branching process
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@ -99,7 +99,7 @@ characteristic functions of Fourier transforms.
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If $\mu$ and $\nu$ have Lebesgue densities $f_\mu$ and $f_\nu$,
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If $\mu$ and $\nu$ have Lebesgue densities $f_\mu$ and $f_\nu$,
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then the convolution has Lebesgue density
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then the convolution has Lebesgue density
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\[
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\[
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f_{\mu \ast \nu}(x) \coloneqq
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f_{\mu \ast \nu}(x) =
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\int_{\R^d} f_\mu(x - t) f_\nu(t) \lambda^d(\dif t).
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\int_{\R^d} f_\mu(x - t) f_\nu(t) \lambda^d(\dif t).
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\]
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\]
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\end{fact}
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\end{fact}
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@ -274,13 +274,12 @@ We have shown, that $\mu_{n_k} \implies \mu$ along a subsequence.
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We still need to show that $\mu_n \implies \mu$.
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We still need to show that $\mu_n \implies \mu$.
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\begin{fact}
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\begin{fact}
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Suppose $a_n$ is a bounded sequence in $\R$,
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Suppose $a_n$ is a bounded sequence in $\R$,
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such that any subsequence has a subsequence
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such that any convergent subsequence converges to $a \in \R$.
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that converges to $a \in \R$.
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Then $a_n \to a$.
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Then $a_n \to a$.
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\end{fact}
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\end{fact}
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\begin{subproof}
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% \begin{subproof}
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\notes
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% \notes
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\end{subproof}
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% \end{subproof}
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Assume that $\mu_n$ does not converge to $\mu$.
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Assume that $\mu_n$ does not converge to $\mu$.
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By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,
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By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,
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such that $F_n(x_0) \not\to F(x_0)$.
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such that $F_n(x_0) \not\to F(x_0)$.
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@ -292,6 +291,11 @@ which converges.
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$G_1, G_2, \ldots$ is a subsequence of $F_1, F_2,\ldots$.
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$G_1, G_2, \ldots$ is a subsequence of $F_1, F_2,\ldots$.
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However $G_1, G_2, \ldots$ is not converging to $F$,
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However $G_1, G_2, \ldots$ is not converging to $F$,
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as this would fail at $x_0$. This is a contradiction.
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as this would fail at $x_0$. This is a contradiction.
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\end{refproof}
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\begin{refproof}{genlevycontinuity}
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% TODO TODO TODO
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\end{refproof}
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\end{refproof}
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% IID is over now
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% IID is over now
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@ -183,9 +183,23 @@ Recall
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\bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
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\bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
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\]
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\]
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\end{theorem}
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\end{theorem}
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\begin{proof}
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% TODO
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\todo{Exercise}
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% \begin{proof}
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\end{proof}
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% Take some $G \in \cG$.
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% We first consider the case of $|X(\omega)|^p, |Y(\omega)|^p > 0$
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% a.s.~for $\omega \in G$.
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% Then
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% \begin{IEEEeqnarray*}{rCl}
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% \int_G\frac{\bE[|XY| ~ |\cG]}%
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% {\bE[\bE[|X|^p | \cG]^{\frac{1}{p}} \bE[|Y|^p| \cG]^{\frac{1}{q}}}
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% \dif \bP
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% &=& \int_G \frac{|X|}{\bE[|X|^p]^{\frac{1}{p}}} \frac{|Y|}{\bE[|Y|^q]^{\frac{1}{q}}}
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% \dif \bP\\
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% &\le& \left(\int_G \frac{|X|^p}{\bE[|X|^p]} \dif \bP\right)^p
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% \left(\int_G \frac{|Y|^q}{\bE[|Y|^q]} \dif \bP\right)^q\\
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% &=& \bE[\One_G]
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% \end{IEEEeqnarray*}
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% \end{proof}
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\begin{theorem}[Tower property]
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\begin{theorem}[Tower property]
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\label{ceprop10}
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\label{ceprop10}
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@ -119,6 +119,42 @@ we need the following theorem, which we won't prove here:
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we get the convergence.
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we get the convergence.
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\end{refproof}
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\end{refproof}
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\begin{example}+[\vocab{Branching Process}; Exercise 10.1, 12.4]
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Let $(Y_{n,k})_{n \in \N_0, k \in \N}$ be i.i.d.~with values in $\N_0$
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such that $0 < \bE[Y_{n,k}] = m < \infty$.
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Define
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\[
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S_0 \coloneqq 1, S_n \coloneqq \sum_{k=1}^{S_{n-1}} Y_{n-1,k}
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\]
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and let $M_n \coloneqq \frac{S_n}{m^n}$.
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$S_n$ models the size of a population.
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\begin{claim}
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$M_n$ is a martingale.
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\end{claim}
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\begin{subproof}
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bE[M_{n+1} - M_n | \cF_n]
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&=& \frac{1}{m^n} \left( \frac{1}{m}\sum_{k=1}^{S_{n}} \bE[X_{n,k}] - S_n\right)\\
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&=& \frac{1}{m^n}(S_n - S_n).
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\end{IEEEeqnarray*}
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\end{subproof}
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\begin{claim}
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$(M_n)_{n \in \N}$ is bounded in $L^2$ iff $m > 1$.
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\end{claim}
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\todo{TODO}
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\begin{claim}
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If $m > 1$ and $M_n \to M_\infty$,
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then
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\[
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\Var(M_\infty) = \sigma^2(m(m-1))^{-1}.
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\]
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\end{claim}
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\todo{TODO}
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\end{example}
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\subsection{Stopping Times}
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\subsection{Stopping Times}
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\begin{definition}[Stopping time]
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\begin{definition}[Stopping time]
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