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\begin{document}
\begin{center}
{\bf Probability 1\\
CEU Budapest, fall semester 2013}\\
Imre Péter Tóth\\
{\bf Homework sheet 2 -- due on 07.10.2013 -- and exercises for practice}\\
\end{center}


\begin{enumerate}[2.1]
%

\item \textbf{(homework)} Exercise 3 of ``Homework sheet 1'', delayed from last week (unless already done)
%
\item \textbf{(homework)} Exercise 4 of ``Homework sheet 1'', delayed from last week (unless already done)
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\item \emph{Continuity of the measure}
  \begin{enumerate}
   \item Prove the following:
     \begin{theorem} (Continuity of the measure)
      \begin{enumerate}
       \item If $(\Omega,\sigalg,\mu)$ is a measure space and $A_1,A_2,\dots$ is an increasing sequence of measurable sets (i.e. $A_i\in\sigalg$ and $A_i\subset A_{i+1}$ for all $i$), then $\mu(\cup_{i=1}^\infty A_i)=\lim_{i\to\infty}\mu(A_i)$ (and both sides of the equation make sense).
       \item If $(\Omega,\sigalg,\mu)$ is a measure space, $A_1,A_2,\dots$ is a decreasing sequence of measurable sets (i.e. $A_i\in\sigalg$ and $A_i\supset A_{i+1}$ for all $i$) and $\mu(A_1)<\infty$, then $\mu(\cap_{i=1}^\infty A_i)=\lim_{i\to\infty}\mu(A_i)$ (and both sides of the equation make sense).
      \end{enumerate}
     \end{theorem}
   \item Show that in the second statement the condition $\mu(A_1)<\infty$ is needed, by constructing a counterexample for the statement when this condition does not hold.
  \end{enumerate}
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\item \emph{Usefulness of the linearity of the expectation.}
A building has $10$ floors, not including the ground floor. On the ground floor, $10$ people get into the elevator, and every one of them chooses a destination at random, uniformly out of the $10$ floors, independently of the others. Let $X$ denote the number of floors \emph{on which the elevator stops} -- i.e. the number of floors that were chosen by at least one person. Calculate the expectation and the variance of $X$.
\emph{(hint: First notice that the distribution of $X$ is hard to calculate. Find a way to calculate the expectation and the variance without that.)}
%
\item Calculate the characteristic function of
   \begin{enumerate}
     \item The Bernoulli distribution $B(p)$ (see Homework sheet 1)
     \item The ``pessimistic geometric distribution with parameter $p$'' -- that is, the distribution $\mu$ on $\{0,1,2\dots\}$ with weights $\mu(\{k\})=(1-p)p^k$ $(k=0,1,2\dots)$.
     \item The ``optimistic geometric distribution with parameter $p$'' -- that is, the distribution $\nu$ on $\{1,2,3,\dots\}$ with weights $\nu(\{k\})=(1-p)p^{k-1}$ $(k=1,2\dots)$.
     \item \textbf{(homework)} The Poisson distribution with parameter $\lambda$ -- that is, the distribution $\eta$ on $\{0,1,2\dots\}$ with weights $\eta(\{k\})=e^{-\lambda}\frac{\lambda^k}{k!}$ $(k=0,1,2\dots)$.
     \item \textbf{(homework)} The exponential distribution with parameter $\lambda$ -- that is, the distribution on $\IR$ with density (w.r.t. Lebesgue measure)
     $$f_{\lambda}(x)=\begin{cases}
                        \lambda e^{-\lambda x},\text{ if $x>0$}\\
                        0, \text{ if not}
                      \end{cases}.$$
   \end{enumerate}
%
\item \textbf{(homework)} Calculate the characteristic function of the normal distribution $\mathcal{N}(m,\sigma^2)$. (Remember the definition from the old times: $\mathcal{N}(m,\sigma^2)$ is the distribution on $\IR$ with density (w.r.t. Lebesgue measure)
$$f_{m,\sigma^2}(x)=\frac{1}{\sqrt{2\pi} \sigma}e^{-\frac{(x-m)^2}{2\sigma^2}}.$$
You can save yourself some paperwork if you only do the calculation for $\mathcal{N}(0,1)$ and reduce the general case to this using the relation between different normal distributions. You can and should use the fact that
$$\int_{-\infty}^\infty f_{m,\sigma^2}(x)\diff x=1$$
for every $m$ and $\sigma$.
%
\item \emph{Dominated convergence and continuous differentiability of the characteristic function.}\\
The Lebesgue dominated convergence theorem is the following
\begin{theorem}[dominated convergence]
Let $(\Omega,\sigalg,\mu)$ be a measure space and $f_1,f_2,\dots$ measurable real valued functions on $\Omega$ which converge to the limit function pointwise, $\mu$-almost everywehere. (That is, $\lim_{n\to\infty}f_n(x)=f(x)$ for every $x\in\Omega$, except possibly for a set of $x$-es with $\mu$-measure zero.)
Assume furthermore that the $f_n$ admit a common \emph{integrable dominating function}: there exists a $g:\Omega\to\IR$ such that
$|f_n(x)|\le g(x)$ for every $x\in\Omega$ and $n\in\IN$, and $\int_\Omega g\diff \mu<\infty$. Then (all the $f_n$ and also $f$ are integrable and)
$$\lim_{n\to\infty} \int_\Omega f_n \diff \mu = \int_\Omega f \diff \mu.$$
\end{theorem}

Use this theorem to prove the following

\begin{theorem}[differentiability of the characteristic function]
Let $X$ be a real valued random variable, $\psi(t)=\expect(e^{itX})$ its characteristic function and $n\in\IN$.
If the $n$-th moment of $X$ exists and is finite (i.e. $\expect(|X|^n)<\infty$), then $\psi$ is $n$ times continuously differentiable and
$$
\psi^{(k)}(0)=i^k \expect(X^k),\quad k=0,1,2,\dots,n.
$$
\end{theorem}
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\end{enumerate}

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