% vim: tw=50
% 08/02/2022 11AM

``$\mathrm{Bin} \left( n, \frac{\lambda}{n}
\right)$ converges to $\mathrm{Po}(\lambda)$''.
(note the ``converges'' is not very meaningful).

\subsubsection*{Expectation}

$(\Omega, \mathcal{F}, \PP)$ and $X$ a discrete
random variable. For now: $X$ only takes
non-negative values. ``$X \ge 0$''

\begin{definition*}
    The \emph{expectation of $X$} (or
    \emph{expected value} of \emph{mean}) is
    \[ \EE[X] = \sum_{x \in \mathrm{Im}(X)} x
    \PP(X = x) = \sum_{\omega \in \Omega}
    X(\omega) \PP(\{\omega\}) \]
    ``average of values taken by $X$, weighted by
    $p_X$''.
\end{definition*}

\setcounter{customexample}{0}
\begin{example}
    $X$ uniform on $\{1, 2, \dots, 6\}$ (i.e.
    dice) then
    \[ \EE[X] = \frac{1}{6} \times 1 + \frac{1}{6}
    \times 2 + \cdots + \frac{1}{6} \times 6 = 3.5 \]
    \begin{note*}
        $\EE[X] \not\in \mathrm{Im}(X)$.
    \end{note*}
\end{example}

\begin{example}
    $X \sim \mathrm{Binomial}(n, p)$.
    \[ \EE[X] = \sum_{k = 0}^n k \PP(X = k) =
    \sum_{k = 0}^n k {n \choose k} p^k (1 - p)^{n
    - k} \]
    Trick:
    \begin{align*}
        k {n \choose k}
        &= \frac{k \times n!}{k! \times (n - k)!}
        \\
        &= \frac{n!}{(k - 1)! (n - k)!} \\
        &= \frac{n \times (n - 1)!}{(k - 1)!
        \times (n - k)!} \\
        &= n {n - 1 \choose k - 1} \\
        \EE[X]
        &= n \sum_{k = 1}^n {n - 1 \choose k - 1}
        p^k (1 - p)^{n - k} \\
        &= np \sum_{k = 1}^n {n - 1 \choose k - 1}
        p^{k - 1} (1 - p)^{(n - 1) - (k - 1)} \\
        &= np \sum_{l = 0}^{n - 1} {n - 1 \choose
        l} p^l (1 - p)^{(n - 1) - l} \\
        &= np (p + (1 - p))^{n - 1} \\
        &= np
    \end{align*}
\end{example}

\begin{note*}
    Would like to say:
    \[ \EE[\mathrm{Bin}(n, p)] =
    \EE[\mathrm{Bern}(p)] + \cdots +
    \EE[\mathrm{Bern}(p)] \]
\end{note*}

\begin{example}
    $X \sim \mathrm{Poisson}(\lambda)$.
    \begin{align*}
        \EE[X]
        &= \sum_{k \ge 0} k \PP(X = k) \\
        &= \sum_{k \ge 0} k \cdot e^{-\lambda}
        \frac{\lambda^k}{k!} \\
        &= \sum_{k \ge 1} e^{-\lambda}
        \frac{\lambda^k}{(k - 1)!} \\
        &= \lambda \sum_{k \ge 0} e^{-\lambda}
        \frac{\lambda^{k - 1}}{(k - 1)!} \\
        &= \lambda \sum_{l \ge 0} e^{-\lambda}
        \frac{\lambda^l}{l!} \\
        &= \lambda
    \end{align*}
\end{example}

\begin{note*}
    Would like to say
    \[ \EE[\mathrm{Poisson}(\lambda)] \approx
    \EE\left[\mathrm{Bin} \left( n, \frac{\lambda}{n}
    \right) \right] = \lambda \]
    Can't say this: not true in general that
    \[ \PP(X_n = k) \approx \PP(\lambda = k)
    \implies \EE[X_n] \approx \EE[X] \]
\end{note*}

\begin{example}
    $X \sim \mathrm{Geometric}(p)$. Exercise.
\end{example}

\noindent
Positive and negative: General $X$ (not
necessarily $X \ge 0$).
\[ \EE[X] = \sum_{x \in \mathrm{Im}(X)} x \PP(X =
x) \]
unless
\[ \sum_{\substack{x > 0 \\x \in \mathrm{Im}(x)}}
x\PP(X = x) = +\infty \]
and
\[ \sum_{\substack{x < 0 \\x \in \mathrm{Im}(x)}}
x\PP(X = x) = -\infty \]
then we say that $\EE[X]$ is not defined. \\
\ul{Summary}:
\begin{itemize}
    \item both infinite: not defined
    \item first infinite, second not: $\EE[X] =
        +\infty$
    \item second infinite, first not: $\EE[X] =
        -\infty$
    \item neither infinite: $X$ is
        \emph{integrable}, i.e.
        \[ \sum_{x \in \mathrm{Im}(X)} |x| \PP(X =
        x) \]
        converges.
\end{itemize}
Note that some people say that in cases 2 and 3,
the expectation is undefined.

\begin{example}
    Most examples in the course are integrable
    \emph{except}:
    \begin{itemize}
        \item $\PP(X = n) = \frac{6}{\pi^2} \times
            \frac{1}{n^2}$ for $n \ge 1$. (Note
            $\sum \PP(X = n) = 1$). Then
            \[ \EE[X] = \sum \frac{6}{\pi^2}
            \times \frac{1}{n} = +\infty \]
        \item $\PP(X = n) = \frac{3}{\pi^2} \times
            \frac{1}{n^2}$ for $n \in \ZZ
            \setminus \{0\}$, then $\EE[X]$ is
            not defined. (``It's symmetric so
            $\EE[X] = 0$'' is considered wrong for
            us).
    \end{itemize}
\end{example}

\begin{example*}
    $\EE[\mathbbm{1}_A = \PP(A)$ Important!
\end{example*}

\subsubsection*{Properties of Expectation}

($X$ discrete).
\begin{enumerate}[(1)]
    \item If $X \ge 0$, then $\EE[X] \ge 0$ with
        equality if and only $\PP(X = 0) = 1$.
        Why?
        \[ \EE[X] = \sum_{\substack{x \in
        \mathrm{Im}(X) \\x \neq 0}} x \PP(X =
        x) \]
    \item If $\lambda, c \in \RR$ then:
        \begin{enumerate}[(i)]
            \item $\EE[X + c] = \EE[X] + c$
            \item $\EE[\lambda X] = \lambda
                \EE[X]$
        \end{enumerate}
    \item \begin{enumerate}[(i)]
            \item $X$, $Y$ random variables (both
                integrable) on same probability space.
                \[ \EE[X + Y] = \EE[X] + \EE[Y] \]
            \item In fact $\lambda, \mu \in \RR$
                \[ \EE[\lambda X + \mu Y] =
                \lambda \EE[X] + \mu \EE[Y] \]
                similarly:
                \[ \EE[\lambda_1 X_1 + \cdots +
                \lambda_n X_n] = \lambda_1
                \EE[X_1] + \cdots + \lambda_n
                \EE[X_n] \]
        \end{enumerate}
\end{enumerate}

\begin{proof}[of (3)(ii)]
    \begin{align*}
        \EE[\lambda X + \mu Y]
        &= \sum_{\omega \in \Omega} (\lambda
        X(\omega) + \mu Y(\omega)) \PP(\{\omega\})
        \\
        &= \lambda \sum_{\omega \in \Omega}
        X(\omega) \PP(\{\omega\}) + \mu
        \sum_{\omega \in \Omega} Y(\omega)
        \PP(\{\omega\}) \\
        &= \lambda \EE[X] + \mu \EE[Y]
    \end{align*}
    Note that this proof only works for countable
    $\Omega$, but there is also a proof for
    general $\Omega$.
\end{proof}

\begin{note*}
    Independence is \emph{not} required for
    linearity of expectation to hold. (This is the
    name for property (3)(ii)).
\end{note*}