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\begin{flashcard}[image-measure]
\begin{definition*}[image / measure]
  \cloze{
  Let $(E, \mathcal{E})$ and $(G, \mathcal{G})$ be
  2 measurable spaces, $f : E \to G$
  measurable, and $\mu$ a measure on $(E,
  \mathcal{E})$. Then $\mu$ induces an \emph{image
  / pull-forward measure} $\mathcal{V}$ on
  $\mathcal{G}$ given by $\mathcal{V} = \mu \circ
  f^{-1}$, i.e. $\mathcal{V}(A) = \mu(f^{-1}(A))
  ~\forall A \in \mathcal{G}$. This is
  well-defined and $\mathcal{V}$ is a measure
  (\es{1}).
  }
\end{definition*}
\end{flashcard}

\begin{note*}
  Starting from Lebesgue measure, we can get all
  probability measures (Radon measures) in this
  way.
\end{note*}

\begin{lemma*}
  Let $g : \RR \to \RR$ be non-constant, right
  continuous, and increasing. Set $G(\pm \infty) =
  \lim_{x \to \pm \infty} g(x)$ and $I =
  (g(-\infty), g(\infty))$. Define $f : I \to \RR$
  by $f(x) = \inf\{y \in \RR : g(y) \ge x\}$. Then
  $f$ is left continuous, increasing and $\forall
  x \in I, y \in \RR$,
  \[ f(x) \le y \iff x \le g(y) \qquad (f(x) > y
  \iff x > g(y)) \]
  $f$ is called a generalised inverse of $g$ (if
  $I = (0, 1)$ then $f$ is the quantile function).
\end{lemma*}

\begin{center}
  \includegraphics[width=0.6\linewidth]
  {images/4a2277de6e6111ee.png}
\end{center}

\begin{proof}
  Fix $x \in I$. Define $J_x = \{y \in \RR : g(y)
  \ge x\}$. Then $J_x$ is non-empty and bounded
  below and hence $f(x) \in \RR$. Since $g$ if
  increasing, if $y \in J_x$ and $y' \ge y$, then
  $g(y') \ge g(y) \ge x$, i.e. $y' \in J_x$. Since
  $g$ is right continuous, if $y_n \in J_x$, $y_n
  \downarrow y$, then $g(y) = \lim_{n \to \infty}
  g(y_n) \ge x$, i.e. $y \in J_x$. So $J_x =
  [f(x), \infty)$. So $x \le g(y) \iff
  y \in J_x \iff f(x) \le y$. If $x \le x'$, we
  have $J_x \supseteq J_{x'}$ (as $y \in J_x
  \Rightarrow y \in J_{x'}$, as $y \in J_x'
  \iff g(y) \ge x' \implies g(y) \ge x \implies y
  \in J_x$). So $[f(x), \infty) \supseteq [f(x'),
  \infty)$, so $f(x) \le f(x')$, i.e. $f$ is
  increasing.

  To show $f$ is left continuous: Let $x_n
  \uparrow x$. Then $J_x = \bigcap_n J_{x_n}$,
  i.e. $[f(x), \infty) = \bigcap_n [f(x_n),
  \infty)$, so $f(x_n) \to f(x)$.
\end{proof}

\begin{theorem*}
  Let $g : \RR \to \RR$ as in the lemma. Then
  there exists a unique Radon measure $\mu_g$ on
  $\RR$ such that $\forall a, b \in \RR$ with $a <
  b$,
  \[ \mu_g((a, b]) = g(b) - g(a) \]
  Also, every Radon measure can be obtained in
  this way.
\end{theorem*}

\begin{remark*}
  The measure $\mu_g$ is called the
  Lebesgre-Stieljtes measure associated with $g$.
\end{remark*}

\begin{proof}
  Define $I, f$ as in the lemma, and let $\lambda$
  be the Lebesgue measure on $I$. $f$ is Borel
  measurable since
  \[ f^{-1}((-\infty, z]) = \{x \in I : f(x) \le
  z\} = \{x \in I : x \le g(z)\} = (g(-\infty),
  g(z)] \in \mathcal{B} \]
  and $\{(-\infty, z] : z \in \RR\}$ generate
  $\mathcal{B}$, hence $f$ is measurable.

  Thus, the induced measure $\mu_g = \lambda \circ
  f^{-1}$ exists on $\mathcal{B}$, where $\mu_g(A)
  = \lambda(f^{-1}(A))$. Then
  \begin{align*}
  \mu_g((a, b])
  &= \lambda(f^{-1}(a, b]) \\
  &= \lambda(\{x : f(x) > a, f(x) \le b\}) \\
  &= \lambda(\{x : x > g(a), x \le g(b)\}) \\
  &= \lambda((g(a), g(b)]) \\
  &= g(b) - g(a)
  \end{align*}
  By the \nameref{uniqueness_thm} for
  $\sigma$-finite measures, $\mu_g$ is uniquely
  determined.

  Conversely, if $\mathcal{V}$ is any Radon
  measure on $\RR$, define
  \[ g : \RR \to \RR \quad \text{as} \quad
  g(y) = \begin{cases}
    \mathcal{V}((0, y]) & y \ge 0 \\
    -\mathcal{V}((y, 0]) & y < 0
  \end{cases} \]
  $\mathcal{V}$ Radon implies $g : \RR \to \RR$.
  Easy to check $g$ is right continuous ($y \ge
  0$, $y_n \downarrow y$, then $(0, y_n]
  \downarrow (0, y]$ and then $\mathcal{V}((0,
  y_n]) \downarrow \mathcal{V}((0, y])$ by
  countably additivity, and for $y < 0$, if $y_n
  \downarrow y$ then use a similar argument). So
  $g$ is increasing. Lastly,
  \[ \mathcal{V}((a, b]) = g(b) - g(a) \]
  (check cases $0 \in (a, b)$ and $0 \not\in (a,
  b$).
\end{proof}

\begin{example*}
  Fix $x \in \RR$. Take $g = \mathbbm{1}_{[x,
  \infty)}$
  \begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0f1da9c06e6511ee.png}
  \end{center}
  Then $\mu_g = \delta_x$: Dirac measure at $x$,
  i.e.
  \[ \delta_x(A) = \begin{cases}
    1 & x \in A \\
    0 & x \not\in A
  \end{cases} \forall A \in \mathcal{B} \]
\end{example*}

\subsection{Random Variables}

\begin{flashcard}[random-variable-defn]
\begin{definition*}[random variable]
  \cloze{
  Let $(\Omega, \mathcal{F}, \PP)$ be a probability
  space and $(E, \mathcal{E})$ a measurable space.
  Let $X : \Omega \to E$ a measurable function. Then
  $X$ is called a \emph{random variable} in $E$.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

$X$ models a ``random'' outcome of an experiment.
For example $\Omega = \{\text{H}, \text{T}\}$, $X
: \text{\# heads} : \Omega \to \{0, 1\}$.
\begin{center}
  \includegraphics[width=0.6\linewidth]
  {images/7216ff186e6511ee.png}
\end{center}

\begin{flashcard}[distribution-func-defn]
\begin{definition*}[distribution]
  \cloze{The image measure $\mu_X = \PP \circ X^{-1}$ is
  called the \emph{law} or \emph{distribution} of
  $X$. It is a measure on $(E, \mathcal{E})$.

  If $E = \RR$, $\mu_X$ is uniquely determined by
  its values on the $\pi$-system $\{(-\infty, x] :
  x \in \RR\}$ given by
  \[ F_X(x) \defeq \mu_X((-\infty, x]) = \PP \circ
  X^{-1}((-\infty, x]) = \PP(\omega \in \Omega :
  X(\omega) \le x) = \PP(X \le x) .\]
  The function $F_X$ is called the distribution
  function of $X$, because it characterises the
  distribution of $X$.}
\end{definition*}
\end{flashcard}
\vspace{-1em}

By properties of probability measure:
\begin{enumerate}[(1)]
  \item $F_X$ is increasing.
  \item $F_X$ is right continuous ($x_n
    \downarrow x \implies (-\infty, x_n]
    \downarrow (-\infty, x]$ hence countability
    additivity of $\PP \circ X^{-1}$)
  \item $F_X(-\infty) = \lim_{x \to -\infty}
    F_X(x) = 0$, $F_X(\infty = \lim_{x \to \infty}
    F(x) = 1$.
\end{enumerate}
Any $F : \RR \to [0, 1]$ satisfying all these
properties is called a \emph{distribution
function}.