s23-probability-theory/inputs/lecture_6.tex

\lecture{6}{}{}
\todo{Large parts of lecture 6 are missing}
\begin{refproof}{lln}
    We want to deduce the SLLN (\autoref{lln}) from \autoref{thm2}.
    W.l.o.g.~let us assume that $\bE[X_i] = 0$ (otherwise define $X'_i \coloneqq  X_i - \bE[X_i]$).
    We will show that $\frac{S_n}{n} \xrightarrow{a.s.} 0$.
    Define $Y_i \coloneqq \frac{X_i}{i}$.
    Then the $Y_i$ are independent and we have $\bE[Y_i] = 0$
    and $\Var(Y_i) = \frac{\sigma^2}{i^2}$.
    Thus $\sum_{i=1}^\infty \Var(Y_i) < \infty$.
    From \autoref{thm2} we obtain that $\sum_{i=1}^\infty Y_i < \infty$ a.s.
    \begin{claim}
        Let $(a_n)$ be a sequence in $\R$ such that $\sum_{n=1}^{\infty} \frac{a_n}{n}$, then $\frac{a_1 + \ldots + a_n}{n} \to  0$.
    \end{claim}
    \begin{subproof}
        Let $S_m \coloneqq \sum_{n=1}^\infty \frac{a_n}{n}$.
        By assumption, there exists $S \in \R$
        such that $S_m \to  S$ as $m \to  \infty$.
        Note that $j \cdot (S_{j} - S_{j-1}) = a_j$.
        Define $S_0 \coloneqq  0$.
        Then $a_1 + \ldots + a_n = (S_1 - S_0) + 2(S_2 - S_1) + 3(S_3 - S_2) +
        \ldots + n (S_n - S_{n-1})$.
        Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO
   
    \end{subproof}
    The SLLN follows from the claim.
\end{refproof}

We need the fol]
\begin{theorem}[Kolmogorov's inequality]
    If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$
    and $\Var(X_i) = \sigma_i^2$, then
    \[
    \bP\left[\max_{1 \le  i \le  n} \left| \sum_{j=1}^{i}  X_j \right| > \epsilon \right] \le  \frac{1}{\epsilon ^2} \sum_{i=1}^m \sigma_i^2 % TODO
    \]
\end{theorem}
\begin{proof}
    Let $A_1 \coloneqq  \{\omega : |X_1(\omega)| > \epsilon\}, \ldots,
    A_i := \{\omega: |X_1(\omega)| \le \epsilon, |X_1(\omega) + X_2(\omega)| \le  \epsilon, \ldots, |X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon,
    |X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}$.
    We are interested in $\bigcup_{1 \le  i \le n} A_i$.

    We have
    \begin{IEEEeqnarray*}{rCl}
        &&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\
        &=& \int_{A_i} C^2 d\bP + \underbrace{\int_{A_i} D^2 d \bP}_{\ge  0} + 2 \int_{A_i} CD d\bP\\
                                                                                                       &\ge & \int_{A_i}  \underbrace{C^2}_{\ge \epsilon^2} d \bP + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\
                                                                                                       &\ge& \int_{A_i} \epsilon^2 d\bP
    \end{IEEEeqnarray*}
    (By the independence of $X_1,\ldots, X_n$ and therefore that of $E$ and $D$ and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.)

    % TODO

\end{proof}

\begin{refproof}{thm2}
    % TODO

\end{refproof}



\subsubsection{Application: Renewal Theorem}

\begin{theorem}[Renewal theorem]
    Let $X_1,X_2,\ldots$ i.i.d.~random variables with $X_i \ge  0$, $\bE[X_i] = m > 0$. The $X_i$ model waiting times.
    Let $S_n \coloneqq  \sum_{i=1}^n X_i$.
    For all $t > 0$ let \[
    N_t \coloneqq  \sup \{n : S_n \le  t\}.
    \]
    Then $\frac{N_t}{t} \xrightarrow{a.s.} \frac{1}{m}$ as $t \to \infty$.
\end{theorem}

    The $X_i$ can be thought of as waiting times.
    $S_i$ models how long you have to wait for $i$ events to occur.

\begin{proof}
    By SLLN, $\frac{S_n}{n} \xrightarrow{a.s.} m$ as $n \to \infty$.
    Note that $N_t \uparrow \infty$ a.s.~as $t \to \infty (\ast\ast)$, since
    $\{N_t \ge  n\} = \{X_1 + \ldots+ X_n \le t\}$ thus $N_t \uparrow \infty$ as $t \uparrow \infty$.

    \begin{claim}
        $\bP[\frac{S_n}{n} \xrightarrow{n \to  \infty} m , N_t \xrightarrow{t \to  \infty}  \infty] = 1$.
    \end{claim}
    \begin{subproof}
    Let $A \coloneqq  \{\omega: \frac{S_n(\omega)}{n} \xrightarrow{n \to  \infty} m\}$ and $B \coloneqq \{\omega : N_t(\omega \xrightarrow{t \to  \infty} \infty\}$.
    By the SLLN, we have $\bP(A^C) = 0$ and $\ast\ast \implies \bP(B^C) = 0$.
    \end{subproof}

    Equivalently, $\bP\left[ \frac{S_{N_t}}{N_t} \xrightarrow{t \to \infty} m, \frac{S_{N_t + 1}}{N_t + 1} \xrightarrow{t \to  \infty} m \right] = 1$.

    By definition, we have $S_{N_t} \le  t \le  S_{N_t + t}$.
    Then $\frac{S_{N_t}}{N_t} \le  \frac{t}{N_t} \le S_{N_t + 1}{N_t} \le  \frac{S_{N_t + 1}}{N_t + 1} \cdot  \frac{N_t + 1}{N_t}$.
    Hence $\frac{t}{N_t} \to  m$.

\end{proof}