If X is uniformly distributed over (−1,1), find (a) P|X|&gt;12 (b) the density function of the random variable |X|.

Step by step solution to "If X is uniformly distributed over (−1,1), find..."

Ayaana Buck

Answered question

2021-09-15

$If X is uniformly distributed over (- 1, 1), find (a) P | X | > \frac{1}{2}$
(b) the density function of the random variable |X|.

Answer & Explanation

i1ziZ

Skilled2021-09-16Added 92 answers

Step 1
Given that $X \sim U [- 1, 1]$ , a uniform random variable on $[- 1, 1]$
(a)
$P [| X | > \frac{1}{2}] = P [X > \frac{1}{2}] + P [X < - \frac{1}{2}]$
$= \int_{{\frac{1}{2}}}^{1} \frac{1}{2} d x + \int_{{- 1}}^{- \frac{1}{2}} \frac{1}{2} d x$
$= \frac{1}{2} {[x]}_{{\frac{1}{2}}}^{1} + \frac{1}{2} {[x]}_{{- 1}}^{- \frac{1}{2}}$
$= \frac{1}{2} [1 - \frac{1}{2}] + \frac{1}{2} [- \frac{1}{2} + 1]$
$= \frac{1}{4} + \frac{1}{4}$
$= \frac{1}{2}$
Step 2
(b) To find the density function of $| X |$ . Let $Y = | X |$ . We first find the distribution function of Y.
$F_{Y} (y) = P [Y \leq y] = P [| X | \leq y]$
$= P [- y \leq X \leq y]$
$= \int_{{- y}}^{y} \frac{1}{2} d x for 0 \leq y \leq 1$
$= \frac{1}{2} [2 y] for 0 \leq y \leq 1$
$= y for 0 \leq y \leq 1$
$Thus, the density function of Y is given as$
$f_{Y} (y) = \frac{d}{d y} F_{Y} (y) = \frac{d y}{d y} = 1 for 0 \leq y \leq 1$
$Thus,$
${\begin{cases} 1 & for 0 \leq y \leq 1 \\ 0 & o . w \end{cases}$
Result
$(a) P [| X | > \frac{1}{2}] = \frac{1}{2}$

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