I'm given the following problem:
Suppose that $X$ is a continuous random variable, with CDF ($F_X$) given by:
\begin{cases} 0 & x < 0\\ 1 - e^{-x} & x \geq 0 \end{cases}
Find the CDF of $(X-10)^2$.
Attempted solution.
Let $Y = (X - 10)^2$. As $X$ is a random variable, $Y$ must also be a random variable.
Let $F_Y$ be the CDF of $Y$. By definition:
$F_Y = P(Y \leq y) = P((X - 10)^2 \leq y ) = P(10 - \sqrt{y} \leq X \leq \sqrt{y} + 10)$
It can be shown that (this is given in the class text):
$= F_X(\sqrt{y} + 10) - \lim_{x\to 10 - \sqrt{y}} F_X)\;\;\;$
But this $= F_X(10 + \sqrt{y}) - F_X(10 - \sqrt{y})$ since $F_X$ is continuous.
Now, we find $F_X(10+\sqrt{y})$ and $F_X(10 - \sqrt{Y})$...
$F_X(10 + \sqrt{y})= \{\;1 - e^{-(10 + \sqrt{y})} \;\;\;|\;\;\; 10 \leq 10 + \sqrt{y}\;\}$
$F_X(10 - \sqrt{y})=$ \begin{cases} 0 & x < 0\\ 1 - e^{-(10 - \sqrt{y})} & 0 \leq 10 + \sqrt{y} < 10 \end{cases}
Unfortunately, I'm not sure where to go from here (it doesn't seem to me that the difference can be taken?). Further, I'm not entirely confident in what I have thus far.
Is this a valid approach, and what's the next step?
Using the support of $X$,
$F_X(10 − \sqrt y) = \begin{cases} 0 & y \gt 100\\ 1 - e^{-(10 - \sqrt{y})} & 0 \leq y \leq 100 \end{cases} $
Also, $F_X(10+\sqrt y) = 1 - e^{-(10 + \sqrt{y})}, y \geq 0$
As you rightly said, $F_Y(y) = F_X(10 + \sqrt{y}) - F_X(10 - \sqrt{y})$. So,
$F_Y(y) = \begin{cases} 0 & y \lt 0 \\ e^{-(10 - \sqrt{y})} - e^{-(10 + \sqrt{y})} & 0 \leq y \leq 100\\ 1 - e^{-(10 + \sqrt{y})} & y \gt 100 \end{cases} $