lecture 18 exercise branching process
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@ -15,31 +15,32 @@ Hence the same holds for submartingales, i.e.
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\subsection{Doob's $L^p$ Inequality}
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\begin{question}
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What about $L^p$ convergence of martingales?
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\end{question}
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\begin{example}[\vocab{Branching process}]
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Fix $u > 0$ and let $p = \frac{1}{1+u}$.
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Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with
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$\bP[Z_n = 1] = p \in (0,1)$.
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$\bP[Z_n = 1] = p$.
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Fix $u > 0$. Let $X_0 = x > 0$.
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Define $X_{n+1} \coloneqq u^{Z_{n+1}} X_{n}$.
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Let $X_0 = x > 0$ and
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define $X_{n+1} \coloneqq u^{Z_{n+1}} X_{n}$.
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Then $(X_n)_n$ is a martingale,
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since
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\begin{IEEEeqnarray*}{rCl}
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\bE[X_{n+1} |\cF_n] &=& X_n \bE[u^{Z_{n+1}}]\\
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&=& X_n \left(p \cdot u + (1-p)\cdot\frac{1}{u}\right)\\
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&=& X_n \left(\frac{p (u^2-1) + 1}{u}\right)\\
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&=& X_n.
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\end{IEEEeqnarray*}
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\begin{exercise}
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Given $u \ge 0$, find $p = p(u)$
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such that $(X_n)_n$ is a martingale w.r.t.~the canonical filtration.
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\end{exercise}
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\todo{TODO}
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By \autoref{doobmartingaleconvergence}, there is an
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a.s.~limit $X_\infty$.
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By \autoref{doobmartingaleconvergence},
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there exists an a.s.~limit $X_\infty$.
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By the SLLN, we have
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\[
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\frac{1}{n} \sum_{k=1}^{n} Z_k \xrightarrow{a.s.} \bE[Z_1] = zp - 1.
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\frac{1}{n} \sum_{k=1}^{n} Z_k \xrightarrow{a.s.} \bE[Z_1] = 2p - 1.
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\]
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Hence
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\[
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