lecture 19

inputs/lecture_18.tex

@@ -51,7 +51,7 @@ Hence the same holds for submartingales, i.e.
     $N_0(\epsilon)$ (possibly random)
     such that for all $n > N_0(\epsilon)$
     \[
-        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le  u^{2p - 1}(1 + \epsilon) \implies x [\underbrace_{< 1}{u^{2p - 1} (1+\epsilon)}]^n \xrightarrow{n \to \infty} 0.
+        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le  u^{2p - 1}(1 + \epsilon) \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
     \]
     Hence it can not converge in $L^1$.
     % TODO Confusion
@@ -132,7 +132,7 @@ First, we need a very important inequality:
     Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
     denote the \vocab{running maximum}.
     \begin{enumerate}[(1)]
-        \item Then \[ \forall  \ell > 0 .~\bP[X_n^\ast \ge  \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ell\} |X_n| \dif \bP \le  \frac{1}{\ell} \bE[|X_n|].  \]
+        \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].
@@ -167,7 +167,7 @@ We first need
 
     We have
     \begin{IEEEeqnarray*}{rCl}
-        \eqref{l18star} &\le & \int_0^\infty \frac{1}{\ell} \int_{\{Y(\omega) \ge \ell\} \ell^p \dif \ell\\
+        \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},

inputs/lecture_19.tex

@@ -0,0 +1,238 @@
+\lecture{19}{2023-06-22}{}
+
+\subsection{Uniform integrability}
+
+\begin{example}
+    Let $\Omega = [0,1]$, $\cF = \cB$
+    and $\bP = \lambda \defon{[0,1]}$.
+    Consider $X_n \coloneqq  n \One_{(0,\frac{1}{n}))}$.
+    We know that $X_n \xrightarrow{n \to \infty} 0$ a.s.,
+    however $\bE[X_n] = \bE[|X_n|] = 1$,
+    hence $X_n$ does not converge in $L^1(\bP)$.
+
+    Let $\mu_n(\cdot ) = \bP[X_n \in \cdot ]$.
+
+    Intuitively, for a series that converges in probability,
+    for $L^1$-convergence to hold we somehow need to make sure
+    that probability measures don't assign mass far away from $0$.
+    This will be made precise in the notion of uniform integrability.
+\end{example}
+\begin{goal}
+    We want to show that uniform integrability and convergence in probability
+    is equivalent to convergence in $L^1$.
+\end{goal}
+
+\begin{definition}
+    A sequence of random variables $(X_n)_n$ is called \vocab{uniformly integrable} (UI),
+    if
+    \[\forall  \epsilon > 0 .~\exists  k > 0 .~ \forall n.~\bE[|X_n| \One_{\{|X_n > k\} }] < \epsilon.\]
+
+    Similarly, we define uniformly integrable for sets of random variables.
+\end{definition}
+\begin{example}
+    $X_n \coloneqq n \One_{(0,\frac{1}{n})}$ is not uniformly integrable.
+\end{example}
+There is no nice description of uniform integrability.
+However, some subsets can be easily described, e.g.
+\begin{fact}\label{lec19f1}
+    If $(X_n)_{n \ge 1}$ is a sequence bounded in $L^{1 + \delta}(\bP)$
+    for some $\delta > 0$ (i.e.~$\sup_n \bE[|X_n|^{1+\delta}] < \infty$),
+    then $(X_n)_n$ is uniformly integrable.
+\end{fact}
+\begin{proof}
+    Let $\epsilon > 0$.
+    Let $p \coloneqq  1 + \delta > 1$.
+    Choose $q$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
+    Then
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[|X_n| \One_{|X_n| > k}] &\le& \bE[|X_n|^p]^{\frac{1}{p}} \bP[|X_n| > k]^{\frac{1}{q}}\\
+    \end{IEEEeqnarray*}
+    i.e.
+    \begin{IEEEeqnarray*}{rCl}
+        \sup_n\bE[|X_n| \One_{|X_n| > k}] &\le& \underbrace{\sup_n\bE[|X_n|^p]^{\frac{1}{p}}}_{< \infty} \sup_n \underbrace{\bP[|X_n| > k]^{\frac{1}{q}}}_{\le k^{\frac{1}{q}} \bE[|X_n|]^{\frac{1}{q}}}\\
+    \end{IEEEeqnarray*}
+    where we have applied Markov's inequality. % TODO REF
+
+    Since $\sup_n \bE[|X_n|^{1+\delta}] < \infty$,
+    we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen. % TODO REF
+    Hence, choose $k$ large enough to make the relevant
+    term less than $\epsilon$.
+\end{proof}
+
+\begin{fact}\label{lec19f2}
+    If $(X_n)_n$ is uniformly integrable,
+    then $(X_n)_n$ is bounded in $L^1$.k:w
+\end{fact}
+
+\begin{fact}\label{lec19f3}
+    Suppose $Y \in  L^1(\bP)$ and $\sup_n |X_n(\cdot )| \le  Y(\cdot )$.
+    Then $(X_n)_n$ is uniformly integrable.
+\end{fact}
+
+\begin{fact}\label{lec19f4}
+    Let $X \in  L^1(\bP)$.
+    \begin{enumerate}[(a)]
+        \item $\forall \epsilon > 0 .~ \exists \delta > 0 .~\forall F \in \cF .~ \bP(F) < \delta \implies\int_F |X| \dif \bP < \epsilon$.
+        \item $\forall  \epsilon > 0 .~ \exists  k \in  (0,\infty) .~ \int_{|X| > k} | X| \dif \bP < \epsilon$.
+    \end{enumerate}
+\end{fact}
+\begin{proof}
+    \begin{enumerate}[(a)]
+        \item Suppose not. Then for $\delta = 1, \frac{1}{2}, \frac{1}{2^2}, \ldots$
+            there exists $F_n$ such that $\bP(F_n) <\frac{1}{2^n}$
+            but $\int_{F_n} |X| \dif \bP \ge  \epsilon$.
+
+            Since $\sum_{n}  \bP(F_n) < \infty$,
+            by \autoref{borelcantelli},
+            \[\bP[\underbrace{\limsup_n F_n}_{\text{\reflectbox{$\coloneqq$}}F}] = 0.\]
+            We have
+            \begin{IEEEeqnarray*}{rCl}
+                \int_F | X| \dif \bP &=& \int |X| \One_F \dif \bP\\
+                                     &=& \int \limsup_n (|X| \One_{F_n}) \dif \bP\\
+                                     &\overset{\text{Reverse Fatou}}{\ge }&
+                                     \limsup_n \int |X| \One_{F_n} \dif \bP\\
+                                     &\ge & \epsilon
+            \end{IEEEeqnarray*}
+            where the assumption that $X$ is in $L^1$ was used to apply
+            the reverse of Fatou's lemma.
+
+            This yields a contradiction since $\bP(F) = 0$.
+        \item We want to apply part (a) to $F = \{ |X| > k\}$.
+            By Markov, $\bP(F) \le  \frac{1}{k} \bE[|X|]$.
+            Since $\bE[|X|] < \infty$, we can choose $k$ large enough
+            to get $\bP(F) \le \delta$.
+    \end{enumerate}
+\end{proof}
+
+\begin{refproof}{lec19f3}
+    Fix $\epsilon > 0$.
+    We have
+    \[
+    \bE[|X_n| \One_{|X_n| > k}] \le  \bE[|Y| \One_{|Y| > k}] < \epsilon
+    \]
+    for $k$ large enough by \autoref{lec19f4} (b).
+\end{refproof}
+
+\begin{fact}\label{lec19f5}
+    Let $X \in  L^1(\bP)$.
+    Then $\bF \coloneqq  \{ \bE[X | \cG] : \cG \subseteq \cF \text{ sub-$\sigma$-algebra}\}$ is uniformly integrable.
+\end{fact}
+\begin{proof}
+    Fix $\epsilon > 0$.
+    Choose $\delta > 0$ such that
+    \[\forall  F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon. (\ast)\]
+    Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
+    Then, by \autoref{condjensen}, $|Y| \le  \bE[ |X| | \cG]$.
+    Hence $\bE[|Y|] \le  \bE[|X|]$.
+    It follows that $\bP[|Y| > k] < \delta$
+    for $k$ suitably large,
+    since $\bE[|X|] \le \infty$.
+    Note that $\{Y > k\}  \in \cG$.
+    We have
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
+    \end{IEEEeqnarray*}
+    by $(\ast)$, since $\bP[|Y| > k] < \delta$.
+\end{proof}
+
+\begin{theorem}
+    Assume that $X_n \in L^1$ for all $n$ and $X \in L^1$.
+    Then the following are equivalent:
+    \begin{enumerate}[(1)]
+        \item $X_n \to X$ in $L^1$.
+        \item $(X_n)_n$ is uniformly integrable and $X_n \to X$ in probability.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (2) $\implies$ (1)
+
+    Define
+    \begin{IEEEeqnarray*}{rCl}
+        \phi(x) &\coloneqq & \begin{cases}
+            -k, & x \le -k\\
+            x, & x \in (-k,k)\\
+            k, & x \ge k.
+        \end{cases}
+    \end{IEEEeqnarray*}
+
+    $\phi$ is $1$-Lipshitz. % TODO
+
+    We have
+    \begin{IEEEeqnarray*}{rCl}
+        \int |X_n - X| \dif \bP
+        &\le & \int |X_n - \phi(X_n)| \dif \bP
+        + \int  |\phi(X) - X| \dif \bP
+        + \int |\phi(X_n) - \phi(X)| \dif \bP\\
+    \end{IEEEeqnarray*}
+    We have $\int_{|X_n| > k} \underbrace{|X_n - \phi(X_n)|}_{\le  |X_n| + | \phi(X_n)| \le  2 |X_n|}  \dif \bP\le  \epsilon$ by uniform integrability and
+    \autoref{lec19f4} part (b).
+    Similarly $\int_{|X| > k} |X - \phi(X)| \dif \bP < \epsilon$.
+
+    Since $\phi$ is Lipshitz,
+    $ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$.
+    By the bounded convergence theorem % TODO
+    $|\phi(X_n)| \le k \implies \int |  \phi(X_n) - \phi(X)| \dif \bP \to  0$.
+
+
+    (1) $\implies$ (2)
+
+    $X_n \xrightarrow{L^1}  X \implies X_n \xrightarrow{\bP} X$ 
+    by Markov's inequality.
+
+    Fix $\epsilon > 0$.
+    We have
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[|X_n|] &=& \bE[|X_n - X + X|]\\
+                   &\le & \epsilon + \bE[|X|]\\
+                   &<& \delta k
+    \end{IEEEeqnarray*}
+    for all $\delta > 0$ and suitable $k$.
+
+    Hence $\bP[|X_n| < k] < \delta$ by Markov's inequality.
+    Then by \autoref{lec19f4} part (a) it follows that
+    \[
+    \int_{|X_n| > k} |X_n| \dif \bP \le  \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le  2 \epsilon.
+    \] 
+\end{proof}
+
+\subsection{Martingale convergence theorems in $L^p, p \ge 1$}
+
+Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
+
+\begin{fact}\label{lec19f6}
+    Suppose that $X \in  L^p$ for some $p \ge 1$.
+
+    Then $(\bE[X | \cF_n])_n$ is an $\cF_n$-martingale.
+\end{fact}
+\begin{proof}
+    It is clear that $(\bE[X | \cF_n])_n$ is adapted to $(\cF_n)_n$.
+
+    Let  $X_n \coloneqq  \bE[X | \cF_n]$.
+    Consider 
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[X_n - X_{n-1} | \cF_{n-1}]
+        &=& \bE[\bE[X | \cF_n] - \bE[X | \cF_{n-1}] | \cF_{n-1}]\\
+        &=& \bE[X | \cF_{n-1}] - \bE[X | \cF_{n-1}]\\
+        &=& 0.
+    \end{IEEEeqnarray*}
+\end{proof}
+
+\begin{theorem}
+    Let $X \in L^p$ for some $p \ge 1$.
+    Then $X_n \coloneqq  \bE[X | \cF_n]$ defines a martingale which converges
+    to $X$ in $L^p$.
+\end{theorem}
+\begin{proof}
+    
+\end{proof}
+
+\begin{theorem}
+    Let $p > 1$.
+    Let $(X_n)_n$ be a martingale bounded in $L^p$.
+    Then there exists a random variable $X \in  L^p$, such that
+    $X_n = \bE[X | \cF_n]$ for all $n$.
+    
+\end{theorem}
+
+
+

probability_theory.tex

@@ -42,6 +42,7 @@
 \input{inputs/lecture_16.tex}
 \input{inputs/lecture_17.tex}
 \input{inputs/lecture_18.tex}
+\input{inputs/lecture_19.tex}
 
 \cleardoublepage