From 29a36b4dbc57618dff463a4f5ba485299c2f9b3f Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 28 Jul 2023 22:06:08 +0200 Subject: [PATCH] some small changes --- inputs/lecture_10.tex | 8 ++++---- inputs/lecture_20.tex | 3 ++- 2 files changed, 6 insertions(+), 5 deletions(-) diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex index d4a1faf..d3afe64 100644 --- a/inputs/lecture_10.tex +++ b/inputs/lecture_10.tex @@ -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)\\ diff --git a/inputs/lecture_20.tex b/inputs/lecture_20.tex index a177cde..4f117b6 100644 --- a/inputs/lecture_20.tex +++ b/inputs/lecture_20.tex @@ -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 $\frac{1}{p} + \frac{1}{q} = 1$, \[ \int X_{n_k} Y \dif \bP \to \int XY \dif \bP \]