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\begin{document}
\begin{center}
{\bf Probability 1\\
CEU Budapest, fall semester 2013}\\
Imre Péter Tóth\\
{\bf Homework sheet 5 -- solutions}\\
\end{center}


\begin{enumerate}[5.1]
%
\item \textbf{(homework)} Let $X_1,X_2,\dots$ be i.i.d. random variables with density (w.r.t. Lebesgue measure)
$f(x)=\frac{1}{\pi}\frac{1}{1+x^2}$. (So they have the Cauchy distribution.) Find the weak limit (as $n\to\infty$)
of the average
$$\frac{X_1+\dots +X_n}{n}.$$
\emph{Warning: this is not hard, but also not as trivial as it may seem. Hint: a possible solution is using characteristic functions. Calculating the characteristic function of the Cauchy distribution is a little tricky, but you can look it up.}

\textbf{Solution:} The characteristic function of the Cauchy distribution is $\psi_{X_k}(t)=e^{-|t|}$ (see e.g. Durrett~\cite{durrett}, Example 3.3.9). So $S_n=X_1+\dots +X_n$ has characteristic function
$\psi_{S_n}(t)=\left(\psi_{X_k}(t)\right)^n=e^{-n|t|}$ and $\frac{S_n}{n}$ has characteristic function
$\psi_{\frac{S_n}{n}}(t)=\psi_{S_n}\left(\frac{t}{n}\right)=e^{-|t|}$. This means that $\frac{S_n}{n}$ has the same Cauchy distribution as the $X_k$ for every $n$, so it also converges to the Cauchy distribution weakly.

Note that this does not contradict the weak law of large numbers, because our $X_k$ do not have an expectation.
%
\item Durrett~\cite{durrett}, Exercise 3.3.20
%
\item \textbf{(homework)} Durrett~\cite{durrett}, Exercise 3.4.4

\textbf{Solution 1:} \emph{My solution, no trickery.} We calculate the limiting distribution function:
$$
F_n(x):=\prob(2(\sqrt{S_n}-\sqrt{n})\le x)=\prob\left(\sqrt{S_n}\le \sqrt{n}+\frac{x}{2}\right).
$$
Note that $S_n\ge 0$, so there is no problem with the square root and for every $x$ the right hand side of the inequality $\sqrt{S_n}\le \sqrt{n} +\frac{x}{2}$ is positive for big enough $n$, so it holds if and only if $S_n\le \left(\sqrt{n}+\frac{x}{2}\right)^2$. So, for $n$ big enough, using $\expect S_n=n$,
$$
F_n(x)=\prob\left(S_n\le n+\sqrt{n}x + \frac{x^2}{4}\right)=\prob\left(\frac{S_n-\expect S_n}{\sqrt{n}}\le x+\frac{x^2}{4\sqrt{n}}\right).
$$
As $n\to\infty$, this converges to $\Phi\left(\frac{x}{\sigma}\right)$ (where $\Phi$ is the standard normal distribution function), because $\frac{S_n-\expect S_n}{\sqrt{n}}$ converges weakly to $\mathcal{N}(0,\sigma^2)=\sigma\chi$ (by the C.L.T.) and $\frac{x^2}{4\sqrt{n}}\to 0$. This is what we had to show.

\emph{(The $\frac{x^2}{4\sqrt{n}}\to 0$ term can be treated rigorously using the continuity of the normal distribution function and some $\eps$-ing, or looking at the $\liminf$ and $\limsup$ separately.)}

\textbf{Solution 2:} \emph{Invented by clever students. More elegant, more tricky and one needs to know more.}
Write the random variable under study in the tricky form
$$
2(\sqrt{S_n}-\sqrt{n})=\frac{2}{\sqrt{\frac{S_n}{n}}+1}\frac{S_n-n}{\sqrt{n}}.
$$
The second factor converges to $\sigma \chi$ by the C.L.T., and the $\frac{S_n}{n}$ under the square root converges weakly to $\expect X_k=1$ by the weak law of large numbers. So Theorem 3.2.4 in Durrett~\cite{durrett} gives that the first factor converges weakly to $1$ (because the map $x\mapsto\frac{2}{\sqrt{x}+1}$ is continuous in $1$), and consequently Exercise 3.2.14 from Durrett~\cite{durrett} gives that the product converges weakly to $\sigma \chi$ as well.



%
\item Durrett~\cite{durrett}, Exercise 3.4.5
%
\item \textbf{(homework)} Durrett~\cite{durrett}, Exercise 3.6.1

\textbf{Solution:}
\begin{enumerate}[(i)]
 \item We check the defining properties of a metric. Let $\mu$, $\nu$ and $\rho$ be any probability measures on $\IZ$.
  For $z\in\IZ$, we abuse notation and write $\mu(z)=\mu(\{z\})$, just like Durrett.
  \begin{itemize}
   \item $d(\mu,\nu)\ge 0$ clearly from the definition.
   \item $d(\mu,\nu)=d(\nu,\mu)$ clearly from the definition.
   \item If $d(\mu,\nu) = 0$, then for every $A\subset\IZ$ we have $|\mu(A)-\nu(A)|=0$ meaning $\mu(A)=\nu(A)$, so $\mu=\nu$.
   \item
   \begin{eqnarray*}
d(\mu,\nu)&=&\frac12\sum_{z\in\IZ}|\mu(z)-\rho(z)+\rho(z)-\nu(z)| \\
&\le& \frac12 \sum_{z\in\IZ}\left(|\mu(z)-\rho(z)|+|\rho(z)-\nu(z)|\right)=d(\mu,\rho)+d(\rho,\nu).
   \end{eqnarray*}
  \end{itemize}
 \item One direction is easy: for every $x\in\IZ$ we have $|\mu_n(x)-\mu(x)|\le 2||\mu_n-\mu||$, so if $||\mu_n-\mu||\to 0$, then $\mu_n(x)\to\mu(x)$.
 
 For the other direction we need to work. Suppose that $\mu_n(x)\to\mu(x)$ for every $x\in\IZ$. Then for every $\eps>0$ one can take $A\subset \IZ$ big enough but still \emph{finite} such that $\mu(\IZ\setminus A)<\frac{\eps}{2}$. Set $K:=|A|<\infty$. We will treat the points inside $A$ one by one, and those outside $A$ together. Now take $n_0$ so big that $|\mu_n(x)-\mu(x)|<\frac{\eps}{2K}$ for every $n\ge n_0$ and every $x\in A$. (This is possible because $A$ is finite.) This implies that $|\mu_n(A)-\mu(A)|<\frac{\eps}{2}$, so $|\mu_n(\IZ\setminus A)-\mu(\IZ\setminus A)|=|\mu_n(A)-\mu(A)|<\frac{\eps}{2}$ as well. With the definition of $A$ this gives $\mu_n(\IZ\setminus A)<\eps$ for every $n\ge n_0$. Now
 \begin{eqnarray*}
 ||\mu_n-\mu||&=&\frac12\sum_{x\in A}|\mu_n(x)-\mu(x)|+\frac12 \sum_{x\in\IZ\setminus A} |\mu_n(x)-\mu(x)| \\
 &\le& \frac12 \sum_{x\in A} \frac{\eps}{2K} + \frac12 \sum_{x\in\IZ\setminus A} (\mu_n(x)+\mu(x))=\frac12 K \frac{\eps}{2K} + \frac12 \mu_n(\IZ\setminus A) + \frac12 \mu(\IZ\setminus A) \\
 &\le& \frac{\eps}{4} + \frac{\eps}{2} + \frac{\eps}{4} =\eps.
 \end{eqnarray*}
\end{enumerate}

%
\item Durrett~\cite{durrett}, Exercise 3.6.2
%
\end{enumerate}

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


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