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Josia Pietsch 2023-07-12 18:24:36 +02:00
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inputs/lecture_10.tex
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@ -133,7 +133,7 @@ However, Fourier analysis is not only useful for continuous probability density
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) \\ 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\\ &\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}{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)\\ &=& \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
\[ \[
@ -141,10 +141,11 @@ However, Fourier analysis is not only useful for continuous probability density
1, &y = x,\\ 1, &y = x,\\
0, &y \neq x. 0, &y \neq x.
\end{cases} \end{cases}
% TODO TODO TODO
\] \]
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)} \bP(\dif 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}
@ -154,7 +155,7 @@ However, Fourier analysis is not only useful for continuous probability density
Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$. Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
Then Then
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item $\phi(0) = 1$, $|\phi(1)| \le t$ and $\phi(\cdot )$ is continuous. \item $\phi(0) = 1$, $|\phi(t)| \le 1$ and $\phi(\cdot )$ is continuous.
\item $\phi$ is a \vocab{positive definite function}, \item $\phi$ is a \vocab{positive definite function},
i.e.~ 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 \[\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
@ -164,8 +165,6 @@ However, Fourier analysis is not only useful for continuous probability density
\end{theorem} \end{theorem}
\begin{refproof}{thm:lec_10thm5} \begin{refproof}{thm:lec_10thm5}
Part (a) is obvious. Part (a) is obvious.
% TODO
For part (b) we have: For part (b) we have:
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}