%! TEX root = PM.tex
% vim: tw=50
% 04/11/2023 10AM

\begin{flashcard}[indep-rv-prop]
\begin{proposition*}
  Let $X_1, \ldots, X_n$ be random variables $X_i
  : (\Omega, \mathcal{F}, \PP) \to (E_i,
  \mathcal{E}_i)$. Set $E = E_1 \times \cdots
  \times E_n$, $\mathcal{E} = \mathcal{E}_1 \pmeas
  \cdots \pmeas \mathcal{E}_n$. Consider $X :
  (\Omega, \mathcal{F}, \PP) \to (E, \mathcal{E})$
  given by $X(\omega) = (X_1(\omega), \ldots,
  X_n(\omega))$. Then \cloze{$X$ is
  $\mathcal{E}$-measurable and the following are
  equivalent:}
  \begin{enumerate}[(i)]
    \item \cloze{$X_1, \ldots, X_n$ are \gls{indep_rv}
      (i.e. $\sigma\{X_i^{-1}(A) : A \in
      \mathcal{E}_i\}$ are independent).}
    \item \cloze{$\mu_X = \mu_{X_1} \pmu \mu_{X_2} \pmu
      \cdots \pmu \mu_{X_n}$.}
    \item \cloze{For all bounded measurable $f_i : E_i
      \to \RR$,
      \[ \EE \left( \prod_{i = 1}^n f_i(X_i)
      \right) = \prod_{i = 1}^n \EE (f_i(X_i))
      .\]}
  \end{enumerate}
\end{proposition*}

\begin{proof}
  \cloze{
  $X$ measurable: $X^{-1}(A) \in \mathcal{F}
  ~\forall A \in \mathcal{E}$. Enough to check
  \begin{align*}
    X^{-1}(A_1 \times \cdots \times A_n)
    &= \{\omega : X_i(\omega) \in A_1, \ldots,
    X_n(\omega) \in A_n\} \\
    &= \bigcap_{i = 1}^n X_i^{-1}(A_i)
  \end{align*}
  is in $\mathcal{F}$. But this is true since $X_i
  : (\Omega, \mathcal{F}) \to (E_i,
  \mathcal{E}_i)$ are all measurable, so
  $X_i^{-1}(A_i) \in \mathcal{F}$ for all $A_i \in
  \mathcal{E}_i$. Now we will show (i) $\implies$
  (ii) $\implies$ (iii) $\implies$ (i).
  \begin{enumerate}[(i) $\implies$ (ii)]
    \item[(i) $\implies$ (ii)] Let $\nu =
      \mu_{X_1} \pmu \cdots \pmu \mu_{X_n}$.
      Enough to show that
      \begin{align*}
        \mu_X(A_1 \times \cdots \times A_n)
        &\stackrel{?}{=} \nu (A_1 \times \cdots
        \times A_n) \\
        &= \PP(X_1 \in A_1, \ldots, X_n \in A_n)
        \\
        &\stackrel{\text{(a)}}{=} \PP(X_1 \in A_1)
        \cdots \PP(X_n \in A_n) \\
        &= \mu_{X_1} (A_1) \cdots \mu_{X_n}(A_n)
        \\
        &= \nu(A)
      \end{align*}
      Now finish since $\mathcal{A}$ is a \pisys{}
      generating $\mathcal{E}$\ldots
    \item[(ii) $\implies$ (iii)]
      \begin{align*}
        \EE \left( \prod_{i = 1}^n f_i(X_i) \right)
        &= \int \prod_{i = 1}^n f_i(x_i) \dd
        \mu_X(x) \\
        &= \prod_{i = 1}^n \int f_i(x_i) \dd
        \mu_{X_i}(x_i) \\
        &= \prod_{i = 1}^n \EE (f_i(X_i))
      \end{align*}
    \item[(iii) $\implies$ (i)] Use $f_i =
      \indicator{A_i}$, $A_i \in \mathcal{E}_i$.
      Then
      \begin{align*}
        \EE \left( \prod_{i = 1}^n
        \indicator{A_i}(X_i) \right)
        &= \EE (\indicator{A_1 \times \cdots
        \times A_n}(X)) \\
        &= \PP(X_1 \in A_1, \ldots, X_n \in A_n)
        \\
        &\stackrel{\text{(a)}}{=} \prod_{i = 1}^n
        \EE(\indicator{A_i}(X_i)) \\
        &= \prod_{i = 1}^n \PP(X_i \in A_i)
        \qedhere
      \end{align*}
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

\newpage
\section{$L^p$ Spaces, Norms, Inequalities}

\begin{flashcard}[norm-defn]
\begin{definition*}[Norm]
  \cloze{
  Recall that a norm on a real vector space $V$ is
  $\|\bullet\| : V \to [0, \infty)$ such that
  \begin{enumerate}[(1)]
    \item $\|\lambda v\| = |\lambda| \|v\|$ for
      all $\lambda \in \RR, v \in V$.
    \item $\|v + w\| \le \|v\| + \|w\|$ for all
      $v, w \in W$.
    \item $\|v\| = 0 \iff v = 0$.
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[Lp-norm-defn]
\begin{definition*}[$L^p$ norm]
  \glssymboldefn{Lp_norm}{$L^p$}{$L^p$}
  \glssymboldefn{Lp_norm_func}{$\|f\|_p$}{$\|f\|_p$}
  \cloze{For $(E, \mathcal{E}, \mu)$ a measure space and
  $1 \le p \le \infty$, define
  \[ L^p = L^p(E, \mathcal{E}, \mu) = \{f : E \to
  \RR \text{ measurable } : \normp\|f\|_p < \infty\} \]
  where
  \[ \normp\|f\|_p = \left( \int |f(x)|^p \dd \mu(x)
  \right)^{1/p} \]
  for $1 \le p < \infty$, and
  \[ \normp\|f\|_\infty = \esssup |f| = \inf\{\lambda
  \ge 0 : |f| \le \lambda \text{
  $\mu$.\gls{al_ev}}\} \]}
\end{definition*}
\end{flashcard}
\vspace{-1em}

By linearity of integral, this satisfies the
properties of a norm:
\begin{enumerate}[(1)]
  \item This holds for $1 \le p < \infty$ and for
    $p = \infty$ it's obvious.
  \item This holds for $p = 1, \infty$ easily. For
    other $p$, we shall show by Minkowski
    inequality.
  \item $f = 0$ implies $\normp\|f\|_p = 0$. But
    $\normp\|f\|_p = 0$ implies $f = 0$ \gls{al_ev}. So
    we fix this by defining equivalence classes
    \[ [f] = \{g : g = f \text{ \gls{al_ev}}\} \]
    and
    \glssymboldefn{Lpcal_norm}{$L^p$}{$L^p$}
    \[ \mathcal{L}^p = \{[f] : f \in \Lp\} \]
\end{enumerate}
Then $\mathcal{L}^p$, $1 \le p \le \infty$ are
normed vector spaces.

\subsection{Inequalities}

\subsubsection*{Markov / Chebyshev's Inequality}

Let $f \ge 0$ measurable, then $\forall \lambda >
0$,
\[ \ub{\mu(\{x \in E : f(x) \ge \lambda\})}_{\mu(f
\ge \lambda)} \le
\frac{\mu(f)}{\lambda} \]

\begin{proof}
  $\lambda \indicator{\{f \ge \lambda\}} \le f$.
  So integrating with respect to $\mu$, we get
  \[ \lambda \mu(f \ge \lambda) \le \mu(f).
  \qedhere \]
\end{proof}

In particular, if $g \in L^p$, $p < \infty$, then
$\mu(|g| \ge \lambda) \le
\frac{\mu(|g|^p)}{\lambda^p} < \infty$. This gives
the tail estimates as $\lambda \to \infty$.

\begin{flashcard}[convex-fn-defn]
\begin{definition*}[Convex function]
  \glsadjdefn{convex_fn}{convex}{function}
  \cloze{
  For $I \subseteq \RR$ an interval, say a
  function $c : I \to \RR$ is \emph{convex} if
  $\forall x, y \in I$, $t \in [0, 1]$,
  \[ c(tx + (1 - t)x) \le tc(x) + (1 - t)c(y) \]
  equivalently:
  \[ \label{convex_defn_2} \frac{c(\tilde{t}) -  c(t)}{\tilde{t} - x}
  \le \frac{f(y) - c(\tilde{t})}{y - \tilde{t}}
  \tag{$*$} \]
  for $x < \tilde{t} < y$ in $I$. (In particular,
  this second definition shows that $c$ is
  continuous on $I$, hence Borel measurable).
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[convex-linear-ineq-lemma]
\begin{lemma*}
  Let $c : I \to \RR$ be \gls{convex_fn}, $m \in
  I$ (an interval). Then there exists $a, b \in
  \RR$ such that $c(x) \ge ax + b$ for all $x \in
  I$, and equality at $x = m$.
\end{lemma*}

\begin{proof}
  \cloze{
  Let
  \[ a = \sup \left\{ \frac{c(m) - c(x)}{m - x} : x
  \in I, x < m\right\} < \infty .\]
  Then by \eqref{convex_defn_2}, $\forall y > m, y
  \in I$,
  \[ \frac{c(m) - c(x)}{m - x}
  \le a \le \frac{c(y) - c(m)}{y - m} \]
  These inequalities imply
  \[ c(y) \ge ay - am + c(m) \qquad \forall y \ge
  m \]
  and
  \[ c(x) \ge ax - am + c(m) \qquad \forall x \le
  m. \qedhere \]
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[jensens-inequality-thm]
\begin{theorem*}[Jensen's Inequality]
  \label{jensens_ineq}
  \cloze{
  Let $X$ be an integrable random variable (i.e.
  $\EE |X| < \infty$), taking values in an
  interval $I \subset \RR$, and let $c : I \to
  \RR$ be convex. Then $\EE(c(x))$ is
  well-defined, and
  \[ \boxed{\EE(c(X)) \ge c(\EE(X))} \]
  }
\end{theorem*}

\begin{proof}
  \cloze{If $X$ is a constant \gls{al_surely}, then
  nothing to prove. Assume otherwise. Then $\EE(x)
  = m \in I$. Using the previous lemma, $\exists
  a, b \in \RR$ such that
  \[ \label{lec14_l236_eq} c(X) \ge aX + b \tag{$*$} \]
  In particular,
  \[ c((X))^- \le |a| |X| + |b| \]
  so $\EE(c(X)^-) < \infty$. Hence $\EE(c(X)) =
  \EE(c(X)^+) - \EE(c(X)^-)$ is well-defined on
  $(-\infty, \infty]$.

  Claim: \eqref{lec14_l236_eq} $\implies$
  $\EE(c(X)) \ge a\EE(X) + b$. If $\EE(c(X)) =
  \infty$, nothing to prove. Otherwise, $c(X)$ and
  $aX + b$ are integrable random variables
  satisfying \eqref{lec14_l236_eq}, so taking
  expectatation,
  \[ \EE(c(X)) \ge a\EE(X) + b = am + b = c(m) =
  c(\EE(X)). \qedhere \]}
\end{proof}
\end{flashcard}

As an application: $(\Omega, \mathcal{F}, \PP)$
and $1 \le p \le \infty$. If $X \in \Lp[\infty(\PP)]$,
then $X \in \Lp^{p(\PP)}$ for all $1 \le p <
\infty$ as $\normp\|X\|_p \le \normp\|X\|_\infty$.

Claim: If $X \in \Lp[q]$ and $q > p \ge 1$, then
$X \in \Lp$.

% $c(X) = |x|^{q / p}$ is convex, hence
\[ \normp\|X\|_p =
(\EE |X|^p)^{1/p}
= (c(\EE(|X|^p)))^{1/q}
\le (\EE(c(|X|^p)))^{1/q}
= (\EE(|X|^q))^{1/q}
= \normp\|X\|_q \]
Hence, $X \in \Lp[q]$ implies $X \in \Lp$ for all
$1 \le p < q$, i.e.
\[ \Lp[\infty](\PP) \subseteq \Lp[q](\PP)
\subseteq \Lp(\PP) \subseteq \Lp[1](\PP) \]
for all $1 \le p \le q \le \infty$.
