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% Submitted:      Dr. Eric Suess,Statistics 6401, Winter 2012
% Submitted Date: January 1st, 2002
% Accepted Date:  January 9th, 2002
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\begin{document}

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\markright{Statistics 6401, Winter 2012}

\large
\begin{center}
\textbf{Model of Convergence}
\end{center}

\normalsize
Let $Z_1, Z_2,...,Z_n$ be a sequence of jointly distributed random variables
defined on the same sample space $\Omega$.  Let $Z$ be another random variable
defined on this sample space $\Omega$.  We now consider what is meant by
$Z_n$ ``tending to'' $Z$.

{\bf Def:  (Convergence in probability or weak convergence)}

$Z_n$ converges to $Z$ in probability, $Z_n \stackrel{p}{\longrightarrow} Z$,
if for every $\epsilon > 0$

\begin{equation}
\lim_{n\rightarrow\infty} P\left(|Z_n - Z| > \epsilon\right) = 0
\end{equation}

{\bf Def:  (Strong Convergence or almost sure convergence)}

$Z_n$ converges to $Z$ almost surely, $Z_n \stackrel{a.s.}{\longrightarrow} Z$, if

\begin{equation}
P\left(\lim_{n\rightarrow\infty} |Z_n - Z|  = 0\right)=1
\end{equation}

{\bf Def:  (Convergence in Mean Square)}

$Z_n$ converges to $Z$ in mean square, $Z_n \stackrel{m.s.}{\longrightarrow} Z$, if

\begin{equation}
\lim_{n\rightarrow\infty} E[(Z_n - Z)^2] = 0
\end{equation}

{\bf Thm: (Weak Law of Large Numbers)}

Let $X_1,X_2,...,X_n$ be a sequence of i.i.d. random variables, each having $E[X_i] = \mu$
and $Var(X_i) = \sigma^2$.  Let $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$.  Then, for
any $\epsilon > 0$


\begin{equation}
 P\left(|\bar{X}_n - \mu| > \epsilon\right) \rightarrow 0
\end{equation}

as ${n\rightarrow\infty}$.

{\bf Proof:}

Use Chebychev's Inequality.

So $\bar{X}_n \stackrel{p}{\longrightarrow} \mu$.

{\bf Thm: (Strong Law of Large Numbers)}

Let $X_1,X_2,...,X_n$ be a sequence of i.i.d. random variables, each having $E[X_i] = \mu$
and $Var(X_i) = \sigma^2$.  Then, for any $\epsilon > 0$
\begin{equation}
 P\left( \lim_{n\rightarrow\infty} \bar{X}_n = \mu \right) = 1.
\end{equation}

So $\bar{X}_n \stackrel{a.s.}{\longrightarrow} \mu$.

\newpage

{\bf Def:  (Convergence in Distributon)}

$Z_n$ converges to $Z$ in distribution, $Z_n \stackrel{d}{\longrightarrow} Z$, if,

\begin{equation}
\lim_{n\rightarrow\infty} F_{Z_n}(t) = F_Z(t)
\end{equation}

for all $t$ where $F_Z(t)$ is continuous.

{\bf Thm: (Continuity Theorem)}

Let $Z_1, Z_2,...,Z_n$ be a sequence of random variables whose moment
generating funtion, $M_{Z_n}(t)$, existis for all $|t| < \epsilon$, for
every $n$.  Then
\begin{equation}
 F_{Z_n}(z) \  {\rightarrow} \  F_Z(z),
\end{equation}

$Z_n \stackrel{d}{\longrightarrow} Z$, if and only if,
\begin{equation}
\lim_{n\rightarrow\infty} M_{Z_n}(t) \stackrel{d}{\longrightarrow} M_Z(t)
\end{equation}

where $M_Z(t)$ is the m.g.f. of $F_Z(z)$.

{\bf Thm: (Central Limit Theorem)}

Let $X_1,X_2,...,X_n$ be a sequence of i.i.d. random variables, each
having mean $\mu$ and variance $\sigma^2$.  The the distribution of
\begin{equation}
 \frac{S_n - n\mu}{\sigma \sqrt{n}},
\end{equation}

where $S_n = \sum_{i=1}^n X_i$, tends to the standard normal distribution

\begin{equation}
 \lim_{n\rightarrow\infty} P\left( \frac{S_n - n\mu}{\sigma \sqrt{n}} \le x \right) = \Phi(x)
\end{equation}

where $-\infty < x < \infty$ and $\Phi(x)$ is the c.d.f. of the standard normal distribution.

{\bf Thm:} Let  $F_1,F_2,...$ be a sequence of c.d.f.'s and the corresponding
p.d.f.'s be $f_1,f_2,...$ .

If
\begin{equation}
 \lim_{n\rightarrow\infty} f_n(x) = f(x)
\end{equation}

where $f(x)$ is a p.d.f., then $F_n$ converges (in distribution) to the
c.d.f. $F$.

{\bf Note: (Stirling Formula)}
\begin{equation}
 \frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2}) \sqrt{\frac{n}{2}}} \  {\rightarrow} \ 1
\end{equation}

which says $\Gamma(n+1) = n! \approx \sqrt{2\pi} \  n^{n+1/2} \  e^{-n} = \sqrt{2\pi} \ n \left(\frac{n}{e}\right)^n$.

\end{document}
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