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\begin{flashcard}[orthogonal-projection-thm]
\begin{theorem*}[Orthogonal projection]
  \cloze{
  If $V$ is a \emph{closed subspace} of $\Lp[2]$,
  then $\forall f \in \Lp[2]$, $f = v + u$ where
  $v \in V$, $u \in V^\perp$. Moreover, $\normp\|f
  - v\|_2 \le \normp\|f - g\|_2$ for all $g \in V$
  with equality if and only if $g = v$
  \gls{al_ev}. In particular, $v$ is unique
  (\gls{al_ev}) and is called the orthogonal
  projection of $f$ on $V$.}
\end{theorem*}
\fcscrap{
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/bc3f3d3aa0fa4d56.png}
\end{center}
}
\begin{proof}
  \cloze{Define $d(f, V) = \inf_{g \in V} \normp\|f -
  g\|_2$. Let $(g_n) \in V$ be a sequence such
  that $\normp\|f - g_n\|_2 \to d(f, V)$. Now by
  parallelogram law,
  \[ 2(\normp\|f - g_n\|_2^2 + \normp\|f -
  g_m\|_2^2) = \normp\|g_n - g_m\|_2^2 + \ub{4 \left\|
  f - \frac{g_n + g_m}{2} \right\|^2}_{\ge 4d(f,
  V)^2} \]
  so by taking $\lim_{n, m \to \infty}$, we deduce
  that $\normp\|g_n - g_m\|_2^2 \to 0$, i.e.
  $(g_n)$ is Cauchy. As $\Lp[2]$ is complete and
  $V$ is closed, $g_n
  \stackrel{\Lp[2]}{\longrightarrow} v \in
  V$. Then $\normp\|f - g_n\|_2^2 \to \normp\|f -
  v\|_2^2$ hence $\normp\|f - v\|_2^2 = d(f, V)^2$,
  i.e. $d(f, V) = \normp\|f - v\|_2$.

  Then for any $h \in V$, $t \in \RR$,
  \[ \label{lec16_l40_eq}
  d(f, v)^2 \le \normp\|f - (v + th)\|_2^2 =
  d(f, v)^2 - 2t \langle f - v, h \rangle + t^2
  \normp\|h\|_2^2 \tag{$*$} \]
  Letting $t \downarrow 0$ and $t \uparrow 0$,
  $\langle f - v, h \rangle = 0$, hence $f - v \in
  V^\perp$. Now
  \[ f = \ub{v}_{\in V} + \ub{f - v}_{\in V^\perp} \]
  as desired. For any $g \in V$,
  \[ f - g = \ub{f - v}_{\in V^\perp} + \ub{v -
  g}_{\in V} \]
  and
  \[ \normp\|f - g\|_2^2 = \normp\|f - v\|_2^2 +
  \normp\|v - g\|_2^2 \]
  Hence $\normp\|f - g\|_2 \ge \normp\|f - v\|_2$ with
  equality if and only if $\normp\|v - g\|_2 = 0$,
  i.e. $v = g$ \gls{al_ev}.}
\end{proof}
\end{flashcard}

$(\Omega, \mathcal{F}, \PP)$ and $X, Y \in
\Lp[2](\Omega, \mathcal{F}, \PP)$ with $\EE X =
\EE Y = 0$. Then $\Cov(X, Y) = \EE(X - \EE X)(Y -
\EE Y) = \EE XY = \langle X, Y \rangle$. $\Var(X)
= \Cov(X, X)$. If $X$ and $Y$ independent,
$\langle X, Y \rangle = 0$, converse not true.

If $\mathcal{G}$ is a sub-\sigalg{} of
$\mathcal{F}$ (i.e. $\mathcal{G} \subseteq
\mathcal{F}$), then $\Lp[2](\Omega, \mathcal{G},
\PP)$ is a closed subspace of $\Lp[2](\Omega,
\mathcal{F}, \PP)$. For $X \in \Lp[2](\Omega,
\mathcal{F}, \PP)$, (a variant of) the conditional
expectation of $X$ given $\mathcal{G}$, $\EE(X
\given \mathcal{G})$ is defined as the orthogonal
projection of $X$ on $\Lp[2](\Omega, \mathcal{F},
\PP)$ ($X$ should be measurable with respect to
$\mathcal{G}$ and $\normp\|X - Y\|_2 \ge \normp\|X
- \EE(X \given \mathcal{G})\|_2$ for all $Y$
$\mathcal{G}$-measurable).

\textbf{Question:} How to define $\EE(X \given
\mathcal{G})$ is $X \in \Lp[1](\Omega,
\mathcal{F}, \PP)$? (advanced probability).

\textbf{Exercise:} Let $(G_i)_{i \in I}$ be a
countable family of disjoint events whose union is
$\Omega$ and set $\mathcal{G} = \sigma(G_i : i \in
I)$. Let $X$ be integrable. Then the conditional
expression of $X$ given $\mathcal{G}$ is given by
\[ Y = \sum_i \EE(X \given G_i) \indicator{G_i},
\qquad \EE(X \given G_i) =
\frac{\EE(x\indicator{G_i})}{\PP(G_i)} ~\forall i
\in I \]
Check:
\begin{enumerate}[(1)]
  \item $Y$ is $\mathcal{G}$-measurable
  \item $Y \in \Lp[2](\Omega, \mathcal{F}, \PP)$
  \item $Y$ is ``the'' orthogonal projection of
    $X$ onto $\Lp[2](\Omega, \mathcal{G}, \PP)$ if
    $X \in \Lp[2](\Omega, \mathcal{F}, \PP)$.
\end{enumerate}

\subsubsection*{$L^p$ Convergence and Uniform
Integrability}

$(\Omega, \mathcal{F}, \PP)$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/790f013e1a114061.png}
\end{center}
Explanations:

$f_n = n\indicator{(0, 1/n)}$ on $((0, 1),
\mathcal{B}, \lambda)$. Then $f_n \to 0$
\gls{al_surely}. But $\EE|f_n| = \EE f_n = 1
~\forall n$, i.e. \gls{al_surely} $\nimplies$
$\Lp$ convergence.

\[ \PP(|X_n - X| > \eps) \le \frac{\EE|X_n -
X|^p}{\eps^p} \to 0 \]
(Markov inequality).

\begin{theorem*}[DCT]
  \label{dct}
  Let $(X_n)$ be such that $X_n \convP X$ and
  $|X_n| \le Y$ for all $n$, for some integrable
  random variable. Then $X_n
  \stackrel{\Lp[1]}{\to} X$, i.e. $\EE|X_n - X|
  \to 0$ as $n \to \infty$.
\end{theorem*}
\vspace{-1em}

\textbf{Question:} What is the ``minimum
condition'' on $(X_n)$ under which $X_n \convP X$
implies $X_n \stackrel{\Lp[1]}{\longrightarrow}
X$? $\implies \EE|X_n| \to \EE|X|$, ``Uniformly
Integrable''.

\begin{flashcard}[IXdelta]
For $X \in \Lp[1](\PP)$ define
\[ I_X(\delta) = \cloze{\sup\{\EE(|X|\indicator{A}) : A
\in \mathcal{F}, \PP(A) \le \delta\}} \]
\end{flashcard}
Then $I_X(\delta) \to 0$ as $\delta \to 0$. (If
not then $\exists \eps > 0$ and $(A_n) \in
\mathcal{F}$ such that $\PP(A_n) \le 2^{-n}$ and
$\EE|X|\indicator{A_n} \ge \eps$. Then $\sum
\PP(A_n) < \infty$ so $\PP(A_n \text{ i.o.}) = 0$
by \nameref{borel_cantelli_1}. Then
\[ \eps \le \EE|X| \indicator{A_n} \le \EE|X|
\indicator{\bigcup_{m = n}^\infty A_m}
\stackrel{\text{\nameref{dct}}}{\longrightarrow} 0 \]
since
\[ \indicator{\bigcup_{m = n}^\infty A_m} \to
\indicator{\bigcap_m \bigcup_{m = n}^\infty
A_m} = \indicator{\{A_n \text{ i.o.}\}} = 0
\text{ \gls{al_surely}} \]
contradiction).

\begin{flashcard}[uniformly-integrable-defn]
\begin{definition*}
  \glsadjdefn{u_int}{uniformly
  integrable}{collection of random variables}
  \glsadjdefn[u_int]{ui}{UI}{collection
  of random variables}
  Let $\chi$ be a collection of random
  variables in $\Lp[1](\PP)$. Define
  \[ I_{\chi}(\delta) = \cloze{
  \sup\{\EE(|X|\indicator{A}) : X \in \chi,
  A \in \mathcal{F}, \PP(A) \le \delta\}} \]
  We say $\chi$ is \emph{uniformly
  integrabl} (UI) if
  \begin{enumerate}[(1)]
    \item \cloze{$\chi$ is bounded in $\Lp[1]$
      (i.e. $\sup_{X \in \chi}
      \normp\|X\|_1 = \sup_{X \in \chi}
      \EE|X| = I_{\chi} < \infty$)}
    \item \cloze{$I_{\chi}(\delta) \to 0$
      as $\delta \to 0$.}
  \end{enumerate}
\end{definition*}
\end{flashcard}

\begin{remark*}
  Note to reader: I didn't follow what the
  lecturer was writing in these remarks so they
  are probably nonsense, but if you rearrange
  things appropriately it should hopefully make
  sense (I've tried thinking about it for a bit
  but haven't figured it out).
  \begin{enumerate}[(1)]
    \item Any single integrable random variable is
      \gls{ui} (so does any finite collection of
      integrable random variables, also if
      \[\chi = \{X : \text{$X$ a random
      variable such that $|X| \le Y$ for some $Y
      \in \Lp[1]$}\}\]) Then $X$ is \gls{ui},
      \[ I_{\chi}(\delta) \le
      I_Y(\delta) \to 0 \]
      as $\delta \to 0$.
    \item If $\chi$ is bounded in $\Lp$ for
      some $p > 1$ then
      \[ \sup \EE(X \indicator{A})
      \stackrel{\text{\nameref{holder_ineq}}}{\le}
      \normp\|X\|_p (\PP(A))^{1/q} \le C\delta^{1/q} \]
      $X \in \chi$ such that $\PP(A) \le
      \delta$ i.e. $I_{\chi}(\delta) \le C
      \delta^{1/q} \to 0$ as $\delta \to 0$ for
      all $A$ such that $\PP(A) \le \delta$ as
      $\sup_{X \in \chi} \EE|X|
      \indicator{A} \le \EE Y\indicator{A}$.
    \item $\Lp[1]$ bounded $\nimplies$ \gls{ui}.
  \end{enumerate}
\end{remark*}
