221 lines
7.9 KiB
TeX
221 lines
7.9 KiB
TeX
\lecture{9}{}{Percolation, Introduction to characteristic functions}
|
|
\subsubsection{Application: Percolation}
|
|
|
|
We will now discuss another application of Kolmogorov's $0-1$-law, percolation.
|
|
|
|
\begin{definition}[\vocab{Percolation}]
|
|
Consider the graph with nodes $\Z^d$, $d \ge 2$, where edges from the lattice are added with probability $p$. The added edges are called \vocab[Percolation!Edge!open]{open};
|
|
all other edges are called
|
|
\vocab[Percolation!Edge!closed]{closed}.
|
|
|
|
More formally, we consider
|
|
\begin{itemize}
|
|
\item $\Omega = \{0,1\}^{\bE_d}$, where $\bE_d$ are all edges in $\Z^d$,
|
|
\item $\cF \coloneqq \text{product $\sigma$-algebra}$,
|
|
\item $\bP \coloneqq \left(p \underbrace{\delta_{\{1\} }}_{\text{edge is open}} + (1-p) \underbrace{\delta_{\{0\} }}_{\text{edge is absent closed}}\right)^{\otimes \bE_d}$.
|
|
\end{itemize}
|
|
\end{definition}
|
|
\begin{question}
|
|
Starting at the origin, what is the probability, that there exists
|
|
an infinite path (without moving backwards)?
|
|
\end{question}
|
|
\begin{definition}
|
|
An \vocab{infinite path} consists of an infinite sequence of distinct points
|
|
$x_0, x_1, \ldots$
|
|
such that $x_n$ is connected to $x_{n+1}$, i.e.~the edge $\{x_n, x_{n+1}\}$ is open.
|
|
\end{definition}
|
|
|
|
Let $C_\infty \coloneqq \{\omega | \text{an infinite path exists}\}$.
|
|
|
|
\begin{exercise}
|
|
Show that changing the presence / absence of finitely many edges
|
|
does not change the existence of an infinite path.
|
|
Therefore $C_\infty$ is an element of the tail $\sigma$-algebra.
|
|
Hence $\bP(C_\infty) \in \{0,1\}$.
|
|
\end{exercise}
|
|
Obviously, $\bP(C_\infty)$ is monotonic with respect to $p$.
|
|
For $d = 2$ it is known that $p = \frac{1}{2}$ is the critical value.
|
|
For $d > 2$ this is unknown.
|
|
|
|
% TODO: more in the notes
|
|
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}.
|
|
|
|
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 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} \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.~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
|
|
\end{IEEEeqnarray*}
|
|
|
|
For the r.h.s.~we get a mean of $0$ and a variance of $1$.
|
|
So, to determine what \eqref{eqn:lec09ast} could mean, it is necessary that $\sqrt{n}$
|
|
is the right scaling.
|
|
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.
|
|
|
|
\subsection{Convolutions${}^\dagger$}
|
|
|
|
\begin{definition}+[Convolution]
|
|
Let $\mu$ and $\nu$ be probability measures on $\R^d$.
|
|
Then the \vocab{convolution} of $\mu$ and $\nu$,
|
|
$\mu \ast \nu$,
|
|
is the probability measure on $\R^d$
|
|
defined by
|
|
\[
|
|
(\mu \ast \nu)(A) = \int_{\R^d} \int_{\R^d} \One_A(x + y) \mu(\dif x) \nu(\dif y).
|
|
\]
|
|
\end{definition}
|
|
\begin{fact}
|
|
If $\mu$ and $\nu$ have Lebesgue densities $f_\mu$ and $f_\nu$,
|
|
then the convolution has Lebesgue density
|
|
\[
|
|
f_{\mu \ast \nu}(x) =
|
|
\int_{\R^d} f_\mu(x - t) f_\nu(t) \lambda^d(\dif t).
|
|
\]
|
|
\end{fact}
|
|
|
|
\begin{fact}+[Exercise 6.1]
|
|
If $X_1,X_2,\ldots$ are independent with
|
|
distributions $X_1 \sim \mu_1$,
|
|
$X_2 \sim \mu_2, \ldots$,
|
|
then $X_1 + \ldots + X_n$
|
|
has distribution
|
|
\[
|
|
\mu_1 \ast \mu_2 \ast \ldots \ast \mu_n.
|
|
\]
|
|
\end{fact}
|
|
\todo{TODO}
|
|
|
|
|
|
\subsection{Characteristic Functions and Fourier Transform}
|
|
|
|
\begin{definition}
|
|
\label{def:characteristicfunction}
|
|
Consider $(\R, \cB(\R), \bP)$.
|
|
The \vocab{characteristic function} of $\bP$ is defined as
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi_{\bP}: \R &\longrightarrow & \C \\
|
|
t &\longmapsto & \int_{\R} e^{\i t x} \bP(\dif x).
|
|
\end{IEEEeqnarray*}
|
|
\end{definition}
|
|
\begin{abuse}
|
|
$\phi_\bP(t)$ will often be abbreviated as $\phi(t)$.
|
|
\end{abuse}
|
|
We have
|
|
\[
|
|
\phi(t) = \int_{\R} \cos(tx) \bP(dx) + \i \int_{\R} \sin(tx) \bP(dx).
|
|
\]
|
|
\begin{itemize}
|
|
\item Since $|e^{\i t x}| \le 1$ the function $\phi(\cdot )$ is always defined.
|
|
\item We have $\phi(0) = 1$.
|
|
\item $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$.
|
|
\end{itemize}
|
|
|
|
\begin{fact}+
|
|
Let $X$, $Y$ be independent random variables
|
|
and $a,b \in \R$.
|
|
Then
|
|
\begin{itemize}
|
|
\item $\phi_{a X + b}(t) = e^{\i t b} \phi_X(\frac{t}{a})$,
|
|
\item $\phi_{X + Y}(t) = \phi_X(t) \cdot \phi_Y(t)$.
|
|
\end{itemize}
|
|
\end{fact}
|
|
\begin{proof}
|
|
We have
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi_{a X + b}(t) &=& \bE[e^{\i t (aX + b)}]\\
|
|
&=& e^{\i t b} \bE[e^{\i t a X}]\\
|
|
&=& e^{\i t b} \phi_X(\frac{t}{a}).
|
|
\end{IEEEeqnarray*}
|
|
Furthermore
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi_{X + Y}(t) &=& \bE[e^{\i t (X + Y)}]\\
|
|
&=& \bE[e^{\i t X}] \bE[e^{\i t Y}]\\
|
|
&=& \phi_X(t) \phi_Y(t).
|
|
\end{IEEEeqnarray*}
|
|
\end{proof}
|
|
|
|
\begin{remark}
|
|
Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and
|
|
$X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable.
|
|
Then we can define
|
|
\[
|
|
\phi_X(t) \coloneqq \bE[e^{\i t X}]
|
|
= \int e^{\i t X(\omega)} \bP(\dif \omega)
|
|
= \int_{\R} e^{\i t x} \mu(dx) = \phi_\mu(t),
|
|
\]
|
|
where $\mu = \bP \circ X^{-1}$.
|
|
\end{remark}
|
|
|
|
\begin{theorem}[Inversion formula] % thm1
|
|
\label{inversionformula}
|
|
Let $(\Omega, \cB(\R), \bP)$ be a probability space.
|
|
Let $F$ be the distribution function of $\bP$
|
|
(i.e.~$F(x) = \bP((-\infty, x])$ for all $x \in \R$ ).
|
|
Then for every $a < b$ we have
|
|
\begin{eqnarray}
|
|
\frac{F(b) + F(b-)}{2} - \frac{F(a) + F(a-)}{2} = \lim_{T \to \infty} \frac{1}{2 \pi} \int_{-T}^T \frac{e^{-\i t b} - e^{- \i t a}}{- \i t} \phi(t) dt
|
|
\label{invf}
|
|
\end{eqnarray}
|
|
where $F(b-)$ is the left limit.
|
|
\end{theorem}
|
|
% TODO!
|
|
We will prove this later.
|
|
|
|
\begin{theorem}[Uniqueness theorem] % thm2
|
|
\label{charfuncuniqueness}
|
|
Let $\bP$ and $\Q$ be two probability measures on $(\R, \cB(\R))$.
|
|
Then $\phi_\bP = \phi_\Q \implies \bP = \Q$.
|
|
|
|
Therefore, probability measures are uniquely determined by their characteristic functions.
|
|
Moreover, \eqref{invf} gives a representation of $\bP$ (via $F$)
|
|
from $\phi$.
|
|
\end{theorem}
|
|
\begin{refproof}{charfuncuniqueness}
|
|
Assume that we have already shown \autoref{inversionformula}.
|
|
Suppose that $F$ and $G$ are the distribution functions of $\bP$ and $\Q$.
|
|
Let $a,b \in \R$ with $a < b$.
|
|
Assume that $a $ and $b$ are continuity points of both $F$ and $G$.
|
|
By \autoref{inversionformula} we have
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
F(b) - F(a) = G(b) - G(a) \label{eq:charfuncuniquefg}
|
|
\end{IEEEeqnarray*}
|
|
|
|
Since $F$ and $G$ are monotonic, \autoref{eq:charfuncuniquefg}
|
|
holds for all $a < b$ outside a countable set.
|
|
|
|
Take $a_n$ outside this countable set, such that $a_n \ssearrow -\infty$.
|
|
Then, \autoref{eq:charfuncuniquefg} implies that
|
|
$F(b) - F(a_n) = G(b) - G(a_n)$ hence $F(b) = G(b)$.
|
|
Since $F$ and $G$ are right-continuous, it follows that $F = G$.
|
|
\end{refproof}
|