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\begin{document}
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\large



\title{Math 533, Spring 2020, HW11} 
\author{Due in class, May 4}
\maketitle

\begin{problem}
 If $X_n \to X$ in probability, then $P_{X_n} \to P_X$ vaguely.
\end{problem}

%\begin{problem}
% Under the hypotheses of the central limit theorem, the sequence $Y_n = (\sigma \sqrt{n})^{-1} \sum_1^n (X_j - \mu)$ does not converge in probability.
%\end{problem}

%\begin{problem}
% Let $X_n, X$ be random variables such that $P_{X_n} \to P_X$ vaguely and $\sup_n \E(|X_n|^r) < \infty$. Then $\E(|X_n|^s) \to \E(|X|^s)$ for all $s \in (0,r)$, and if also $s \in \bN$, then $\E(X_n^s) \to \E(X^s)$.
%\end{problem}

%\begin{problem}
% Let $\{X_n\}$ be a sequence of independent random variables, each having the Poisson distribution with mean 1.  Let
% \[
%  S_n = X_1 + \cdots + X_n, \qquad Y_n = \max\left(\frac{n-S_n}{\sqrt{n}} , 0\right).
% \]
%\begin{enumerate}
% \item $\E(Y_n) = e^{-n} \sum_{k=0}^n \frac{n-k}{\sqrt{n}} \frac{n^k}{k!} = \frac{n^{n+1/2}e^{-n}}{n!}.$
% \item $P_{Y_n}$ converges vaguely to $\frac{1}{2} \delta_0 + \chi_{(0,\infty)} \nu_0^1$.
% \item $\int_0^\infty t^2 dP_{Y_n}(t) = \E(Y_n^2) \leq 1$ for all $n$.
% \item $\E(Y_n) = \int_0^\infty t dP_{Y_n}(t) \to \int_0^\infty t d\nu_0^1(t) = (2\pi)^{-1/2}$.
%\end{enumerate}

%\end{problem}

\begin{problem}
 Identify $\bT^1$ with $\{z \in \bC: |z| = 1\}$.  
 \begin{enumerate}
  \item If $X_1, ..., X_n$ are independent, then $P_{X_1X_2\cdots X_n} = P_{X_1}* \cdots * P_{X_n}$.
  \item If $\{X_j\}$ is a sequence of independent random variables with common distribution $\lambda$, the distribution of $\prod_1^n X_j$ converges vaguely to the uniform distribution on $\bT^1$ unless $X_1$ is supported on a finite subgroup of $\bT^1$.
 \end{enumerate}

\end{problem}

\begin{problem}
 Given $b \in \bN \setminus \{1\}$, let $B = \{0, 1, ..., b-1\}$ and $\Omega = B^\bN$.  Put the discrete topology on $B$ and the product topology on $\Omega$, and let $P$ be the product measure on $\Omega$, where each $P_n$ is $b^{-1}$ times counting measure on $B$.  Let $\{X_n\}_1^\infty$ be the coordinate functions on $\Omega$.  Then if $A_1, ..., A_n \subset B$,
 \[
  \Prob\left(\bigcap_1^n X_j^{-1}(A_j)\right) = b^{-n} \prod_1^n |A_j|
 \]
 and $P(\{\omega\}) = 0$ for all $\omega \in \Omega$.

\end{problem}

\begin{problem} Prove the following.
 Let \[\Omega' = \{\omega \in \Omega: X_n(\omega) \neq 0 \text{ for infinitely many $n$}\}.\]  Then $\Omega \setminus \Omega'$ is countable and $P(\Omega') = 1$.  Define $F: \Omega \to[0,1]$ by $F(\omega) = \sum_1^\infty X_n(\omega)b^{-n}$.  Then $F|_{\Omega'}$ is a bijection from $\Omega'$ to $(0,1]$ which maps $\sB_{\Omega'}$ bijectively onto $\sB_{(0,1]}$.
\end{problem}

%\begin{problem}
% Prove that the image measure of $P$ under $F$ is Lebesgue measure.
%\end{problem}

\begin{problem}(Borel's normal number theorem)
 A number $x \in (0,1]$ is called \emph{normal} in base $b$ if the digits $0, 1, ..., b-1$ occur with equal frequency in its base $b$ decimal expansion, that is, if $n^{-1} X_j^{-1}(F^{-1}(x)) \to b^{-1}$ as $n \to \infty$.  Almost every $x \in (0,1]$ (with respect to Lebesgue measure) is normal in base $b$ for every $b$.
\end{problem}



\end{document}

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