lecture 18 exercise branching process

inputs/lecture_18.tex

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