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@ -9,7 +9,7 @@ We consider $(\R, \cB(\R))$.
By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$. By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
\end{notation} \end{notation}
For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}\dif\bP(x)$. For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}\bP(\dif x)$.
If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
$\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$, $\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$,
where $\mu = \bP X^{-1}$. where $\mu = \bP X^{-1}$.
@ -20,22 +20,22 @@ where $\mu = \bP X^{-1}$.
exists and is equal to the LHS. exists and is equal to the LHS.
Note that the term on the RHS is integrable, as Note that the term on the RHS is integrable, as
\[ \[
\lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \pi(t) = a - b \lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \phi(t) = a - b
\] \]
and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$. and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
% TODO think about this % TODO think about this
We have We have
\begin{IEEEeqnarray*}{rCl} \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} dt \dif\bP(x)\\ &&\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)\\
&\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} dt \dif\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} \dif t \bP(\dif 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} dt \dif\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} \dif t \bP(\dif 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] dt \dif\bP(x)}_{=0 \text{, as the function is odd}} &=& \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 \bP(\dif 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 \dif\bP(x)\\ + \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)\\
&=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt \dif\bP(x)\\ &=& \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)\\
&\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 } &\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}) \dif\bP(x)\\ - (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) \bP(\dif x)\\
&=& \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\ &=& \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} &=& \frac{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
@ -44,13 +44,13 @@ where $\mu = \bP X^{-1}$.
\begin{fact} \begin{fact}
\label{fact:intsinxx} \label{fact:intsinxx}
\[ \[
\int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2} \int_0^\infty \frac{\sin x}{x} \dif x = \frac{\pi}{2}
\] \]
where the LHS is an improper Riemann-integral. where the LHS is an improper Riemann-integral.
Note that the LHS is not Lebesgue-integrable. Note that the LHS is not Lebesgue-integrable.
It follows that It follows that
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{x} dt &=& \lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{x} \dif t &=&
\begin{cases} \begin{cases}
- \frac{\pi}{2}, &x < a,\\ - \frac{\pi}{2}, &x < a,\\
0, &x = a,\\ 0, &x = a,\\
@ -64,7 +64,7 @@ where $\mu = \bP X^{-1}$.
Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$. Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
Then $\bP$ has a continuous probability density given by Then $\bP$ has a continuous probability density given by
\[ \[
f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) dt. f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) \dif t.
\] \]
\end{theorem} \end{theorem}
@ -83,7 +83,7 @@ where $\mu = \bP X^{-1}$.
\end{itemize} \end{itemize}
\end{example} \end{example}
\begin{refproof}{thm:lec10_3} \begin{refproof}{thm:lec10_3}
Let $f(x) \coloneqq \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) dt$. Let $f(x) \coloneqq \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) \dif t$.
\begin{claim} \begin{claim}
If $x_n \to x$, then $f(x_n) \to f(x)$. If $x_n \to x$, then $f(x_n) \to f(x)$.
\end{claim} \end{claim}
@ -101,17 +101,17 @@ where $\mu = \bP X^{-1}$.
We'll show that for all $a < b$ we have We'll show that for all $a < b$ we have
\[ \[
\bP\left( (a,b] \right) = \int_a^b (x) dx.\label{thm10_3eq1} \bP\left( (a,b] \right) = \int_a^b (x) \dif x.\label{thm10_3eq1}
\] \]
Let $F$ be the distribution function of $\bP$. Let $F$ be the distribution function of $\bP$.
It is enough to prove \autoref{thm10_3eq1} It is enough to prove \autoref{thm10_3eq1}
for all continuity points $a $ and $ b$ of $F$. for all continuity points $a $ and $ b$ of $F$.
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
RHS &\overset{\text{Fubini}}{=}& \frac{1}{2 \pi} \int_{\R} \int_{a}^b e^{-\i t x} \phi(t) dx dt\\ RHS &\overset{\text{Fubini}}{=}& \frac{1}{2 \pi} \int_{\R} \int_{a}^b e^{-\i t x} \phi(t) \dif x \dif t\\
&=& \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) \int_a^b e^{-\i t x} \dif x \dif t\\
&=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) dt\\ &=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) \dif t\\
&\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 &\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
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
By \autoref{inversionformula}, the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$. By \autoref{inversionformula}, the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
\end{refproof} \end{refproof}
@ -122,17 +122,17 @@ However, Fourier analysis is not only useful for continuous probability density
Let $\bP \in M_1(\lambda)$. Let $\bP \in M_1(\lambda)$.
Then Then
\[ \[
\forall x \in \R ~ \bP\left( \{x\} \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) dt. \forall x \in \R ~ \bP\left( \{x\} \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) \dif t.
\] \]
\end{theorem} \end{theorem}
\begin{refproof}{bochnersformula} \begin{refproof}{bochnersformula}
We have We have
\begin{IEEEeqnarray*}{rCl} \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} \dif\bP(y) \\ 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) \\
&\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\\ &\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))} \dif t\\
&=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} \dif\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\ &=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} \bP(\dif y) \int_{-T}^T \cos(t(y - x)) \dif t\\
&=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \dif\bP(y)\\ &=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \bP(\dif y)\\
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Furthermore Furthermore
\[ \[
@ -143,7 +143,7 @@ However, Fourier analysis is not only useful for continuous probability density
\] \]
Hence Hence
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \dif\bP(y) &=& \bP\left( \{x\}\right) \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \bP(\dif y) &=& \bP\left( \{x\}\right)
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
% TODO by dominated convergence? % TODO by dominated convergence?
\end{refproof} \end{refproof}
@ -168,9 +168,9 @@ However, Fourier analysis is not only useful for continuous probability density
For part (b) we have: For part (b) we have:
\begin{IEEEeqnarray*}{rCl} \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} \dif\bP(x)\\ \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)\\
&=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} \dif\bP(x)\\ &=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} \bP(\dif x)\\
&=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} \dif\bP(x)\\ &=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} \bP(\dif x)\\
&=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0 &=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\end{refproof} \end{refproof}