%! TEX root = PM.tex
% vim: tw=50
% 18/11/2023 10AM

\begin{proof}
  First assume $f \in \Lp[1]$ with $\ft{f} \in
  \Lp[1]$. Then $(x, u) \mapsto f(x)\ft{f}(u)$ is
  $\dd x \dd u$-integrable. So
  \begin{align*}
    (2\pi)^d \|f\|_2^2
    &= (2\pi)^d \int f(x)
    \ol{f}(x) \dd x \\
    &= \iint \ft{f}(u) e^{-i\langle x, u\rangle}
    \ol{f}(x) \dd u \dd x < \infty
    &&\text{(\gls{FI}, and $f \in \Lp[2]$)} \\
    &= \int \ft{F}(u) \ol{\left( \int
    \ub{f(x) e^{i\langle x, u\rangle}}_{\ft{f}(u)}
    \right)} \dd u
    &&\text{(\nameref{fubini_tonelli})} \\
    &= \int \ft{f} (u) \ol{\ft{f}(u)} \dd u
    &&\text{(\gls{FT})} \\
    &= \normp\|\ft{f}\|_2^2 \label{lec20_l22_eq} \tag{$*$}
  \end{align*}
  (Note: $f \in \Lp[1]$ and $\ft{f} \in \Lp[1]$
  implies $f \in \Lp[2] \cap \Lp[\infty]$). Now,
  let $f \in \Lp[1] \cap \Lp[2]$. For $t > 0$,
  take $f_t = f \conv \gt \stackrel{t \to
  \infty}{\longrightarrow} f$ in $\Lp[2]$ and so
  \[ \normp\|f_t\|_2 \stackrel{t \to
  0}{\longrightarrow} \normp\|f\|_2
  \label{lec20_l30_eq} \tag{$**$} .\]
  Also,
  \[ |\ft{f_t}(u)| = |\ft{f}(u) \ft{\gt}(u)| =
  |\ft{f}(u)| e^{-t \frac{|u|^2}{2}} \uparrow
  |\ft{f}(u)| \]
  as $t \to 0$.
  \[ \normp\|\ft{f_t}\|_2^2 = \int |\ft{f_t}(u)|^2
  \dd u \stackrel{t \to 0}{\longrightarrow} \int
  |\ft{f}(u)|^2 \dd u = \normp\|\ft{f}\|_2^2
  \label{lec20_l40_eq} \tag{\dag} \]
  by \nameref{mconv_thm} But, $f_t = f \conv \gt
  \in \Lp[1]$, and $\ft{f_t} \in \Lp[1]$. So by
  \eqref{lec20_l22_eq}, $(2\pi)^d
  \normp\|f_t\|_2^2 = \normp\|\ft{f_t}\|_2^2$. Let
  $t \to 0$, then $\LHS \to (2\pi)^d
  \normp\|f\|_2^2$ by \eqref{lec20_l30_eq}, and
  $\RHS \to \normp\|\ft{f}\|_2^2$ by
  \eqref{lec_20_l40_eq}. Hence $(2\pi)^d
  \normp\|f\|_2^2 = \normp\|\ft{f}\|_2^2$. Similar
  proof for $\langle f, g \rangle$.
\end{proof}

\subsubsection*{Characteristic functions, weak
convergence and the CLT}

\[ \cf_X(t) = \EE(e^{itX}) = \ft{\mu_X} = \int
e^{i\langle t, x\rangle} \dd \mu_X(x) \]
For dirac measure $\delta_0$, $\ft{\delta_0} =
\int e^{itx} \dd \delta_0(x) = 1$ not integrable
on $\RR$ so \gls{FI} does not make sense. To
circumvent this, we `test' $\mu$ on nice test
functions $f$.

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \setcounter{enumi}{-1}
    \item 2 probability measures $\mu$ and $\nu$
      on $\RR^d$ coincide if and only if
      \[ \int f \dd \mu = \int f
      \dd \nu \label{lec20_l71_eq} \tag{$*$} \]
      for all $f ;: \RR^d \to \RR$ bounded
      continuous (\es{2}). In fat, enough to have
      \eqref{lec20_l71_eq} holds for all $f \in
      \cptinf$
      \glssymboldefn{cptinf}{$C_c^\infty$}{$C_c^\infty$}
      (space of infinitely
      differentiable functions with compact
      support). ($\mu : \cptinf \to \RR, f \mapsto
      \mu(f)$ linear, continuous (Lf top), hence
      $\mu$ is ``Schwarz distribution''
      $\mathcal{A} \circ f$, $\mu \in (\cptinf)^*$).
  \end{enumerate}
\end{remark*}

\begin{flashcard}[conv-wly-defn]
\begin{definition*}[Converges weakly]
  \glsverbdefn{conv_wly}{weakly}{probability measure}
  \cloze{
  Let $(\mu_n)$, $\mu$ be Borel probability
  measures on $\RR^d$. Then \emph{$\mu_n$ converges to
  $\mu$ weakly} if
  \[ \int f \dd \mu_n \stackrel{n \to
  \infty}{\longrightarrow} \int f \dd \mu
  \label{lec20_l92_eq} \tag{$*$} \]
  for all $f : \RR^d \to \RR$ bounded and
  continuous.
  }
\end{definition*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item
      \glsadjdefn{vc_wly}{weakly}{random variables
      converging}
      For a sequence of random variables
      $(X_n)$ and $X$ another random variable,
      $X_n \to X$ weakly if $\mu_{X_n}
      \to \mu_X$ \gls{conv_wly}.
    \item A sequence $(\mu_n)$ can have at most
      one weak limit (by Remark (0)).
    \item If $X_n \to X$ \gls{vc_wly}, and $h :
      \RR^d \to \RR^k$ is continuous, then $h(X_n)
      \to h(X)$ \gls{vc_wly} (as random variables
      in $\RR^k$) (continuous mapping theorem)
      (from definition as $f \circ h$ bounded
      continuous if $f$ bounded continuous).
    \item \refsteplabel[Remark (4)]{lec20_Cc_infty_remark}
      Sufficient to check \eqref{lec20_l92_eq}
      for all $f \in \cptinf$ (``tightness''
      argument, i.e. there exists $K$ compact such
      that $\mu_n(K^c) < \eps$ for all $n$ if
      $\mu_n \to \mu$ \gls{conv_wly}, see \es{4}).
    \item When $d = 1$, this is equivalent to $X_n
      \to \convdist X$ (i.e. $F_{X_n}(x) \to F_X(x)$
      at all points where $x \mapsto F_X(x)$ is
      continuous). (\es{4}) $\RR^d$, $F(x_1,
      \ldots, x_d) = \PP(x \le (x_1, \ldots,
      x_d))$.
  \end{enumerate}
\end{remark*}

\begin{flashcard}[ft-of-rv-meas-determined-by-cf-thm]
\begin{theorem*}
  Let $X$ be a random variable on $\RR^d$. Then
  $\mu_X$ is uniquely determined by $\ft{\mu_X} =
  \cf_X$. Further, if $\cf_X \in \Lp[1]$, then
  $\mu_X$ has a bounded continuous pdf given by
  \[ f(x) \defeq \frac{1}{(2\pi)^d} \int \cf_X(u)
  e^{-i\langle x, u\rangle} \dd u .\]
\end{theorem*}

\begin{proof}
  \cloze{
  Take $Z \sim \normaldist(0, I_d)$ independent of
  $X$. Thus $\sqrt{t} Z$ has pdf $\gt$ and $X +
  \sqrt{t} Z$ has pdf $\mu_X \conv \gt \eqdef
  f_t$. Then
  \[ \ft{f_t}(u) = \ft{\mu_X}(u) \ft{\gt}(u) =
  \cf_X(u) e^{-t \frac{|u|^2}{2}} .\]
  So by \gls{FI} of \gls{gconv},
  \[ f_t(x) = \frac{1}{(2\pi)^d} \int \cf_X(u)
  e^{-t \frac{|u|^2}{2}} e^{-i\langle u, x\rangle}
  \dd u \qquad \forall x \]
  i.e. $f_t$ is uniquely determined by $\cf_X$.
  Now for any $g$ bounded continuous, $g : \RR^d
  \to \RR$, as $t \to 0$,
  \[ \int g(x) f_t(x) \dd x = \EE(g(x + \sqrt{t}
  Z)) \stackrel{\text{BCT}}{\longrightarrow}
  \EE(g(X)) = \int g(x) \mu_X(\dd x)
  \label{l20_l160_eq} \tag{$*$} \]
  i.e. $\int g(x) \dd \mu_X$ is uniquely
  determined by $\cf_X$. Hence $\mu_X$ is uniquely
  determined by $\cf_X$ (Remark (0)). If $\cf_X
  \in \Lp[1]$, then
  \[ \cf_X(u) e^{-t \frac{|u|}{2}} e^{-i\langle u,
  x\rangle} \stackrel{t \to 0}{\longrightarrow}
  \cf_X(u) e^{-i\langle u, x\rangle} \]
  By \nameref{dct}, $f_t(x) \stackrel{t \to
  0}{\longrightarrow} f_X(t)$ for all $x$. In
  particular, $f_X(x) \ge 0$ for all $x$ and
  $|f_t(x)| \le \frac{1}{(2\pi)^d}
  \normp\|\cf_X\|_1$. Then for
  any $g$ bounded continuous with compact support,
  \[ \int \ub{g(x) f_t(x)}_{\to g(x) f_X(x)} \dd x
  \stackrel{\text{\nameref{dct}}}{\longrightarrow}
  \int g(x) f_X(x) \dd x \]
  Also, $\LHS \to \int f(x) \mu_X(\dd x)$ from
  \eqref{l20_l160_eq}. So
  \[ \int g(x) \mu_X(\dd x) = \int g(x) f_X(x) \dd
  x \qquad \]
  for all $g$ bounded continuous with compat
  support. Hence $\mu_X$ has density $f_X$ (Remark
  (0)).}
\end{proof}
\end{flashcard}

\begin{flashcard}[levy-thm]
\begin{theorem*}[Levy]
  \label{levy_thm}
  \cloze{
  Let $(X_n)$, $X$ be random variables on $\RR^d$
  with $\cf_{X_n}(u) \to \cf_X(u)$ for all $u$ as
  $n \to \infty$. Then $X_n \to X$ \gls{vc_wly}.}
\end{theorem*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Levy's continuous theorem states that if
      $\cf_{X_n}(u) \to \phi(u)$ for all $u$ for
      some function $\phi$ that is continuous in a
      neighbourhood of 0, then $\phi$ is the
      \gls{char_func} of some random variable $X$
      and $X_n \to X$ \gls{vc_wly}.
    \item \refsteplabel[Cram\'er Wold device]{Cramer_Wold}
      Cram\'er Wold device: Let $(X_n)$, $X$ be
      random variables on $\RR^d$, then $X_n \to
      X$ \gls{vc_wly} if and only if
      \[ \langle u, X_n \rangle
      \stackrel{\text{\gls{vc_wly}}}{\longrightarrow}
      \langle u, X \rangle \qquad \forall u \in
      \RR^d \]
      (hence $\cf_{X_n}(u) \to \cf_X(u)$ by BCT
      and Levy).
  \end{enumerate}
\end{remark*}
