s23-probability-theory/inputs/prerequisites.tex

This section provides a short recap of things that should be known
from the lecture on stochastic.

\subsection{Notions of Convergence}
\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}
                ($X_n \xrightarrow{\text{a.s.}} X$)
                iff
                \[
                \bP(\{\omega | X_n(\omega) \to X(\omega)\}) = 1.
                \]
        \item We say that $X_n$ converges to $X$
            \vocab[Convergence!in probability]{in probability}
            ($X_n \xrightarrow{\bP} X$)
            iff
            \[
                \lim_{n \to \infty}\bP[|X_n - X| > \epsilon] = 0
            \]
            for all $\epsilon > 0$.
        \item We say that $X_n$ converges to $X$
            \vocab[Convergence!in mean]{in the $p$-th mean}
            ($X_n \xrightarrow{L^p} X$ )
            iff
            \[
                \bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
            \]
        \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

\pagebreak
\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) { $X_n \xrightarrow{\text{a.s.}} X$};
            \node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
            \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] (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 (apart from the transitive closure).
\end{theorem}
\begin{refproof}{thm:convergenceimplications}
    \begin{claim}
        $X_n \xrightarrow{\text{a.s.}} X \implies X_n \xrightarrow{\bP} X$.
    \end{claim}
    \begin{subproof}
        Let $\Omega_0 \coloneqq
        \{\omega \in  \Omega : \lim_{n\to \infty} X_n(\omega) = X(\omega)\}$.
        Fix some $\epsilon > 0$ 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 $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
        Since $X_n \xrightarrow{a.s.} X$ we have that
        \[\forall  \omega \in \Omega_0 .~ \exists  n \in \N .~
          \forall m \ge n.~ |X_m(\omega) - X(\omega)| < \epsilon.\]
        We have $A \subseteq  \Omega_0^{c}$, hence $\bP[A_n] \to 0$.
        Thus \[
        \bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to  0.
        \]
    \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}
        Suppose $\bE[|X_n - X|] \to  0$.
        Then for every $\epsilon > 0$
        \begin{IEEEeqnarray*}{rCl}
            \bP[|X_n - X| \ge \epsilon]
              &\overset{\text{Markov}}{\le}& \frac{\bE[|X_n - X|]}{\epsilon}\\
              &\xrightarrow{n \to \infty} & 0,
        \end{IEEEeqnarray*}
        hence $X_n \xrightarrow{\bP} X$.
    \end{subproof}
    \begin{claim} %+
        $X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{dist}} X$.
    \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
        \begin{IEEEeqnarray*}{rCl}
            |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,
        \end{IEEEeqnarray*}
        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}
    Take $([0,1], \cB([0,1 ]), \lambda)$
    and define $X_n \coloneqq  n \One_{[0, \frac{1}{n}]}$.
    We have $\bP[|X_n| > \epsilon] = \frac{1}{n}$
    for $n$ large enough.

    However $\bE[|X_n|] = 1$.
    \end{subproof}

    \begin{claim}
    $X_n \xrightarrow{\text{a.s.}} X \notimplies X_n\xrightarrow{L^1} X$.
    \end{claim}
    \begin{subproof}
        We can use the same counterexample as in \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}
    $X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{\text{a.s.}} X$.
    \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
    \[
    \bE[|X_n|] = \int_{\Omega}|X_n| \dif\bP = \frac{1}{2^k} \to 0.
    \]
    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}

    \begin{claim}
        $X_n \xrightarrow{\text{dist}} X \notimplies X_n \xrightarrow{\bP} X$.
    \end{claim}
    \begin{subproof}
        Note that $X_n \xrightarrow{\text{dist}} X$ only makes a statement
        about the distributions of $X$ and $X_1,X_2,\ldots$
        For example, take some $p \in (0,1)$ and let
        $X$, $X_1, X_2,\ldots$ 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}

\begin{theorem}[Bounded convergence theorem]
    \label{thm:boundedconvergence}
    Suppose that $X_n \xrightarrow{\bP} X$ and there exists
    some $K$ such that $|X_n| \le K$ for all $n$.
    Then $X_n \xrightarrow{L^1} X$.
\end{theorem}
\begin{proof}
    Note that $|X| \le K$ a.s.~since
    \[\bP[|X| \ge  K + \epsilon] \le \bP[|X_n - X| > \epsilon]
        \xrightarrow{n \to \infty} 0.\]
    Hence
    \begin{IEEEeqnarray*}{rCl}
        \int |X_n - X| \dif \bP &\le& \int_{|X_n - X| \ge \epsilon} |X_n - X| \dif \bP + \epsilon\\
                                &\le &2K\bP[|X_n - X| \ge \epsilon] +\epsilon.
    \end{IEEEeqnarray*}
\end{proof}


\subsection{Some Facts from Measure Theory}
\begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]
    Let $\mu$ be a finite measure on $(\R, \cB(\R))$.
    Then for all $\epsilon > 0$,
    there exists a compact set $K \in \cB(\R)$ such that
    $\mu(K) > \mu(\R) - \epsilon$.
\end{fact}
\begin{proof}
    We have $[-k,k] \uparrow \R$, hence $\mu([-k,k]) \uparrow \mu(\R)$.
\end{proof}

\begin{theorem}[Riemann-Lebesgue]
    \label{riemann-lebesgue}
    Let $f: \R \to  \R$ be integrable.
    Then
    \[
    \lim_{n \to \infty} \int_{\R} f(x) \cos(n x) \lambda(\dif x) = 0.
    \]
\end{theorem}



\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}