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fixed typos

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Josia Pietsch 2023-07-12 16:09:48 +02:00
parent 6972a481bb
commit 5013048fbf
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2 changed files with 7 additions and 7 deletions
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inputs/lecture_09.tex
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@ -113,7 +113,7 @@ We have
$X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable. $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable.
Then we can define Then we can define
\[ \[
\phi_X(t) \coloneqq \bE[e^{\i t x}] \phi_X(t) \coloneqq \bE[e^{\i t X}]
= \int e^{\i t X(\omega)} \bP(\dif \omega) = \int e^{\i t X(\omega)} \bP(\dif \omega)
= \int_{\R} e^{\i t x} \mu(dx) = \phi_\mu(t), = \int_{\R} e^{\i t x} \mu(dx) = \phi_\mu(t),
\] \]

8
inputs/lecture_10.tex
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@ -30,11 +30,11 @@ where $\mu = \bP X^{-1}$.
&&\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)\\ &&\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{\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} \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} \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}{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)\\ &=& \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}) \bP(\dif 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}