|
Probability with Martingales by David Williams\[ \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\Z}{\mathbb Z} \newcommand{\N}{\mathbb N} \newcommand{\abs}[1]{\lvert #1 \rvert} \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\abs}[1]{\lvert #1 \rvert} \newcommand{\Norm}[1]{\left \lVert #1 \right \rVert} \newcommand{\Abs}[1]{\left \lvert #1 \right \rvert} \newcommand{\pind}{\bot \!\!\! \bot} \newcommand{\probto}{\buildrel P\over \to} \newcommand{\vect}[1]{\boldsymbol #1} \DeclareMathOperator{\EE}{\mathbb E} \DeclareMathOperator{\PP}{\mathbb P} \DeclareMathOperator{\E}{E} \DeclareMathOperator{\dnorm}{\mathcal N} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Leb}{Leb} \DeclareMathOperator{\Bin}{Bin} \newcommand{\wto}{\buildrel w\over \to} \] Algebra's, etc.Problem 1.1
Let \(V\subset \N\), and say \(V\) has Cesaro density \(\gamma(V)\) whenever \[ \begin{equation} \gamma(V) = \lim_{n\to \infty} \frac{ \sharp (V\cap \{1,2,\dots,n\})} n \end{equation} \] exists. Let \(\mathcal C\) be the set of all sets with Cesaro density. Give an example of \(V_1,V_2\in \mathcal C\) such that \(V_1\cap V_2 \not \in \mathcal C\) Take \(V_1\) to be 1 and \(n>10\) whose base 10 representation has an equal first and last digit. Take \(V_2\) to be the natural numbers whose last digit in its base 10 representation is 1. Then \(0\leq \sharp (V_i\cap \{1,2,\dots,n\}) - \lfloor n/10 \rfloor \leq 1\) since in every subset \(\{ q\cdot 10, q\cdot 10+1,q\cdot 10+2,\dots, q\cdot 10+9\}\) each of \(V_1\) and \(V_2\) have exactly one representative. Therefore \(\gamma(V_1) = \gamma(V_2)=1\). Note \(V_1\cap V_2\) consists of 1 and all numbers of the form \(1b_2 b_3 \dots b_{n-1} 1\) where \(b_i\) can take any digit from 0 to 9. Thus between \(10^k\) and \(10^{k+1}\) there are \(10^{k-1}\) members of \(V_1\cap V_2\). Therefore if \(n=10^k\), \[ \begin{equation} \sharp( V_1\cap V_2 \cap \{1,\dots,n\}) = 1 + 1 + 10 + 100 + \dots + 10^{k-2} = \frac{10^{k-1}-1} 9 + 1 \end{equation} \] and \(\frac{\sharp( V_1\cap V_2 \cap \{1,\dots,n\})} n < \frac 1 {89}\). However if \(n=2 \cdot 10^k\) then \[ \begin{equation} \sharp( V_1\cap V_2 \cap \{1,\dots,n\}) = 1 + 1 + 10 + 100 + \dots + 10^{k-1} = \frac{10^k-1} 9 + 1 \end{equation} \] and \(\frac{\sharp( V_1\cap V_2 \cap \{1,\dots,n\})} n > \frac 1 {18}\). Therefore the limit diverges since the \(\limsup\) is greater than the \(\liminf\). ContactFor comments or corrections please contact Ryan McCorvie at ryan@martingale.group |