0
$\begingroup$

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.

  • 0
    Are you using the two letters $\omega$ and $w$ synonymously?2012-11-10
  • 0
    . . . . and did you intend $\omega$ to be uniformly distributed in $[0,1]$?2012-11-10
  • 0
    no just one, thanks for the editing2012-11-10
  • 0
    yeah w is uniformly distributed in [0,1]2012-11-10
  • 1
    Maybe you should try the following question first. Let $Y \sim U[0,1]$ and $Z = 2(1-|2X-1|)$. What is the CDF of $Z$? When solving this problem you might want to consider drawing a sketch which will help you express $Z$ as a piecewise linear function of $Y$ and to discover that $Z$ takes on values in $[0,2]$.2012-11-11
  • 0
    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

1 Answers 1

2

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 $z\gt2$.

  • 0
    thanks for the solution, good logic :).2012-11-11