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some small changes

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Josia Pietsch 2023-07-28 22:06:08 +02:00
parent a3b5b209ce
commit 29a36b4dbc
Signed by: josia
GPG key ID: E70B571D66986A2D
2 changed files with 6 additions and 5 deletions
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8
inputs/lecture_10.tex
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@ -26,12 +26,12 @@ where $\mu = \bP X^{-1}$.
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} \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)\\
&\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)\\
&=& \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}_{=0 \text{, as the function is odd}} \bP(\dif 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{\yaref{fact:sincint},\text{dominated convergence}}}{=}&
&\overset{\substack{\yaref{fact:sincint},\text{DCT}}}{=}&
\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}) \bP(\dif x)\\
&=& \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
@ -107,7 +107,7 @@ where $\mu = \bP X^{-1}$.
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) \dif x \dif t\\
\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
@ -130,7 +130,7 @@ However, Fourier analysis is not only useful for continuous probability density
\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} \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 \int_{-T}^T
e^{-\i t (y - x)} \dif t \bP(\dif y)\\

3
inputs/lecture_20.tex
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@ -98,7 +98,8 @@ we need the following theorem, which we won't prove here:
\begin{refproof}{martingaleisce}
Since $(X_n)_n$ is bounded in $L^p$, by \yaref{banachalaoglu},
there exists $X \in L^p$ and a subsequence
$(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
$(X_{n_k})_k$ such that for all $Y \in L^q$,
where as always $\frac{1}{p} + \frac{1}{q} = 1$,
\[
\int X_{n_k} Y \dif \bP \to \int XY \dif \bP
\]