branching process

inputs/lecture_09.tex

@@ -99,7 +99,7 @@ characteristic functions of Fourier transforms.
     If $\mu$ and $\nu$ have Lebesgue densities $f_\mu$ and $f_\nu$,
     then the convolution has Lebesgue density
     \[
-        f_{\mu \ast \nu}(x) \coloneqq
+        f_{\mu \ast \nu}(x) =
             \int_{\R^d} f_\mu(x - t) f_\nu(t) \lambda^d(\dif t).
    \]
 \end{fact}

inputs/lecture_13.tex

@@ -274,13 +274,12 @@ 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 has a subsequence
-    that converges to $a \in \R$.
+    such that any convergent subsequence converges to $a \in \R$.
     Then $a_n \to  a$.
 \end{fact}
-\begin{subproof}
-    \notes
-\end{subproof}
+% \begin{subproof}
+%     \notes
+% \end{subproof}
 Assume that $\mu_n$ does not converge to $\mu$.
 By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,
 such that $F_n(x_0) \not\to F(x_0)$.
@@ -292,6 +291,11 @@ 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}
+\begin{refproof}{genlevycontinuity}
+    % TODO TODO TODO
+
+    
 \end{refproof}
 
 % IID is over now

inputs/lecture_15.tex

@@ -183,9 +183,23 @@ Recall
         \bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
     \]
 \end{theorem}
-\begin{proof}
-    \todo{Exercise}
-\end{proof}
+% TODO
+% \begin{proof}
+%     Take some $G \in \cG$.
+%     We first consider the case of $|X(\omega)|^p, |Y(\omega)|^p > 0$
+%     a.s.~for $\omega \in G$.
+%     Then
+%      \begin{IEEEeqnarray*}{rCl}
+%          \int_G\frac{\bE[|XY| ~ |\cG]}%
+%             {\bE[\bE[|X|^p | \cG]^{\frac{1}{p}} \bE[|Y|^p| \cG]^{\frac{1}{q}}}
+%             \dif \bP
+%         &=& \int_G \frac{|X|}{\bE[|X|^p]^{\frac{1}{p}}} \frac{|Y|}{\bE[|Y|^q]^{\frac{1}{q}}}
+%         \dif \bP\\
+%         &\le& \left(\int_G \frac{|X|^p}{\bE[|X|^p]} \dif \bP\right)^p
+%         \left(\int_G \frac{|Y|^q}{\bE[|Y|^q]} \dif \bP\right)^q\\
+%         &=& \bE[\One_G]
+%     \end{IEEEeqnarray*}
+% \end{proof}
 
 \begin{theorem}[Tower property]
     \label{ceprop10}

inputs/lecture_20.tex

@@ -119,6 +119,42 @@ we need the following theorem, which we won't prove here:
     we get the convergence.
 \end{refproof}
 
+\begin{example}+[\vocab{Branching Process}; Exercise 10.1, 12.4]
+    Let $(Y_{n,k})_{n \in \N_0, k \in \N}$ be i.i.d.~with values in $\N_0$
+    such that $0 < \bE[Y_{n,k}] = m < \infty$.
+    Define
+    \[
+    S_0 \coloneqq  1, S_n \coloneqq \sum_{k=1}^{S_{n-1}} Y_{n-1,k}
+    \]
+    and let $M_n \coloneqq  \frac{S_n}{m^n}$.
+    $S_n$ models the size of a population.
+
+    \begin{claim}
+        $M_n$ is a martingale.
+    \end{claim}
+    \begin{subproof}
+        We have
+        \begin{IEEEeqnarray*}{rCl}
+            \bE[M_{n+1} - M_n | \cF_n]
+            &=& \frac{1}{m^n} \left( \frac{1}{m}\sum_{k=1}^{S_{n}}  \bE[X_{n,k}] - S_n\right)\\
+            &=& \frac{1}{m^n}(S_n - S_n).
+        \end{IEEEeqnarray*}
+    \end{subproof}
+
+    \begin{claim}
+        $(M_n)_{n \in \N}$ is bounded in $L^2$ iff $m > 1$.
+    \end{claim}
+    \todo{TODO}
+    \begin{claim}
+        If $m > 1$ and $M_n \to M_\infty$,
+        then
+        \[
+        \Var(M_\infty) = \sigma^2(m(m-1))^{-1}.
+        \]
+    \end{claim}
+    \todo{TODO}
+\end{example}
+
 \subsection{Stopping Times}
 
 \begin{definition}[Stopping time]