% scribe: Daisy Yan Huang
% lastupdate: Oct. 3, 2005
% lecture: 7
% title: Borel-Cantelli lemmas and almost sure convergence
% references: Durrett, section 1.6
% keywords: Borel-Cantelli lemmas, almost sure convergence, a.s. convergence
% end

\documentclass[letterpaper,12pt]{article}

\include{macros}

\newcommand{\conv}{\longrightarrow}

\begin{document}

\lecture{7}{Borel-Cantelli lemmas and 
almost sure convergence}
{Daisy Yan Huang}{yanhuang@stat.berkeley.edu}

This set of notes is a revision of the work of Jin Kim, 2002.

\section{Borel-Cantelli Lemmas}
% keywords: Borel-Cantelli lemmas
% end

Recall that for real valued random variables $X_n$ and $X$,

$$\begin{array}{ll}
\{X_n\to X\} &=\{\omega: X_n(\omega)\to X(\omega)\} \\
&=\{\forall \epsilon > 0, |X_n-X|\leq\epsilon\ \mbox{ eventually}\}
\end{array}
$$

Thus,

$$\begin{array}{ll}
\P(X_n\to X)=1
&\Leftrightarrow \forall \epsilon>0,\ \P(|X_n-X|\leq \epsilon \ ev.)=1\\
&\Leftrightarrow\forall \epsilon>0,\ \P(|X_n-X|>\epsilon \ i.o.)=0
\end{array}
$$

Let the event $A_n :=\{|X_n-X|>\epsilon\}$.  Then, we are motivated by
consideration of a.s. convergence to find useful conditions for
$\P(A_n \mbox{ i.o.})=0$.

Recall that $\{A_n \mbox{ i.o.}\}=\bigcap_n\ \bigcup_{m\geq n} A_m$.

\begin{theorem}[Borel-Cantelli Lemmas]
Let $(\Omega,F, \P)$ be a probability space and let $(A_n)$ be a sequence of events in F. Then,
\begin{enumerate}
\item If $\sum_n \P(A_n) < \infty$, then $\P(A_n \mbox{ i.o.}) = 0$. \\
\item If $\sum_n \P(A_n) = \infty$ and $A_n$ are independent, then $\P(A_n \mbox{ i.o.}) = 1$.
\end{enumerate}
\end{theorem}

There are many possible substitutes for independence in BCL II, see
Kochen-Stone Lemma.

Before proving BCL, notice that
\begin{itemize}
\item $ \1(A_n \mbox{ i.o.}) = \limsup_{n \to \infty} \1( A_n ) $
\item $ \1(A_n \mbox{ ev.}) = \liminf_{n \to \infty}  \1( A_n ) $
\item $  \{A_n \mbox{ i.o.}\} = \lim_{m \to \infty} ( \cup_{n > m} A_n )
         \hspace{.53cm} (\mbox{note: as } m \uparrow, \;  \cup_{n \ge m} A_n \downarrow \; )$
\item $ \{A_n \mbox{ ev.}\} = \lim_{m \to \infty} ( \cap_{n > m} A_n )
         \hspace{.53cm} (\mbox{note: as } m \uparrow, \;  \cap_{n \ge m} A_n \uparrow \; )$.
\end{itemize}
Therefore,
\begin{align*}
  \P(A_n \mbox{ ev.}) & \le  \liminf_{n \to \infty} \P(A_n)  \hspace{.62cm} \mbox{ by Fatou's lemma }\\
   & \le  \limsup_{n \to \infty} \P(A_n)  \hspace{.62cm} \mbox{ obvious from definition }\\
   & \le  \P(A_n \mbox{ i.o.})  \hspace{.62cm} \mbox{ dual of Fatou's lemma (i.e. apply to $-\P$)}
\end{align*}

\begin{proof} (Of BCL I)
\begin{align*}
  \P(A_n \mbox{ i.o.}) & =  \lim_{m \to \infty} \P(\cup_{n \ge m} A_n)  \\
   & \le  \lim_{m \to \infty} \sum_{n \ge m}^\infty \P(A_n)  \; = \; 0
        \hspace{.32cm} \mbox{ since } \sum_{i=1}^\infty \P(A_n) < \infty .
\end{align*}
\end{proof}

%{\bf Pf of BCL I} (Alternative method)\\
%Consider a random variable $N := \sum_\1(A_n)$, i.e. the number of events that occur. Then
%$\E[N] = \sum_{n=1}^\infty \P(A_n)$ by the Monotone Convergence Theorem, and
%\begin{align*}
  %\sum_{n=1}^\infty \P(A_n) < \infty  & \Longrightarrow    \E[N] < \infty\\
  %      & \Longrightarrow       \P(N < \infty)=1 \\
  %      & \Longrightarrow       \P(N = \infty)=0 \\
  %     & \Longrightarrow       \P(A_n \mbox{ i.o.})=0  \hspace{.63cm} \mbox{ because }(N=\infty) \equiv ( A_n \mbox{ i.o.})  . \; \qedsymbol
%\end{align*}
\bigskip

\begin{proof} (Of BCL II)
Assume that $\Sigma \P(A_n)=\infty$ and the
$A_n$'s are independent.
We will show that $\P(A_n^c \mbox{ ev.}) =0$. \\
\begin{align}
  \P(A_n^c \mbox{ ev.}) &=  \lim_{n \to \infty} \P(\cap_{m \ge n} A_m^c )
 \; = \; \lim_{n \to \infty} \prod_{m \ge n} \P( A_m^c )  \label{eq:BCL2a}\\
&= \lim_{n \to \infty} \prod_{m \ge n} \left( 1-\P( A_m ) \right)
\; \le \; \lim_{n \to \infty} \prod_{m \ge n} \exp{(-\P( A_m^c ))}  \label{eq:BCL2b} \\
&= \lim_{n \to \infty} \exp{\left(-\sum_{m \ge n} \P( A_m^c )\right)} = 0 \nonumber
\end{align}
since $(-\sum_{m \ge n} \P( A_m^c ))\rightarrow \infty, \mbox{ as
} n\rightarrow \infty$

For~\eqref{eq:BCL2a}, we used the following fact (due to the
independence of $A_n$):
\begin{align*}
                \P(\cap_{m \ge n} A_m^c ) = \lim_{N \to \infty} \P( \cap_{n \le m \le N} A_m^c) =
         \lim_{N \to \infty} \prod_{n \le m \le N} \P( A_m^c) =  \prod_{n \le m} \P(A_m^c).
\end{align*}
For ~\eqref{eq:BCL2b}, $1-x \le \exp(-x)$ was used.
\end{proof}

\bigskip

For an example in which the theorem cannot be applied, consider $A_n
= (0,1/n)$ in $(0,1)$. Then, $\P(A_n) = 1/n$, $\sum \P(A_n) =
\infty$, but $\P(A_n \mbox{ i.o.}) = \P( \emptyset ) =0$.

\bigskip

\begin{example} 
Consider random walk  in $\Z^d$, $d=0, 1, \cdots$
$S_n=X_1+\cdots+X_n,, \; n=0, 1, \cdots$ where $X_i$ are independent in $\Z^d$. In the simplest case, each $X_i$ has uniform distribution on $2^d$ possible strings. i.e., if $d=3$,  we have $2^3=8$ neighbors
\begin{align*}
   \left\{
   \begin{array}{c}
        (+1, +1, +1)\\ \vdots \\ (-1, -1, -1)
   \end{array}
   \right\} \; .
\end{align*}
Note that each coordinate of $S_n$ does a simple coin-tossing walk independently. We can prove that
\begin{align}
  \P(S_n=0 \mbox{ i.o.}) = \left\{
  \begin{array}{cll}
        1 & \mbox{if } d=1 \mbox{ or } 2 & \mbox{(recurrent)}  \\
        0 & \mbox{if } d \ge 3 & \mbox{(transient)} \; . \label{eq:randwalk:tran}
  \end{array}   \right.
\end{align}
\end{example}

\begin{proofsketch} (of \eqref{eq:randwalk:tran})\\
Let us start with $d=1$, then
\begin{align}
\P(S_{2n}=0)
&= \P(\mbox{$n$ ``$+$'' signs and $n$ ``$-$'' signs})\\
&= \left( {2n \atop n} \right) 2^{-2n}\\
&\sim \frac{c}{\sqrt{n}} \mbox{ as } n \conv \infty .
\end{align}

where we used the facts that $ n ! \sim  \left( {n \atop e}
\right)^n \sqrt{2 \pi n}$, and that $a_n\thicksim b_n$ iff
${{a_n}\over {b_n}} \rightarrow 1 \mbox{ as }n\rightarrow \infty$.

Note
\begin{align}
  \sum \left( \frac{1}{\sqrt{n}} \right) ^d \left\{
  \begin{array}{rl}
        = \infty & \hspace{.61cm} d=1, 2  \\
        < \infty & \hspace{.61cm} d= 3, 4, \cdots \label{eq:randwalk:tran2}
  \end{array}   \right.
\end{align}
Thus, $\sum_n{\P(S_{2n}=0)=\infty}$, and BC II and
\eqref{eq:randwalk:tran2} together gives \eqref{eq:randwalk:tran}.
\end{proofsketch}


\begin{example}[for the case $d=1$]

$\{S_2=0\}$ is the event of ending up back to the origin at step 2 when we 
started at the origin.  $\P(S_{2}=0)=1/2$.  Note:

$$\P(S_{10,000}=0)\thicksim {c\over \sqrt{n}}\thickapprox 1/100,$$

$$\P(S_{10,002}=0)\thickapprox 1/100,$$

$$\P(S_{10,000}=0, S_{10,002}=0)
=\P(S_{10,000}=0)\P(S_{10,002}=0| S_{10,000}=0)
\thickapprox 1/100\cdot 1/2,$$

Later in the course, we will show that for the case $d=1$, even when the $(S_{2n}=0)$
are dependent, it is still true that $\P(S_{2n}=0 \mbox{ i.o. })=1$.
\end{example}

The same result holds for the case $d=2$.

In general,
%
$$\P(S_{2n}=0)={{ { {2n} \choose n  }\over {2^ {2n}} }^d} \approx {{c^d} \over {n^{d/2}}}.$$
%
For $d=2$, this is $\sim {{c^2} \over {n}}$ which is not summable.  Thus, $\P(S_{2n}=0 \mbox{ i.o.})=1$.
For $d\geq3$, this is $\sim {{c^3} \over {n^{3/2}}}$ which is summable.  Then, by {\bf BCL I},
$\P(S_{2n}=0 \mbox{ i.o.})=0$.



\section{Almost sure convergence}
% keywords: almost sure convergence, a.s. convergence
% end

Because
\[X_n \conv X \mbox{ a.s.}
        \hspace{.81cm} \iff \hspace{.81cm}
  X_n-X \conv 0 \mbox{ a.s.} \; , \]
it is enough to prove for the case of convergence to $0$.

\begin{proposition} 
The following are equivalent:
\begin{enumerate}
  \item $ X_n \ascv 0$
  \item $ \forall \epsilon > 0,\ \P( | X_n | >\epsilon \mbox{ i.o.} ) =0 $
  \item $ M_n \pcv 0$ where $M_n := \sup_{n \le k} | X_k |$
  \item $ \forall\  \epsilon_n  \downarrow 0 \; : \;  \P( | X_n | > \epsilon_n \mbox{ i.o.} ) =0 $
\end{enumerate}
\end{proposition}
Note: ``$\forall$'' in Proposition 4 cannot be replaced by ``$\exists$''.  For example,
Let $X_n=(1/\sqrt{n})U_n$, where $U_1, U_2,...$ are independent $U[0,1]$.

Take $\epsilon_n= {1/2}/{\sqrt{n}}$.  Then,
$\P(X_n>\epsilon_n)=\P(U_n>1/2)=1/2$.
So, $\P(X_n>\epsilon_n \mbox{ i.o.})=1$.

But if we take $\epsilon_n= {{1} \over {\sqrt{n}}}$.  Then,
$\P(X_n>\epsilon_n)=\P(U_n>1)=0$.

\bigskip

\begin{proof} (only for the equivalence of 1 and 3)

Suppose Proposition 1 holds.  If $X_n(\omega)\rightarrow 0$ a.s., then
$\sup_{n \le k} | X_k(\omega) |\rightarrow 0$ a.s.  But this implies that $M_n\rightarrow 0$ a.s.  Thus,
$M_n\pcv 0$.

Conversely, if $M_n\ \downarrow \mbox{ as } n\uparrow$, then we know in advance
that $M_n$ has a almost-surely-limit in $[0, \infty]$.
\end{proof}



\begin{lemma}
If $X_n \pcv X$, then there exists a subsequence $n_k$ such that $X_{n_k} \rightarrow X$ a.s.
\end{lemma}

\begin{proof}
It is enough to show that there exists $\epsilon_k \downarrow 0$
such that $ \sum_k{\P(|X_{n_k} - X|> \epsilon_k)}<\infty$.  We can take $\epsilon_k = 1/k \mbox{ and choose } n_k$
 so that $\P(|X_{n_k} - X|>1/k)\leq 1/{2^k}$.  Then, $ \sum_k{\P(|X_{n_k} - X|> \epsilon_k)}<\infty$,
 and by {\bf BCL I} we can conclude that $X_{n_k} \rightarrow X$ a.s.
\end{proof}






\end{document}
