Question

If$X$is uniformly distributed over $(-1,1),$find

$\left(a\right)P\{\left|X\right|>\frac{1}{2}\}$

$\left(b\right)$the density function of the random variable$\left|X\right|$.

Short answer

Therefore,

$\left(a\right)P\left(\left|X\right|>\frac{1}{2}\right)=\frac{1}{2}$

$\left(b\right)$We have found that$Y=\left|X\right|~UNIF(0,1)$

Step-by-step solution

Given information:

If $X$is uniformly distributed over$(-1,1)$

Part (a) step 2 Explanation:

We have that $X~UNIF(-1,1)$and wish to calculate

$$\begin{gathered} P\left(\left|X\right|>\frac{1}{2}\right)=P\left(-\frac{1}{2}>X>\frac{1}{2}\right) \\ =P\left(X<-\frac{1}{2}\cup X>\frac{1}{2}\right) \\ =P\left(X<-\frac{1}{2}\right)+P\left(X>\frac{1}{2}\right) \\ ={\int }_{-1}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{2}dx+{\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\frac{1}{2}dx \\ =\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \end{gathered}$$

Part (b) step 3 Explanation:

The CDF of$Y=\left|X\right|$, where

$$\begin{gathered} F_x\left(\left|X\right|\right)=P\left(\left|X\right|\le x\right)=P\left(-x\le X\le x\right) \\ ={\int }_{-x}^x\frac{1}{2}dt={\left[\frac{1}{2}t\right]}_{-x}^x=\frac{2x}{2}=x \end{gathered}$$

for all$x\in (0,1)$. We then differentiate the CDF to obtain the PDF as$f_{\left|x\right|}\left(x\right)=\frac{d}{dx}F\left(\left|X\right|\right)=\frac{d}{dx}\left(x\right)=1$for $x\in (0,1)$. Therefore, we have found that $Y=\left|X\right|~UNIF(0,1).$