\}} X(\omega) \bP(\dif \omega)\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_{\omega} 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.
@@ -225,3 +239,4 @@ In order to prove \autoref{dooblp}, we first need
     For the second part, we apply the first part and
     \autoref{dooplplemma} (choose $Y \coloneqq  X_n^\ast$).
 \end{refproof}
+\todo{Branching process}