s23-probability-theory/inputs/prerequisites.tex

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

\subsection{Notions of convergence}
\begin{definition}
    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}
                ($X_n \xrightarrow{a.s.} X$)
                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
            \[
                \bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
            \]
    \end{itemize}
\end{definition}
% TODO Connect to ANaIII

\begin{theorem}
    \vspace{10pt}
    Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
    Then
    \begin{figure}[H]
        \centering
        \begin{tikzpicture}
            \node at (0,1.5) (as) { $X_n \xrightarrow{a.s.} X$};
            \node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
            \node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
            \draw[double equal sign distance, -implies] (as) -- (p);
            \draw[double equal sign distance, -implies] (L1) -- (p);
        \end{tikzpicture}
    \end{figure}
    and none of the other implications hold.
\end{theorem}
\begin{proof}
    \begin{claim}
        $X_n \xrightarrow{a.s.} X \implies X_n \xrightarrow{\bP} X$.
    \end{claim}
    \begin{subproof}
        $\Omega_0 \coloneqq  \{\omega \in  \Omega : \lim_{n\to \infty} X_n(\omega) = X(\Omega)\} $.
        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 $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
        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}
    $X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
    \end{claim}
    \begin{subproof}
    We have $\bE[|X_n - X|] \to  0$.
    Suppose there exists an $\epsilon > 0$ such that
    $\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
    We have
    \begin{IEEEeqnarray*}{rCl}
        \bE[|X_n - X|] &=& \int_\Omega |X_n - X | d\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}\\
                       &\ge& \epsilon \int_{|X_n -X | > \epsilon} d\bP\\
                       &=& \epsilon \cdot c > 0 \lightning
    \end{IEEEeqnarray*}
    \todo{Improve this with Markov}
    \end{subproof}
    \begin{claim}
    $X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$
    \end{claim}
    \begin{subproof}
    Take $([0,1], \cB([0,1 ]), \lambda)([0,1], \cB([0,1 ]), \lambda)$
    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}
    $X_n \xrightarrow{a.s.} X \notimplies X_n\xrightarrow{L^1} X$.
    \end{claim}
    \begin{subproof}
    We can use the same counterexample as in c).
    
    $\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
    We have already seen, that $X_n$ does not converge in $L_1$.
    \end{subproof}

    \begin{claim}
    $X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{a.s.} X$.
    \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
    \[
    \bE[|X_n|] = \int_{\Omega}|X_n| d\bP = \frac{1}{2^k} \to 0.
    \]
    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?
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}



\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 $X_i$.



Modes of covergence: $L^p$, in probability, a.s.
\fi