222 lines
8.0 KiB
TeX
222 lines
8.0 KiB
TeX
\lecture{9}{}{Percolation, Introduction to characteristic functions}
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\subsubsection{Application: Percolation}
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We will now discuss another application of \yaref{kolmogorov01}, percolation.
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\begin{definition}[\vocab{Percolation}]
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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};
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all other edges are called
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\vocab[Percolation!Edge!closed]{closed}.
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More formally, we consider
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\begin{itemize}
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\item $\Omega = \{0,1\}^{\bE_d}$, where $\bE_d$ are all edges in $\Z^d$,
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\item $\cF \coloneqq \text{product $\sigma$-algebra}$,
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\item $\bP \coloneqq \left(p \underbrace{\delta_{\{1\} }}_{\text{edge is open}} + (1-p) \underbrace{\delta_{\{0\} }}_{\text{edge is absent closed}}\right)^{\otimes \bE_d}$.
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\end{itemize}
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\end{definition}
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\begin{question}
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Starting at the origin, what is the probability, that there exists
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an infinite path (without moving backwards)?
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\end{question}
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\begin{definition}
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An \vocab{infinite path} consists of an infinite sequence of distinct points
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$x_0, x_1, \ldots$
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such that $x_n$ is connected to $x_{n+1}$, i.e.~the edge $\{x_n, x_{n+1}\}$ is open.
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\end{definition}
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Let $C_\infty \coloneqq \{\omega | \text{an infinite path exists}\}$.
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\begin{exercise}
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Show that changing the presence / absence of finitely many edges
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does not change the existence of an infinite path.
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Therefore $C_\infty$ is an element of the tail $\sigma$-algebra.
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Hence $\bP(C_\infty) \in \{0,1\}$.
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\end{exercise}
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Obviously, $\bP(C_\infty)$ is monotonic with respect to $p$.
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For $d = 2$ it is known that $p = \frac{1}{2}$ is the critical value.
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For $d > 2$ this is unknown.
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% TODO: more in the notes
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We'll get back to percolation later.
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\section{Characteristic Functions, Weak Convergence and the Central Limit Theorem}
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% Characteristic functions are also known as the \vocab{Fourier transform}.
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%Weak convergence is also known as \vocab{convergence in distribution} / \vocab{convergence in law}.
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So far we have dealt with the average behaviour,
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\[
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\frac{\overbrace{X_1 + \ldots + X_n}^{\text{i.i.d.}}}{n} \to \bE(X_1).
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\]
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We now want to understand fluctuations from the average behaviour,
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i.e.\[
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X_1 + \ldots + X_n - n \cdot \bE(X_1).
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\]
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% TODO improve
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The question is, what happens on other timescales than $n$?
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An example is
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\begin{equation}
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\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }
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\xrightarrow{n \to \infty} \cN(0, \Var(X_i))
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\label{eqn:lec09ast}
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\end{equation}
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Why is $\sqrt{n}$ the right order?
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Handwavey argument:
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Suppose $X_1, X_2,\ldots$ are i.i.d.~with $X_1 \sim \cN(0,1)$.
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The mean of the l.h.s.~is $0$ and for the variance we get
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\begin{IEEEeqnarray*}{rCl}
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\Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} })
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&=& \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right)\\
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&=& \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
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\end{IEEEeqnarray*}
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For the r.h.s.~we get a mean of $0$ and a variance of $1$.
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So, to determine what \eqref{eqn:lec09ast} could mean, it is necessary that $\sqrt{n}$
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is the right scaling.
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To make \eqref{eqn:lec09ast} precise,
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we need another notion of convergence.
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This will be the weakest notion of convergence, hence it is called
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\vocab{weak convergence}.
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This notion of convergence will be defined in terms of
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characteristic functions of Fourier transforms.
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\subsection{Convolutions${}^\dagger$}
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\begin{definition}+[Convolution]
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Let $\mu$ and $\nu$ be probability measures on $\R^d$.
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Then the \vocab{convolution} of $\mu$ and $\nu$,
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$\mu \ast \nu$,
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is the probability measure on $\R^d$
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defined by
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\[
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(\mu \ast \nu)(A) = \int_{\R^d} \int_{\R^d} \One_A(x + y) \mu(\dif x) \nu(\dif y).
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\]
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\end{definition}
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\begin{fact}
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If $\mu$ and $\nu$ have Lebesgue densities $f_\mu$ and $f_\nu$,
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then the convolution has Lebesgue density
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\[
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f_{\mu \ast \nu}(x) =
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\int_{\R^d} f_\mu(x - t) f_\nu(t) \lambda^d(\dif t).
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\]
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\end{fact}
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\begin{fact}+[Exercise 6.1]
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If $X_1,X_2,\ldots$ are independent with
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distributions $X_1 \sim \mu_1$,
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$X_2 \sim \mu_2, \ldots$,
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then $X_1 + \ldots + X_n$
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has distribution
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\[
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\mu_1 \ast \mu_2 \ast \ldots \ast \mu_n.
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\]
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\end{fact}
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\todo{TODO}
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\subsection{Characteristic Functions and Fourier Transform}
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\begin{definition}
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\label{def:characteristicfunction}
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Consider $(\R, \cB(\R), \bP)$.
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The \vocab{characteristic function} of $\bP$ is defined as
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\begin{IEEEeqnarray*}{rCl}
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\phi_{\bP}: \R &\longrightarrow & \C \\
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t &\longmapsto & \int_{\R} e^{\i t x} \bP(\dif x).
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\end{IEEEeqnarray*}
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\end{definition}
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\begin{abuse}
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$\phi_\bP(t)$ will often be abbreviated as $\phi(t)$.
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\end{abuse}
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We have
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\[
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\phi(t) = \int_{\R} \cos(tx) \bP(dx) + \i \int_{\R} \sin(tx) \bP(dx).
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\]
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\begin{itemize}
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\item Since $|e^{\i t x}| \le 1$ the function $\phi(\cdot )$ is always defined.
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\item We have $\phi(0) = 1$.
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\item $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$.
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\end{itemize}
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\begin{fact}+
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Let $X$, $Y$ be independent random variables
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and $a,b \in \R$.
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Then
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\begin{itemize}
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\item $\phi_{a X + b}(t) = e^{\i t b} \phi_X(\frac{t}{a})$,
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\item $\phi_{X + Y}(t) = \phi_X(t) \cdot \phi_Y(t)$.
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\end{itemize}
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\end{fact}
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\begin{proof}
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We have
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\begin{IEEEeqnarray*}{rCl}
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\phi_{a X + b}(t) &=& \bE[e^{\i t (aX + b)}]\\
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&=& e^{\i t b} \bE[e^{\i t a X}]\\
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&=& e^{\i t b} \phi_X(\frac{t}{a}).
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\end{IEEEeqnarray*}
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Furthermore
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\begin{IEEEeqnarray*}{rCl}
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\phi_{X + Y}(t) &=& \bE[e^{\i t (X + Y)}]\\
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&=& \bE[e^{\i t X}] \bE[e^{\i t Y}]\\
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&=& \phi_X(t) \phi_Y(t).
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{remark}
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Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and
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$X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable.
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Then we can define
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\[
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\phi_X(t) \coloneqq \bE[e^{\i t X}]
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= \int e^{\i t X(\omega)} \bP(\dif \omega)
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= \int_{\R} e^{\i t x} \mu(dx) = \phi_\mu(t),
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\]
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where $\mu = \bP \circ X^{-1}$.
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\end{remark}
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\begin{theorem}[Inversion formula] % thm1
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\yalabel{Inversion Formula}{Inversion Formula}{inversionformula}
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Let $(\Omega, \cB(\R), \bP)$ be a probability space.
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Let $F$ be the distribution function of $\bP$
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(i.e.~$F(x) = \bP((-\infty, x])$ for all $x \in \R$ ).
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Then for every $a < b$ we have
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\begin{eqnarray}
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\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
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\label{invf}
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\end{eqnarray}
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where $F(b-)$ is the left limit.
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\end{theorem}
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% TODO!
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We will prove this later.
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\begin{theorem}[Uniqueness theorem] % thm2
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\yalabel{Uniqueness Theorem}{Uniqueness}{charfuncuniqueness}
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Let $\bP$ and $\Q$ be two probability measures on $(\R, \cB(\R))$.
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Then $\phi_\bP = \phi_\Q \implies \bP = \Q$.
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Therefore, probability measures are uniquely determined by their characteristic functions.
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Moreover, \eqref{invf} gives a representation of $\bP$ (via $F$)
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from $\phi$.
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\end{theorem}
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\begin{refproof}{charfuncuniqueness}
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Assume that we have already shown the \yaref{inversionformula}.
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Suppose that $F$ and $G$ are the distribution functions of $\bP$ and $\Q$.
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Let $a,b \in \R$ with $a < b$.
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Assume that $a $ and $b$ are continuity points of both $F$ and $G$.
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By the \yaref{inversionformula} we have
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\begin{IEEEeqnarray*}{rCl}
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F(b) - F(a) = G(b) - G(a) \label{eq:charfuncuniquefg}
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\end{IEEEeqnarray*}
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Since $F$ and $G$ are monotonic, \yaref{eq:charfuncuniquefg}
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holds for all $a < b$ outside a countable set.
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Take $a_n$ outside this countable set, such that $a_n \ssearrow -\infty$.
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Then, \yaref{eq:charfuncuniquefg} implies that
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$F(b) - F(a_n) = G(b) - G(a_n)$ hence $F(b) = G(b)$.
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Since $F$ and $G$ are right-continuous, it follows that $F = G$.
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\end{refproof}
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