some small changes

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Josia Pietsch 2023-07-10 23:56:21 +02:00
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commit e6d50be72e
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2 changed files with 18 additions and 12 deletions

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@ -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)\\

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@ -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
\[
\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} } \xrightarrow{n \to \infty} hv \cN(0, \Var(X_i)) (\ast)
\]
Why is $\sqrt{n}$ the right order? (Handwavey argument)
\begin{equation}
\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }
\xrightarrow{n \to \infty} \cN(0, \Var(X_i))
\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)\\
&=& \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
\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}