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Given any distribution $F$, there exists a
function $X$ such that $F = F_X$, i.e. $F(x) =
F_X(x) = \PP(X \le x) ~\forall x$ ($\PP(\omega \in
\Omega : X(\omega) \le x)$.

\begin{proof}
  Let $\Omega = (0, 1)$ and $\PP$ the Lebesgue
  measure $\lambda|_{(0, 1)}$. Let $F$ be any
  distribution function. Then $F$ is $\uparrow$,
  right continuous, so we can define
  \[ X(\omega) = \inf\{x : \omega \le F(x)\} :
  (0,1) \to \RR \]
  Since $X$ is a measurable function, $X$ is a
  random variable.
  \[ \forall x, F_X(x) = \PP( \le x) = \PP(\omega
  \in \Omega : X(\omega) \le x) = \PP(\omega \in
  \Omega : \omega \le F(x)) = \PP((0, F(x)]) =
  F(x) \qedhere \]
\end{proof}

\begin{flashcard}[independent-rvs-defn]
\begin{definition*}[Independent variables]
  \glsadjdefn{indep_rv}{independent}{random
  variable}
  \cloze{
  A (countable) family of random variables ($X_i$,
  $i \in I$) is said to be \emph{independent}, if
  the family of \sigalgpl\ $(\sigma(X_i), i \in I)$
  is independent (where recall $\sigma(X) =
  \sigma\{X^{-1}(A) : A \in \mathcal{E}\}$
  for $X : \Omega \to (E, \mathcal{E})$).
  }
\end{definition*}
\end{flashcard}

\begin{proposition*}
  For a sequence of random variables $(X_n, n \in
  \NN)$, this sequence is \emph{independent} if
  \[ \PP(X_1 \le x_1, \ldots, X_n \le x_n) =
  \PP(X_1 \le x_1) \cdots \PP(X_n \le x_n) \qquad
  \forall x_1, \ldots, x_n \in \RR, n \in \NN .\]
\end{proposition*}

\begin{proof}
  \es{1}
\end{proof}

\subsection{Rademacher functions}

\textbf{Question:} Given a distribution function
$F$, we know there exists a random variable $X$
corresponding to it. But given an infinite
sequence of distribution functions $F_1, F_2,
\ldots$ does there exist \gls{indep_rv} random
variables $(X_1, X_2, \ldots)$ corresponding to
them?

Let $\probspace = ((0, 1), \mathcal{B}(0. 1),
\lambda|_{(0, 1)})$. Any $\omega \in \Omega$ has a
binary expansion:
\[ \omega = 0 \cdot \ub{\omega_1}_{\half}
\ub{\omega_2}_{\quarter} \ub{\omega_3}_{\frac{1}{8}}
\cdots, \qquad \omega_i \in \{0 ,1\} \]
If we exclude representations ending in an
infinite sequence of $0s$, then the representation
is unique.

Define $R_n : \Omega \to \{0, 1\}$ by $R_n(\omega)
= \omega_n$, i.e. $R_n = \indicator{\{\omega_n =
1\}}$ So
\[ R_1 = \indicator{(1/2, 1]}, \quad R_2 =
\indicator{\{\omega_2 = 1\}} =
\indicator{\{\omega_1 = 0, \omega_2 = 1\}} +
\indicator{\{\omega_1 = 1, \omega_2 = 1\}} =
\indicator{(1/4, 1/2)} + \indicator{(3/4, 1]},
\quad R_3 = \cdots \]
\glsnoundefn{rad_fun}{Rademacher
function}{Rademacher functions}
So $R_n$s are finite sums of indicators of
intervales, hence measurable, i.e. they are random
variables. They are called \emph{Rademacher
functions}.

\textbf{Claim:} $R_i$ are \IID\ $\Ber \left( \half \right)$.
$\PP(R_n = 1) = \half = \PP(R_n = 0) ~\forall
n$.
\[ \PP(R_1 = x_1, R_2 = x_2, \ldots, R_n = x_n)
= 2^{-n} = \PP(R_1 = x_1) \cdots \PP(R_n = x_n) \]
and hence $(R_i)_{i \in \NN}$ \gls{indep_rv}.

Now, choose a bijection $m : \NN^2 \to \NN$ and
define $Y_{k, n} = R_{m(k, n)}$ and set $Y_n =
\sum_{k = 1}^\infty 2^{-k} Y_{k, n}$ (converges on
$|Y_{k, m}| \le 1$).

\textbf{Claim:} $(Y_n)_n$ are \IID\ $U(0, 1)$
(i.e. $\mu_{Y_n} = \lambda|_{(0, 1)}$ and $(Y_n)$
are \gls{indep_rv}). \Gls{indep_rv} is easy
($Y_1$ is measurable function of $Y_{1, 1}, Y_{2,
1}, \ldots$, similarly $Y_2$ is a measurable
function of $Y_{1, 2}, Y_{2, 2}, \ldots$, but note
that these two lists are independent). Any
measurable functions of \gls{indep_rv} random
variables are independent (check!).

The law of $Y_n$ is identified on the \pisys
of intervals $\left( \frac{i}{2^m}, \frac{i +
1}{2^m} \right]$, $i = 0, 1, \ldots, 2^m - 1, m
\in \NN$. And
\begin{align*}
  \PP \left( \frac{i}{2^m} < Y_n \le \frac{i +
  1}{2^m} \right)
  &= \PP \left( \frac{i}{2^m} < \sum_{K =
  1}^\infty 2^{-k} Y_{k, n} \le \frac{i + 1}{2^m}
  \right) \\
  &= \PP \left( Y_{1, n} = y_1, \ldots, Y_{m, n} =
  y_m \right)
  &&\text{where $\frac{i}{2^m} = 0 \cdot y_1 y_2
  \cdots y_m$} \\
  &= \prod_{i = 1}^m \PP (Y_{i, n} = y_i) \\
  &= 2^{-m} \\
  &= \lambda \left( \frac{i}{2^m}, \frac{i +
  1}{2^m} \right]
\end{align*}
Hence $\mu_{Y_n}$ is $\lambda|_{(0, 1)}$, i.e.
$Y_1$ are \IID\ $U(0, 1)$. Then, as before, set
\[ G_n(x) = F_n^-(x) = \inf\{y : x \le F_n(y)\} \]
then $G_n$'s are Borel functions. Set $X_n =
G_n(Y_n)$, $n = 1, 2, \ldots$. Then as before
$F_{X_n} = F_n$ and $(X_n)$ are \gls{indep_rv} (as
$(Y_n)_N$ are \gls{indep_rv}).

\subsection{Convergence of Random Variables}

\begin{flashcard}[almost-everywhere-defn]
\begin{definition*}[almost everywhere]
  \glspropdefn{al_ev}{almost everywhere}{property}
  \glspropdefn{al_surely}{almost surely}{event}
  \cloze{
  $(E, \mathcal{E}, \mu)$ be a measure space. Let $A
  \in \mathcal{E}$ be defined by some property. We
  say the property holds almost everywhere (a.e /
  $\mu$.a.e) if $\mu(A^c) = 0$. If $\mu$ is a
  probability measure, we say almost surely (a.s)
  if $\PP(A^c) = 0$, i.e. $\PP(A) = 1$ (w.p.1).
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

Thus if $(f_n)$, $f$, $(E, \mathcal{E}, \mu) \to
(\RR, \mathcal{B})$ measurable, we say
\begin{itemize}
  \item $f_n \to f$ \gls{al_ev} if
    \[ \mu(\{x \in E : f_n(x) \not\to f(x)\}) = 0 \]
    for $\PP$, \gls{al_surely} $\PP(\{\omega \in
    \Omega : X_n(\omega) \to X(\omega)\}) = 1$.
  \item $f_n \to f$ in ($\mu$-)measure if $\forall
    \eps > 0$,
    \[ \mu(\{x \in E : |f_n(x) - f(x)| > \eps\})
    \to 0 \]
    as $n \to \infty$, and in ($\PP$)-probability
    if $\PP(|X_n - X| > \eps) \to 0$.
\end{itemize}

\begin{theorem*}
  Let $(f_n)$ be a sequence of measurable
  functions. Then if $\mu(E) < \infty$, then $f_n
  \to 0$ \gls{al_ev} $\implies f_n \to 0$ in
  $\mu$-measure. ($f_n = \indicator{(n, \infty)}$
  and the Lebesgue measure then $f_n \to 0$
  \gls{al_ev} but $\mu(|f_n| > \eps) = \infty
  ~\forall n$).
\end{theorem*}

\begin{proof}
  Fix $\eps > 0$. Suppose $f_n \to 0$ \gls{al_ev}.
  Then
  \begin{align*}
    \mu(|f_n| \le \eps) \ge \mu \left( \bigcap_{m
    = 1}^\infty \{|f_m| \le \eps\}\right) \\
    &\uparrow \mu (|f_n| \le \text{ eventually})
    \\
    &\ge \mu(f_n \stackrel{n \to
    \infty}{\longrightarrow} 0) \\
    &= \mu(E) \\
    &<\infty
  \end{align*}
  \[ A_n = \bigcap_{m = n}^\infty \{|f_m| \le
  \eps\} \uparrow \bigcup_{n = 1}^\infty
  \bigcap_{m = n}^\infty \{|f_m| \le \eps\} \]
  Hence
  \[ \lim_{n \to \infty} \mu(|f_n| \le \eps) =
  \mu(E) \]
  So, $\lim_{n \to \infty} \mu(|f_n| > \eps) = 0$.
\end{proof}