Friday, 29 August 2014

Monotonic Sequence

Analysis with Programming has recently been accepted as a contributing blog on Mathblogging.org, a blogosphere aiming to be the best place to discover mathematical writing on the web. And as a first post, being a member of the said site, I will do proving on the theory of probability. This problem by the way, is part of my first homework on my masteral. This is my solution and if you find errors, do let me know.

Problem

  1. If $\{A_k\}$ is either expanding or contracting, we say that it is monotone, and for monotone sequence $\{A_k\}$, $\displaystyle\lim_{n\to \infty} A_n$ is defined as follows: \begin{equation}\nonumber \lim_{n\to \infty} A_n = \begin{cases} \displaystyle\bigcup_{k=1}^\infty A_k&\text{if}\;\{A_k\}\;\text{is expanding}\\[0.3cm] \displaystyle\bigcap_{k=1}^\infty A_k&\text{if}\;\{A_k\}\;\text{is contracting} \end{cases}. \end{equation} Prove the above equation.

Solution

  1. Proof. If $\{A_k\}$ is either expanding or contracting, then for an infinite sequence $A_1,A_2,\cdots$ one can define two events from $\displaystyle\lim_{n\to \infty}A_n$, i.e. \begin{equation} \label{eq:limAn} \lim_{n\to \infty} A_n = \begin{cases} \displaystyle\lim_{n\to\infty}\sup_{k\in [n,\infty)}\{A_k\}\\[0.3cm] \displaystyle\lim_{n\to\infty}\inf_{k\in [n, \infty)}\{A_k\} \end{cases}. \end{equation} Now the $\displaystyle\sup_{k\in [n,\infty)} \{A_k\}$ and $\displaystyle\inf_{k\in [n,\infty)} \{A_k\}$ are defined as follows: \begin{equation}\nonumber \displaystyle\sup_{k\in [n,\infty)} \{A_k\} = \displaystyle\bigcup_{k=n}^{\infty} A_k,\quad\mathrm{and}\quad \displaystyle\inf_{k\in [n,\infty)} \{A_k\} = \displaystyle\bigcap_{k=n}^{\infty} A_k. \end{equation} Hence, \begin{equation}\nonumber \displaystyle\lim_{n\to \infty}\sup_{k\in [n,\infty)} \{A_k\} = \displaystyle\lim_{n\to \infty}\bigcup_{k=n}^{\infty} A_k,\quad\mathrm{and}\quad \displaystyle\lim_{n\to \infty}\inf_{k\in [n,\infty)} \{A_k\} = \displaystyle\lim_{n\to \infty}\bigcap_{k=n}^{\infty} A_k. \end{equation} Since $\displaystyle\bigcup_{k=n}^{\infty} A_k$ is an event that "at least one $A_k$ occurs", then $\displaystyle\bigcup_{k=n}^{\infty}A_k$ occurs for all $n$. This statement simply defines the intersection, and thus we have \begin{equation} \label{eq:sup} \displaystyle\lim_{n\to \infty}\sup_{k\in [n,\infty)} \{A_k\} = \displaystyle\lim_{n\to \infty}\bigcup_{k=n}^{\infty} A_k = \displaystyle\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty} A_k. \end{equation} Now for $\displaystyle\bigcap_{k=n}^{\infty} A_k$, it can be observed that, for every $n$, $\{A_k\}$ must occur for all $k$, $k \in[n,\infty)$. This statement is sometimes not satisfied on some $n$, wherein an empty set is obtained if no common values between the sequence $\{A_k\}$ are observed. Therefore the event, $\displaystyle\bigcap_{k=n}^{\infty} A_k$ occurs for some $n$, that is, \begin{equation} \label{eq:inf} \displaystyle\lim_{n\to \infty}\inf_{k\in [n,\infty)} \{A_k\} = \displaystyle\lim_{n\to \infty}\bigcap_{k=n}^{\infty} A_k = \displaystyle\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty} A_k. \end{equation} Finally, Equations (\ref{eq:sup}) and (\ref{eq:inf}) can now be equated to Equation (\ref{eq:limAn}), \begin{equation} \lim_{n\to \infty} A_n = \begin{cases} \displaystyle\lim_{n\to\infty}\sup_{k\in [n,\infty)}\{A_k\} = \displaystyle\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty} A_k \\[0.3cm] \displaystyle\lim_{n\to\infty}\inf_{k\in [n, \infty)}\{A_k\} = \displaystyle\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty} A_k \end{cases}.\nonumber \end{equation} Now if the sequence $\{A_k\}$ is expanding, then the inner union $\left(\displaystyle\bigcup_{k=n}^{\infty} A_k\right)$ in $\displaystyle\lim_{n\to\infty}\sup_{k\in [n, \infty)}\{A_k\}$ remains the same independently of $n$, so that $\displaystyle\lim_{n\to\infty}\sup_{k\in [n, \infty)}\{A_k\} = \displaystyle\bigcup_{k=1}^{\infty} A_k$. On the other hand, the inner intersection $\left(\displaystyle\bigcap_{k=n}^{\infty} A_k\right)$ in $\displaystyle\lim_{n\to\infty}\inf_{k\in [n, \infty)}\{A_k\}$, is equal to $A_n$, implying that the limit is $\displaystyle\bigcup_{n=1}^{\infty} A_n$. Since both $\displaystyle\lim_{n\to\infty}\sup_{k\in [n, \infty)}\{A_k\}$ and $\displaystyle\lim_{n\to\infty}\inf_{k\in [n, \infty)}\{A_k\}$ converge to the same event, then that proves the first part of the problem.

    If the sequence $\{A_k\}$ is contracting. Then the inner union $\left(\displaystyle\bigcup_{k=n}^{\infty} A_k\right)$ in $\displaystyle\lim_{n\to\infty}\sup_{k\in [n, \infty)}\{A_k\}$ is equal to $A_n$, so that $\displaystyle\lim_{n\to\infty}\sup_{k\in [n, \infty)}\{A_k\} = \displaystyle\bigcap_{n=1}^{\infty} A_n$. On the other hand, the inner intersection $\left(\displaystyle\bigcap_{k=n}^{\infty} A_k\right)$ in $\displaystyle\lim_{n\to\infty}\inf_{k\in [n, \infty)}\{A_k\}$, remains the same independently of $n$, and thus the limit is $\displaystyle\bigcap_{k=1}^{\infty} A_k$. Since both $\displaystyle\lim_{n\to\infty}\sup_{k\in [n, \infty)}\{A_k\}$ and $\displaystyle\lim_{n\to\infty}\inf_{k\in [n, \infty)}\{A_k\}$ converge to the same event, then that proves the second part of the problem.$\hspace{12cm}\blacksquare$


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