more details on example from lecture 05
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@ -26,18 +26,26 @@ We fix a probability space $(\Omega, \cF, \bP)$ once and for all.
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\begin{enumerate}[(a)]
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\begin{enumerate}[(a)]
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\item Given $\epsilon > 0$, we need to show that
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\item Given $\epsilon > 0$, we need to show that
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\[
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\[
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\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] \to 0 \]
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\bP\left[
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as $n \to 0$.
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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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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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Then $\bE[S_n] = \bE[X_1] + \ldots + \bE[X_n] = nm$.
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We have
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We have
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] &=& \bP\left[\left|\frac{S_n}{n}-m\right| > \epsilon\right]\\
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\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n} - m\right| > \epsilon\right]
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&\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
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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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\end{IEEEeqnarray*}
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since
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since
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\[\Var(\frac{S_n}{n}) = \frac{1}{n^2} \Var(S_n) = \frac{1}{n^2} n \Var(X_i).\]
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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{enumerate}
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\end{refproof}
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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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For the proof of (b) we need the following general result:
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@ -54,4 +62,17 @@ We'll prove this later\todo{Move proof}
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Does the converse hold? I.e.~does $\sum_{n \ge 1} X_n < \infty$ a.s.~
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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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then $\sum_{n \ge 1} \Var(X_n) < \infty$.
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\end{question}
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\end{question}
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This does not hold. Consider for example $X_n = \frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$.
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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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