% lecture 10 - 2023-05-09

% RECAP

First, we will prove some of the most important facts about Fourier transforms.

We consider $(\R, \cB(\R))$.
\begin{notation}
   By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
\end{notation}

For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}d\bP(x)$.
If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
$\phi_X(t) \coloneqq  \bE[e^{\i t X}] = \phi_{\mu}(t)$,
where $\mu = \bP X^{-1}$.


\begin{refproof}{inversionformula}
    We will prove that the limit in the RHS of \autoref{invf}
    exists and is equal to the LHS.
    Note that the term on the RHS is integrable, as
    \[
    \lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \pi(t) = a - b
    \] 
    and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
    % TODO think about this

    We have
    \begin{IEEEeqnarray*}{rCl}
        &&\lim_{T \to  \infty} \frac{1}{2 \pi} \int_{-T}^T \int_{\R} \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} d  \bP(x)\\
        &\overset{\text{Fubini for  $L^1$}}{=}& \lim_{T \to  \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} d  \bP(x)\\
        &=& \lim_{T \to  \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{\i t (b-x)}- e^{\i t (x-a)}}{-\i t} d  \bP(x)\\
        &=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \underbrace{\int_{-T}^T \left[ \frac{\cos(t (x-b)) - \cos(t(x-a))}{-\i t}\right] d \bP(x)}_{=0 \text{, as the function is odd}}
      \\&&
        + \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} dt d\bP(x)\\
        &=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt d\bP(x)\\
        &\overset{\substack{\text{\autoref{fact:intsinxx},}\\\text{dominated convergence}}}{=}& \frac{1}{\pi} \int -\frac{\pi}{2} \One_{x < a} + \frac{\pi}{2} \One_{x > a }
                                                                - (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) d\bP(x)\\
        &=&  \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
        &=& \frac{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
    \end{IEEEeqnarray*}
\end{refproof}

\begin{fact}
    \label{fact:intsinxx}
    \[
    \int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2}
    \] 
    where the LHS is an improper Riemann-integral.
    Note that the LHS is not Lebesgue-integrable.
    It follows that
    \begin{IEEEeqnarray*}{rCl}
        \lim_{T \to  \infty} \int_0^T \frac{\sin(t(x-a))}{x} dt &=&
    \begin{cases}
        - \frac{\pi}{2}, &x < a,\\
        0, &x = a,\\
        \frac{\pi}{2}, & \frac{\pi}{2}
    \end{cases}
    \end{IEEEeqnarray*}
\end{fact}

\begin{theorem} % Theorem 3
    \label{thm:lec10_3}
    Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
    Then $\bP$ has a continuous probability density given by
    \[
    f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R(t) dt}.
    \] 
\end{theorem}

\begin{example}
    \begin{itemize}
        \item Let $\bP = \delta_{\{0\}}$.
          Then
          \[
          \phi_{\R}(t) = \int e^{\i t x} d \delta_0(x) = e^{\i t 0 } = 1
          \] 
        \item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
            Then
            \[
            \phi_{\R}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
            \] 
    \end{itemize}
\end{example}
\begin{refproof}{thm:lec10_3}
    Let $f(x) \coloneqq  \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) dt$.
    \begin{claim}
        If $x_n \to  x$, then $f(x_n) \to  f(x)$.
    \end{claim}
    \begin{subproof}
        If $e^{-\i t x_n} \phi(t) \xrightarrow{n \to  \infty}  e^{-\i t x } \phi(t) $ for all $t$.

        Then
        \[
        |e^{-\i t x} \phi(t)| \le  |\phi(t)|
        \] 
        and $\phi \in L^1$, hence $f(x_n) \to  f(x)$
        by the dominated convergence theorem.
        
    \end{subproof}

    We'll show that for all  $a < b$ we have
    \[
    \bP\left( (a,b] \right)  = \int_a^b (x) dx.\label{thm10_3eq1}
    \] 
    Let $F$ be the distribution function of $\bP$.
    It is enough to prove \autoref{thm10_3eq1}
    for all continuity points $a $ and $ b$ of $F$.
    We have
    \begin{IEEEeqnarray*}{rCl}
        RHS &\overset{\text{Fubini}}{=}& \frac{1}{2 \pi} \int_{\R} \int_{a}^b e^{-\i t x} \phi(t) dx dt\\
            &=& \frac{1}{2 \pi} \int_\R \phi(t) \int_a^b e^{-\i t x} dx dt\\
            &=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) dt\\
            &\overset{\text{dominated convergence}}{=}& \lim_{T \to \infty} \frac{1}{2\pi} \int_{-T}^{T} \phi(t) \left( \frac{e^{-\i t b} - e^{- \i t a}}{- \i t} \right) dt
    \end{IEEEeqnarray*}
    By \autoref{inversionformula}, the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
\end{refproof}

However, Fourier analysis is not only useful for continuous probability density functions:

\begin{theorem}[Bochner's formula for the mass at a point]\label{bochnersformula} % Theorem 4
    Let $\bP \in M_1(\lambda)$.
    Then
    \[
    \forall x \in  \R ~ \bP\left( \{x\}  \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) dt.
    \] 
    
\end{theorem}
\begin{refproof}{bochnersformula}
    We have
    \begin{IEEEeqnarray*}{rCl}
        RHS &=& \lim_{T \to  \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} d \bP(y) \\
            &\overset{\text{Fubini}}{=}& \lim_{T \to  \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} dt\\
            &=& \lim_{T \to  \infty} \frac{1}{2T} \int_{\R} d\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\
            &=& \lim_{T \to  \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} d \bP(y)\\
    \end{IEEEeqnarray*}
    Furthermore
    \[
    \lim_{T \to  \infty} \frac{\sin(T(x-y)}{T (y- x)} = \begin{cases}
        1, &y = x,\\
        0, &y \neq  x.
    \end{cases}
    \] 
    Hence
    \begin{IEEEeqnarray*}{rCl}
            \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} d \bP(y) &=& \bP\left( \{x\}\right) 
    \end{IEEEeqnarray*}
    % TODO by dominated convergence?
\end{refproof}

\begin{theorem} % Theorem 5
    \label{thm:lec_10thm5}
    Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
    Then
    \begin{enumerate}[(a)]
        \item $\phi(0) = 1$, $|\phi(t)| \le t$ and $\phi(\cdot )$ is continuous.
        \item $\phi$ is a \vocab{positive definite function},
            i.e.~
            \[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge  0
            \]
            (equivalently, the matix $(\phi(t_j- t_k))_{j,k}$ is positive definite.
    \end{enumerate}
\end{theorem}
\begin{refproof}{thm:lec_10thm5}
    Part (a) is obvious.
    % TODO


    For part (b) we have:
    \begin{IEEEeqnarray*}{rCl}
        \sum_{j,k}  c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k}  c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} d \bP(x)\\
                                                       &=& \int_{\R} \sum_{j,k}  c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} d\bP(x)\\
                                                       &=& \int_{\R}\sum_{j,k}  c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} d\bP(x)\\
                                                       &=& \int_{\R} \left| \sum_{l}  c_l e^{\i t_l x}\right|^2 \ge  0
    \end{IEEEeqnarray*}
\end{refproof}
\begin{theorem}[Bochner's theorem]\label{bochnersthm}
    The converse to \autoref{thm:lec_10thm5} holds, i.e.~ any
    $\phi: \R \to  \C$ satisfying (a) and (b) of \autoref{thm:lec_10thm5}
    must be the Fourier transform of a probability measure $\bP$ 
    on $(\R, \cB(\R))$.
\end{theorem}
Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.


\begin{definition}[Convergence in distribution / weak convergence]
    We say that $\bP_n \subseteq  M_1(\R)$ \vocab[Convergence!weak]{converges weakly} towards $\bP \in M_1(\R)$ (notation: $\bP_n \implies \bP$), iff
    \[
    \forall f \in C_b(\R)~  \int f d\bP_n \to \int f d\bP.
    \]
    Where
    \[
    C_b(\R) \coloneqq  \{ f: \R \to  \R \text{ continuous and bounded}\}
    \]

    In analysis, this is also known as  $\text{weak}^\ast$ convergence.
\end{definition}
\begin{remark}
    This notion of convergence makes $M_1(\R)$ a separable metric space. We can construc a metric on $M_1(\R)$ that turns $M_1(\R)$ into a complete
    and separable metric space:

    Consider the sets
    \[
    \{\bP \in M_1(\R): \forall i=1,\ldots,n ~ \int f d \bP - \int f_i d\bP < \epsilon \}
    \]
    for any $f,f_1,\ldots, f_n \in C_b(\R)$.
    These sets form a basis for the topology on $M_1(\R)$.
    More of this will follow later.
\end{remark}
\begin{example}
    \begin{itemize}
        \item Let $\bP_n = \delta_{\frac{1}{n}}$.
            Then $\int f d \bP_n = f(\frac{1}{n}) \to f(0) = \int f d \delta_0$
            for any continuous, bounded function $f$.
            Hence $\bP_n \to \delta_0$.
        \item $\bP_n \coloneqq  \delta_n$ does not converge weakly,
            as for example
            \[
            \int \cos(\pi x) d\bP_n(x)
            \]
            does not converge.

        \item $\bP_n \coloneqq  \frac{1}{n} \delta_n + (1- \frac{1}{n}) \delta_0$.
            Let $f \in C_b(\R)$ arbitrary.
            Then
            \[
            \int f d\bP_n = \frac{1}{n}(n) + (1 - \frac{1}{n}) f(0) \to f(0)
            \]
            since $f$ is bounded.
            Hence $\bP_n \implies \delta_0$.
        \item $\bP_n \coloneqq  \frac{1}{\sqrt{2 \pi n}} e^{-\frac{x^2}{2n}}$.
            This ``converges'' towards the $0$-measure, which is not a probability measure. Hence $\bP_n$ does not converge weakly.
            (Exercise) % TODO
    \end{itemize}
\end{example}
\begin{definition}
    We say that a series of random variables $X_n$
     \vocab[Convergence!in distribution]{converges in distribution}
     to $X$ (notation: $X_n \xrightarrow{\text{dist}} X$), iff
     $\bP_n \implies \bP$, where $\bP_n$ is the distribution of $X_n$
     and  $\bP$ is the distribution of $X$.
\end{definition}
\begin{example}
    Let $X_n \coloneqq  \frac{1}{n}$
    and $F_n$ the distribution function, i.e.~$F_n = \One_{[\frac{1}{n},\infty)}$.
    Then $\bP_n = \delta_{\frac{1}{n}} \implies \delta_0$
    which is the distribution of $X \equiv 0$.
    But $F_n(0) \centernot\to F(0)$.
\end{example}
\begin{theorem}
    $X_n \xrightarrow{\text{dist}} X$ iff
    $F_n(t) \to  F(t)$ for all continuity points $t$ of $F$.
\end{theorem}
\begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
    $X_n \xrightarrow{\text{dist}} X$ iff
    $\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$.
\end{theorem}
We will assume these two theorems for now and derive the central limit theorem.
The theorems will be proved later.