339 lines
14 KiB
TeX
339 lines
14 KiB
TeX
\lecture{10}{2023-05-09}{}
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First, we will prove some of the most important facts about Fourier transforms.
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We consider $(\R, \cB(\R))$.
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\begin{notation}
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By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
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\end{notation}
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For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}\bP(\dif x)$.
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If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
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$\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$,
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where $\mu = \bP X^{-1}$.
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\begin{refproof}{inversionformula}
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We will prove that the limit in the RHS of \yaref{invf}
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exists and is equal to the LHS.
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Note that the term on the RHS is integrable, as
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\[
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\lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \phi(t) = a - b
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\]
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and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
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% TODO think about this
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We have
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\begin{IEEEeqnarray*}{rCl}
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&&\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} \dif t \bP(\dif x)\\
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&\overset{\yaref{thm:fubini}}{=}& \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} \dif t \bP(\dif x)\\
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&=& \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} \dif t \bP(\dif x)\\
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&=& \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] \dif t}_{=0 \text{, as the function is odd}} \bP(\dif x) \\
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&& + \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} \dif t \bP(\dif x)\\
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&=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} \dif t \bP(\dif x)\\
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&\overset{\substack{\yaref{fact:sincint},\text{DCT}}}{=}&
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\frac{1}{\pi} \int -\frac{\pi}{2} \One_{x < a} + \frac{\pi}{2} \One_{x > a}
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- (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) \bP(\dif x)\\
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&=& \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
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&=& \frac{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
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\end{IEEEeqnarray*}
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\end{refproof}
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\begin{fact}
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\label{fact:sincint}
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\[
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\int_0^\infty \frac{\sin x}{x} \dif x = \frac{\pi}{2}
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\]
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where the LHS is an improper Riemann-integral.
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Note that the LHS is not Lebesgue-integrable.
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It follows that
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\begin{IEEEeqnarray*}{rCl}
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\lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{t} \dif t &=&
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\begin{cases}
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- \frac{\pi}{2} &\text{if }x < a,\\
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0 &\text{if }x = a,\\
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\frac{\pi}{2}&\text{if } x > a.
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\end{cases}
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\end{IEEEeqnarray*}
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\end{fact}
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\begin{theorem} % Theorem 3
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\label{thm:lec10_3}
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Let $\bP \in M_1(\R)$ such that $\phi_\bP \in L^1(\lambda)$.
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Then $\bP$ has a continuous probability density given by
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\[
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f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\bP}(t) \dif t.
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\]
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\end{theorem}
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\begin{example}
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\begin{itemize}
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\item Let $\bP = \delta_{0}$.
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Then
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\[
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\phi_{\bP}(t) = \int e^{\i t x} \delta_0(\dif x) = e^{\i t 0 } = 1
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\]
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\item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
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Then
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\[
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\phi_{\bP}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
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\]
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\end{itemize}
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\end{example}
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\begin{refproof}{thm:lec10_3}
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Let $f(x) \coloneqq \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) \dif t$.
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\begin{claim}
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If $x_n \to x$, then $f(x_n) \to f(x)$.
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\end{claim}
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\begin{subproof}
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Suppose that
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$e^{-\i t x_n} \phi(t) \xrightarrow{n \to \infty} e^{-\i t x} \phi(t)$
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for all $t$.
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Since
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\[
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|e^{-\i t x} \phi(t)| \le |\phi(t)|
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\]
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and $\phi \in L^1$,
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we get $f(x_n) \to f(x)$
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by the dominated convergence theorem.
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\end{subproof}
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We'll show that for all $a < b$ we have
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\[
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\bP\left( (a,b] \right) = \int_a^b f(x) \dif x.\label{thm10_3eq1}
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\]
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Let $F$ be the distribution function of $\bP$.
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It is enough to prove \yaref{thm10_3eq1}
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for all continuity points $a $ and $ b$ of $F$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\rhs &\overset{\text{Fubini}}{=}& \frac{1}{2 \pi} \int_{\R} \int_{a}^b e^{-\i t x} \phi(t) \dif x \dif t\\
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&=& \frac{1}{2 \pi} \int_\R \phi(t) \int_a^b e^{-\i t x} \dif x \dif t\\
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&=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) \dif t\\
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&\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) \dif t
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\end{IEEEeqnarray*}
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By the \yaref{inversionformula},
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the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
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\end{refproof}
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However, Fourier analysis is not only useful for continuous probability density functions:
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\begin{theorem}[Bochner's formula for the mass at a point]
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\yalabel{Bochner's Formula for the Mass at a Point}{Bochner}{bochnersformula} % Theorem 4
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Let $\bP \in M_1(\lambda)$.
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Then
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\[
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\forall x \in \R .~ \bP\left( \{x\} \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) \dif t.
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\]
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\end{theorem}
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\begin{refproof}{bochnersformula}
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We have
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\begin{IEEEeqnarray*}{rCl}
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\rhs &=& \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \bP(\dif y) \\
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&\overset{\text{Fubini}}{=}&
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\lim_{T \to \infty} \frac{1}{2 T} \int_\R \int_{-T}^T
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e^{-\i t (y - x)} \dif t \bP(\dif y)\\
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&=& \lim_{T \to \infty} \frac{1}{2 T} \int_\R \int_{-T}^T
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\cos(t(y-x)) + \underbrace{\i \sin(t (y-x))}_{\text{odd}}
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\dif t \bP(\dif y)\\
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&=& \lim_{T \to \infty} \frac{1}{2T}\int_{\R}
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\int_{-T}^T \cos(t(y - x)) \dif t \bP(\dif y)\\
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&=& \lim_{T \to \infty} \frac{1}{2T}\int_{\R}
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2T \sinc(T(y-x))
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\footnote{$\sinc(x) = \begin{cases}
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\frac{\sin(x)}{x} &\text{if } x \neq 0,\\
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1 &\text{otherwise.}
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\end{cases}$} \bP(\dif y)\\
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&\overset{\text{DCT}}{=}& \int_{\R}\lim_{T \to \infty}
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\sinc(T(y-x)) \bP(\dif y)\\
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&=& \bP(\{x\}).
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\end{IEEEeqnarray*}
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\end{refproof}
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\begin{theorem} % Theorem 5
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\label{thm:lec_10thm5}
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Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
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Then
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\begin{enumerate}[(a)]
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\item $\phi(0) = 1$, $|\phi(t)| \le 1$, $\phi(-t) = \overline{\phi(t)}$
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and $\phi(\cdot )$ is continuous.
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\item $\phi$ is a \vocab{positive definite function},
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i.e.~
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\[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge 0
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\]
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Equivalently, the matrix $(\phi(t_j- t_k))_{j,k}$ is positive definite.
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\end{enumerate}
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\end{theorem}
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\begin{refproof}{thm:lec_10thm5}
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Part (a) is obvious.
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For part (b) we have:
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\begin{IEEEeqnarray*}{rCl}
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\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} \bP(\dif x)\\
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&=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} \bP(\dif x)\\
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&=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} \bP(\dif x)\\
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&=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0
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\end{IEEEeqnarray*}
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\end{refproof}
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\begin{theorem}[Bochner's theorem]
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\yalabel{Bochner's Theorem for Positive Definite Functions}{Bochner's Theorem}{thm:bochner}%
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The converse to \yaref{thm:lec_10thm5} holds, i.e.~any
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$\phi: \R \to \C$ satisfying (a) and (b) of \yaref{thm:lec_10thm5}
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must be the Fourier transform of a probability measure $\bP$
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on $(\R, \cB(\R))$.
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\end{theorem}
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Unfortunately, we won't prove \yaref{thm:bochner} in this lecture.
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\begin{definition}[Convergence in distribution / weak convergence]
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\label{def:weakconvergence}
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We say that $\bP_n \in M_1(\R)$ \vocab[Convergence!weak]{converges weakly} towards $\bP \in M_1(\R)$ (notation: $\bP_n \implies \bP$), iff
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\[
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\forall f \in C_b(\R)~ \int f \dif\bP_n \to \int f \dif\bP.
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\]
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Where
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\[
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C_b(\R) \coloneqq \{ f: \R \to \R \text{ continuous and bounded}\}
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\]
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In analysis, this is also known as $\text{weak}^\ast$ convergence.
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\end{definition}
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\begin{remark}
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This notion of convergence makes $M_1(\R)$ a separable metric space.
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We can construct a metric on $M_1(\R)$ that turns $M_1(\R)$ into a complete
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and separable metric space:
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Consider the sets
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\[
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\{\bP \in M_1(\R): \forall i=1,\ldots,n ~ \int f \dif\bP - \int f_i \dif\bP < \epsilon \}
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\]
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for any $f,f_1,\ldots, f_n \in C_b(\R)$.
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These sets form a basis for the topology on $M_1(\R)$.
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More of this will follow later.
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\end{remark}
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\begin{example}
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\begin{itemize}
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\item Let $\bP_n = \delta_{\frac{1}{n}}$.
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Then $\int f \dif\bP_n = f(\frac{1}{n}) \to f(0) = \int f d \delta_0$
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for any continuous, bounded function $f$.
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Hence $\bP_n \to \delta_0$.
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\item $\bP_n \coloneqq \delta_n$ does not converge weakly,
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as for example
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\[
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\int \cos(\pi x) \dif\bP_n(x)
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\]
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does not converge.
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\item $\bP_n \coloneqq \frac{1}{n} \delta_n + (1- \frac{1}{n}) \delta_0$.
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Let $f \in C_b(\R)$ arbitrary.
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Then
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\[
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\int f \dif\bP_n = \frac{1}{n}(n) + (1 - \frac{1}{n}) f(0) \to f(0)
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\]
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since $f$ is bounded.
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Hence $\bP_n \implies \delta_0$.
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\item $\bP_n \coloneqq \frac{1}{\sqrt{2 \pi n}} e^{-\frac{x^2}{2n}}$.
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This ``converges'' towards the $0$-measure,
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which is not a probability measure.
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Hence $\bP_n$ does not converge weakly.
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(Exercise) % TODO
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\end{itemize}
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\end{example}
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\begin{definition}
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We say that a series of random variables $X_n$
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\vocab[Convergence!in distribution]{converges in distribution}
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to $X$ (notation: $X_n \xrightarrow{\text{d}} X$), iff
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$\bP_n \implies \bP$, where $\bP_n$ is the distribution of $X_n$
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and $\bP$ is the distribution of $X$.
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\end{definition}
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It is easy to see, that this is equivalent to $\bE[f(X_n)] \to \bE[f(X)]$
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for all $f \in C_b(\R)$.
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\begin{example}
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Let $X_n \coloneqq \frac{1}{n}$
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and $F_n$ the distribution function, i.e.~$F_n = \One_{[\frac{1}{n},\infty)}$.
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Then $\bP_n = \delta_{\frac{1}{n}} \implies \delta_0$
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which is the distribution of $X \equiv 0$.
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But $F_n(0) \centernot\to F(0)$.
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\end{example}
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\begin{theorem} % Theorem 1
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\label{lec10_thm1}
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$X_n \xrightarrow{\text{d}} X$ iff
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$F_n(t) \to F(t)$ for all continuity points $t$ of $F$.
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\end{theorem}
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% \begin{proof}\footnote{This proof was not done in the lecture,
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% but can be found in the official notes from lecture 13}
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% ``$\implies$''
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% Suppose $\mu_n \implies \mu$.
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% Let $F_n$ and $F$ denote the respective density functions.
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% Fix a continuity point $x_0 \in \R$ of $F$.
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% We'll show
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% \[
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% \limsup_{n \to \infty} F_n(x_0) \le F(x_0) + \epsilon
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% \]
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% and
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% \[
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% \liminf_{ \to \infty} F_n(x_0) \ge F(x_0) - \epsilon
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% \]
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% for all $\epsilon > 0$.
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% Fix some $\epsilon > 0$.
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% Choose $\delta > 0$ such that $F(x_0 + \delta) < F(x_0) + \epsilon$
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% and define
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% \[
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% g(x) \coloneqq \begin{cases}
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% 1 &\text{if } x \le x_0,\\
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% 1 - \frac{1}{\delta}(x - x_0)&
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% \text{if } x \in (x_0, x_0 + \delta],\\
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% 0 &\text{if } x \ge x_0 + \delta.
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% \end{cases}
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% \]
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% Since $g$ is continuous and bounded, we have
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% \[
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% \int g \dif \mu_n \to \int g \dif \mu.
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% \]
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% It is clear that $\One_{(-\infty, x_0]} \le g$.
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% Hence
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% \[
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% F_n(x_0) = \int \One_{(-\infty, x_0]} \dif \mu_n \le \int g \dif \mu_n.
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% \]
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% It follows that
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% \begin{IEEEeqnarray*}{rCl}
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% \limsup_{n} F_n(x_0)
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% &\le& \limsup_n \int g \dif \mu_n\\
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% &=& \lim_n \int g \dif \mu_n\\
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% &=& \int g \dif \mu\\
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% &\overset{g \le \One_{(-\infty, x + \delta]}}{=}& F(x + \delta)\\
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% &=& F(x) + \epsilon.
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% \end{IEEEeqnarray*}
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% The assertion about $\liminf_{n \to \infty} F_n(x_0)$
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% follows by a similar argument.
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%
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% ``$\impliedby$''
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% Assume that $F_n(x) \to F(x)$ at all continuity points of $F$.
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% We need to show
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% \[
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% \fgrall g \in C_b(\R) .~\int g \dif \mu_n \to \int g \dif \mu.
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% \]
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% Let $C$ denote the set of continuity points of $f$.
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% We apply measure theoretic induction:
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% \begin{itemize}
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% \item For $g = \One_{(a,b]}$, $a< b \in C$,
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% we have
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% \[\int g \dif \mu_n = F_n(b) - F_n(a) \to F(b) - F(a) = \int g \dif \mu.\]
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% \item For $g = \sum_{i} \alpha_i \One_{(a_i, b_i]}$,
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% $a_i < b_i \in C$,
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% we get $\int g \dif \mu_n \to \int g \dif \mu$
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% by the same argument.
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% \item % TODO continue from Lec13 page 21 (iii)
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% \end{itemize}
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%
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% \end{proof}
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\begin{theorem}[Levy's continuity theorem]
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\yalabel{Levy's Continuity Theorem}{Levy}{levycontinuity}
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% Theorem 2
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$X_n \xrightarrow{\text{d}} X$ iff
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$\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$.
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\end{theorem}
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We will assume these two theorems for now and derive the central limit theorem.
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The theorems will be proved later.
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