s23-probability-theory/inputs/lecture_09.tex

\lecture{9}{}{Percolation, Introduction to characteristic functions}
\subsubsection{Application: Percolation}

We will now discuss another application of \yaref{kolmogorov01}, 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
    \yalabel{Inversion Formula}{Inversion Formula}{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
    \yalabel{Uniqueness Theorem}{Uniqueness}{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 the \yaref{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 the \yaref{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, \yaref{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, \yaref{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}