diff --git a/inputs/lecture_1.tex b/inputs/lecture_1.tex index e68ca7d..10f049f 100644 --- a/inputs/lecture_1.tex +++ b/inputs/lecture_1.tex @@ -72,6 +72,16 @@ The converse to this fact is also true: 1 & x \in (1,\infty).\\ \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}: \[ F(x) = \begin{cases} @@ -79,10 +89,19 @@ The converse to this fact is also true: 0 & x < 0. \end{cases} \] + \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}: \[ \Phi(x) \coloneqq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{y^2}{2}} dy. \] + \item $\bP[X = 1] = \bP[X = -1] = \frac{1}{2}$ : \[ F(x) = \begin{cases} @@ -91,5 +110,14 @@ The converse to this fact is also true: 1 & x \in [1, \infty). \end{cases} \] + \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{example} diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 3c8be54..d7e0670 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -70,8 +70,8 @@ First, we need to prove some properties of characteristic functions. \begin{subproof} For $y \ge 0$, we have \begin{IEEEeqnarray*}{rCl} - |e^{\i y} - 1| &=& |\int_0^y \cos(s) \d s + \i \int_0^y \sin(s) \d s|\\ - &=& |\int_0^y e^{\i 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} \dif s|\\ &\overset{\text{Jensen}}{\le}& \int_0^y |e^{\i s}| ds = y. \end{IEEEeqnarray*} 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} 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}} \d 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(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}} \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 $0$. \begin{IEEEeqnarray*}{rCl} \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}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \d 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)\\ - &=& \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\\ + &=& \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\\ &=& \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) \end{IEEEeqnarray*} Thus, for all $t \in \R$ @@ -172,18 +172,20 @@ Now, we can finally prove the CLT: Let $t \in \R$. 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[e^{\i t \frac{Y_1}{\sqrt{n}}}] \cdot \ldots \cdot \bE[e^{\i t \frac{Y_n}{\sqrt{n} }}]\\ - &=& \left( \phi(\frac{t}{\sqrt{n} } \right)^n. + \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. \end{IEEEeqnarray*} where $\phi(t) \coloneqq \phi_{Y_1}(t)$. We have \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} - - \bE[Y_1^2] \frac{s^2}{2} + o(s^2)\\ - &=& 1 - \frac{s^2}{2} + o(s^2), \text{as $s \to $} + - \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 $s \to 0$} \end{IEEEeqnarray*} Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain @@ -194,7 +196,7 @@ Now, we can finally prove the CLT: \[ \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n = - (1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}}, + 1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}}, \] where we have used the following: diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index 77dae54..4d8c8df 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -15,8 +15,10 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$, \label{lindebergclt} Assume $X_1, X_2, \ldots,$ are independent (but not necessarily identically distributed) with $\mu_i = \bE[X_i] < \infty$ and $\sigma_i^2 = \Var(X_i) < \infty$. 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 $\epsilon > 0$ - (\vocab{Lindeberg condition}, ``The truncated variance is negligible compared to the variance.''). + 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 $\epsilon > 0$ + (\vocab{Lindeberg condition}\footnote{``The truncated variance is negligible compared to the variance.''}). Then the CLT holds, i.e.~ \[ @@ -94,16 +96,26 @@ A generalized version of \autoref{levycontinuity} is the following: \end{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 $f_C(x) = \frac{1}{\pi} \frac{1}{1 + x^2}$. + \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 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 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} We will prove \autoref{levycontinuity} assuming @@ -229,7 +241,7 @@ We still need to show that $\mu_n \implies \mu$. \begin{subproof} \todo{in the notes} \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 $F$, such that $F_n(x_0) \not\to F(x_0)$. Pick $\delta > 0$ and a subsequence $F_{n_1}(x_0), F_{n_2}(x_0), \ldots$ diff --git a/inputs/lecture_6.tex b/inputs/lecture_6.tex index fd19744..ed4dd95 100644 --- a/inputs/lecture_6.tex +++ b/inputs/lecture_6.tex @@ -25,7 +25,7 @@ The SLLN follows from the claim. \end{refproof} -We need the following inequality: +We need the fol] \begin{theorem}[Kolmogorov's inequality] If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$ and $\Var(X_i) = \sigma_i^2$, then @@ -41,7 +41,8 @@ We need the following inequality: We have \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} \epsilon^2 d\bP \end{IEEEeqnarray*} diff --git a/inputs/lecture_8.tex b/inputs/lecture_8.tex index 7b67105..03d7f59 100644 --- a/inputs/lecture_8.tex +++ b/inputs/lecture_8.tex @@ -85,7 +85,7 @@ for any $k \in \N$. This follows from the independence of the $X_i$. 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 \begin{enumerate}[(i)] diff --git a/inputs/lecture_9.tex b/inputs/lecture_9.tex index 90ab3b8..f936429 100644 --- a/inputs/lecture_9.tex +++ b/inputs/lecture_9.tex @@ -64,9 +64,11 @@ Why is $\sqrt{n}$ the right order? (Handwavey argument) Suppose $X_1, X_2,\ldots$ are i.i.d. $\cN(0,1)$. The mean of the l.h.s.~is $0$ and for the variance we get -\[ -\Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) = \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right) = \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1 -\] +\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 +\end{IEEEeqnarray*} + 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 $\sqrt{n}$ 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} -Consider $(\R, \cB(\R), \bP)$. -For every $t \in \R$ define a function $\phi(t) \coloneqq \phi_\bP(t) \coloneqq \int_{\R} e^{\i t x} \bP(dx)$. +\begin{definition} + Consider $(\R, \cB(\R), \bP)$. + 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 \[ \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 $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$. \end{itemize} -We call $\phi_{\bP}$ the \vocab{characteristic function} of $\bP$. \begin{remark} Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and diff --git a/wtheo.sty b/wtheo.sty index 728d857..e42ca30 100644 --- a/wtheo.sty +++ b/wtheo.sty @@ -23,6 +23,10 @@ \usepackage{float} %\usepackage{algorithmicx} +\usepackage{pgfplots} +\pgfplotsset{compat = newest} + + \newcounter{subsubsubsection}[subsubsection] \renewcommand\thesubsubsubsection{\thesubsubsection.\arabic{subsubsubsection}} \newcommand\subsubsubsection[1] @@ -97,3 +101,5 @@ \DeclareSimpleMathOperator{Bin} \DeclareSimpleMathOperator{Ber} \DeclareSimpleMathOperator{Exp} + +\newcommand*\dif{\mathop{}\!\mathrm{d}}