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Recall for a collection $\chi$ of random
variables, $\chi$ is \gls{ui} if
\begin{enumerate}[(1)]
  \item $\chi$ is $\Lp[1]$ bounded
  \item $I_{\chi}(\delta) = \sup_{X \in
    \chi} \{\EE(|X|\indicator{A}) : A \in
    \mathcal{F}, \PP(A) \le \delta\} \downarrow 0$ as $\delta
    \downarrow 0$.
\end{enumerate}

\begin{flashcard}[alt-def-of-ui-lemma]
\begin{lemma*}[Alternative definition of \gls{ui}]
  \cloze{
  $\chi$ is \gls{ui} if and only if
  \[ \sup_{X \in \chi} \EE (|X|
  \indicator{(|X| \ge K)}) \to 0 \]
  as $K \to \infty$.
  }
\end{lemma*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] \cloze{Fix any $\eps > 0$. Then
      $\exists \delta > 0$ such that
      $I_{\chi}(\delta) < \eps$ (as
      $\chi$ is \gls{ui}). Choose $K <
      \infty$ such that
      \[ \frac{I_\chi(1)}{\delta} = \frac{\sup_{X
      \in \chi} \EE|X|}{\delta} \le K .\]
      Then for any $X \in \chi$,
      \[ \PP(|X| \ge K) \stackrel{\text{Markov}}{\le} \frac{\EE|X|}{K} \le
      \frac{I_\chi(1)}{K} \le \delta .\]
      So, with $A = \{|X| \ge K\}$,
      \[ \EE(|X|\indicator{|X| \ge K}) \le
      I_\chi(\delta) < \eps .\]}
    \item[$\Leftarrow$] \cloze{\[ \EE|X|
      = \EE|X|\indicator{(|X| \ge K)} +
      \EE\ub{|X|\indicator{(|X| \le K)}}_{\le K} \]
      So,
      \[ \sup_{X \in \chi} \EE|X| \le K + \sup_{X
      \in \chi} \EE|X|\indicator{(|X| \le K)} \]
      Choose $K$ large so that the second term on
      the $\RHS$ is $\le 1$. Then $\chi$ is
      $\Lp[1]$ bounded. Fix any $\eps > 0$. Choose
      $K$ so that $\EE(|X|\indicator{|X| \ge K}) <
      \eps/2$ for all $X \in \chi$.
      Then choose $\delta > 0$ such that $\delta
      \le \frac{\eps}{2K}$.
      Then $\forall X \in \mathcal{X}$ and $A \in
      \mathcal{F}$ with $\PP(A) \le \delta$,
      \begin{align*}
        \EE(|X| \indicator{A})
        &= \EE(|X|\indicator{A} \indicator{|X| \ge K})
        + \EE(|X|\indicator{A} \indicator{|X| \le
        K}) \\
        &\le \ub{\EE(|X| \indicator{|X| \ge
        K})}_{< \eps} + \ub{K\PP(A)}_{\le K\delta
        \le \eps/2} \\
        &\le \eps \qedhere
      \end{align*}
      }
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{theorem*}
  Let $(X_n)$, $X$ be random variables on
  $(\Omega, \mathcal{F}, \PP)$. Then the following
  are equivalent:
  \begin{enumerate}[(a)]
    \item $X, X_n \in \Lp[1]$ for all $n$ and $X_n
      \stackrel{\Lp[1]}{\longrightarrow} X$.
    \item $\chi = (X_n : n \in \NN)$ is \gls{ui}
      and $X_n \convP X$.
  \end{enumerate}
\end{theorem*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(a) $\implies$ (b)]
    \item[(a) $\implies$ (b)] By Markov,
      \[ \PP(|X_n - X| > \eps) \le \frac{\EE|X_n -
      X|)}{\eps} \to 0 \]
      as $n \to \infty$, so $X_n \convP X$. Choose
      $N$ such that $\EE|X_n - X| <
      \frac{\eps}{2}$ for all $n \ge N$. Choose
      $\delta$ so that $\EE|X|\indicator{A} \le
      \frac{\eps}{2}$ whenever $\PP(A) < \delta$.
      \[ \EE|X_n| \indicator{A}
      \le \ub{\EE|X_n - X|}_{\le \delta/2} +
      \ub{\EE|X|\indicator{A}}_{\le \eps/2} \le
      \eps \qquad \forall n \ge N .\]
      For all $n = 1, 2, \ldots, N - 1$,
      \[ \EE|X_n| \indicator{A} \le \eps \]
      by the choice of $\delta$. Hence $\chi$ is
      \gls{ui}.
    \item[(b) $\implies$ (a)] $X_n \convP X$. So
      take a subsequence $(n_k)$ such that
      $X_{n_k}
      \stackrel{\text{\gls{al_surely}}}{\longrightarrow}
      X$. Then
      \[ \EE|X| = \EE\liminf |X_{n_k}|
      \stackrel{\text{\nameref{fatous_lemma}}}{\le}
      \liminf \EE|X_{n_K}|
      \le \sup_{X_n \in \chi} \EE|X_n| < \infty .\]
      (as $\chi$ is \gls{ui}, hence $\Lp[1]$
      bounded). So $X \in \Lp[1]$. Define the
      truncated random variables
      \begin{align*}
        X_n^K
        &= (-K) \vee X_n \wedge K \\
        X^k (-k) \vee X \wedge K
      \end{align*}
      Then $X_n^K \convP X^K$ (as $\PP(|X_n^K -
      X^K| > \eps) \le \PP(|X_n - X| > \eps)$).

      Aside: If $X_n \convP X$ and $f$ is a
      continuous function, then $f(x_n) \convP
      f(x)$. Also $|X_n^K| \le K$ for all $n$.
      Hence by BCT, $X_n^K
      \stackrel{\Lp[1]}{\longrightarrow} X^k$.
      Now,
      \[ \EE|X_n - X| \le \EE\ob{|X_n -
      X_n^K|}^{\le |X_n|\indicator{(|X_n| \ge K)}
      ~~(1)}
      + \EE \ob{|X - X^K|}^{\le
      \EE|X|\indicator{(|X| \le K)} ~~(2)} + \EE|X_n^K -
      X^K| \le \eps \qquad \]
      for all $n \ge N$.
      By \gls{ui} choose $K$ large so that (1) and
      (2) are $\le \frac{\eps}{3}$. Then choose
      $N$ large so that the last term is $\le
      \frac{\eps}{3}$ for all $n \ge N$. \qedhere
  \end{enumerate}
\end{proof}

\newpage
\section{Fourier Transforms}

For $g$ measurable
such that $\int |G| \dd x < \infty$, define
\[ \int g(x) \mu (\dd x)
= \int \Re(g(x)) \mu (\dd x) + i\int \Im(g(x))
\mu(\dd x) \]

Here $\Lp = \Lp(\RR^d)$ is the space of
\emph{complex valued} Borel measurable functions
on $\RR^d$, i.e. $f : \RR^d \to \CC$ for which
\[ \ub{\left( \int_{\RR^d} |f(x)|^p \mu(\dd x)
\right)^{1/p}}_{= \normp\|f\|_p} < \infty \]
for all $1 \le p < \infty$.
and
\[ \left| \int g(x) \mu (\dd x) \right| \le \int
|g(x)| \mu(\dd x) \]
(\es{3}).
We also define for $f, g \in \Lp[2]$,
\[ \langle f, g \rangle = \int f(x) \ol{g(x)} \dd
\mu(x) \]
which is an inner product on $\Lp[2](\mu)$. For
any $y \in \RR^d$,
\begin{align*}
  \int f(x - y) \dd x
  &= \int f(y - x) \dd x - \int f(x) \dd x \\
  &= \int f(-x) \dd x
\end{align*}
(translation invariance and $x \mapsto -x$
symmetry of $\lambda$, see \es{3}). Also for $a
\in \RR$, $a \neq 0$,
\[ \int f(ax) \dd x = \frac{1}{a^d} \int f(x) \dd
x\]

\begin{flashcard}[fourier-transform-defn]
\begin{definition*}[Fourier Transform]
  \glsnoundefn{FT}{Fourier transform}{Fourier
  transforms}
  \glssymboldefn{finv}{FT of $f$}{FT of $f$}
  \cloze{The \emph{Fourier transform} $\hat{f}$ of $f \in
  \Lp[1](\RR^d)$ is defined as
  \[ \hat{f}(u) = \int_{\RR^d} f(x) e^{i\langle u,
  x\rangle} \dd x \]
  for all $u \in \RR^d$ and $\langle u, x \rangle
  = \sum_{i = 1}^d u_i x_i$.}
\end{definition*}
\end{flashcard}
\vspace{-1em}

For all $u \in \RR^d$,
\[ \sup_u |\hat{f}(u)| \le \int |f(x)| \dd x =
\normp\|f\|_1 < \infty \]
i.e. $\hat{f} \in \Lp[\infty]$. Also, for $u_n \to
u$,
\[ f(x) e^{i\langle u_n, x \rangle} \to f(x)
e^{i\langle u, x \rangle} \]
and $|f(x) e^{i\langle u_n, x \rangle}| \le
|f(x)|$ and $f \in \Lp[1]$, so by \nameref{dct},
$\hat{f}(u_n) \to \hat{f}(u)$. Moreover,
\[ \lim_{\|u\| \to \infty} \hat{f}(u) = 0 \]
(Riemann-Lebesgue Lemma, \es{3}). Thus
\[ \hat{f} \in C_0(\RR^d) = \{\text{$f$ bounded
continuous vanishing at $\infty$}\} \]
The map is $1-1$ (but not onto).

For a finite / probability measure $\mu$ on
$\RR^d$, define similarly,
\[ \hat{\mu}(u) = \int e^{i\langle u x\rangle} \dd
\mu(x) \qquad u \in \RR^d \]
Then $\hat{\mu}$ is a bounded continuous function
on $\RR^d$ and $|\hat{\mu}(u)| \le \mu(\RR^d) <
\infty$. If $\mu$ has density $f$ (with respect to
$\lambda$), then
\[ \hat{\mu}(u) = \int e^{i\langle u, x\rangle}
f(x) \dd x = \hat{f}(u) .\]

\begin{flashcard}[char-func-defn]
\begin{definition*}[Characteristic function]
  \glsnoundefn{char_func}{characteristic
  function}{characteristic functions}
  \glssymboldefn{cf}{$\phi$}{$\phi$}
  \cloze{The \emph{characteristic function} (c.f.)
  $\phi_X$ of a random variable $X$ on $\RR^d$ is
  the \gls{FT} of its law $\mu_X = \PP \circ
  X^{-1}$. So
  \[ \phi_X(u) = \hat{\mu}_X(u) = \int e^{i\langle
  u, x\rangle} \dd \mu_X(x) = \int e^{i\langle u,
  x \rangle} \dd\PP = \EE e^{i\langle u, X\rangle} \]
  ($\nu \circ f^{-1}(g) = \nu(f \circ g)$). In
  particular if $X$ has pdf $f$, then $\phi_X(u) =
  \hat{f}(u)$.}
\end{definition*}
\end{flashcard}