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% 03/03/2022 11AM

\subsubsection*{Studying $\mathrm{N}(\mu, \sigma^2)$ via linear transformations}

Facts about $X \sim \mathrm{N}(\mu, \sigma^2)$:
\begin{enumerate}[(i)]
    \item $X$ has the same distribution as $\mu +
        \sigma Z$ where $Z \sim \mathrm{N}(0, 1)$.
    \item $X$ has CDF
        \[ F_X(x) = \Phi \left( \frac{x -
        \mu}{\sigma} \right) \]
        \begin{notation*}
            $\Phi$ is the CDF of $\mathrm{N}(0,
            1)$
        \end{notation*}
    \item $\EE[X] = \mu$, $\mathrm{Var}(X) =
        \sigma^2$.
\end{enumerate}

\begin{proof}
    \begin{enumerate}[(i)]
        \item $g(z) = \mu + \sigma z$ so
            $g^{-1}(x) = \frac{x - \mu}{\sigma}$.
            Then $g(Z)$ has density
            \begin{align*}
                &= \frac{1}{\sigma} f_Z \left(
                \frac{x - \mu}{\sigma} \right) \\
                &= \frac{1}{\sigma \sqrt{2\pi}}
                e^{-\frac{(x - \mu)^2}{2\sigma^2}}
            \end{align*}
        \item $F_{g(Z)}(x) = \PP(g(Z) \le x) =
            \PP \left( Z \le \frac{x -
            \mu}{\sigma} \right) = \Phi \left(
            \frac{x - \mu}{\sigma} \right)$.
        \item Use part (i):
            \[ \EE[X] = \EE[\mu + \sigma Z] = \mu
            + \sigma \EE[Z] = \mu \]
            \[ \mathrm{Var}(\mu + \sigma Z) =
            \sigma^2 \mathrm{Var}(Z) = \sigma^2 \]
    \end{enumerate}
\end{proof}

\begin{remark*}
    Reduces to $\Phi$: lookup in book / table /
    Wolfram Alpha.
\end{remark*}

\noindent
\ul{Usage}: $X \sim \mathrm{N}(\mu, \sigma^2)$
\begin{align*}
    \PP(a \le X \le b)
    &= \PP \left( \frac{a - \mu}{\sigma} \le
    \frac{X - \mu}{\sigma} \le \frac{b -
    \mu}{\sigma} \right) \\
    &= \PP \left( \frac{a - \mu}{\sigma} \le Z \le
    \frac{b - \mu}{\sigma} \right) \\
    &= \Phi \left( \frac{b - \mu}{\sigma} \right)
    - \Phi \left( \frac{a - \mu}{\sigma} \right)
\end{align*}
\ul{Special case}:
\[ a = \mu - k\sigma, \qquad b = \mu + k\sigma \]
($k \in \{1, 2, \dots\}$). Recall: $\sigma$ is the
\emph{standard deviation}.
\[ \PP(a \le X \le b) = \Phi(k) - \Phi(-k) \]
``within $k$ standard deviations of the mean''.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/019f955ea30211ec.png}
\end{center}

\begin{definition*}
    $X$ a continuous random variable. The
    \emph{median} of $X$ is the number $m$ such
    that $\PP(X \le m) = \PP(X \ge m) = \half$, i.e.
    \[ \int_{-\infty}^m f_X(x) \dd x =
    \int_m^\infty f_X(x) \dd x = \half \]
\end{definition*}

\subsubsection*{Comments}

\begin{itemize}
    \item For $X \sim \mathrm{N}(\mu, \sigma^2)$
        and other distributions symmetric about
        mean, we have median $m = \EE[X]$.
    \item Sometimes $|X - m|$ better than $|X -
        \mu|$ for interpretation.
\end{itemize}

\subsubsection*{More than one continuous Random Variables}

Allow random variables to take values in $\RR^n$.
For example
\[ X = (X_1, \dots, X_n) \in \RR^n \]
is a random variable. Say $X$ has density $f :
\RR^n \to [0, \infty)$ if
\[ \PP(X_1 \le x_1, \dots, x_n \le x_n) =
\int_{-\infty}^{x_1} \cdots \int_{-\infty}^{x_n}
f(u_1, \dots, u_n) \prod_i \dd u_i \]
(integrate over $(-\infty, x_1] \times \cdots
\times (-\infty, x_n]$) % ))

\myskip
\ul{Consequence}:
\[ \PP((X_1, \dots, X_n) \in A) = \int_A f(u) \dd
u \]
for all ``measurable'' $A \subset \RR^n$.

\begin{definition*}
    $f$ is called a \emph{multivariate density
    function} or (especially $n = 2$) a
    \emph{joint density}.
\end{definition*}

\begin{definition*}
    Random variables $X_1, \dots, X_n$
    \emph{independent} if
    \[ \PP(X_1 \le x_1, \dots, X_n \le x_n) =
    \PP(X_1 \le x_1) \cdots \PP(X_n \le x_n)
    \tag{$*$} \]
\end{definition*}

\noindent
\ul{Goal}: convert to statement about densities.

\begin{definition*}
    $X = (X_1, \dots, X_n)$ has density $f$. The
    \emph{marginal density} $f_{X_i}$ of $X_i$ is
    \[ f_{X_i}(x_i) = \int_{-\infty}^\infty \cdots
    \int_{-\infty}^\infty f(x_1, \dots, x_n)
    \prod_{j \neq i} \dd x_j \]
    ``density of $X_i$ viewed as a random variable
    by itself''.
\end{definition*}

\begin{theorem}
    $X = (X_1, \dots, X_n)$ has density $f$.
    \begin{enumerate}[(a)]
        \item if $X_1, \dots, X_n$ independent,
            with marginals $f_{X_1}, \dots,
            x_{X_n}$. Then
            \[ f(X_1, \dots, X_n) = f_{X_1}(x_1)
            \cdots f_{X_n}(x_n) \]
        \item Suppose $f$ factorises as
            \[ f(X_1, \dots, X_n) = g_1(x_1)
            \cdots g_n(x_n) \]
            for non-negative functions $(g_i)$.
            Then $X_1, \dots, X_n$ are independent
            and marginal $f_{X_i} \propto g_i$.
    \end{enumerate}
\end{theorem}

\begin{proof}
    \begin{enumerate}[(a)]
        \item \eqnoskip
            \begin{align*}
                \PP(X_1 \le x_1, \dots, X_n \le
                x_n)
                &= \PP(X_1 \le x_1) \cdots \PP(X_n
                \le x_n) \\
                &= \left[ \int{-\infty}^\infty
                f_{X_1}(u_1) \dd u_1 \right]
                \cdots \left[
                \int_{-\infty}^\infty f_{X_n}(u_n)
                \dd u_n \right] \\
                &= \int_{-\infty}^{x_1}
                \int_{-\infty}^{x_n} \prod f_{X_i}
                (u_i) \prod \dd u_i
            \end{align*}
            which matches with definition of $f$.
        \item Idea:
            \begin{itemize}
                \item Replace $g_i(x)$ with
                    $h_i(x) = \frac{g_i(x)}{\int
                    g_i(u) \dd u}$. $h_i$
                    \emph{is} a density.
                \item compute integral at ($*$)
            \end{itemize}
    \end{enumerate}
\end{proof}

\subsubsection*{Transformation of Multiple Random Variables}

Key Example 1: $X, Y$ independent with densities
$f_X, f_Y$. \\
Goal: density of $Z = X + Y$.
\begin{enumerate}[Step 1:]
    \item Declare the joint density
        \[ f_{X, Y}(x, y) = f_X(x) f_Y(y) .\]
    \item CDF of $Z$:
        \begin{align*}
            \PP(X + Y \le z)
            &= \iint_{\{x + y \le z\}} f_{X, Y}(x,
            y) \dd x \dd y \\
            &= \int_{x = -\infty}^\infty \int_{y =
            -\infty}^{z - x} f_X(x) f_Y(y) \dd x
            \dd y \\
            &= \int_{x = -\infty}^\infty \int_{y' =
            -\infty}^z f_Y(y' - x) f_X(x) \dd y'
            \dd x
            &\text{substitute $y' = y + x$} \\
            &= \int_{y = -\infty}^x \dd y \left(
            \int{x = -\infty}^\infty f_Y(y - x)
            f_X(x) \dd x \right) 
        \end{align*}
        So density of $Z$:
        \[ f_Z(z) = \ub{\int_{x = -\infty}^\infty
        f_Y(z - x) f_X(x) \dd
        x}_{\text{Convolution of $f_X$ and $f_Y$}} \]
\end{enumerate}