Convergence

inputs/intro.tex

@@ -22,3 +22,7 @@ please send me a message:\\
     \item Martingales,
     \item Markov chains.
 \end{enumerate}
+
+This notes follow the way the material was presented in the lecture rather
+closely. Additions (e.g.~from exercise sheets)
+and slight modifications have been marked with $\dagger$.

inputs/prerequisites.tex

@@ -2,14 +2,14 @@ This section provides a short recap of things that should be known
 from the lecture on stochastic.
 
 \subsection{Notions of Convergence}
-\begin{definition}
+\begin{definition}+
     \label{def:convergence}
     Fix a probability space $(\Omega,\cF,\bP)$.
     Let $X, X_1, X_2,\ldots$ be random variables.
     \begin{itemize}
         \item We say that $X_n$ converges to $X$
             \vocab[Convergence!almost surely]{almost surely}
-                ($X_n \xrightarrow{a.s.} X$)
+                ($X_n \xrightarrow{\text{a.s.}} X$)
                 iff
                 \[
                 \bP(\{\omega | X_n(\omega) \to X(\omega)\}) = 1.
@@ -29,34 +29,58 @@ from the lecture on stochastic.
             \[
                 \bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
             \]
+        \item We say that $X_n$ converges to $X$
+            \vocab[Convergence!in distribution]{in distribution}%
+            \footnote{
+                This notion of convergence was actually
+                defined during the course of the lecture,
+                but has been added here for completeness;
+                see \autoref{def:weakconvergence}.
+            }
+            ($X_n \xrightarrow{\text{dist}} X$)
+            iff for every continuous, bounded $f: \R \to  \R$
+            \[
+                \bE[f(X_n)] \xrightarrow{n \to \infty} \bE[f(X)].
+            \]
     \end{itemize}
 \end{definition}
 % TODO Connect to AnaIII
 
-\begin{theorem}
+\begin{theorem}+
     \label{thm:convergenceimplications}
     \vspace{10pt}
     Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
+    Let $1 \le p < q < \infty$.
     Then
     \begin{figure}[H]
         \centering
         \begin{tikzpicture}
-            \node at (0,1.5) (as) { $X_n \xrightarrow{a.s.} X$};
+            \node at (0,1.5) (as) { $X_n \xrightarrow{\text{a.s.}} X$};
             \node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
-            \node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
+            \node at (1.5,-1.5) (w) { $X_n \xrightarrow{\text{dist}} X$};
+            %\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
+            \node at (3,1.5) (Lp) { $X_n \xrightarrow{L^p} X$};
+            \node at (3,3) (Lq) { $X_n \xrightarrow{L^q} X$};
             \draw[double equal sign distance, -implies] (as) -- (p);
-            \draw[double equal sign distance, -implies] (L1) -- (p);
+            \draw[double equal sign distance, -implies] (p) -- (w);
+            % \draw[double equal sign distance, -implies] (L1) -- (p);
+            % \draw[double equal sign distance, -implies] (Lp) -- (L1);
+            \draw[double equal sign distance, -implies] (Lp) -- (p);
+            \draw[double equal sign distance, -implies] (Lq) -- (Lp);
         \end{tikzpicture}
     \end{figure}
-    and none of the other implications hold.
+    and none of the other implications hold (apart from the transitive closure).
 \end{theorem}
-\begin{proof}
+\begin{refproof}{thm:convergenceimplications}
     \begin{claim}
-        $X_n \xrightarrow{a.s.} X \implies X_n \xrightarrow{\bP} X$.
+        $X_n \xrightarrow{\text{a.s.}} X \implies X_n \xrightarrow{\bP} X$.
     \end{claim}
     \begin{subproof}
-        $\Omega_0 \coloneqq  \{\omega \in  \Omega : \lim_{n\to \infty} X_n(\omega) = X(\Omega)\} $.
+        $\Omega_0 \coloneqq
-        Let $\epsilon > 0$ and consider $A_n \coloneqq  \bigcup_{m \ge n} \{\omega \in \Omega: |X_m(\omega) - X(\Omega)| > \epsilon\}$.
+        \{\omega \in  \Omega : \lim_{n\to \infty} X_n(\omega) = X(\omega)\}$.
+        Let $\epsilon > 0$ and consider
+        $A_n \coloneqq  \bigcup_{m \ge n}
+        \{\omega \in \Omega: |X_m(\omega) - X(\omega)| > \epsilon\}$.
         Then  $A_n \supseteq A_{n+1} \supseteq \ldots$
         Define $A \coloneqq  \bigcap_{n \in \N} A_n$.
         Then $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
@@ -67,28 +91,68 @@ from the lecture on stochastic.
         \bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to  0.
         \]
     \end{subproof}
+    \begin{claim}
+        Let $1 \le p < q < \infty$.
+        Then $X_n \xrightarrow{L^q}  X  \implies X_n \xrightarrow{L^p} X$.
+    \end{claim}
+    \begin{subproof}
+        Take $r$ such that $\frac{1}{p} = \frac{1}{q} + \frac{1}{r}$.
+        We have
+        \begin{IEEEeqnarray*}{rCl}
+            \|X_n - X\|_{L^p} &=& \|1 \cdot (X_n - X)\|_{L^p}\\
+                              &\overset{\text{Hölder}}{\le }& \|1\|_{L^r} \|X_n - X\|_{L^q}\\
+                              &=& \|X_n - X\|_{L^q}
+        \end{IEEEeqnarray*}
+        Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$.
+    \end{subproof}
     \begin{claim}
     $X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
     \end{claim}
     \begin{subproof}
-    We have $\bE[|X_n - X|] \to  0$.
+    Let $\bE[|X_n - X|] \to  0$.
     Suppose there exists an $\epsilon > 0$ such that
     $\limsup\limits_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
     W.l.o.g.~$\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c$,
     otherwise choose an appropriate subsequence.
     We have
     \begin{IEEEeqnarray*}{rCl}
-        \bE[|X_n - X|] &=& \int_\Omega |X_n - X | d\bP\\
+        \bE[|X_n - X|] &=& \int_\Omega |X_n - X | \dif\bP\\
-                       &=& \int_{|X_n - X| > \epsilon} |X_n - X| d\bP + \underbrace{\int_{|X_n - X| \le  \epsilon} |X_n - X | d\bP}_{\ge 0}\\
+                       &=& \int_{|X_n - X| > \epsilon} |X_n - X| \dif\bP
-                       &\ge& \epsilon \int_{|X_n -X | > \epsilon} d\bP\\
+                        + \underbrace{\int_{|X_n - X| \le \epsilon}|X_n-X|\dif\bP}_{\ge 0}\\
+                       &\ge& \epsilon \int_{|X_n -X | > \epsilon} \dif\bP\\
                        &=& \epsilon \cdot c > 0 \lightning
     \end{IEEEeqnarray*}
     \todo{Improve this with Markov}
-    \todo{Counter examples}
+    \end{subproof}
-    \todo{weak convergence}
+    \begin{claim} %+
-    \todo{$L^p$ convergence}
+        $X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{dist}} X$.
+    \end{claim}
+    \begin{subproof}
+        Let $F$ be the distribution function of $X$
+        and $(F_n)_n$ the distribution functions of $(X_n)_n$.
+        By \autoref{lec10_thm1}
+        it suffices to show that $F_n(t) \to F(t)$ for all continuity
+        points $t$ of $F$.
+        Let $t$ be a continuity point of $F$.
+        Take some $\epsilon > 0$.
+        Then there exists $\delta > 0$ such that
+        $|F(t) - F(t')| < \frac{\epsilon}{2}$
+        for all $t'$ with $|t - t'| \le \delta$.
+        For all $n$ large enough,
+        we have
+        $\bP[|X_n - X| > \delta] < \frac{\epsilon}{2}$.
+        It is
+        \[|F_n(t) - F(t)|
+            = |\bP[X_n \le t] - F(t)|
+            \le \max(|\frac{\epsilon}{2} + \bP[X \le t + \delta] - F(t)|,
+            |\bP[X \le  t -\delta] - F(t)|)\\
+            \le \max(|\frac{\epsilon}{2} + F(t + \delta) - F(t)|, |F(t-\delta) -F(t)|
+            \le \epsilon,
+        \]
+        hence $F_n(t) \to F(t)$.
     \end{subproof}
     \begin{claim}
+        \label{claim:convimplpl1}
     $X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$
     \end{claim}
     \begin{subproof}
@@ -101,60 +165,58 @@ from the lecture on stochastic.
     \end{subproof}
 
     \begin{claim}
-    $X_n \xrightarrow{a.s.} X \notimplies X_n\xrightarrow{L^1} X$.
+    $X_n \xrightarrow{\text{a.s.}} X \notimplies X_n\xrightarrow{L^1} X$.
     \end{claim}
     \begin{subproof}
-    We can use the same counterexample as in c).
+        We can use the same counterexample as in \autoref{claim:convimplpl1}
 
     $\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
     We have already seen, that $X_n$ does not converge in $L_1$.
     \end{subproof}
 
     \begin{claim}
-    $X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{a.s.} X$.
+    $X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{\text{a.s.}} X$.
     \end{claim}
     \begin{subproof}
     Take $\Omega = [0,1], \cF = \cB([0,1]), \bP = \lambda$.
     Define $A_n \coloneqq  [j 2^{-k}, (j+1) 2^{-k}]$ where $n = 2^k + j$.
     We have
     \[
-    \bE[|X_n|] = \int_{\Omega}|X_n| d\bP = \frac{1}{2^k} \to 0.
+    \bE[|X_n|] = \int_{\Omega}|X_n| \dif\bP = \frac{1}{2^k} \to 0.
     \]
     However $X_n$ does not converge a.s.~as for all $\omega \in [0,1]$
     the sequence $X_n(\omega)$ takes the values $0$ and $1$ infinitely often.
     \end{subproof}
-\end{proof}
 
-How do we prove that something happens almost surely?
+    \begin{claim}
-The first thing that should come to mind is:
+        $X_n \xrightarrow{\text{dist}} X \notimplies X_n \xrightarrow{\bP} X$.
-\begin{lemma}[Borel-Cantelli]
+    \end{claim}
-    \label{borelcantelli}
+    \begin{subproof}
-    If we have a sequence of events  $(A_n)_{n \ge 1}$
+        Note that $X_n \xrightarrow{\text{dist}} X$ only makes a statement
-    such that $\sum_{n \ge 1}  \bP(A_n) < \infty$,
+        about the distributions of $X$ and $X_1,X_2,\ldots$
-    then $\bP[ A_n \text{for infinitely many $n$}] = 0$
+        For example, take some $p \in (0,1)$ and let
-    (more precisely: $\bP[\limsup_{n \to  \infty} A_n] = 0$).
+        $X$, $X_1, X_2,\ldots$ be i.i.d.~with $X \sim \Bin(1,p)$.
+        Trivially $X_n \xrightarrow{\text{dist}} X$.
+        However
+        \[
+            \bP[|X_n - X| = 1]
+            = \bP[X_n = 0]\bP[X = 1] + \bP[X_n = 1]\bP[X = 0]
+            = 2p(1-p).
+        \]
+    \end{subproof}
+    \begin{claim}
+        Let $1 \le p < q < \infty$. Then
+        $X_n \xrightarrow{L^p} X \notimplies X_n \xrightarrow{L^q} X$.
+    \end{claim}
+    \begin{subproof}
+        Consider $\Omega = [0,1]$, $\cF = \cB([0,1])$, $\bP = \lambda\upharpoonright [0,1]$
+        and $X_n(\omega) = \frac{1}{n \sqrt[q]{\omega}}$.
+        Then $\|X_0(\omega)\|_{L^p} < \infty$, since $p < q$.
+        Thus $X_n \xrightarrow{L_p} 0$.
+        However $\|X_n(\omega)\|_{L^q} = \infty$ for all $n$.
+    \end{subproof}
+\end{refproof}
 
-    For independent events $A_n$ the converse holds as well.
-\end{lemma}
-
-
-
-\iffalse
-\todo{Add more stuff here}
-\subsection{Some inequalities}
-% TODO: Markov
-
-
-\begin{theorem}[Chebyshev's inequality] % TODO Proof
-    Let $X$ be a r.v.~with  $\Var(x) < \infty$.
-    Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$.
-
-\end{theorem}
-
-
-We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and $\Var(X_1 +\ldots + X_n) = \Var(X_1) + \ldots + \Var(X_n)$ for independent $X_i$.
-
-\fi
 
 \subsection{Some Facts from Measure Theory}
 \begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]
@@ -178,5 +240,51 @@ We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and
 
 
 
+\subsection{Inequalities}
+This is taken from section 6.1 of the notes on Stochastik.
+
+\begin{theorem}[Markov's inequality]
+    Let $X$ be a random variable and $a > 0$.
+    Then
+    \[
+        \bP[|X| \ge  a] \le  \frac{\bE[|X|]}{a}.
+    \]
+\end{theorem}
+\begin{proof}
+    We have
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[|X|] &\ge & \int_{|X| \ge a} |X| \dif \bP\\
+                 &=& a \int_{|X| \ge a} \dif \bP = a\bP[|X| \ge a].
+    \end{IEEEeqnarray*}
+\end{proof}
+
+\begin{theorem}[Chebyshev's inequality]
+    Let $X$ be a random variable and $a > 0$.
+    Then
+    \[
+        \bP[|X - \bE(X)| \ge  a] \le \frac{\Var(X)}{a}.
+    \]
+\end{theorem}
+\begin{proof}
+    We have
+    \begin{IEEEeqnarray*}{rCl}
+        \bP[|X-\bE(X)| \ge a]
+        &=& \bP[|X - \bE(X)|^2 \ge a^2]\\
+        &\overset{\text{Markov}}{\ge}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
+    \end{IEEEeqnarray*}
+\end{proof}
+
+How do we prove that something happens almost surely?
+The first thing that should come to mind is:
+\begin{lemma}[Borel-Cantelli]
+    \label{borelcantelli}
+    If we have a sequence of events  $(A_n)_{n \ge 1}$
+    such that $\sum_{n \ge 1}  \bP(A_n) < \infty$,
+    then $\bP[ A_n \text{for infinitely many $n$}] = 0$
+    (more precisely: $\bP[\limsup_{n \to  \infty} A_n] = 0$).
+
+    For independent events $A_n$ the converse holds as well.
+\end{lemma}
+