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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.

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    Thank you for the answer, if I try this way. Fw(y) = 0, y <= 0 | y, y in [0, 1] | 1 , y > 1 and from the definition Fw = P(w < y). But when I try to solve I get this P(4w < x) and P(4 - 4w < x) => P(-w < (4 + x)/4) but the definition say P(w < y)2012-11-11

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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 $z\gt2$.

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    thanks for the solution, good logic :).2012-11-11