Lecture 11

inputs/lecture_10.tex

@@ -175,8 +175,8 @@ However, Fourier analysis is not only useful for continuous probability density
     \end{IEEEeqnarray*}
 \end{refproof}
 \begin{theorem}[Bochner's theorem]\label{bochnersthm}
-    The converse to \autoref{thm:lec_10thm5} holds, i.e.~
+    The converse to \autoref{thm:lec_10thm5} holds, i.e.~ any
-    any $\phi: \R \to  \C$ satisfying (a) and (b) of \autoref{thm:lec_10thm5}
+    $\phi: \R \to  \C$ satisfying (a) and (b) of \autoref{thm:lec_10thm5}
     must be the Fourier transform of a probability measure $\bP$ 
     on $(\R, \cB(\R))$.
 \end{theorem}

inputs/lecture_11.tex

@@ -0,0 +1,130 @@
+\subsection{The central limit theorem}
+
+For $X_1, X_2,\ldots$ i.i.d.~we were looking
+at $S_n \coloneqq  \sum_{i=1}^n X_i$.
+Then the LLN basically states, that $S_n$ can be approximated by $n \bE[X_1]$.
+\begin{question}
+    What is the error of this approximation?
+\end{question}
+
+We set $\mu\coloneqq \bE[X_1]$ and $\sigma^2 \coloneqq \Var(X_1) \in (0,\infty)$.
+We know that $\bE[S_n] = n \mu$ and  $\Var(S_n) = n\sigma^2$.
+
+The central limit theorem basically states, that the distribution of  $S_n$
+can be approximated by a normal distribution with mean $n \mu$ and
+variance $n \sigma^2$,
+i.e.~$S_n \approx n \mu + \sigma \sqrt{n} N$ for $N \sim \cN(0,1)$,
+where $\approx$ is to be made precise.
+
+For intuition, watch \url{https://3blue1brown.com/lessons/clt}.
+
+\begin{example}
+    We throw a fair die $n = 100$ times and denote the sum of the faces
+    by $S_n \coloneqq  X_1 + \ldots + X_n$, where $X_1,\ldots, X_n$ 
+    are i.i.d.~and uniformly distributed on $\{1,\ldots,6\}$.
+    Then $\bE[S_n] = 350$ and $\sqrt{\Var(S_n)} = \sigma \approx 17.07$.
+    \todo{Missing pictures}
+\end{example}
+
+\begin{question}
+    Why do statisticians care about $\sigma$ instead of $\sigma^2$?
+\end{question}
+By definition, $\Var(X) = \bE[(X- \bE(X))^2]$, hence $\sqrt{\Var(X)}$
+can be interpreted as a distance.
+One could also define $\Var(X)$ to be $\bE[|X - \bE(X)|]$ but this is not 
+well behaved.
+
+
+\begin{example}
+    Let $X_1,\ldots,X_n$  be i.i.d.~and $X_1\sim \Exp(1)$.
+    We knot that for $n \in \N$, $\bE[S_n] = n$
+    and $\sqrt{\Var(S_n)}  = \sqrt{n}$.
+    For $n = 100, 300, 500$, we get the following picture
+    \todo{Missing picture}
+\end{example}
+
+In order to make things nicer, we do the following:
+\begin{enumerate}[1.]
+    \item center: $S_n - \bE[S_n]$,
+    \item normalize: $\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)} }$.
+\end{enumerate}
+Then $\bE[\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}] = 0$
+and $\Var(\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}) = 1$.
+
+\begin{theorem}[Central limit theorem, 1920s, Lindeberg and Levy]\label{clt}
+    Let $X_1,X_2,\ldots$ be i.i.d.~random variables
+    with $\bE[X_1] = \mu$ and $\Var(X_1) = \sigma^2 \in (0, \infty)$.
+
+    Let $S_n \coloneqq  \sum_{i=1}^n X_i$.
+    Then
+    \[
+    \frac{S_n - n \nu}{\sigma \sqrt{n} } \xrightarrow{dist.} \cN(0,1),
+   \]
+   i.e.~$\forall  x \in \R:$
+   \[
+       \lim_{n \to \infty} \bP\left[\frac{S_n - n \mu}{\sigma \sqrt{n} } \le x\right] = \Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2 \pi} } e^{\frac{-t^2}{2}}dt.
+   \]
+\end{theorem}
+
+There exists a special case of this theorem, which was proved earlier:
+\begin{theorem}[de-Moivre (1730, $p =  0.5$), Laplace (1812, general $p$ )]
+    \label{preclt}
+    Let $S_n = \Bin(n,p)$, where $p \in (0,1)$ is constant.
+    Then, for all $x \in \R$ :
+    \[
+    \lim_{n \to \infty} \bP\left[ \frac{ S_n - np}{\sqrt{n p(1-p)}} \le x\right] = \Phi(x).
+    \]
+\end{theorem}
+\begin{proof}
+    Let $X_1, X_2,\ldots$ i.i.d.~with $X_1 \sim \Ber(p)$.
+    Then $\bE[X_1] = p$ and $\Var(X_1) = p(1-p )$.
+    Furthermore $\sum_{i=1}^n X_i \sim \Bin(n,p)$,
+    and the special case follows from \autoref{clt}.
+\end{proof}
+\autoref{preclt} is a useful tool for approximating the Binomial distribution with the normal distribution.
+If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq  \R$, we have
+\[\bP[a \le S_n \le b] = \bP\left[\frac{a - np}{\sqrt{np(1-p)}}  \le \frac{S_n -np}{\sqrt{n p (1-p)}} \le  \frac{b - np}{\sqrt{n p (1-p)} }\right] \approx \Phi(b') - \Phi(a').\]
+
+\begin{example}
+    We consider a $n=40$-times Bernoulli trial with success probability $p = \frac{1}{2}$.
+    Then $0.9597 = \bP[S \le 25] \approx \Phi(\frac{5}{\sqrt{10}} \approx 0.9431$.
+
+    However, $S$ takes only integer values, which means $\bP[S \le 25] = \bP[S  26]$.
+    With this in mind, a better approximation is
+    \[
+    \bP[S \le  25] = \bP[S \le  25.5] \approx \Phi\left( \frac{5.5}{\sqrt{10} } \right) \approx 0.9541.
+    \] 
+\end{example}
+
+\begin{example}
+    Consider a particle that start at $0$ and moves on the lattice $\Z$.
+    In every step, takes a step $+ 1$ with probability $\frac{1}{2}$
+    or $-1$ with probability $\frac{1}{2}$.
+
+    More formally: Let $X_1,X_2,\ldots$ be i.i.d.~with $\bP[X_1=1] = \bP[X_1=-1] = \frac{1}{2}$ and consider $S_n \coloneqq \sum_{i=1}^n X_i$.
+
+    Then \autoref{clt} states, that $S_n \approx \cN(0,n)$.
+\end{example}
+
+\begin{example}
+    Consider an election with two candidates $A$ and $B$.
+    The relative number of votes for $A$ is $p \in (0,1)$ (constatn, but unknown)
+    How many ballots do we need to count to make sure that the probability of erring more than $1\%$ is not bigger than $5\%$?
+
+    Each ballot is a vote for $A$ with probability $p$.
+    We have $S_n \sim \Bin(n,p)$ and we want to find $n$ such that
+    $\bP[|S_n - np| \le  0.01 n] \le  0.05$.
+    We have that
+    \begin{IEEEeqnarray*}{rCl}
+        &&\bP[|S_n - nü| \le  0.01n] \\
+        &=& \bP[ -0.01 n \le  S_n - np \le  0.01n]\\
+        &=& \bP[-\frac{0.01 n}{\sqrt{n p (1-p)} } \le  \frac{S_n - np}{\sqrt{n p (1-p)} } \le \frac{0.01 n}{\sqrt{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\\
+    \end{IEEEeqnarray*}
+    Hence, we want $\Phi(0.01 \sqrt{\frac{n}{p(1-p}}) \approx \frac{1.95}{2}$,
+    i.e.~$n = (1.96)^2 100^2 p\cdot (1-p)$
+    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.
+\end{example}
+

probability_theory.tex

@@ -34,6 +34,7 @@
 \input{inputs/lecture_8.tex}
 \input{inputs/lecture_9.tex}
 \input{inputs/lecture_10.tex}
+\input{inputs/lecture_11.tex}
 
 \cleardoublepage