Cumulative distribution of mixed random variable function

Question

Random variable $X$ has probability density function:
$$
f_X(x) =
\begin{cases}
2e^{−2x}, & \text{for } x\geq 0,\\
0, & \text{otherwise.}\\
\end{cases}
$$

Furthermore, the random variable Y is defined as
$$
Y =
\begin{cases}
2X, & \text{if } X < 1/2,\\
0, & \text{if } X\geq 1/2.\\
\end{cases}
$$
Determine the cumulative distribution $F_Y(y)$ of the random variable $Y$.

How do I solve this question? If $Y$ was simply defined as $Y=2X$ then it would simply be finding $P[Y\leq y]=P[2X\leq y]=P[X\leq y/2]$ and thus solving the integral
$$
\int^{y/2}_0{2e^{-2x}}dx
$$

Answer 1

You can rely on basic principles. Note that $Y$ is of the form

$$ Y = g(X), \qquad \text{where} \quad g(x) = \begin{cases} 2x, & \text{if $x < \frac{1}{2}$,} \\ 0, & \text{if $x \geq \frac{1}{2}$.} \end{cases} $$

From this and using the definition of CDF,

\begin{align*} F_Y(y) &= \mathbf{P}(Y \leq y) = \mathbf{P}(g(X) \leq y). \end{align*}

So the problem boils down to solving the inequality $g(x) \leq y$ for each given $y$. To this end, you may study the behavior of the function $g(x)$. Note that the graph of $g(x)$ looks like:

From this, we can solve the inequality $g(x) \leq y$ for $x$ for each given value of $y$:

Range of $y$	Sol. of $g(x) \leq y$	$F_Y(y)$	Graph
$y < 0$	$x \leq y/2$	$\begin{aligned} \mathbf{P}(X \leq y/2) = 0 \end{aligned}$
$0 \leq y < 1$	$x \leq y/2$ or $x \geq \frac{1}{2}$	$\begin{aligned} &\mathbf{P}(X \leq y/2) + \mathbf{P}(X \geq \tfrac{1}{2}) \\ &= 1 - e^{-y} + e^{-1} \end{aligned}$
$y \geq 1$	$x \in \mathbb{R}$	$1$

Summarizing, the CDF of $Y$ is given by

$$ F_Y(y) = \begin{cases} 0, & y < 0, \\ 1 - e^{-y} + e^{-1}, & 0 \leq y < 1, \\ 1, & y \geq 1. \end{cases} $$

For fun, I also included the graph of $F_Y$ below: