From 4e3c50102297069f420838cfdfe814c486482666 Mon Sep 17 00:00:00 2001
From: Josia Pietsch <josia.pietsch@uni-bonn.de>
Date: Tue, 11 Jul 2023 23:37:14 +0200
Subject: [PATCH] more details on example from lecture 05

---
 inputs/lecture_05.tex | 33 +++++++++++++++++++++++++++------
 1 file changed, 27 insertions(+), 6 deletions(-)

diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex
index 76fc1ef..e927114 100644
--- a/inputs/lecture_05.tex
+++ b/inputs/lecture_05.tex
@@ -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}