\lecture{7}{}{Kolmogorov's three series theorem}
\begin{goal}
    We want to drop our assumptions on finite mean or variance
    and say something about the behaviour of $ \sum_{n \ge 1}  X_n$
    when the $X_n$ are independent.
\end{goal}
\begin{theorem}[Kolmogorov's three-series theorem] % Theorem 3
    \label{thm3}
    Let $X_n$ be a family of independent random variables.
    \begin{enumerate}[(a)]
        \item Suppose for some $C \ge 0$, the following three series
            of numbers converge:
            \begin{itemize}
                \item  $\sum_{n \ge 1} \bP(|X_n| > C)$,
                \item $\sum_{n \ge 1}  \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$,
                \item $\sum_{n \ge 1}  \underbrace{\int_{|X_n| \le  C} X_n^2 d\bP - \left( \int_{|X_n| \le  C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
            \end{itemize}
            Then $\sum_{n \ge  1}  X_n$ converges almost surely.
        \item Suppose $\sum_{n \ge 1}  X_n$ converges almost surely.
            Then all three series above converge for every $C > 0$.
    \end{enumerate}
\end{theorem}
For the proof we'll need a slight generalization of \autoref{thm2}:
\begin{theorem} %[Theorem 4]
    \label{thm4}
    Let $\{X_n\}_n$ be independent and \vocab{uniformly bounded}
    (i.e. $\exists M < \infty : \sup_n \sup_\omega |X_n(\omega)| \le M$).
    Then $\sum_{n \ge 1} X_n$ converges almost surely
    $\iff$ $\sum_{n \ge 1} \bE(X_n)$ and $\sum_{n \ge 1}  \Var(X_n)$
    converge.
\end{theorem}
\begin{refproof}{thm3}
    Assume, that we have already proved \autoref{thm4}.
    We prove part (a) first.
    Put $Y_n = X_n \cdot \One_{\{|X_n| \le  C\}}$.
    Since the $X_n$ are independent, the $Y_n$ are independent as well.
    Furthermore, the $Y_n$ are uniformly bounded.
    By our assumption, the series
    $\sum_{n \ge  1}  \int_{|X_n| \le  C} X_n d\bP = \sum_{n \ge 1}  \bE[Y_n]$
    and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le  C} X_n d\bP \right)^2 = \sum_{n \ge 1}  \Var(Y_n)$
    converges.
    By \autoref{thm4} it follows that $\sum_{n \ge 1}  Y_n < \infty$
    almost surely.
    Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
    Since the first series $\sum_{n \ge 1}  \bP(A_n) < \infty$,
    by Borel-Cantelli, $\bP[\text{infinitely many $A_n$ occcur}] = 0$.


    For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
    for almost every $\omega$.
    Fix an arbitrary $C > 0$.
    Define
    \[
    Y_n(\omega) \coloneqq  \begin{cases}
        X_n(\omega) & \text{if} |X_n(\omega)| \le  C,\\
        C &\text{if } |X_n(\omega)| > C.
    \end{cases}
    \]
    Then the $Y_n$ are independent and $\sum_{n \ge 1} Y_n(\omega) < \infty$
    almost surely and the $Y_n$ are uniformly bounded.
    By \autoref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$
    converge.
    Define
    \[
    Z_n(\omega) \coloneqq \begin{cases}
        X_n(\omega) &\text{if } |X_n| \le  C,\\
        -C &\text{if } |X_n| > C.
    \end{cases}
    \]
    Then the $Z_n$ are independent, uniformly bounded and  $\sum_{n \ge 1}  Z_n(\omega) < \infty$
    almost surely.
    By \autoref{thm4} we have
    $\sum_{n \ge 1} \bE(Z_n) < \infty$
    and $\sum_{n \ge 1} \Var(Z_n) < \infty$.

    We have
    \begin{IEEEeqnarray*}{rCl}
        \bE(Y_n) &=& \int_{|X_n| \le  C} X_n d \bP + C \bP(|X_n| \ge C),\\
        \bE(Z_n) &=& \int_{|X_n| \le  C} X_n d \bP - C \bP(|X_n| \ge C).
    \end{IEEEeqnarray*}
    Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$
    the second series converges,
    and since
    $\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
    For the third series, we look at
    $\sum_{n \ge 1} \Var(Y_n)$ and
    $\sum_{n \ge 1} \Var(Z_n)$ to conclude that this series converges
    as well.
\end{refproof}
Recall \autoref{thm2}.
We will see, that the converse of \autoref{thm2} is true if the $X_n$ are uniformly bounded.
More formally:
\begin{theorem}[Theorem 5]
    \label{thm5}
    Let $X_n$ be a series of independent variables with mean $0$,
    that are uniformly bounded.
    If $\sum_{n \ge 1} X_n < \infty$ almost surely,
    then $\sum_{n \ge 1} \Var(X_n) < \infty$.
\end{theorem}
\begin{refproof}{thm4}
    Assume we have proven \autoref{thm5}.

    ``$\impliedby$'' Assume $\{X_n\} $ are independent, uniformly bounded
    and $\sum_{n \ge 1}  \bE(X_n) < \infty$ as well as $\sum_{n \ge 1}  \Var(X_n) < \infty$.
    We need to show that $\sum_{n \ge 1} X_n < \infty$ a.s.
    Let $Y_n \coloneqq  X_n - \bE(X_n)$.
    Then the $Y_n$ are independent, $\bE(Y_n) = 0$ and $\Var(Y_n) = \Var(X_n)$.
    By \autoref{thm2} $\sum_{n \ge 1} Y_n < \infty$ a.s.
    Thus $\sum_{n \ge 1}  X_n < \infty$ a.s.

    ``$\implies$'' We assume that $\{X_n\}$ are independent, uniformly bounded
    and $\sum_{n \ge 1}  X_n(\omega) < \infty$ a.s.
    We have to show that $\sum_{n \ge 1}  \bE(X_n) < \infty$
    and $\sum_{n \ge 1} \Var(X_n) < \infty$.

    Consider the product space $(\Omega, \cF, \bP) \otimes (\Omega, \cF, \bP)$.
    On this product space, we define
    $Y_n \left( (\omega, \omega') \right) \coloneqq  X_n(\omega)$
    and $Z_n \left( (\omega, \omega') \right) \coloneqq  X_n(\omega')$.
    \begin{claim}
        For every fixed $n$, $Y_n$ and $Z_n$ are independent.
    \end{claim}
    \begin{subproof}
        This is obvious, but well prove it carefully here.
        \begin{IEEEeqnarray*}{rCl}
            &&(\bP \otimes \bP) [Y_n \in (a,b) , Z_n \in (a',b') ]\\
            &=& (\bP\otimes\bP) \left( (\omega, \omega') : X_n(\omega) \in (a,b) \land X_n(\omega') \in (a',b') \right)\\
            &=& (\bP\otimes\bP)(A \times A') \text{where }
            A \coloneqq X_n^{-1}\left( (a,b)\right) \text{ and } A' \coloneqq  X_n^{-1}\left( (a',b') \right)\\
            &=& \bP(A)\bP(A')
        \end{IEEEeqnarray*}
    \end{subproof}
    Now $\bE[Y_n - Z_n] = 0$ (by definition) and $\Var(Y_n - Z_n) = 2\Var(X_n)$.
    Obviously, $(Y_n - Z_n)_{n \ge 1}$ is also uniformly bounded.
    \begin{claim}
        $\sum_{n \ge 1} (Y_n - Z_n) < \infty$ almost surely
        on $(\Omega \otimes \Omega, \cF \otimes\cF, \bP \otimes\bP)$.
    \end{claim}
    \begin{subproof}
        Suppose $\Omega_0 = \{\omega: \sum_{n \ge 1}  X_n(\omega) < \infty\}$.
        Then $\bP(\Omega_0) = 1$.
        Thus $(\bP\otimes\bP)(\Omega_0 \otimes \Omega_0) = 1$.
        Furthermore
        $\sum_{n \ge 1} \left(Y_n(\omega, \omega') - Z_n(\omega, \omega') \right)= \sum_{n \ge 1} \left(X_n(\omega) - X_n(\omega')\right)$.
            Thus $\sum_{n \ge 1} \left( Y_n(\omega, \omega') - Z_n(\omega, \omega') \right) < \infty$ a.s.~on $\Omega_0\otimes\Omega_0$.
    \end{subproof}
    By \autoref{thm5}, $\sum_{n} \Var(X_n) = \frac{1}{2}\sum_{n \ge 1}  \Var(Y_n - Z_n) < \infty$ a.s.
    Define $U_n \coloneqq  X_n - \bE(X_n)$.
    Then $\bE(U_n) = 0$ and the $U_n$ are independent
    and uniformly bounded.
    We have $\sum_{n} \Var(U_n) = \sum_{n}  \Var(X_n) < \infty$.
    Thus $\sum_{n} U_n$ converges a.s.~by \autoref{thm2}.
    Since by assumption $\sum_{n} X_n < \infty$ a.s.,
    it follows that $\sum_{n} \bE(X_n) < \infty$.
\end{refproof}
\begin{remark}
    In the proof of \autoref{thm4}
    ``$\impliedby$'' is just a trivial application of \autoref{thm2}
    and uniform boundedness was not used.
    The idea of `` $\implies$ '' will lead to coupling. % TODO ?
\end{remark}

 % TODO Proof of thm5 in the notes
\begin{example}[Application of \autoref{thm5}]
    The series $\sum_{n}  \frac{1}{n^{\frac{1}{2} + \epsilon}}$
    does not converge for $\epsilon < \frac{1}{2}$.
    However
    \[
    \sum_{n}  X_n \frac{1}{n^{\frac{1}{2} + \epsilon}}
    \]
    where $\bP[X_n = 1] = \bP[X_n = -1] = \frac{1}{2}$
    converges almost surely for all $\epsilon > 0$.
    And
    \[
    \sum_{n} X_n \frac{1}{n^{\frac{1}{2} - \epsilon}}
    \]
    does not converge.

\end{example}
