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Josia Pietsch 2023-07-12 15:25:25 +02:00
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inputs/lecture_10.tex
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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 $\left( \R, \cB(\R) \right)$.
\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
$\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$,
where $\mu = \bP X^{-1}$.
@ -20,22 +20,22 @@ where $\mu = \bP X^{-1}$.
exists and is equal to the LHS.
Note that the term on the RHS is integrable, as
\[
\lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \pi(t) = a - b
\lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \phi(t) = a - b
\]
and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
% 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} 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} 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} dt \dif\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] dt \dif\bP(x)}_{=0 \text{, as the function is odd}}
&&\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} \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} \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] \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}{\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}{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} \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 }
- (- \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{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
\end{IEEEeqnarray*}
@ -44,13 +44,13 @@ where $\mu = \bP X^{-1}$.
\begin{fact}
\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.
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 &=&
\lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{x} \dif t &=&
\begin{cases}
- \frac{\pi}{2}, &x < a,\\
0, &x = a,\\
@ -64,7 +64,7 @@ where $\mu = \bP X^{-1}$.
Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
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}
@ -83,7 +83,7 @@ where $\mu = \bP X^{-1}$.
\end{itemize}
\end{example}
\begin{refproof}{thm:lec10_3}
Let $f(x) \coloneqq \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) dt$.
Let $f(x) \coloneqq \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) \dif t$.
\begin{claim}
If $x_n \to x$, then $f(x_n) \to f(x)$.
\end{claim}
@ -101,17 +101,17 @@ where $\mu = \bP X^{-1}$.
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 $\bP$.
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
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} \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) \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) \dif t
\end{IEEEeqnarray*}
By \autoref{inversionformula}, the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
\end{refproof}
@ -122,17 +122,17 @@ However, Fourier analysis is not only useful for continuous probability density
Let $\bP \in M_1(\lambda)$.
Then
\[
\forall x \in \R ~ \bP\left( \{x\} \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) dt.
\forall x \in \R ~ \bP\left( \{x\} \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) \dif t.
\]
\end{theorem}
\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} \dif\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} \dif\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)} \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))} \dif t\\
&=& \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)} \bP(\dif y)\\
\end{IEEEeqnarray*}
Furthermore
\[
@ -143,7 +143,7 @@ However, Fourier analysis is not only useful for continuous probability density
\]
Hence
\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*}
% TODO by dominated convergence?
\end{refproof}
@ -168,9 +168,9 @@ However, Fourier analysis is not only useful for continuous probability density
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} \dif\bP(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 e^{\i t_j x} \overline{c_k e^{\i 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}} \bP(\dif 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
\end{IEEEeqnarray*}
\end{refproof}