exercise08.02.tex

\paragraph{Exercise 8.2} Let $X$ and $Y$ be independent, exponentially distributed
random variables with parameter 1. Let $Z$ be a random variable,
such that $Z = X + Y$. For all $z \in \mathbb{R}$,
\[
    F_Z(z) = \int_{- \infty}^{z} f_Z(u) \, du,
\]
and
\[
    f_Z(z)
        = \int_{- \infty}^{\infty} f_{X,Y}(x,z-x) \, dx
        = \int_{- \infty}^{\infty} f_X(x)\cdot f_Y(z-x) \, dx.
\]
Clearly, $f_Z(z) = 0$ for $z < 0$. Consider the case $z \geq 0$. In order to have
$f_Y(z-x) \not= 0$, we need $z-x \geq 0$, that is, $x \leq z$. Consequently, it
suffices to integrate from $x=0$ to $x = z$,
\[
    f_z(z)
        = \int_{0}^{z} f_X(x)\cdot f_Y(z-x) \, dx
        = \int_{0}^{z} e^{-x} \cdot e^{-(z-x)} \, dx
        = \int_{0}^{z} e^{-x-(z-x)} \, dx
        = e^{-z} \int_{0}^{z} 1 \, dx
        = z \cdot e^{-z}.
\]
Thus, the density function of $Z$ is defined as
\[
    f_Z: z \mapsto
    \begin{cases}
        z \cdot e^{-z}, &\text{if }z  \geq 0; \\
        0,  &\text{else.}
    \end{cases}
\]
Let us now consider the distribution function of $Z$. $F_Z(z) = 0$ for $z < 0$.
In regard to the case $z \geq 0$, one has
\[
    F_Z(z)
        = \int_{- \infty}^{z} f_Z(u) \, du
        = \int_{0}^{z} z \cdot e^{-z} \, du
        = z \cdot e^{-z} \int_{0}^{z} 1 \, du
        = z^2 \cdot e^{-z}.
\]
Thus, the distribution function of $Z$ is given by
\[
    F_Z: z \mapsto
    \begin{cases}
        z^2 \cdot e^{-z},   &\text{if }z \geq 0; \\
        0,                  &\text{else.}
    \end{cases}
\]