lecture 13 part 2

inputs/lecture_13.tex

@@ -40,7 +40,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
     The Lyapunov condition implies the Lindeberg condition.
     (Exercise).
 \end{remark}
-We will not prove the \autoref{linebergclt} or \autoref{lyapunovclt}
+We will not prove the \autoref{lindebergclt} or \autoref{lyapunovclt}
 in this lecture. However, they are quite important.
 
 We will now sketch the proof of \autoref{levycontinuity},
@@ -106,4 +106,154 @@ A generalized version of \autoref{levycontinuity} is the following:
     Exercise: $\phi_{S_n}(t) = e^{-|t|} = \phi_{C_1}(t)$, thus $S_n \sim  C$.
 \end{example}
 
+We will prove \autoref{levycontinuity} assuming
+\autoref{lec10_thm1}.
+\autoref{lec10_thm1} will be shown in the notes.\todo{TODO}
+We will need the following:
+\begin{lemma}
+    \label{lec13_lem1}
+    Given a sequence $(F_n)_n$ of probability distribution functions,
+    there is a subsequence $(F_{n_k})_k$ of $F_n$
+    and a right continuous, non-decreasing function $F$,
+    such that $F_{n_k} \to  F$ at all continuity points of $F$.
+    (We do not yet claim, that $F$ is a probability distribution function,
+    as we ignore $\lim_{x \to \infty} F(x)$ and $\lim_{x \to -\infty} F(x)$ for now).
+\end{lemma}
+\begin{lemma}
+    \label{s7e1}
+    Let $\mu \in M_1(\R)$, $A > 0$ and $\phi$ the characteristic function of $\mu$.
+    Then $\mu\left( (-A,A) \right) \ge  \frac{A}{2} \left| \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi(t) d t \right| - 1$.
+\end{lemma}
+\begin{refproof}{s7e1}
+    Exercise.\todo{TODO}
+\end{refproof}
+
+
+
+\begin{refproof}{levycontinuity}
+``$\implies$ '' If $\mu_n \implies \mu$, then
+$\int f d \mu_n \to  \int f d \mu$
+for all $f \in C_b$ and $x \to  e^{\i t x}$ is continuous and bounded.
+
+
+``$ \impliedby$''
+
+% Step 1:
+\begin{claim}
+    \label{levyproofc1}
+    Given $\epsilon > 0$ there exists $A > 0$ such that
+    $\liminf_n \mu_n\left( (-A,A) \right) \ge  1 - 2 \epsilon$.
+\end{claim}
+\begin{refproof}{levyproofc1}
+    If $f$ is continuous, then
+    \[
+    \frac{1}{\eta} \int_{x - \eta}^{x + \eta} f(t) d t \xrightarrow{\eta \downarrow 0}  f(x).
+    \]
+    Applying this to $\phi$ at $t = 0$, one obtains:
+    \begin{equation}
+    \left| \frac{A}{4} \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi(t) dt - 1 \right| < \frac{\epsilon}{2}
+    \label{levyproofc1eqn1}
+    \end{equation}
+
+    \begin{claim}
+        For $n$ large enough, we have
+        \begin{equation}
+        \left| \frac{A}{4} \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) d t - 1\right| < \epsilon.
+        \label{levyproofc1eqn2}
+        \end{equation}
+    \end{claim}
+    \begin{subproof}
+        Apply dominated convergence.
+    \end{subproof}
+    So to prove $\mu_n\left( (-A,A) \right) \ge 1 - 2 \epsilon$,
+    apply \autoref{s7e1}.
+    It suffices to show that
+    \[
+        \frac{A}{2} \left| \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt\right| - 1 \ge  1 - 2\epsilon
+    \] 
+    or
+    \[
+    1 - \frac{A}{4} \left|\int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt \right| \le \epsilon,
+    \]
+    which follows from \autoref{levyproofc1eqn2}.
+\end{refproof}
+
+% Step 2
+By \autoref{lec13_lem1}
+there exists a right continuous, non-decreasing $F $
+and a subsequence $(F_{n_k})_k$ of $(F_n)_n$ where $F_n$ is
+the probability distribution function of $\mu_n$,
+such that $F_{n_k}(x) \to F(x)$ for all $x$ where $F$ is continuous.
+\begin{claim}
+    \[
+    \lim_{n \to -\infty} F(x) = 0
+    \]
+    and
+    \[
+    \lim_{n \to  \infty} F(x) = 1,
+    \]
+    i.e.~$F$ is a probability distribution function.\footnote{This does not hold in general!}
+\end{claim}
+\begin{subproof}
+    We have
+    \[
+    \mu_{n_k}\left( (- \infty, x] \right) = F_{n_k}(x) \to F(x).
+    \]
+    Again, given $\epsilon > 0$, there exists $A > 0$, such that
+    $\mu_{n_k}\left( (-A,A) \right) > 1 - 2 \epsilon$ (\autoref{levyproofc1}).
+
+    Hence $F(x) \ge  1 - 2 \epsilon$ for $x > A $
+    and $F(x) \le  2\epsilon$ for $x < -A$.
+    This proves the claim.
+\end{subproof}
+
+Since $F$ is a probability distribution function, there exists
+a probability measure $\nu$ on $\R$ such that $F$ is the distribution
+function of $\nu$.
+Since $F_{n_k}(x) \to F_n(x)$ at all continuity points $x$ of $F$.
+By \autoref{lec10_thm1} we obtain that
+$\mu_{n_k} \overset{k \to \infty}{\implies} \nu$.
+Hence
+$\phi_{\mu_{n_k}}(t) \to \phi_\nu(t)$, by the other direction of that theorem.
+But by assumption,
+$\phi_{\mu_{n_k}}(\cdot ) \to  \phi_n(\cdot )$ so $\phi_{\mu}(\cdot) = \phi_{\nu}(\cdot )$.
+By \autoref{charfuncuniqueness}, we get $\mu = \nu$.
+
+We have shown, that $\mu_{n_k} \implies \mu$ along a subsequence.
+We still need to show that $\mu_n \implies \mu$.
+\begin{fact}
+    Suppose $a_n$ is a bounded sequence in $\R$,
+    such that any subsequence converges to $a \in \R$.
+    Then $a_n \to  a$.
+\end{fact}
+\begin{subproof}
+    \todo{in the notes}
+\end{subproof}
+Assume $\mu_n$ 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 $F_{n_1}(x_0), F_{n_2}(x_0), \ldots$
+which are all outside $(F(x_0) - \delta, F(x_0) + \delta)$.
+Then $\phi_{n_1}, \phi_{n_2}, \ldots \to  \phi$.
+Now, there exists a further subsequence $G_1, G_2, \ldots$ of $F_{n_i}$,
+which converges.
+$G_1, G_2, \ldots$ is a subsequence of $F_1, F_2,\ldots$.
+However $G_1, G_2, \ldots$ is not converging to $F$,
+as this would fail at $x_0$. This is a contradiction.
+
+\end{refproof}
+
+
+
+% IID is over now
+\subsection{Summary}
+What did we learn:
+
+\begin{itemize}
+    \item How to construct product measures
+    \item WLLN and SLLN
+    \item Kolmogorov's three series theorem
+    \item Fourier transform, weak convergence and CLT
+\end{itemize}
+
 

probability_theory.tex

@@ -36,6 +36,7 @@
 \input{inputs/lecture_10.tex}
 \input{inputs/lecture_11.tex}
 \input{inputs/lecture_12.tex}
+\input{inputs/lecture_13.tex}
 
 \cleardoublepage