some diagrams

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Josia Pietsch 2023-05-25 00:33:14 +02:00
parent 0930ed6c95
commit 19f81a9a3c
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7 changed files with 90 additions and 31 deletions

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@ -72,6 +72,16 @@ The converse to this fact is also true:
1 & x \in (1,\infty).\\ 1 & x \in (1,\infty).\\
\end{cases} \end{cases}
\] \]
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[samples=1000, xmin=-1, xmax=2, width=10cm, height=5cm]
\addplot[] {and(x>0,x<=1) * x + (x>1)};
\end{axis}
\end{tikzpicture}
\end{figure}
\item \vocab{Exponential distribution}: \item \vocab{Exponential distribution}:
\[ \[
F(x) = \begin{cases} F(x) = \begin{cases}
@ -79,10 +89,19 @@ The converse to this fact is also true:
0 & x < 0. 0 & x < 0.
\end{cases} \end{cases}
\] \]
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[samples=1000, smooth, width=10cm, height=5cm, xmin=-2, xmax=5]
\addplot[] {(x > 0) * (1 - exp( - 5 * x))};
\end{axis}
\end{tikzpicture}
\end{figure}
\item \vocab{Gaussian distribution}: \item \vocab{Gaussian distribution}:
\[ \[
\Phi(x) \coloneqq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{y^2}{2}} dy. \Phi(x) \coloneqq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{y^2}{2}} dy.
\] \]
\item $\bP[X = 1] = \bP[X = -1] = \frac{1}{2}$ : \item $\bP[X = 1] = \bP[X = -1] = \frac{1}{2}$ :
\[ \[
F(x) = \begin{cases} F(x) = \begin{cases}
@ -91,5 +110,14 @@ The converse to this fact is also true:
1 & x \in [1, \infty). 1 & x \in [1, \infty).
\end{cases} \end{cases}
\] \]
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[samples=1000, width=10cm, height=5cm]
\addplot[] {and(x >= -1, x < 1) * 0.5 + (x >= 1)};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{enumerate} \end{enumerate}
\end{example} \end{example}

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@ -70,8 +70,8 @@ First, we need to prove some properties of characteristic functions.
\begin{subproof} \begin{subproof}
For $y \ge 0$, we have For $y \ge 0$, we have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
|e^{\i y} - 1| &=& |\int_0^y \cos(s) \d s + \i \int_0^y \sin(s) \d s|\\ |e^{\i y} - 1| &=& |\int_0^y \cos(s) \dif s + \i \int_0^y \sin(s) \dif s|\\
&=& |\int_0^y e^{\i s} \d s|\\ &=& |\int_0^y e^{\i s} \dif s|\\
&\overset{\text{Jensen}}{\le}& \int_0^y |e^{\i s}| ds = y. &\overset{\text{Jensen}}{\le}& \int_0^y |e^{\i s}| ds = y.
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
For $y < 0$, we have $|e^{\i y} - 1| = |e^{-\i y} - 1|$ For $y < 0$, we have $|e^{\i y} - 1| = |e^{-\i y} - 1|$
@ -122,21 +122,21 @@ First, we need to prove some properties of characteristic functions.
\begin{refproof}{lec12_2} \begin{refproof}{lec12_2}
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\phi_X(t) &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \d 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}} \d 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}} \d x,\\ &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \dif x,\\
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
since, as $x \mapsto \sin(tx)$ is odd and $x \mapsto e^{-\frac{x^2}{2}}$ 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 $0$. is even, their product is odd, wich gives that the integral is $0$.
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\phi'_X(t) &=& \bE[\i X e^{\i t X}] \\ \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}} \d 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}} \d 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}} \d x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \d x\right)\\ &=& \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)} \d x\\ &=& \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} &=& \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}} \d x\\ - \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \dif x\\
&=& -t \phi_X(t) &=& -t \phi_X(t)
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Thus, for all $t \in \R$ Thus, for all $t \in \R$
@ -172,9 +172,10 @@ Now, we can finally prove the CLT:
Let $t \in \R$. Let $t \in \R$.
Then Then
\begin{IEEEeqnarray*}{rCl} \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) }] \\ \phi_{V_n}(t) &=& \bE[e^{\i t Y_n}]\\
&=& \bE[e^{\i t \frac{Y_1}{\sqrt{n}}}] \cdot \ldots \cdot \bE[e^{\i t \frac{Y_n}{\sqrt{n} }}]\\ &=& \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n} } \right) }] \\
&=& \left( \phi(\frac{t}{\sqrt{n} } \right)^n. &=& \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*} \end{IEEEeqnarray*}
where $\phi(t) \coloneqq \phi_{Y_1}(t)$. where $\phi(t) \coloneqq \phi_{Y_1}(t)$.
@ -182,8 +183,9 @@ Now, we can finally prove the CLT:
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{ as $s \to 0$}\\ \phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{ as $s \to 0$}\\
&=& 1 - \underbrace{\i \bE[Y_1] s}_{=0} &=& 1 - \underbrace{\i \bE[Y_1] s}_{=0}
- \bE[Y_1^2] \frac{s^2}{2} + o(s^2)\\ - \bE[Y_1^2] \frac{s^2}{2} + o(s^2), \text{ as $s \to 0$}
&=& 1 - \frac{s^2}{2} + o(s^2), \text{as $s \to $} \\
&=& 1 - \frac{s^2}{2} + o(s^2), \text{ as $s \to 0$}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain
@ -194,7 +196,7 @@ Now, we can finally prove the CLT:
\[ \[
\phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n = \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
(1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}}, 1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
\] \]
where we have used the following: where we have used the following:

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@ -15,8 +15,10 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
\label{lindebergclt} \label{lindebergclt}
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 $\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0$ for all $\epsilon > 0$ and assume that
(\vocab{Lindeberg condition}, ``The truncated variance is negligible compared to the variance.''). \[\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0\]
for all $\epsilon > 0$
(\vocab{Lindeberg condition}\footnote{``The truncated variance is negligible compared to the variance.''}).
Then the CLT holds, i.e.~ Then the CLT holds, i.e.~
\[ \[
@ -94,16 +96,26 @@ A generalized version of \autoref{levycontinuity} is the following:
\end{example} \end{example}
\begin{example} \begin{example}
Suppose $C$ is a random variable which is Cauchy distributed, i.e.~$C$ Suppose $C$ is a random variable which is \vocab[Cauchy distribution]{Cauchy distributed}, i.e.~$C$
has probability distribution $f_C(x) = \frac{1}{\pi} \frac{1}{1 + x^2}$. has probability distribution $f_C(x) = \frac{1}{\pi} \frac{1}{1 + x^2}$.
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[samples=100, smooth]
\addplot[] { (1/3.14159265358979323846) * (1 / ( 1 + x * x))};
\end{axis}
\end{tikzpicture}
\caption{Probability density function of $C$}
\end{figure}
We know that $\bE[|C|] = \infty$. We know that $\bE[|C|] = \infty$.
We have $\phi_C(t) = \bE[e^{\i t C}] = e^{-|t|}$. We have $\phi_C(t) = \bE[e^{\i t C}] = e^{-|t|}$.
Suppose $C_1, C_2, \ldots, C_n$ are i.i.d.~Cauchy distributed Suppose $C_1, C_2, \ldots, C_n$ are i.i.d.~Cauchy distributed
and let $S_n \coloneqq C_1 + \ldots + C_n$. and let $S_n \coloneqq C_1 + \ldots + C_n$.
Exercise: $\phi_{S_n}(t) = e^{-|t|} = \phi_{C_1}(t)$, thus $S_n \sim C$. Exercise: $\phi_{\frac{S_n}{n}}(t) = e^{-|t|} = \phi_{C_1}(t)$, thus $\frac{S_n}{n} \sim C$.
\end{example} \end{example}
We will prove \autoref{levycontinuity} assuming We will prove \autoref{levycontinuity} assuming
@ -229,7 +241,7 @@ We still need to show that $\mu_n \implies \mu$.
\begin{subproof} \begin{subproof}
\todo{in the notes} \todo{in the notes}
\end{subproof} \end{subproof}
Assume $\mu_n$ does not converge to $\mu$. Assume that $\mu_n$ does not converge to $\mu$.
By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$, By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,
such that $F_n(x_0) \not\to F(x_0)$. such that $F_n(x_0) \not\to F(x_0)$.
Pick $\delta > 0$ and a subsequence $F_{n_1}(x_0), F_{n_2}(x_0), \ldots$ Pick $\delta > 0$ and a subsequence $F_{n_1}(x_0), F_{n_2}(x_0), \ldots$

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@ -25,7 +25,7 @@
The SLLN follows from the claim. The SLLN follows from the claim.
\end{refproof} \end{refproof}
We need the following inequality: We need the fol]
\begin{theorem}[Kolmogorov's inequality] \begin{theorem}[Kolmogorov's inequality]
If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$ If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$
and $\Var(X_i) = \sigma_i^2$, then and $\Var(X_i) = \sigma_i^2$, then
@ -41,7 +41,8 @@ We need the following inequality:
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP &=& \int_{A_i} C^2 d\bP + \underbrace{\int_{A_i} D^2 d \bP}_{\ge 0} + 2 \int_{A_i} CD d\bP\\ &&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\
&=& \int_{A_i} C^2 d\bP + \underbrace{\int_{A_i} D^2 d \bP}_{\ge 0} + 2 \int_{A_i} CD d\bP\\
&\ge & \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} d \bP + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\ &\ge & \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} d \bP + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\
&\ge& \int_{A_i} \epsilon^2 d\bP &\ge& \int_{A_i} \epsilon^2 d\bP
\end{IEEEeqnarray*} \end{IEEEeqnarray*}

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@ -85,7 +85,7 @@ for any $k \in \N$.
This follows from the independence of the $X_i$. This follows from the independence of the $X_i$.
It is It is
\[ \[
\sigma\left( X_1,\ldots,X_n \right) = \sigma\left(\underbrace{\{X_{1}^{-1}(B_1) \cap \ldots \cap X_n^{-1}(B_n)\} | B_1,\ldots,B_n \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}}\cA} \right). \sigma\left( X_1,\ldots,X_n \right) = \sigma\left(\underbrace{\{X_{1}^{-1}(B_1) \cap \ldots \cap X_n^{-1}(B_n) | B_1,\ldots,B_n \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}}\cA} \right).
\] \]
$\cA$ is a semi-algebra, since $\cA$ is a semi-algebra, since
\begin{enumerate}[(i)] \begin{enumerate}[(i)]

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@ -64,9 +64,11 @@ Why is $\sqrt{n}$ the right order? (Handwavey argument)
Suppose $X_1, X_2,\ldots$ are i.i.d. $\cN(0,1)$. Suppose $X_1, X_2,\ldots$ are i.i.d. $\cN(0,1)$.
The mean of the l.h.s.~is $0$ and for the variance we get The mean of the l.h.s.~is $0$ and for the variance we get
\[ \begin{IEEEeqnarray*}{rCl}
\Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) = \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right) = \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1 \Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) &=& \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right)\\
\] &=& \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
\end{IEEEeqnarray*}
For the r.h.s.~we get a mean of $0$ and a variance of $1$. For the r.h.s.~we get a mean of $0$ and a variance of $1$.
So, to determine what $(\ast)$ could mean, it is necessary that $\sqrt{n}$ So, to determine what $(\ast)$ could mean, it is necessary that $\sqrt{n}$
is the right scaling. is the right scaling.
@ -77,8 +79,17 @@ This notion of convergence will be defined in terms of characteristic functions
\subsection{Characteristic functions and Fourier transform} \subsection{Characteristic functions and Fourier transform}
\begin{definition}
Consider $(\R, \cB(\R), \bP)$. Consider $(\R, \cB(\R), \bP)$.
For every $t \in \R$ define a function $\phi(t) \coloneqq \phi_\bP(t) \coloneqq \int_{\R} e^{\i t x} \bP(dx)$. The \vocab{characteristic function} of $\bP$ is defined as
\begin{IEEEeqnarray*}{rCl}
\phi_{\bP}: \R &\longrightarrow & \C \\
t &\longmapsto & \int_{\R} e^{\i t x} \bP(\dif x).
\end{IEEEeqnarray*}
\end{definition}
\begin{abuse}
$\phi_\bP(t)$ will often be abbreviated as $\phi(t)$.
\end{abuse}
We have We have
\[ \[
\phi(t) = \int_{\R} \cos(tx) \bP(dx) + \i \int_{\R} \sin(tx) \bP(dx). \phi(t) = \int_{\R} \cos(tx) \bP(dx) + \i \int_{\R} \sin(tx) \bP(dx).
@ -88,7 +99,6 @@ 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}
We call $\phi_{\bP}$ the \vocab{characteristic function} of $\bP$.
\begin{remark} \begin{remark}
Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and

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@ -23,6 +23,10 @@
\usepackage{float} \usepackage{float}
%\usepackage{algorithmicx} %\usepackage{algorithmicx}
\usepackage{pgfplots}
\pgfplotsset{compat = newest}
\newcounter{subsubsubsection}[subsubsection] \newcounter{subsubsubsection}[subsubsection]
\renewcommand\thesubsubsubsection{\thesubsubsection.\arabic{subsubsubsection}} \renewcommand\thesubsubsubsection{\thesubsubsection.\arabic{subsubsubsection}}
\newcommand\subsubsubsection[1] \newcommand\subsubsubsection[1]
@ -97,3 +101,5 @@
\DeclareSimpleMathOperator{Bin} \DeclareSimpleMathOperator{Bin}
\DeclareSimpleMathOperator{Ber} \DeclareSimpleMathOperator{Ber}
\DeclareSimpleMathOperator{Exp} \DeclareSimpleMathOperator{Exp}
\newcommand*\dif{\mathop{}\!\mathrm{d}}