203 lines
7.6 KiB
TeX
203 lines
7.6 KiB
TeX
\lecture{6}{}{Proof of SLLN}
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\begin{refproof}{lln}
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We want to deduce the SLLN (\yaref{lln}) from \yaref{thm2}.
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W.l.o.g.~let us assume that $\bE[X_i] = 0$
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(otherwise define $X'_i \coloneqq X_i - \bE[X_i]$).
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We will show that $\frac{S_n}{n} \xrightarrow{a.s.} 0$.
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Define $Y_i \coloneqq \frac{X_i}{i}$.
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Then the $Y_i$ are independent and we have $\bE[Y_i] = 0$
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and $\Var(Y_i) = \frac{\sigma^2}{i^2}$.
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Thus $\sum_{i=1}^\infty \Var(Y_i) < \infty$.
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From \yaref{thm2} we obtain that $\sum_{i=1}^\infty Y_i$ converges a.s.
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\begin{claim}
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Let $(a_n)$ be a sequence in $\R$
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such that $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges,
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then $\frac{a_1 + \ldots + a_n}{n} \to 0$.
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\end{claim}
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\begin{subproof}
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Let $S_m \coloneqq \sum_{n=1}^\infty \frac{a_n}{n}$.
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By assumption, there exists $S \in \R$
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such that $S_m \xrightarrow{m \to \infty} S$.
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Note that $j \cdot (S_{j} - S_{j-1}) = a_j$.
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Define $S_0 \coloneqq 0$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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a_1 + \ldots + a_n &=&
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(S_1 - S_0) + 2(S_2 - S_1) + \ldots + n(S_n - S_{n-1})\\
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&=& n S_n - (S_1 + S_2 + \ldots + S_{n-1}).
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\end{IEEEeqnarray*}
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Thus
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\begin{IEEEeqnarray*}{rCl}
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\frac{a_1 + \ldots + a_n}{n}
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&=& S_n - \frac{S_1 + \ldots + S_{n-1}}{n}\\
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&=& \underbrace{S_n}_{\to S}
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- \underbrace{\left( \frac{n-1}{n} \right)}_{\mathclap{\to 1}}
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\cdot \underbrace{\frac{S_1 + \ldots + S_{n-1}}{n-1}}_{\to S}\\
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&\to & 0,
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\end{IEEEeqnarray*}
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where we have used
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\begin{fact}
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\[
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\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n} S_i
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\].
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\end{fact}
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\end{subproof}
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The SLLN follows from the claim.
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\end{refproof}
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In order to prove \yaref{thm2}, we need the following:
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\begin{theorem}[Kolmogorov's inequality]
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\yalabel{Kolmogorov's Inequality}{Kolmogorov}{thm:kolmogorovineq}
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If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$
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and $\Var(X_i) = \sigma_i^2$, then
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\[
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\bP\left[\max_{1 \le i \le n} \left| \sum_{j=1}^{i} X_j \right| > \epsilon \right]
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\le \frac{1}{\epsilon^2} \sum_{i=1}^m \sigma_i^2.
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\]
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\end{theorem}
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\begin{proof}
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Let
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\begin{IEEEeqnarray*}{rCl}
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A_1 &\coloneqq& \{\omega : |X_1(\omega)| > \epsilon\},\\
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A_2 &\coloneqq & \{\omega: |X_1(\omega)| \le \epsilon,
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|X_1(\omega) + X_2(\omega)| > \epsilon \},\\
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\ldots\\
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A_i &\coloneqq& \{\omega: |X_1(\omega)| \le \epsilon,
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|X_1(\omega) + X_2(\omega)| \le \epsilon, \ldots, %
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|X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon,\\
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&& ~ ~|X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}.
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\end{IEEEeqnarray*}
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It is clear, that the $A_i$ are disjoint.
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We are interested in $\bigcup_{1 \le i \le n} A_i$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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&&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C
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+ \underbrace{X_{i+1} + \ldots + X_n}_D)^2 \dif\bP\\
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&=& \int_{A_i} C^2 \dif\bP
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+ \underbrace{\int_{A_i} D^2 \dif\bP}_{\ge 0}
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+ 2 \int_{A_i} CD \dif\bP\\
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&\ge& \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} \dif\bP
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+ 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D \dif\bP\\
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&\ge& \int_{A_i} \epsilon^2 \dif\bP,
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\end{IEEEeqnarray*}
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since by the independence of $E$ and $D$,
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and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E \dif\bP = 0$.
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Hence
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\[
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\bP(A_i)
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\le \frac{1}{\epsilon^2} \int_{A_i} (X_1 + \ldots + X_n)^2 \dif \bP.
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\]
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Since the $A_i$ are disjoint, we obtain
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\begin{IEEEeqnarray*}{rCl}
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\bP\left( \bigcup_{i \in \N} A_i \right)
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&\le & \frac{1}{\epsilon^2}
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\int_{\bigcup_{i \in \N} A_i} (X_1 + \ldots + X_n)^2 \dif \bP\\
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&\le & \frac{1}{\epsilon^2}
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\int_{\Omega} (X_1 + \ldots + X_n)^2 \dif \bP\\
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&\overset{\text{independence}}{=}&
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\frac{1}{\epsilon^2}(\bE[X_1^2] + \ldots + \bE[X_n^2])\\
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&\overset{\bE[X_i] = 0}{=}& \frac{1}{\epsilon^2}
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\left( \Var(X_1) + \ldots + \Var(X_n)\right).
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{refproof}{thm2}
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Let $S_n \coloneqq x_1 + \ldots + x_n$.
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We'll show that $\{S_n(\omega)\}_{n \in \N}$
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is a Cauchy sequence
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for almost every $\omega$.
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Let
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\[
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a_m(\omega) \coloneqq \sup_{k \in \N}
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\{ | S_{ m+k}(\omega) - S_m(\omega)|\}
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\]
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and
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\[
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a(\omega) \coloneqq \inf_{m \in \N} a_m(\omega).
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\]
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Then $\{S_n(\omega)\}_{n \in \R}$
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is a Cauchy sequence iff $a(\omega) = 0$.
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We want to show that $\bP[a(\omega) > 0] = 0$.
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For this, it suffices to show that $\bP[a(\omega) > \epsilon] = 0$
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for all $\epsilon > 0$.
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For a fixed $\epsilon > 0$, we obtain:
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\begin{IEEEeqnarray*}{rCl}
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\bP[a_m > \epsilon]
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&=& \bP[ \sup_{k \in \N} | S_{m+k} - S_m| > \epsilon]\\
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&=& \lim_{l \to \infty} \bP[%
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\underbrace{\sup_{k \le l} |S_{m+k} - S_m| > \epsilon}_{%
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\text{\reflectbox{$\coloneqq$}} B_l \uparrow%
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B \coloneqq \{\sup_{k \in \N} |S_{m+k} - S_m| > \epsilon\}}%
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]
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\end{IEEEeqnarray*}
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Now,
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\begin{IEEEeqnarray*}{rCl}
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&&\max \{|S_{m+1} - S_m|, |S_{m+2} - S_m|, \ldots, |S_{m+l} - S_m|\}\\
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&=& \max \{|X_{m+1}|, |X_{m+1} + X_{m+2}|, \ldots, |X_{m+1} + X_{m+2} + \ldots + X_{m+l}|\}\\
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&\overset{\yaref{thm:kolmogorovineq}}{\le}&
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\frac{1}{\epsilon^2} \sum_{i=m}^{l} \Var(X_i)\\
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&\le & \frac{1}{\epsilon^2} \sum_{i=m}^\infty \Var(X_i)
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\xrightarrow{m \to \infty} 0,
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\end{IEEEeqnarray*}
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since by our assumption, $\sum_{n \in \N} \Var(X_i) < \infty$.
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Hence
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\[
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\bP[a_m > \epsilon] \xrightarrow{m \to \infty} 0.
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\]
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It follows that $\bP[a > \epsilon] = 0$,
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as claimed.
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\end{refproof}
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\subsubsection{Application: Renewal Theorem}
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\begin{theorem}[Renewal theorem]
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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.
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Let $S_n \coloneqq \sum_{i=1}^n X_i$.
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For all $t > 0$ let \[
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N_t \coloneqq \sup \{n : S_n \le t\}.
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\]
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Then $\frac{N_t}{t} \xrightarrow{a.s.} \frac{1}{m}$ as $t \to \infty$.
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\end{theorem}
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The $X_i$ can be thought of as waiting times.
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$S_i$ models how long you have to wait for $i$ events to occur.
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\begin{proof}
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By SLLN, $\frac{S_n}{n} \xrightarrow{a.s.} m$ as $n \to \infty$.
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Note that
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\begin{equation}
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N_t \uparrow \infty \text{ a.s.~as } t \to \infty, \label{eqn:renewalnt}
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\end{equation}
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since $\{N_t \ge n\} = \{X_1 + \ldots+ X_n \le t\}$.
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\begin{claim}
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$\bP[\frac{S_n}{n} \xrightarrow{n \to \infty} m
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\land N_t \xrightarrow{t \to \infty} \infty] = 1$.
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\end{claim}
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\begin{subproof}
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Let $A \coloneqq \{\omega: \frac{S_n(\omega)}{n} \xrightarrow{n \to \infty} m\}$
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and $B \coloneqq \{\omega : N_t(\omega) \xrightarrow{t \to \infty} \infty\}$.
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By the SLLN we have $\bP(A^C) = 0$
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and by \eqref{eqn:renewalnt} it holds that $\bP(B^C) = 0$.
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\end{subproof}
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Equivalently,
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$\bP\left[ \frac{S_{N_t}}{N_t} \xrightarrow{t \to \infty} m
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\land \frac{S_{N_t + 1}}{N_t + 1} \xrightarrow{t \to \infty} m \right] = 1$.
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By definition, we have $S_{N_t} \le t \le S_{N_t + 1}$.
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Thus
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\[\frac{S_{N_t}}{N_t} \le \frac{t}{N_t} \le \frac {S_{N_t + 1}}{N_t}
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\le \frac{S_{N_t + 1}}{N_t + 1} \cdot \frac{N_t + 1}{N_t}.\]
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Hence $\frac{t}{N_t} \to m$.
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\end{proof}
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