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Let X be a random variable uniformly distributed on the interval [−1, 1], and Y = |X|. Find the density function and the distribution function of Y .

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closed as too localized by cardinal, t.b., Benjamin Lim, Did, J. M. Apr 24 '12 at 12:49

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What have you tried? –  Brett Frankel Apr 18 '12 at 3:21
Hint: You can solve this with no calculation whatsoever. No pen, no paper. –  cardinal Apr 18 '12 at 3:26

1 Answer 1

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Let $Y=|X|$. We want to find the cumulative distribution function $F_Y(y)$ of $Y$. By definition, $F_Y(y)=P(Y \le y)=P(|X|\le y)$.

It is clear that $F_Y(y)=0$ if $y\le 0$, and that $F_Y(y)=1$ if $y>1$. For $0<y<1$, we have $$F_Y(y)=P(|X|\le y)=P(-y \le X\le y)=\frac{2y}{2}.$$ This is because the interval $[-y,y]$ has length $2y$, and the density function of $X$ is $\frac{1}{2}$ on the interval $(-1,1)$. So on $(0,1)$, we have $F_Y(y)=y$, and therefore $f_Y(y)=1$. The random variable $Y$ has uniform distribution on $(0,1)$. This is intuitively almost obvious. We went through some formal calculations as preparation for handling more complicated problems of a similar type.

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