Square of Bernoulli Random Variable I was wondering about the distribution of the square of a Bernoulli RV. My background in statistics is not too good, so maybe this doesn't even make sense, or it is a trivial problem.
Let, $Z\sim X^2$, where $X\sim \text{Ber}(p)$.
$F_Z(z)=\Pr(X^2\leq z)$
$=\Pr(-z^{1/2}\leq X\leq z^{1/2})$
$=F_X(z^{1/2})-F_X(-z^{1/2})$
At this point I'm pretty confused I mean the CDF is right-continuous while I know $Z$ is a discrete RV.
$\implies P_Z(z)=\frac{ d}{dz}\{F_X(z^{1/2})-F_X(-z^{1/2})\}$
I guess you can define the derivative to be:
$\frac{d}{dx}f(x)=\frac{f(x+1)-f(x)}{1}$ or something... and we have 
$F_X(x)=\begin{cases}
0, & \text{if }x<0 \\
1-p, & \text{if }0\leq x\lt 1 \\ 1, & \text{if } x\geq1
\end{cases}$
Any help is appreciated (is my approach correct?) 

If $X$ is Bernoulli, then $X^2=X$.${}{}{}{}{}{}$
