78 lines
3 KiB
TeX
78 lines
3 KiB
TeX
\lecture{5}{2023-04-21}{Laws of large numbers}
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\subsection{The Laws of Large Numbers}
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We want to show laws of large numbers:
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The LHS is random and represents ``sane'' averaging.
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The RHS is constant, which we can explicitly compute from the distribution of the RHS.
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We fix a probability space $(\Omega, \cF, \bP)$ once and for all.
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\begin{theorem}
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\label{lln}
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Let $X_1, X_2,\ldots$ be i.i.d.~random variables on $(\R, \cB(\R))$
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and $m = \bE[X_i] < \infty$
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and $\sigma^{2} = \Var(X_i) = \bE[ (X_i - \bE(X_i))^2] = \bE[X_i^2] - \bE[X_i]^2 < \infty$.
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Then
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\begin{enumerate}[(a)]
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\item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$
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in probability (\vocab{weak law of large numbers}, WLLN),
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\item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$
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almost surely (\vocab{strong law of large numbers}, SLLN).
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\end{enumerate}
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\end{theorem}
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\begin{refproof}{lln}
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\begin{enumerate}[(a)]
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\item Given $\epsilon > 0$, we need to show that
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\[
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\bP\left[
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\left| \frac{X_1 + \ldots + X_n}{n} - m\right| > \epsilon
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\right] \xrightarrow{n \to 0} 0.
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\]
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Let $S_n \coloneqq X_1 + \ldots + X_n$.
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Then $\bE[S_n] = \bE[X_1] + \ldots + \bE[X_n] = nm$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n} - m\right| > \epsilon\right]
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&=& \bP\left[\left|\frac{S_n}{n}-m\right| > \epsilon\right]\\
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&\overset{\text{Chebyshev}}{\le }&
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\frac{\Var\left( \frac{S_n}{n} \right) }{\epsilon^2}
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= \frac{1}{n} \frac{\Var(X_1)}{\epsilon^2}
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\xrightarrow{n \to \infty} 0
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\end{IEEEeqnarray*}
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since
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\[\Var\left(\frac{S_n}{n}\right)
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= \frac{1}{n^2} \Var\left(S_n\right)
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= \frac{1}{n^2} n \Var(X_i).\]
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\end{enumerate}
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\end{refproof}
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For the proof of (b) we need the following general result:
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\begin{theorem}
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\label{thm2}
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Let $X_1, X_2, \ldots$ be independent (but not necessarily identically distributed) random variables with $\bE[X_i] = 0$ for all $i$
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and
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\[\sum_{i=1}^n \Var(X_i) < \infty.\]
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Then $\sum_{n \ge 1} X_n$ converges almost surely.
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\end{theorem}
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We'll prove this later\todo{Move proof}
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\begin{question}
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Does the converse hold? I.e.~does $\sum_{n \ge 1} X_n < \infty$ a.s.~
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then $\sum_{n \ge 1} \Var(X_n) < \infty$.
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\end{question}
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This does not hold.
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Consider the following:
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\begin{example}
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Let $X_1,X_2,\ldots$ be independent random variables,
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where $X_n$ has distribution
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$\frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$.
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We have $\bP[X_n \neq 0] = \frac{2}{n^2}$.
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Since this is summable, Borel-Cantelli yields
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\[
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\bP[X_{n} \neq 0 \text{ for infinitely many $n$}] = 0.
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\]
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In particular, $X_n$ is summable almost surely.
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However $\Var(X_n) = 2$ is not summable.
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\end{example}
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