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Let $X$ be a random uniform distribution on $[-2,1]$. What is the cumulative distribution of $|X|$, i.e. $G(x) = p(|X| \leq x)$ ?

The cumulative distribution of $X$ is $F(x) = \dfrac{x+2}{3} $. $$G(x) = p(|X| \leq x) = p (-x \leq X \leq x) = p(X \leq x) - p(X \leq -x) = F(x) - F(-x) = \dfrac{2}{3} x$$ But what is the domain of $G$? It can't be $[-2,1]$ since $G(-2) < 0$. Is it $[0, \dfrac{3}{2}]$? Is there something wrong?

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1 Answer 1

up vote 3 down vote accepted

The error that you meade, is that the cdf of X is NOT $F(x) = \frac{x+2}{3}$.

The CDF is

$$F(x) = \begin{cases} 0 & x \leq -2 \\ \frac{x+2}{3} & -2 < x < 1 \\ 1 & x \geq 1 \\ \end{cases}$$


The domain of $G$ is all real numbers.

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