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Josia Pietsch 2023-07-13 00:06:30 +02:00
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49
inputs/lecture_10.tex
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@ -50,11 +50,11 @@ where $\mu = \bP X^{-1}$.
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} \dif t &=&
\lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{t} \dif t &=&
\begin{cases}
- \frac{\pi}{2}, &x < a,\\
0, &x = a,\\
\frac{\pi}{2}, & \frac{\pi}{2}
- \frac{\pi}{2} &\text{if }x < a,\\
0 &\text{if }x = a,\\
\frac{\pi}{2}&\text{if } x > a.
\end{cases}
\end{IEEEeqnarray*}
\end{fact}
@ -73,7 +73,7 @@ where $\mu = \bP X^{-1}$.
\item Let $\bP = \delta_{0}$.
Then
\[
\phi_{\bP}(t) = \int e^{\i t x} \dif \delta_0(x) = e^{\i t 0 } = 1
\phi_{\bP}(t) = \int e^{\i t x} \delta_0(\dif x) = e^{\i t 0 } = 1
\]
\item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
Then
@ -131,23 +131,24 @@ However, Fourier analysis is not only useful for continuous probability density
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) \\
&\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)\\
&\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)\\
&=& \lim_{T \to \infty} \frac{1}{2 T} \int_\R \int_{-T}^T
\cos(t(y-x)) + \underbrace{\i \sin(t (y-x))}_{\text{odd}}
\dif t \bP(\dif y)\\
&=& \lim_{T \to \infty} \frac{1}{2T}\int_{\R}
\int_{-T}^T \cos(t(y - x)) \dif t \bP(\dif y)\\
&=& \lim_{T \to \infty} \frac{1}{2T}\int_{\R}
2T \sinc(T(y-x))
\footnote{$\sinc(x) = \begin{cases}
\frac{\sin(x)}{x} &\text{if } x \neq 0,\\
1 &\text{otherwise.}
\end{cases}$} \bP(\dif y)\\
&\overset{\text{DCT}}{=}& \int_{\R}\lim_{T \to \infty}
\sinc(T(y-x)) \bP(\dif y)\\
&=& \bP(\{x\}).
\end{IEEEeqnarray*}
Furthermore
\[
\lim_{T \to \infty} \frac{\sin(T(x-y)}{T (y- x)} = \begin{cases}
1, &y = x,\\
0, &y \neq x.
\end{cases}
% TODO TODO TODO
\]
Hence
\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)
\end{IEEEeqnarray*}
% TODO by dominated convergence?
\end{refproof}
\begin{theorem} % Theorem 5
@ -160,7 +161,7 @@ However, Fourier analysis is not only useful for continuous probability density
i.e.~
\[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge 0
\]
(equivalently, the matix $(\phi(t_j- t_k))_{j,k}$ is positive definite.
Equivalently, the matrix $(\phi(t_j- t_k))_{j,k}$ is positive definite.
\end{enumerate}
\end{theorem}
\begin{refproof}{thm:lec_10thm5}
@ -231,7 +232,9 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
since $f$ is bounded.
Hence $\bP_n \implies \delta_0$.
\item $\bP_n \coloneqq \frac{1}{\sqrt{2 \pi n}} e^{-\frac{x^2}{2n}}$.
This ``converges'' towards the $0$-measure, which is not a probability measure. Hence $\bP_n$ does not converge weakly.
This ``converges'' towards the $0$-measure,
which is not a probability measure.
Hence $\bP_n$ does not converge weakly.
(Exercise) % TODO
\end{itemize}
\end{example}

31
inputs/lecture_12.tex
View file

@ -94,7 +94,7 @@ First, we need to prove some properties of characteristic functions.
For arbitrary $h \in \R$, we have
\begin{IEEEeqnarray*}{rCl}
|e^{\i t X} \frac{e^{\i h X}}{h}| &\le & \left| \frac{1}{h} \left( e^{\i h X} - 1 \right)\right|\\
&\overset{\text{\autoref{charfprop:c1}}}{\le }& \left|\frac{1}{h} \i h X\right| = |X|.
&\overset{\text{\autoref{charfprop:c1}}}{\le}& \left|\frac{1}{h} \i h X\right| = |X|.
\end{IEEEeqnarray*}
Thus the dominated convergence theorem can be applied and we obtain
\[
@ -123,9 +123,9 @@ First, we need to prove some properties of characteristic functions.
\begin{refproof}{lec12_2}
We have
\begin{IEEEeqnarray*}{rCl}
\phi_X(t) &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \dif x\\
&=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \dif x\\
&=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \dif x,\\
\phi_X(t) &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \dif x\\
&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \dif x\\
&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \dif x,\\
\end{IEEEeqnarray*}
since, as $x \mapsto \sin(tx)$ is odd and $x \mapsto e^{-\frac{x^2}{2}}$
is even, their product is odd, wich gives that the integral is $0$.
@ -134,15 +134,15 @@ First, we need to prove some properties of characteristic functions.
\phi'_X(t) &=& \bE[\i X e^{\i t X}] \\
&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \i x \left( \cos(t x) + \i \sin(tx) \right) e^{-\frac{x^2}{2}} \dif x\\
&=& \frac{1}{\sqrt{2 \pi}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \dif x\\
&=& \frac{1}{\sqrt{2 \pi} } \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}} \dif x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \dif x\right)\\
&=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi} }(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \dif x\\
&=& \frac{1}{\sqrt{2 \pi}} \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}} \dif x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \dif x\right)\\
&=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi}}(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \dif x\\
&=& \underbrace{[ \sin(tx) \frac{1}{\sqrt{2 \pi} e^{-\frac{x^2}{2}}}]_{x=-\infty}^\infty}_{=0}
- \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \dif x\\
- \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} \dif x\\
&=& -t \phi_X(t)
\end{IEEEeqnarray*}
Thus, for all $t \in \R$
\[
(\log(\phi_X(t))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t.
(\log(\phi_X(t)))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t.
\]
Hence there exists $c \in \R$, such that
\[
@ -174,9 +174,9 @@ Now, we can finally prove the CLT:
Then
\begin{IEEEeqnarray*}{rCl}
\phi_{V_n}(t) &=& \bE[e^{\i t Y_n}]\\
&=& \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n} } \right) }] \\
&=& \bE\left[e^{\i t \frac{Y_1}{\sqrt{n}}}\right] \cdot \ldots \cdot \bE\left[e^{\i t \frac{Y_n}{\sqrt{n} }}\right]\\
&=& \left( \phi\left(\frac{t}{\sqrt{n} }\right) \right)^n.
&=& \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n}} \right)}] \\
&=& \bE\left[e^{\i t \frac{Y_1}{\sqrt{n}}}\right] \cdot \ldots \cdot \bE\left[e^{\i t \frac{Y_n}{\sqrt{n}}}\right]\\
&=& \left( \phi\left(\frac{t}{\sqrt{n}}\right) \right)^n.
\end{IEEEeqnarray*}
where $\phi(t) \coloneqq \phi_{Y_1}(t)$.
@ -191,13 +191,14 @@ Now, we can finally prove the CLT:
Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain
\[
\phi\left(\frac{t}{ \sqrt{n} }\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$}
\phi\left(\frac{t}{ \sqrt{n}}\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$}
\]
\[
\phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
\phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n}} \right) \right)^n
= \left(1 - \frac{t^2}{2 n} + o\left( \frac{t^2}{n} \right)\right)^n
\xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
\]
where we have used the following:
@ -209,7 +210,7 @@ Now, we can finally prove the CLT:
We have shown that
\[
\phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_N(t).
\phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_{\cN(0,1)}(t).
\]
Using \autoref{levycontinuity}, we obtain \autoref{clt}.
\end{refproof}