lecture 13 part 1

inputs/lecture_10.tex

@@ -64,7 +64,7 @@ where $\mu = \bP X^{-1}$.
     Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
     Then $\bP$ has a continuous probability density given by
     \[
-    f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R(t) dt}.
+    f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) dt.
     \] 
 \end{theorem}
 
@@ -247,11 +247,13 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
     which is the distribution of $X \equiv 0$.
     But $F_n(0) \centernot\to F(0)$.
 \end{example}
-\begin{theorem}
+\begin{theorem} % Theorem 1
+    \label{lec10_thm1}
     $X_n \xrightarrow{\text{dist}} X$ iff
     $F_n(t) \to  F(t)$ for all continuity points $t$ of $F$.
 \end{theorem}
 \begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
+    % Theorem 2
     $X_n \xrightarrow{\text{dist}} X$ iff
     $\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$.
 \end{theorem}

inputs/lecture_12.tex

@@ -139,19 +139,19 @@ First, we need to prove some properties of characteristic functions.
         - \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \d x\\
         &=& -t \phi_X(t)
     \end{IEEEeqnarray*}
-    Thus, for all $t \in \R$ 
+    Thus, for all $t \in \R$
     \[
         (\log(\phi_X(t))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t.
-    \] 
+    \]
     Hence there exists $c \in \R$, such that
     \[
     \log(\phi_X(t)) = -\frac{t^2}{2} + c.
-    \] 
+    \]
     Since $\phi_X(0) = 1$, we obtain $c = 0$.
     Thus
     \[
     \phi_X(t) = e^{-\frac{t^2}{2}}.
-    \] 
+    \]
 \end{refproof}
 
 
@@ -163,12 +163,12 @@ Now, we can finally prove the CLT:
     Let
     \[
     Y_i \coloneqq \frac{X_i - \mu}{\sigma}
-    \] 
+    \]
     i.e.~we normalize to $\bE[Y_1] = 0$ and $\Var(Y_1) = 1$.
     We need to show that
     \[
     V_n \coloneqq \frac{S_n - n \mu}{ \sigma \sqrt{n}} = \frac{Y_1+ \ldots + Y_n}{\sqrt{n}} \xrightarrow{\omega, n\to \infty}  \cN(0,1) % TODO
-    \] 
+    \]
     Let $t \in \R$.
     Then
     \begin{IEEEeqnarray*}{rCl}
@@ -181,7 +181,7 @@ Now, we can finally prove the CLT:
     We have
     \begin{IEEEeqnarray*}{rCl}
         \phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{as $s \to 0$}\\
-                &=& 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)\\
                 &=& 1 - \frac{s^2}{2} + o(s^2), \text{as $s \to $}
     \end{IEEEeqnarray*}
@@ -189,13 +189,13 @@ Now, we can finally prove the CLT:
     Setting $s \coloneqq  \frac{t}{\sqrt{n}}$ we obtain
     \[
     \phi\left(\frac{t}{ \sqrt{n} }\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to  \infty$}
-    \] 
+    \]
 
 
     \[
     \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
     (1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
-    \] 
+    \]
     where we have used the following:
 
     \begin{claim}
@@ -207,7 +207,7 @@ Now, we can finally prove the CLT:
     We have shown that
     \[
     \phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_N(t).
-    \] 
+    \]
     Using \autoref{levycontinuity}, we obtain \autoref{clt}.
 \end{refproof}
 
@@ -221,6 +221,3 @@ Now, we can finally prove the CLT:
     where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$.
 \end{remark}
 Exercise: Find out, which properties also hold for $d > 1$.
-
-
-

inputs/lecture_13.tex

@@ -0,0 +1,109 @@
+% Lecture 13 2023-05
+%The difficult part is to show \autoref{levycontinuity}.
+%This is the last lecture, where we will deal with independent random variables.
+
+We have seen, that
+if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
+    $\sigma^2 = \Var(X_1)$,
+    then $\frac{\sum_{i=1}^{n} (X_i - \mu)}{\sigma \sqrt{n} } \xrightarrow{(d)} \cN(0,1)$.
+
+\begin{question}
+    What happens if $X_1, X_2,\ldots$ are independent, but not identically distributed? Do we still have a CLT?
+\end{question}
+
+\begin{theorem}[Lindeberg CLT]
+    \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.'').
+
+    Then the CLT holds, i.e.~
+    \[
+    \frac{\sum_{i=1}^n (X_i - \mu_i)}{S_n} \xrightarrow{(d)}  \cN(0,1).
+    \] 
+\end{theorem}
+
+\begin{theorem}[Lyapunov condition]
+    \label{lyapunovclt}
+    Let $X_1, X_2,\ldots$ be independent, $\mu_i = \bE[X_i] < \infty$,
+    $\sigma_i^2 = \Var(X_i) < \infty$
+    and $S_n \coloneqq  \sqrt{\sum_{i=1}^n \sigma_i^2}$.
+    Then, assume that, for some $\delta > 0$,
+    \[
+    \lim_{n \to \infty} \sum_{i=1}^{n} \bE[(X_i - \mu_i)^{2 + \delta}] = 0
+    \] 
+    (\vocab{Lyapunov condition}).
+    Then the CLT holds.
+\end{theorem}
+\begin{remark}
+    The Lyapunov condition implies the Lindeberg condition.
+    (Exercise).
+\end{remark}
+We will not prove the \autoref{linebergclt} or \autoref{lyapunovclt}
+in this lecture. However, they are quite important.
+
+We will now sketch the proof of \autoref{levycontinuity},
+details can be found in the notes.\todo{Complete this}
+A generalized version of \autoref{levycontinuity} is the following:
+\begin{theorem}[A generalized version of Levy's continuity \autoref{levycontinuity}]
+    \label{genlevycontinuity}
+    Suppose we have random variables $(X_n)_n$ such that
+    $\bE[e^{\i t X_n}] \xrightarrow{n \to \infty}  \phi(t)$ for all $t \in \R$
+    for some function $\phi$ on $\R$.
+    Then the following are equivalent:
+    \begin{enumerate}[(a)]
+        \item The distribution of $X_n$ is \vocab[Distribution!tight]{tight} (dt. ``straff''),
+            i.e.~$\lim_{a \to \infty} \sup_{n \in \N} \bP[|X_n| > a] = 0$.
+        \item $X_n \xrightarrow{(d)} X$ for some real-valued random variable $X$.
+        \item  $\phi$ is the characteristic function of $X$.
+        \item $\phi$ is continuous on all of $\R$.
+        \item $\phi$ is continuous at $0$.
+    \end{enumerate}
+\end{theorem}
+\begin{example}
+    Let $Z \sim \cN(0,1)$ and $X_n \coloneqq  n Z$.
+    We have $\phi_{X_n}(t) = \bE[[e^{\i t X_n}] = e^{-\frac{1}{2} t^2 n^2} \xrightarrow{n \to \infty} \One_{\{t = 0\} }$.
+    $\One_{\{t = 0\}}$ is not continuous at $0$.
+    By \autoref{genlevycontinuity}, $X_n$ can not converge to a real-valued
+    random variable.
+
+    Exercise: $X_n \xrightarrow{(d)} \overline{X}$,
+    where $\bP[\overline{X} = \infty] = \frac{1}{2} = \bP[\overline{X} = -\infty]$.
+
+    Similar examples are $\mu_n \coloneqq  \delta_n$ and
+    $\mu_n \coloneqq  \frac{1}{2} \delta_n + \frac{1}{2} \delta_{-n}$.
+\end{example}
+
+\begin{example}
+    Suppose that $X_1, X_2,\ldots$ are i.d.d.~with $\bE[X_1] = 0$.
+    Let $\sigma^2 \coloneqq  \Var(X_i)$.
+    Then the distribution of $\frac{S_n}{\sigma \sqrt{n}}$ is tight:
+
+    \begin{IEEEeqnarray*}{rCl}
+    \bE\left[ \left( \frac{S_n}{\sqrt{n} }^2 \right)^2 \right] &=&
+        \frac{1}{n} \bE[ (X_1+ \ldots + X_n)^2]\\
+                                                          &=& \sigma^2
+    \end{IEEEeqnarray*}
+    For $a > 0$, by Chebyshev's inequality, % TODO
+    we have
+    \[
+    \bP\left[ \left| \frac{S_n}{\sqrt{n}} \right| > a \right] \leq \frac{\sigma^2}{a^2} \xrightarrow{a \to \infty} 0.
+    \]
+    verifying \autoref{genlevycontinuity}.
+\end{example}
+
+\begin{example}
+    Suppose $C$ is a random variable which is Cauchy distributed, i.e.~$C$
+    has probability distribution $f_C(x) = \frac{1}{\pi} \frac{1}{1 + x^2}$.
+
+    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 $S_n \sim  C$.
+\end{example}
+
+