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more details on example from lecture 05

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Josia Pietsch 2023-07-11 23:37:14 +02:00
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@ -26,18 +26,26 @@ We fix a probability space $(\Omega, \cF, \bP)$ once and for all.
\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 $n \to 0$.
\bP\left[
\left| \frac{X_1 + \ldots + X_n}{n} - m\right| > \epsilon
\right] \xrightarrow{n \to 0} 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
\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n} - m\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
\[\Var(\frac{S_n}{n}) = \frac{1}{n^2} \Var(S_n) = \frac{1}{n^2} n \Var(X_i).\]
\[\Var\left(\frac{S_n}{n}\right)
= \frac{1}{n^2} \Var\left(S_n\right)
= \frac{1}{n^2} n \Var(X_i).\]
\end{enumerate}
\end{refproof}
For the proof of (b) we need the following general result:
@ -54,4 +62,17 @@ We'll prove this later\todo{Move proof}
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 $X_n = \frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$.
This does not hold.
Consider the following:
\begin{example}
Let $X_1,X_2,\ldots$ be independent random variables,
where $X_n$ has distribution
$\frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$.
We have $\bP[X_n \neq 0] = \frac{2}{n^2}$.
Since this is summable, Borel-Cantelli yields
\[
\bP[X_{n} \neq 0 \text{ for infinitely many $n$}] = 0.
\]
In particular, $X_n$ is summable almost surely.
However $\Var(X_n) = 2$ is not summable.
\end{example}