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\newtheorem{thm}{Aufgabe H}
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\title{Übungsblatt 9 W-Theorie}
\author{Max Klinger}
 
\begin{document} 
\section*{Wahrscheinlichkeitstheorie Übungsblatt 9}
\begin{thm} \hfill \\
\end{thm}
\begin{proof}
$x_{k,n}:= \frac{x_k}{\sqrt{n}}$\\
$\rightarrow \EE{X_{k,n}}=0$\\
$\rightarrow \sum^n_{k=1}\EE{X^2_{k,n}} = \frac{1}{n}\sum^n_{k=1} 1=1$\\
$\rightarrow$ unabhängig\\
$\rightarrow$ L-B-Bedingung; $\forall \epsilon >0$\\
$\sum^n_{k=1} \EE{\frac{X^2_k}{n} \1_{\{\abs{X_k} > \epsilon \sqrt{n}\}}} \leq\footnote{da $\EE{X^2_k \epsilon^\alpha \sqrt{n}^\alpha \1_{\{\abs{X_k}^\alpha> \epsilon^\alpha \sqrt{n}^\alpha \}}} \leq \EE{\abs{X_k}^{2+\alpha} \1}$} \sum^n_{k=1} \frac{1}{n \epsilon^\alpha \sqrt{n}^\alpha} \EE{\abs{X}^{2+\alpha}}$\\
$\leq \sum^n_{k=1} \frac{1}{n \epsilon^\alpha \sqrt{n}^\alpha} \sup_{k} \EE{\abs{X_k}^{2+\alpha}} \xrightarrow {n\rightarrow \infty}0 \Rightarrow 0$
\end{proof}

\begin{thm}
\end{thm}

\begin{proof}[Beweis]
\begin{itemize}
 \item \begin{align*}
\phi_X(t) &= \EE{e^{itX}}= \EE{e^{it\frac{X+Y}{\sqrt{2}}}} = \EE{e^{i\frac{t}{\sqrt 2} X}} \EE{e^{i\frac{t}{\sqrt{2}}Y}} = \phi_X\left(\frac{t}{\sqrt{2}}\right) \phi_Y\left(\frac{t}{\sqrt 2}\right) \\
&= \phi_X \left(\frac{t}{\sqrt{2}}\right)^2 = \dotsc = \phi_X\left(\frac{X}{\sqrt{2}}\right)^{(2^n)}        
       \end{align*}
\item $X_i $ iid $X_i \sim X$ $\frac{S_m}{\sqrt{m}}\xrightarrow w \mathcal{N}(0,\sigma)$
\item $m = 2^n \Rightarrow \phi_{\frac{S_m}{\sqrt{n}}}(t) = \phi_{X_1} \left(\frac{t}{\sqrt m} \right) \dotsm \phi_{X_m} \left(\frac{t}{\sqrt{m}}\right) = \phi_{X}\left( \frac{t}{\sqrt{m}}\right)^m = \phi_X\left( \frac{t}{\sqrt{2^n}}\right)\xrightarrow{2^n} \phi_{\mathcal{N}(0,\sigma)(t)}$
\item $\phi_X(t) = \phi_{\frac{S_m}{\sqrt{m}}}\xrightarrow{m\rightarrow \infty} \phi_{\mathcal{N(0,1)}}(t)$
\end{itemize}
$\Rightarrow X\sim \mathcal{N}(0, \sigma)$
\end{proof}


\begin{thm}

\end{thm}

\begin{proof}
\begin{itemize}
 \item $W(S_n) = \sum^n_{i=1} 2^{i2}= \sum^n_{i=1} 4^i -1 = \frac{4^{n-1}-1}{4-1} -1 0= \frac{4^{n+1}-4}{3}$
$\Rightarrow \frac{2^n}{\sqrt{W(S_n)}} = \frac{2^n \sqrt{3}}{\sqrt{4^{n+1}-4}} = \frac{2^n}{\sqrt{4^{n+1} -1}} \frac{\sqrt{3}}{2} \rightarrow \frac{\sqrt{3}}{2} \approx 0,86$
\item $\prob{S_n >2^n} \geq \prob{S_{n-2} + X_{n-1} + X_n > 2^n, X_{n+1}>0, X_n > 0} \geq \prob{X_{n-1} > 0, X_n>0} = \frac{1}{4}$
\item $\prob{\frac{S_n}{\sqrt{W(S_n)}} \geq \frac{5}{6}} = \prob{S_n \geq \frac{5}{6} \sqrt{W(S_n)} \frac{2^n}{2^n}}= \prob{S_n \geq \frac{5}{6} \frac{\sqrt{W(S_N)}}{2^n}2^n} \geq \prob{S_n \geq 2^} \geq \frac{1}{4}$
\item $\mathcal{N}_{0,1}([\frac{5}{6}, \infty]) = 1-\mathcal{N}_{(0,1)}(]-\infty, \frac{5}{6}] < 1-\mathcal{N}_{(0,1)(]-\infty, 0,83]}< 1-0,79 = 0,21 < 0,25$
\end{itemize}

\end{proof}

\end{document}
