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lecture 12

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inputs/lecture_11.tex
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@ -65,6 +65,7 @@ and $\Var(\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}) = 1$.
\lim_{n \to \infty} \bP\left[\frac{S_n - n \mu}{\sigma \sqrt{n} } \le x\right] = \Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2 \pi} } e^{\frac{-t^2}{2}}dt.
\]
\end{theorem}
We will abbreviate the central limit theorem by \vocab{CLT}.
There exists a special case of this theorem, which was proved earlier:
\begin{theorem}[de-Moivre (1730, $p = 0.5$), Laplace (1812, general $p$ )]

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% Lecture 12 2023-05-16
We now want to prove \autoref{clt}.
The plan is to do the following:
\begin{enumerate}[1.]
\item Identify the characteristic function of a standard normal
\item Show that the characteristic functions of the $V_n$ converge pointwise
to that of $\cN$.
\item Apply \autoref{levycontinuity}
\end{enumerate}
First, we need to prove some properties of characteristic functions.
\begin{lemma}
\label{charfprops}
For every real random variable $X$, we have
\begin{enumerate}[(i)]
\item $\phi_X(0) = 1$ and $|\phi_X(t)| \le 1$ for all $t \in \R$.
\item $\phi_X$ is uniformly continuous.
\item If $\bE[|X|^n] < \infty$ for any $n \in \N$, then $\phi_X$ i
$n$-times continuously differentiable
and $\bE[X^n] = (-\i)^n \phi_X^{(n)}(0)$.
\item For independent random variables $X$ and $Y$, we have
\[
\phi_{X + Y}(t) = \phi_X(t) \cdot \phi_Y(t).
\]
\end{enumerate}
\end{lemma}
\begin{refproof}{charfprops}
\begin{enumerate}[(i)]
\item $\phi_X(0) = \bE[e^{\i 0 X}] = \bE[1] = 1$.
For $t \in \R$, we have $|\phi_X(t)| = |\bE[e^{\i t X}]| \overset{\text{Jensen}}{\le} \bE|e^{\i t X}|] = 1$.
\item Let $t, h \in \R$.
Then
\begin{IEEEeqnarray*}{rCl}
|\phi_X(t+h) - \phi_X(t)| &=& |\bE[e^{\i (t+h) X} - e^{\i t X}]|\\
&=& |\bE[e^{\i t X} (e^{\i h X} - 1)]|\\
&\overset{\text{Jensen}}{\le}&
\bE[|e^{\i t X}| \cdot |e^{\i h X} -1|]\\
&=& \bE[|e^{\i h X} - 1|]\\
\end{IEEEeqnarray*}
Hence $\sup_{t \in \R} |\phi_X(t + h) - \phi_X(t) | \le g(h)$.
We show that $\lim_{h \to 0} g(h) = 0$.
For all $\omega \in \Omega$, we realize
\[
\lim_{h \to 0} |e^{\i h X(\omega)}- 1| = 0.
\]
Thus $|e^{\i h X} - 1| \xrightarrow{h \to 0} 0$ almost surely.
Since also for all $h \in \R$ we have $|e^{\i h X} - 1| \le 2$,
it follow that $|e^{\i h X} - 1|$ is dominated for all $h \in \R$.
Thus, we can apply the dominated convergence theorem % TODO REF
and obtain
\[
\lim_{h \to 0} g(h) = \lim_{h \to 0} \bE[|e^{\i h X} - 1|]
= \bE[\lim_{h \to 0} |e^{\i h X} - 1|] = 0.
\]
It follows that
\[
\lim_{n \to 0} \sup_{t \in \R} | \phi_X(t + h) - \phi_X(t)| = 0,
\]
which means that $\phi_X$ is uniformly continuous.
\item
\begin{claim}
\label{charfprop:c1}
For $y \in \R$, we have $|e^{\i y} - 1| \le |y|$.
\end{claim}
\begin{subproof}
For $y \ge 0$, we have
\begin{IEEEeqnarray*}{rCl}
|e^{\i y} - 1| &=& |\int_0^y \cos(s) \d s + \i \int_0^y \sin(s) \d s|\\
&=& |\int_0^y e^{\i s} \d s|\\
&\overset{\text{Jensen}}{\le}& \int_0^y |e^{\i s}| ds = y.
\end{IEEEeqnarray*}
For $y < 0$, we have $|e^{\i y} - 1| = |e^{-\i y} - 1|$
and we can apply the above to $-y$.
\end{subproof}
First, we look at $n = 1$. Then $\bE[|X|] < \infty$.
Consider
\[
\frac{\phi_X(t + h) - \phi_X(t)}{h} = \bE\left[e^{\i t X} \frac{e^{\i h X} - 1}{h}\right].
\]
We have $e^z = \sum_{n = 0}^\infty \frac{z^k}{n!}$.
Hence
\begin{IEEEeqnarray*}{rCl}
\lim_{n \to \infty} e^{\i t X} \left( \frac{1 + \i h X + \frac{(\i h X)^2}{2} + o(h^2) - 1}{h} \right)
&=& e^{\i t X} \i X \text{ almost surely.}
\end{IEEEeqnarray*}
For arbitrary $h \in \R$, we have
\begin{IEEEeqnarray*}{rCl}
|e^{\i t X} \frac{e^{\i h X}}{h}| &\le & \left| \frac{1}{h} \left( e^{\i h X} - 1 \right)\right|\\
&\overset{\text{\autoref{charfprop:c1}}}{\le }& \left|\frac{1}{h} \i h X\right| = |X|.
\end{IEEEeqnarray*}
Thus the dominated convergence theorem can be applied and we obtain
\[
\lim_{h \to 0} \frac{\phi_X(t + h) - \phi_X(t)}{h} = \lim_{h \to 0} \bE\left[ e^{\i t X} \left( \frac{e^{\i h X}-1}{h} \right) \right]
= \bE[e^{\i t X} \i X].
\]
It follows that $\phi_X$ is differentiable and $\phi_X(t) = \bE[e^{\i t X} \i X]$.
For $t = 0$ we get $\phi'_X(0) = \i \bE[X]$, i.e.~
-$\i \phi'_X(0) = \bE[X]$.
Adapting the proof of (ii) gives that
$\phi'_X(t)$ is continuous.
Adapting the proof of (iii) gives
the statement for arbitrary $n \in \N$.
\item Easy exercise.
\end{enumerate}
\end{refproof}
\begin{lemma}\label{lec12_2} % Lemma 2
For $X \sim \cN(0,1)$, we have $\phi_X(t) = e^{-\frac{t^2}{2}}$
for all $t \in \R$.
\end{lemma}
\begin{refproof}{lec12_2}
We have
\begin{IEEEeqnarray*}{rCl}
\phi_X(t) &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \d x\\
&=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \d x\\
&=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \d x,\\
\end{IEEEeqnarray*}
since, as $x \mapsto \sin(tx)$ is odd and $x \mapsto e^{-\frac{x^2}{2}}$
is even, their product is odd, wich gives that the integral is $0$.
\begin{IEEEeqnarray*}{rCl}
\phi'_X(t) &=& \bE[\i X e^{\i t X}] \\
&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \i x \left( \cos(t x) + \i \sin(tx) \right) e^{-\frac{x^2}{2}} \d x\\
&=& \frac{1}{\sqrt{2 \pi}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \d x\\
&=& \frac{1}{\sqrt{2 \pi} } \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}} \d x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \d x\right)\\
&=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi} }(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \d x\\
&=& \underbrace{[ \sin(tx) \frac{1}{\sqrt{2 \pi} e^{-\frac{x^2}{2}}}]_{x=-\infty}^\infty}_{=0}
- \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \d x\\
&=& -t \phi_X(t)
\end{IEEEeqnarray*}
Thus, for all $t \in \R$
\[
(\log(\phi_X(t))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t.
\]
Hence there exists $c \in \R$, such that
\[
\log(\phi_X(t)) = -\frac{t^2}{2} + c.
\]
Since $\phi_X(0) = 1$, we obtain $c = 0$.
Thus
\[
\phi_X(t) = e^{-\frac{t^2}{2}}.
\]
\end{refproof}
Now, we can finally prove the CLT:
\begin{refproof}{clt}
Let $X_1,X_2,\ldots$ be i.i.d.~random variables
with $\bE[X_1] = \mu_1$, $\Var(X_1) = \sigma^2$.
Let
\[
Y_i \coloneqq \frac{X_i - \mu}{\sigma}
\]
i.e.~we normalize to $\bE[Y_1] = 0$ and $\Var(Y_1) = 1$.
We need to show that
\[
V_n \coloneqq \frac{S_n - n \mu}{ \sigma \sqrt{n}} = \frac{Y_1+ \ldots + Y_n}{\sqrt{n}} \xrightarrow{\omega, n\to \infty} \cN(0,1) % TODO
\]
Let $t \in \R$.
Then
\begin{IEEEeqnarray*}{rCl}
\phi_{V_n}(t) = \bE[e^{\i t Y_n}] = \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n} } \right) }] \\
&=& \bE[e^{\i t \frac{Y_1}{\sqrt{n}}}] \cdot \ldots \cdot \bE[e^{\i t \frac{Y_n}{\sqrt{n} }}]\\
&=& \left( \phi(\frac{t}{\sqrt{n} } \right)^n.
\end{IEEEeqnarray*}
where $\phi(t) \coloneqq \phi_{Y_1}(t)$.
We have
\begin{IEEEeqnarray*}{rCl}
\phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{as $s \to 0$}\\
&=& 1 - \underbrace{\i \bE[Y_1] s}_{=0}
- \bE[Y_1^2] \frac{s^2}{2} + o(s^2)\\
&=& 1 - \frac{s^2}{2} + o(s^2), \text{as $s \to $}
\end{IEEEeqnarray*}
Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain
\[
\phi\left(\frac{t}{ \sqrt{n} }\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$}
\]
\[
\phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
(1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
\]
where we have used the following:
\begin{claim}
For a sequence $a_n, n\in \N$ with $\lim_{n \to \infty} n a_n = \lambda$,
it holds that $\lim_{n \to \infty} (1 + a_n)^n = e^{\lambda}$.
\end{claim}
We have shown that
\[
\phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_N(t).
\]
Using \autoref{levycontinuity}, we obtain \autoref{clt}.
\end{refproof}
\begin{remark}
If $X: \Omega \to \R^d$ with distribution $\nu$,
we define
\begin{IEEEeqnarray*}{rCl}
\phi_X: \R^d &\longrightarrow & \C \\
t &\longmapsto & \bE[e^{\i \langle t , X\rangle}]
\end{IEEEeqnarray*}
where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$.
\end{remark}
Exercise: Find out, which properties also hold for $d > 1$.

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inputs/lecture_9.tex
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@ -43,9 +43,8 @@ We'll get back to percolation later.
\section{Characteristic functions, weak convergence and the central limit theorem}
Characteristic functions are also known as the \vocab{Fourier transform}.
Weak convergence is also known as \vocab{convergence in distribution} / \vocab{convergence in law}.
We will abbreviate the central limit theorem by \vocab{CLT}.
% Characteristic functions are also known as the \vocab{Fourier transform}.
%Weak convergence is also known as \vocab{convergence in distribution} / \vocab{convergence in law}.
So far we have dealt with the average behaviour,
\[
@ -79,7 +78,7 @@ This notion of convergence will be defined in terms of characteristic functions
\subsection{Characteristic functions and Fourier transform}
Consider $(\R, \cB(\R), \bP)$.
For every $t \in \R$ define a function $\phi(t) \coloneqq \phi_\bP(t) \coloneqq \int_{\R} e^{\i t x} \bP(dx)$.
For every $t \in \R$ define a function $\phi(t) \coloneqq \phi_\bP(t) \coloneqq \int_{\R} e^{\i t x} \bP(dx)$.
We have
\[
\phi(t) = \int_{\R} \cos(tx) \bP(dx) + \i \int_{\R} \sin(tx) \bP(dx).

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@ -35,6 +35,7 @@
\input{inputs/lecture_9.tex}
\input{inputs/lecture_10.tex}
\input{inputs/lecture_11.tex}
\input{inputs/lecture_12.tex}
\cleardoublepage