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\begin{document}
\begin{center}
{\bf Probability 1\\
CEU Budapest, fall semester 2013}\\
Imre Péter Tóth\\
{\bf Homework sheet 4 -- solutions}\\
\end{center}


\begin{enumerate}[4.1]
%
\item \textbf{(homework)} For real numbers $a_1,a_2,a_3,\dots$ define the infinite product $\prod\limits_{k=1}^{\infty}a_k$ as
$$
\prod_{k=1}^{\infty}a_k:=\lim_{n\to\infty}\prod_{k=1}^{n}a_k,
$$
whenever this limit exists.

Let $p_1,p_2,p_3,\dots$ satisfy $0\le p_k<1$ for all $k$. Show that $\prod\limits_{k=1}^{\infty}(1-p_k)>0$ if and only if $\sum\limits_{k=1}^{\infty}p_k <\infty$.

\emph{(Hint: estimate the logarithm of $(1-p)$ with $p$.)}

\textbf{Solution:} For $0\le p_k \lneqq 1$ we have that $\prod\limits_{k=1}^{\infty}(1-p_k)>0$ if and only if
\begin{equation}\label{eq:limsumlog}
 \lim_{n\to\infty} \sum_{k=1}^n \ln(1-p_k) >-\infty.
\end{equation}
Now if $p_k\nrightarrow 0$, then this is clearly false. If $p_k\to 0$, then we get from the linear approximation of $x\mapsto \ln (1+x)$ near $x_0=0$ that -- except possibly for finitely many $k$-s --
$$
-p_k \ge \ln(1-p_k) \ge -2p_k.
$$
This implies that
$$
C-\sum_{k=1}^n p_k \ge \sum_{k=1}^n \ln(1-p_k) \ge C-2 \sum_{k=1}^n p_k,
$$
which means that (\ref{eq:limsumlog}) holds if and only if $\lim_{n\to\infty} \sum_{k=1}^n p_k <\infty$.
%
\item Durrett~\cite{durrett}, Exercise 3.3.1
%
\item Durrett~\cite{durrett}, Exercise 3.3.3
%
\item Durrett~\cite{durrett}, Exercise 3.3.9
%
\item \textbf{(homework)} Durrett~\cite{durrett}, Exercise 3.3.10. Show also that independence is needed.

\textbf{Solution:}
\begin{enumerate}[a.)]
 \item Denote the characteristic functions of $X_n$, $Y_n$ and $X_n+Y_n$ by $\psi_n$, $\phi_n$ and $\rho_n$, respectively. Then the assumptions about independence give $\rho_n(t)=\psi_n(t) \phi_n(t)$ for every $t\in\IR$ and $1\le n\le\infty$, and the continuity theorem gives $\psi_n(t)\to\psi_\infty (t)$ and $\phi_n(t)\to\phi_\infty (t)$, so we get $\rho_n(t)\to\rho_\infty (t)$. Using the continuity theorem again gives that $X_n+Y_n \Rightarrow X_\infty +Y_\infty$.
 \item To see that independence is needed, consider the following example. For $1\le n<\infty$ let $X_n\sim B(\frac12)$ and $Y_n=1-X_n$, so $Y_n\sim B(\frac12)$ also. For $n=\infty$ let $X_\infty\sim B(\frac12)$ again, but set $Y_\infty=X_\infty$. Again, this implies $Y_\infty\sim B(\frac12)$. Clearly $X_n \Rightarrow X_\infty$ and $Y_n \Rightarrow Y_\infty$, but $X_n+Y_n \equiv 1 \nRightarrow X_\infty + Y_\infty$, because e.g. $\prob(X_\infty + Y_\infty=1)=0$.
\end{enumerate}
%
\item Durrett~\cite{durrett}, Exercise 3.3.11
%
\item \textbf{(homework)} Durrett~\cite{durrett}, Exercise 3.3.12

\textbf{Solution:} Let $\xi_1,\xi_2,\dots$ be independent and uniform on the two-element set $\{-1;1\}$, and set
$X_n=\sum_{m=1}^n \frac{\xi_m}{2^m}$. Then the characteristic function of the $\xi_m$ is
$$\psi_\xi (t)=\frac12 e^{it(-1)}+\frac12 e^{it1}=cos(t)$$
and the characteristic function of $X_n$ is
$$\psi_{X_n} (t)=\prod_{m=1}^n \psi_\xi \left(\frac{t}{2^m}\right)=\prod_{m=1}^n \cos \left(\frac{t}{2^m}\right).$$
But notice that $X_n$ is uniform on the $2^n$-element set
$$\left\{\frac{k}{2^n}\,:\,k=-2^n+1;-2^n+3;-2^n+5;\dots ;2^n-3;2^n-1\right\},$$
so $X_n$ converges weakly to some $X$ with the (continuous) uniform distribution on $[-1;1]$. (This can easily be seen e.g. from the pointwise convergence of the distribution functions.) So the characteristic function of $X$ is
$$
\psi_{X} (t)=\int_{-1}^1 e^{itx}\frac12 \diff x = \frac{\sin t}{t},
$$
so the continuity theorem states that
$$\frac{\sin t}{t}=\lim_{n\to\infty} \psi_{X_n}(t)=\prod_{m=1}^\infty \cos \left(\frac{t}{2^m}\right).$$
%
\item Durrett~\cite{durrett}, Exercise 3.3.13
%
\end{enumerate}

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


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