CDF for random variable $X(\omega) = 2(1-|2\omega - 1|)$

Question

I don't know how to calculate this cdf, the modulus is very annoying, because the cdf definition is $P(X< x)$ in my case $P(\omega < x)$. But in the modulus equality I get this $P(-\omega < (4+x)/4)$

$Ω = [0, 1]$ probability space
Random variable $X_2(ω) = 2(1-|2ω-1|)$

Thanks for the help.

Answer 1

Forget CDFs, in this case they only make things more difficult. Instead, start from the two points below:

• If $X$ is uniform on $(0,1)$, then $2X$ is uniform on $(0,2)$ hence $2X-1$ is uniform on $(-1,1)$ hence $|2X-1|$ is uniform on $(0,1)$.

• If $Y$ is uniform on $(0,1)$, then $1-Y$ is uniform on $(0,1)$ hence $2(1-Y)$ is uniform on $(0,2)$.

Putting these together yields the fact that $Z=2(1-|2X-1|)$ is uniform on $(0,2)$ for every $X$ uniform on $(0,1)$.

A (not very interesting) by-product is the CDF you know: $F_Z(z)=0$ for $z\leqslant0$, $F_Z(z)=\frac12z$ for $0\lt z\leqslant2$, $F_Z(z)=1$ for $z\gt2$.