This repository has been archived on 2023-10-20. You can view files and clone it, but cannot push or open issues or pull requests.
probability-theory/inputs/lecture_5.tex

67 lines
2.6 KiB
TeX
Raw Normal View History

2023-05-10 18:56:36 +02:00
% Lecture 5 2023-04-21
\subsection{The laws of large numbers}
2023-04-27 17:07:50 +02:00
We want to show laws of large numbers:
The LHS is random and represents ``sane'' averaging.
The RHS is constant, which we can explicitly compute from the distribution of the RHS.
We fix a probability space $(\Omega, \cF, \bP)$ once and for all.
\begin{theorem}
\label{lln}
Let $X_1, X_2,\ldots$ be i.i.d.~random variables on $(\R, \cB(\R))$
and $m = \bE[X_i] < \infty$
and $\sigma^{2} = \Var(X_i) = \bE[ (X_i - \bE(X_i))^2] = \bE[X_i^2] - \bE[X_i]^2 < \infty$.
Then
\begin{enumerate}[(a)]
\item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$
in probability (\vocab{weak law of large numbers}, WLLN),
\item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$
almost surely (\vocab{strong law of large numbers}, SLLN).
\end{enumerate}
\end{theorem}
\begin{refproof}{lln}
\begin{enumerate}[(a)]
\item Given $\epsilon > 0$, we need to show that
\[
\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] \to 0 \]
as $n \to 0$.
Let $S_n \coloneqq X_1 + \ldots + X_n$.
Then $\bE[S_n] = \bE[X_1] + \ldots + \bE[X_n] = nm$.
We have
\begin{IEEEeqnarray*}{rCl}
\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] &=& \bP\left[\left|\frac{S_n}{n}-m\right| > \epsilon\right]\\
&\overset{\text{Chebyshev}}{\le }& \frac{\Var\left( \frac{S_n}{n} \right) }{\epsilon^2} = \frac{1}{n} \frac{\Var(X_1)}{\epsilon^2} \xrightarrow{n \to \infty} 0
\end{IEEEeqnarray*}
since
\[\Var(\frac{S_n}{n}) = \frac{1}{n^2} \Var(S_n) = \frac{1}{n^2} n \Var(X_i).\]
\end{enumerate}
\end{refproof}
For the proof of (b) we need the following general result:
\begin{theorem}
\label{thm2}
Let $X_1, X_2, \ldots$ be independent (but not necessarily identically distributed) random variables with $\bE[X_i] = 0$ for all $i$
and $\sum_{i=1}^n \Var(X_i) < \infty$.
Then $\sum_{n \ge 1} X_n$ converges almost surely.
\end{theorem}
\begin{proof}
\end{proof}
\begin{question}
Does the converse hold? I.e.~does $\sum_{n \ge 1} X_n < \infty$ a.s.~
then $\sum_{n \ge 1} \Var(X_n) < \infty$.
\end{question}
This does not hold. Consider for example $X_n = \frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$.
\begin{refproof}{lln}
\begin{enumerate}
\item[(b)]
\end{enumerate}
\end{refproof}