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\begin{note*}
    The discrete equivalent is $X, Y \ge 0$
    independent,
    \[ \PP(X + Y) = k) = \sum_{l = 0}^k \PP(X = l)
    \PP(Y = k - l) \]
\end{note*}

\begin{example*}
    $X, Y \stackrel{\text{IID}}{\sim}
    \mathrm{Exp}(\lambda)$. $Z = X + Y$.
    \begin{align*}
        f_Z(z)
        &= \int_{x = 0}^z \lambda^2 e^{-\lambda x}
        e^{-\lambda (z - x)} \dd x \\
        &= \lambda^2 \int_{x = 0}^z e^{-\lambda z}
        \dd z \\
        &= \lambda^2 z e^{-\lambda z}
    \end{align*}
\end{example*}

\begin{definition*}
    $X \sim \mathrm{J}(n, \lambda)$ \emph{Gamma
    distribution}. $\lambda > 0$, $n \in \{1, 2,
    \dots\}$. Range is $[0, \infty)$. Density:
    \[ f_X(x) = e^{-\lambda x} \frac{\lambda^n
    x^{n - 1}}{(n - 1)!} \]
    \[ n = 1 \mapsto \mathrm{Exp}(\lambda) \]
    \[ n = 2 \mapsto \lambda^2 x e^{-\lambda x} \]
    So $X + Y \sim J(2, \lambda)$. (and in fact:
    $X_1 + \cdots + X_n \sim J(n, \lambda)$).
\end{definition*}

\begin{example*}
    $X_1 \sim \mathrm{N}(\sigma_1, \sigma_1^2)$,
    $X_2 \sim \mathrm{N}(\mu_2, \sigma_2^2)$
    independent. Then: $X_1 + X_2 \sim
    \mathrm{N}(\mu_1 + \mu_2, \sigma_1^2 +
    \sigma_2^2)$.
    \begin{note*}
        Already know that
        \[ \EE[X_1 + X_2] = \mu_1 + \mu_2 \qquad
        \mathrm{Var}(X_1 + X_2) = \sigma_1^2 +
        \sigma_2^2 \]
    \end{note*}
    \begin{proof}
        \begin{itemize}
            \item Calculation exercise
            \item Generating functions?? Coming
                up.
        \end{itemize}
    \end{proof}
\end{example*}

\begin{theorem*}
    Let $X = (X_1, \dots, X_n)$ on $D$. $g : \RR^n
    \to \RR^n$ well-behaved.
    \[ U = g(X) = (U_1, \dots, U_n) \]
    Joint density $f_X(x)$ is continuous. Then
    joint density
    \[f_U(\bf{u}) = f_X(g^{-1}(\bf{u})) |J(\bf{u})| \]
    where
    \[ J = \det \left( \left(
    \pfrac{[g^{-1}]_i}{u_j} \right)_{i, j = 1}^n
    \right) \]
    ``Jacobean'' ($d \times d$ matrix)
\end{theorem*}

\begin{customproof}{``Proof''}
    Definition of multivariate integration by
    substitution.
\end{customproof}

\begin{example*}[Radial Symmetry]
    $X, Y \stackrel{\text{IID}}{\sim}
    \mathrm{N}(0, 1)$. Write $(X, Y) =
    (R\cos\theta, R\sin\theta)$. Range: $R > 0$,
    $\theta \in [0, 2\pi)$.
    \begin{align*}
        f_{X, Y}(x, y)
        &= \frac{1}{\sqrt{2\pi}}
        e^{-\frac{x^2}{2}} \frac{1}{\sqrt{2\pi}}
        e^{-\frac{y^2}{2}} \\
        &= \frac{1}{2\pi} e^{-\frac{x^2 + y^2}{2}}
    \end{align*}
    \begin{note*}
        \[ |\text{Jacobean of $g^{-1}$}| =
        \frac{1}{|\text{Jacobean of $g$}|} \]
    \end{note*}
    \[ J = \left|
    \begin{matrix}
        \cos\theta & \sin\theta \\
        -R\sin\theta & R\cos\theta
    \end{matrix}
    \right| = R(\cos^2\theta + \sin^2\theta) = R \]
    So $f_{R, \theta}(r, \theta) = \frac{1}{2\pi}
    e^{-\frac{r^2}{2}} \times r$. Marginal:
    \[ f_\theta(\theta) = \frac{1}{2\pi} \]
    \[ f_R(r) = e^{-\frac{r^2}{2}} \times r \]
    Conclusion: $\theta, R$ are independent.
    $\theta$ is uniform on $[0, 2\pi)$.
\end{example*}

\begin{note*}
    Change of range: for example $X, Y \ge 0$, $Z
    = X + Y$.
    \[ f_{X, Z}(x, z) = ?(x, z)\mathbbm{1}_{(Z \ge
    x)} \]
    so $X, Z$ \emph{not} independent, even if $?$
    splits as a product.
\end{note*}

\subsubsection*{Moment Generating Function}

\begin{definition*}
    Let $X$ have density $f$. The \emph{MGF} of
    $X$ is:
    \[ m_X(\theta) := \EE[e^{\theta X}] =
    \int_{-\infty}^\infty e^{\theta x} f(x) \dd x \]
    whenever this is finite.
\end{definition*}

\begin{note*}
    $m_X(0) = 1$.
\end{note*}

\begin{theorem*}
    The MGF uniquely determines distribution of a
    random variable whenever it exists for all
    $\theta \in (-\eps, \eps)$ for some $\eps >
    0$.
\end{theorem*}

\begin{theorem*}
    Suppose $m(\theta)$ exists for all $\theta \in
    (-\eps, \eps)$. Then
    \[ m^{(n)}(0) = \dfrac[n]{}{\theta} m(\theta)
    \big|_{m = 0} = \EE[X^n] \]
    ($\EE[X^n]$ is the ``$n$-th moment'')
\end{theorem*}

\noindent
Proof comment: $\pfrac[n]{e^{\theta x}}{\theta} =
x^n e^{\theta x}$.

\begin{claim*}
    $X_1, \dots, X_n$ independent.
    \[ X = X_1 + \cdots + X_n \]
    Then
    \begin{align*}
        m_X(\theta)
        &= \EE[e^{\theta(X_1 + \cdots + X_n)}] \\
        &= \EE[e^{\theta X_1}] \cdots
        \EE[e^{\theta X_n}] \\
        &= \prod m_{X_i}(\theta)
    \end{align*}
\end{claim*}

\begin{example*}
    Gamma distribution: $X \sim \mathrm{J}(n,
    \lambda)$.
    \[ f_X(x) = e^{-\lambda x} \frac{\lambda^n
    x^{n - 1}}{(n - 1)!} \]
    \begin{align*}
        m(\theta)
        &= \int_0^\infty e^{\theta x} e^{-\lambda
        x} \frac{\lambda^n x^{n - 1}}{(n - 1)!}
        \dd x \\
        &= \int_0^\infty e^{-(\lambda - \theta)x}
        x^{n - 1} \frac{\lambda^n}{(n - 1)!} \dd x
        \\
        &= \left( \frac{\lambda}{\lambda - \theta}
        \right)^n \int_0^\infty e^{-(\lambda -
        \theta) x} x^{n - 1} \frac{(\lambda -
        \theta)^n}{(n - 1)!} \dd x \\
        &= \left( \frac{\lambda}{\lambda - \theta}
        \right)^n
    \end{align*}
    ($\theta < \lambda$ (and infinite if $\theta
    \ge \lambda$))
    \[ \mathrm{Exp}(\lambda) \to \left(
    \frac{\lambda}{\lambda - \theta} \right)
    \text{MGF} \]
    We've proved
    \[ X_1 + \cdots + X_n \sim J(n, \lambda) \]
\end{example*}