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\begin{document}
\begin{center}
{\bf Probability 1\\
CEU Budapest, fall semester 2013}\\
Imre Péter Tóth\\
{\bf Homework sheet 9 -- due on 02.12.2013 -- and exercises for practice}\\
\end{center}


\begin{enumerate}[9.1]
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\item \textbf{(homework)} Durrett~\cite{durrett}, Exercise 8.1.3
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\item \textbf{(homework)} Durrett~\cite{durrett}, Exercise 8.2.3
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\item \label{ex:Levylemma} It is not hard to show that if $\xi$ is a standard Gaussian random variable and $x\ge 1$, then
$$\prob(|X|\ge x)\le\sqrt{\frac{2}{\pi}}e^{-\frac{x^2}{2}}.$$
Use this to show that if $\xi_1,\xi_2,\dots$ are i.i.d. standard Gaussian, then, with probability $1$, the event $\{|\xi_n|>2\ln n\}$ occurs for at most finitely many $n$-s.
%
\item \emph{Paul Lévy construction of the Wiener process.} In a possible construction of the Wiener process (or Brownian motion) on $[0,1]$ -- discussed in class -- we define a sequence of piecewise linear continuous random functions so that we first define $f_n$ at dyadic rationals that are multiples of $\frac{1}{2^n}$, inheriting every second value (at multiples of $\frac{1}{2^{n-1}}$) form $f_{n-1}$, and setting the values at the remaining points (of the form $\frac{2k-1}{2^n}$) to be the average of the two neighbouring values, plus an independent Gaussian random value with mean $0$ and variance $\frac{1}{4^n}$. Then we extend $f_n$ to $[0,1]$ piecewise linearly.

Formally: we take independent standard Gaussian random variables $\xi_0$ and $\xi_{n,k}$ where $n=1,2,\dots$ and $k=1,2,\dots,2^{n-1}$. Then 
\begin{itemize}
 \item In the $0$th step we fix $f_0(0)=0$ and $f_0(1)=\xi_0$. We connect these two values linearly.
 \item In the $1$st step we leave $f_1(0)=f_0(0)$ and $f_1(1)=f_0(1)$, but also set $f_1(\frac12)=f_0(\frac12)+\frac12 \xi_{1,1}$. We connect these three values linearly.
 \item \dots in the $n$th step we leave $f_n\left(\frac{k}{2^{n-1}}\right)=f_{n-1}\left((\frac{k}{2^{n-1}}\right)$ for $k=0,1,\dots,2^{n-1}$, but also set $f_n\left(\frac{k-\frac12}{2^{n-1}}\right)=f_{n-1}\left(\frac{k-\frac12}{2^{n-1}}\right)+\frac{1}{2^n}\xi_{n,k}$ for $k=1,\dots,2^{n-1}$. We connect these $2^n+1$ values linearly.
\end{itemize}
Notice that, in this construction, the difference $g_n:=f_{n+1}-f_n$ is the sum of $2^n$ ``tent'' maps with disjoint supports and i.i.d. Gaussian ``heights''.

Use the statement of Exercise~\ref{ex:Levylemma} to show that, with probability $1$, the series $$\lim_{n\to\infty}f_n=f_0+\sum_{n=0}^\infty g_n$$
is uniformly absolutely convergent.

\end{enumerate}

\begin{thebibliography}{1}
 \bibitem{durrett}Durrett, R. \emph{Probability: Theory and Examples}. \textbf{4th} edition, Cambridge University Press (2010)
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