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2
inputs/lecture_02.tex
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@ -87,7 +87,7 @@ to an infinite number of random variables.
Let $\bP_n, n \in \N$ be probability measures on $(\R^n, \cB(\R^n))$ Let $\bP_n, n \in \N$ be probability measures on $(\R^n, \cB(\R^n))$
which are \vocab{consistent}, which are \vocab{consistent},
then there exists a unique probability measure $\bP^{\otimes}$ then there exists a unique probability measure $\bP^{\otimes}$
on $(\R^\infty, B(R^\infty))$ (where $B(R^{\infty}$ has to be defined), on $(\R^\infty, B(R^\infty))$ (where $B(R^{\infty})$ has to be defined),
such that such that
\[ \[
\forall n \in \N, B_1,\ldots, B_n \in B(\R): \forall n \in \N, B_1,\ldots, B_n \in B(\R):

3
inputs/lecture_03.tex
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@ -43,7 +43,8 @@ Now, for any $C \subseteq \R^n$ let $C^\ast \coloneqq C \times \R^{\infty}$.
Note that $C \in \cB_n \implies C^\ast \in \cF_n$. Note that $C \in \cB_n \implies C^\ast \in \cF_n$.
Thus $\cF_n = \{C^\ast : C \in \cB_n\}$. Thus $\cF_n = \{C^\ast : C \in \cB_n\}$.
Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq (\mu_1 \otimes \ldots \otimes \mu_n)(C)$. Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq (\mu_1 \otimes \ldots \otimes \mu_n)(C)$.
It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$ (\vocab{consistency}). It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$,
i.e. the $\lambda_n$ form a consistent family.
Recall the following theorem from measure theory: Recall the following theorem from measure theory:

20
inputs/lecture_06.tex
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@ -56,7 +56,7 @@ In order to prove \autoref{thm2}, we need the following:
\] \]
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
Let Let
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
A_1 &\coloneqq& \{\omega : |X_1(\omega)| > \epsilon\},\\ A_1 &\coloneqq& \{\omega : |X_1(\omega)| > \epsilon\},\\
A_2 &\coloneqq & \{\omega: |X_1(\omega)| \le \epsilon, A_2 &\coloneqq & \{\omega: |X_1(\omega)| \le \epsilon,
@ -73,16 +73,16 @@ In order to prove \autoref{thm2}, we need the following:
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
&&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C &&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C
+ \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\ + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 \dif\bP\\
&=& \int_{A_i} C^2 d\bP &=& \int_{A_i} C^2 \dif\bP
+ \underbrace{\int_{A_i} D^2 d \bP}_{\ge 0} + \underbrace{\int_{A_i} D^2 \dif\bP}_{\ge 0}
+ 2 \int_{A_i} CD d\bP\\ + 2 \int_{A_i} CD \dif\bP\\
&\ge& \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} d \bP &\ge& \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} \dif\bP
+ 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\ + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D \dif\bP\\
&\ge& \int_{A_i} \epsilon^2 d\bP, &\ge& \int_{A_i} \epsilon^2 \dif\bP,
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
since by the independence of $E$ and $D$, since by the independence of $E$ and $D$,
and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$. and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E \dif\bP = 0$.
Hence Hence
\[ \[
@ -122,7 +122,7 @@ In order to prove \autoref{thm2}, we need the following:
is a Cauchy sequence iff $a(\omega) = 0$. is a Cauchy sequence iff $a(\omega) = 0$.
We want to show that $\bP[a(\omega) > 0] = 0$. We want to show that $\bP[a(\omega) > 0] = 0$.
For this, it suffices to show that $\bP(a(\omega) > \epsilon] = 0$ For this, it suffices to show that $\bP[a(\omega) > \epsilon] = 0$
for all $\epsilon > 0$. for all $\epsilon > 0$.
For a fixed $\epsilon > 0$, we obtain: For a fixed $\epsilon > 0$, we obtain:
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}

17
inputs/lecture_07.tex
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@ -13,8 +13,8 @@
of numbers converge: of numbers converge:
\begin{itemize} \begin{itemize}
\item $\sum_{n \ge 1} \bP(|X_n| > C)$, \item $\sum_{n \ge 1} \bP(|X_n| > C)$,
\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$, \item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n \dif\bP}_{\text{\vocab{truncated mean}}}$,
\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$. \item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 \dif\bP - \left( \int_{|X_n| \le C} X_n \dif\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
\end{itemize} \end{itemize}
Then $\sum_{n \ge 1} X_n$ converges almost surely. Then $\sum_{n \ge 1} X_n$ converges almost surely.
\item Suppose $\sum_{n \ge 1} X_n$ converges almost surely. \item Suppose $\sum_{n \ge 1} X_n$ converges almost surely.
@ -37,8 +37,8 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
Since the $X_n$ are independent, the $Y_n$ are independent as well. Since the $X_n$ are independent, the $Y_n$ are independent as well.
Furthermore, the $Y_n$ are uniformly bounded. Furthermore, the $Y_n$ are uniformly bounded.
By our assumption, the series By our assumption, the series
$\sum_{n \ge 1} \int_{|X_n| \le C} X_n d\bP = \sum_{n \ge 1} \bE[Y_n]$ $\sum_{n \ge 1} \int_{|X_n| \le C} X_n \dif\bP = \sum_{n \ge 1} \bE[Y_n]$
and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$ and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 \dif\bP - \left( \int_{|X_n| \le C} X_n \dif\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$
converges. converges.
By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$ By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
almost surely. almost surely.
@ -76,10 +76,10 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\bE(Y_n) &=& \int_{|X_n| \le C} X_n d \bP + C \bP(|X_n| \ge C),\\ \bE(Y_n) &=& \int_{|X_n| \le C} X_n \dif\bP + C \bP(|X_n| \ge C),\\
\bE(Z_n) &=& \int_{|X_n| \le C} X_n d \bP - C \bP(|X_n| \ge C). \bE(Z_n) &=& \int_{|X_n| \le C} X_n \dif\bP - C \bP(|X_n| \ge C).
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$ Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n \dif\bP$
the second series converges, the second series converges,
and since and since
$\bE(Y_n) - \bE(Z_n)$ converges, the first series converges. $\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
@ -160,8 +160,7 @@ More formally:
and uniform boundedness was not used. and uniform boundedness was not used.
The idea of `` $\implies$ '' will lead to coupling. % TODO ? The idea of `` $\implies$ '' will lead to coupling. % TODO ?
\end{remark} \end{remark}
A proof of \autoref{thm5} can be found in the notes.\notes
% TODO Proof of thm5 in the notes
\begin{example}[Application of \autoref{thm5}] \begin{example}[Application of \autoref{thm5}]
The series $\sum_{n} \frac{1}{n^{\frac{1}{2} + \epsilon}}$ The series $\sum_{n} \frac{1}{n^{\frac{1}{2} + \epsilon}}$
does not converge for $\epsilon < \frac{1}{2}$. does not converge for $\epsilon < \frac{1}{2}$.

1
inputs/lecture_09.tex
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@ -106,6 +106,7 @@ We have
\item We have $\phi(0) = 1$. \item We have $\phi(0) = 1$.
\item $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$. \item $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$.
\end{itemize} \end{itemize}
\todo{Properties of characteristic functions}
\begin{remark} \begin{remark}
Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and

43
inputs/lecture_10.tex
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@ -9,7 +9,7 @@ We consider $(\R, \cB(\R))$.
By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$. By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
\end{notation} \end{notation}
For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}d\bP(x)$. For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}\dif\bP(x)$.
If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
$\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$, $\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$,
where $\mu = \bP X^{-1}$. where $\mu = \bP X^{-1}$.
@ -27,15 +27,15 @@ where $\mu = \bP X^{-1}$.
We have We have
\begin{IEEEeqnarray*}{rCl} \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} dt d \bP(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} dt \dif\bP(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} dt d \bP(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} dt \dif\bP(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} dt d \bP(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} dt \dif\bP(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] dt d \bP(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] dt \dif\bP(x)}_{=0 \text{, as the function is odd}}
\\&& \\&&
+ \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} dt d\bP(x)\\ + \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} dt \dif\bP(x)\\
&=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt d\bP(x)\\ &=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt \dif\bP(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}) d\bP(x)\\ - (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) \dif\bP(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}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
@ -129,10 +129,10 @@ However, Fourier analysis is not only useful for continuous probability density
\begin{refproof}{bochnersformula} \begin{refproof}{bochnersformula}
We have We have
\begin{IEEEeqnarray*}{rCl} \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} d \bP(y) \\ RHS &=& \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \dif\bP(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))} dt\\ &\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))} dt\\
&=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} d\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\ &=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} \dif\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\
&=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} d \bP(y)\\ &=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \dif\bP(y)\\
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Furthermore Furthermore
\[ \[
@ -143,7 +143,7 @@ However, Fourier analysis is not only useful for continuous probability density
\] \]
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)} d \bP(y) &=& \bP\left( \{x\}\right) \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \dif\bP(y) &=& \bP\left( \{x\}\right)
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
% TODO by dominated convergence? % TODO by dominated convergence?
\end{refproof} \end{refproof}
@ -168,9 +168,9 @@ However, Fourier analysis is not only useful for continuous probability density
For part (b) we have: For part (b) we have:
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\sum_{j,k} c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k} c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} d \bP(x)\\ \sum_{j,k} c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k} c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} \dif\bP(x)\\
&=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} d\bP(x)\\ &=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} \dif\bP(x)\\
&=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} d\bP(x)\\ &=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} \dif\bP(x)\\
&=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0 &=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\end{refproof} \end{refproof}
@ -187,7 +187,7 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
\label{def:weakconvergence} \label{def:weakconvergence}
We say that $\bP_n \subseteq M_1(\R)$ \vocab[Convergence!weak]{converges weakly} towards $\bP \in M_1(\R)$ (notation: $\bP_n \implies \bP$), iff We say that $\bP_n \subseteq M_1(\R)$ \vocab[Convergence!weak]{converges weakly} towards $\bP \in M_1(\R)$ (notation: $\bP_n \implies \bP$), iff
\[ \[
\forall f \in C_b(\R)~ \int f d\bP_n \to \int f d\bP. \forall f \in C_b(\R)~ \int f \dif\bP_n \to \int f \dif\bP.
\] \]
Where Where
\[ \[
@ -197,12 +197,13 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
In analysis, this is also known as $\text{weak}^\ast$ convergence. In analysis, this is also known as $\text{weak}^\ast$ convergence.
\end{definition} \end{definition}
\begin{remark} \begin{remark}
This notion of convergence makes $M_1(\R)$ a separable metric space. We can construc a metric on $M_1(\R)$ that turns $M_1(\R)$ into a complete This notion of convergence makes $M_1(\R)$ a separable metric space.
We can construct a metric on $M_1(\R)$ that turns $M_1(\R)$ into a complete
and separable metric space: and separable metric space:
Consider the sets Consider the sets
\[ \[
\{\bP \in M_1(\R): \forall i=1,\ldots,n ~ \int f d \bP - \int f_i d\bP < \epsilon \} \{\bP \in M_1(\R): \forall i=1,\ldots,n ~ \int f \dif\bP - \int f_i \dif\bP < \epsilon \}
\] \]
for any $f,f_1,\ldots, f_n \in C_b(\R)$. for any $f,f_1,\ldots, f_n \in C_b(\R)$.
These sets form a basis for the topology on $M_1(\R)$. These sets form a basis for the topology on $M_1(\R)$.
@ -211,13 +212,13 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
\begin{example} \begin{example}
\begin{itemize} \begin{itemize}
\item Let $\bP_n = \delta_{\frac{1}{n}}$. \item Let $\bP_n = \delta_{\frac{1}{n}}$.
Then $\int f d \bP_n = f(\frac{1}{n}) \to f(0) = \int f d \delta_0$ Then $\int f \dif\bP_n = f(\frac{1}{n}) \to f(0) = \int f d \delta_0$
for any continuous, bounded function $f$. for any continuous, bounded function $f$.
Hence $\bP_n \to \delta_0$. Hence $\bP_n \to \delta_0$.
\item $\bP_n \coloneqq \delta_n$ does not converge weakly, \item $\bP_n \coloneqq \delta_n$ does not converge weakly,
as for example as for example
\[ \[
\int \cos(\pi x) d\bP_n(x) \int \cos(\pi x) \dif\bP_n(x)
\] \]
does not converge. does not converge.
@ -225,7 +226,7 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
Let $f \in C_b(\R)$ arbitrary. Let $f \in C_b(\R)$ arbitrary.
Then Then
\[ \[
\int f d\bP_n = \frac{1}{n}(n) + (1 - \frac{1}{n}) f(0) \to f(0) \int f \dif\bP_n = \frac{1}{n}(n) + (1 - \frac{1}{n}) f(0) \to f(0)
\] \]
since $f$ is bounded. since $f$ is bounded.
Hence $\bP_n \implies \delta_0$. Hence $\bP_n \implies \delta_0$.

2
inputs/lecture_11.tex
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@ -124,7 +124,7 @@ If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq \R$, we have
&\approx& \Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - \Phi(-0.01 \sqrt{\frac{n}{p(1-p)}})\\ &\approx& \Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - \Phi(-0.01 \sqrt{\frac{n}{p(1-p)}})\\
&=& 2\Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - 1\\ &=& 2\Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - 1\\
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Hence, we want $\Phi(0.01 \sqrt{\frac{n}{p(1-p}}) \approx \frac{1.95}{2}$, Hence, we want $\Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) \approx \frac{1.95}{2}$,
i.e.~$n = (1.96)^2 100^2 p\cdot (1-p)$ i.e.~$n = (1.96)^2 100^2 p\cdot (1-p)$
We have $p\cdot (1-p) \le \frac{1}{4}$, We have $p\cdot (1-p) \le \frac{1}{4}$,
thus $n \approx (1.96)^2 \cdot 100^2 \cdot \frac{1}{4} = 9600$ suffices. thus $n \approx (1.96)^2 \cdot 100^2 \cdot \frac{1}{4} = 9600$ suffices.

1
inputs/lecture_12.tex
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@ -223,3 +223,4 @@ Now, we can finally prove the CLT:
where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$. where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$.
\end{remark} \end{remark}
Exercise: Find out, which properties also hold for $d > 1$. Exercise: Find out, which properties also hold for $d > 1$.
\todo{TODO}

2
inputs/lecture_13.tex
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@ -16,7 +16,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
Assume $X_1, X_2, \ldots,$ are independent (but not necessarily identically distributed) with $\mu_i = \bE[X_i] < \infty$ and $\sigma_i^2 = \Var(X_i) < \infty$. Assume $X_1, X_2, \ldots,$ are independent (but not necessarily identically distributed) with $\mu_i = \bE[X_i] < \infty$ and $\sigma_i^2 = \Var(X_i) < \infty$.
Let $S_n = \sqrt{\sum_{i=1}^{n} \sigma_i^2}$ Let $S_n = \sqrt{\sum_{i=1}^{n} \sigma_i^2}$
and assume that and assume that
\[\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0\] \[\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon S_n}\right] = 0\]
for all $\epsilon > 0$ for all $\epsilon > 0$
(\vocab{Lindeberg condition}\footnote{``The truncated variance is negligible compared to the variance.''}). (\vocab{Lindeberg condition}\footnote{``The truncated variance is negligible compared to the variance.''}).

2
inputs/lecture_20.tex
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@ -87,7 +87,7 @@ we need the following theorem, which we won't prove here:
via via
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
L^p &\longrightarrow & (L^q)^\ast \\ L^p &\longrightarrow & (L^q)^\ast \\
f &\longmapsto & (g \mapsto \int g f \dif d\bP) f &\longmapsto & (g \mapsto \int g f \dif\bP)
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
We also have $(L^1)^\ast \cong L^\infty$, We also have $(L^1)^\ast \cong L^\infty$,