some small changes

inputs/lecture_07.tex

@@ -122,7 +122,7 @@ More formally:
         For every fixed $n$, $Y_n$ and $Z_n$ are independent.
     \end{claim}
     \begin{subproof}
-        This is obvious, but well prove it carefully here.
+        This is obvious, but we will prove it carefully here.
         \begin{IEEEeqnarray*}{rCl}
             &&(\bP \otimes \bP) [Y_n \in (a,b) , Z_n \in (a',b') ]\\
             &=& (\bP\otimes\bP) \left( (\omega, \omega') : X_n(\omega) \in (a,b) \land X_n(\omega') \in (a',b') \right)\\

inputs/lecture_09.tex

@@ -50,32 +50,38 @@ So far we have dealt with the average behaviour,
 \[
 \frac{\overbrace{X_1 + \ldots + X_n}^{\text{i.i.d.}}}{n} \to  \bE(X_1).
 \]
-We now want to understand \vocab{fluctuations} from the average behaviour,
+We now want to understand fluctuations from the average behaviour,
 i.e.\[
 X_1 + \ldots + X_n - n \cdot \bE(X_1).
 \]
 % TODO improve
 The question is, what happens on other timescales than $n$?
 An example is
-\[
+\begin{equation}
-\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} } \xrightarrow{n \to \infty} hv \cN(0, \Var(X_i)) (\ast)
+    \frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }
-\]
+    \xrightarrow{n \to \infty} \cN(0, \Var(X_i))
-Why is $\sqrt{n}$ the right order? (Handwavey argument)
+    \label{eqn:lec09ast}
+\end{equation}
+Why is $\sqrt{n}$ the right order?
+Handwavey argument:
 
-Suppose $X_1, X_2,\ldots$ are i.i.d. $\cN(0,1)$.
+Suppose $X_1, X_2,\ldots$ are i.i.d.~with $X_1 \sim \cN(0,1)$.
 The mean of the l.h.s.~is $0$ and for the variance we get
 \begin{IEEEeqnarray*}{rCl}
-    \Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) &=& \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right)\\
+    \Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} })
+        &=& \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right)\\
         &=& \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
 \end{IEEEeqnarray*}
 
 For the r.h.s.~we get a mean of  $0$ and a variance of $1$.
-So, to determine what $(\ast)$ could mean, it is necessary that $\sqrt{n}$
+So, to determine what \eqref{eqn:lec09ast} could mean, it is necessary that $\sqrt{n}$
 is the right scaling.
-To define $(\ast)$ we need another notion of convergence.
+To make \eqref{eqn:lec09ast} precise,
+we need another notion of convergence.
 This will be the weakest notion of convergence, hence it is called
 \vocab{weak convergence}.
-This notion of convergence will be defined in terms of characteristic functions of Fourier transforms.
+This notion of convergence will be defined in terms of
+characteristic functions of Fourier transforms.
 
 \subsection{Characteristic Functions and Fourier Transform}