Question

The random variabel Y is defined by $Y=\frac{1}{2}(X+|X|)$ where ${}^{\prime }{X}^{\prime }$ is another random variable. Then the density funciton of ${}^{\prime }{Y}^{\prime }$ for $y>0$ is equal to

• A. $f_Y\left(y\right)=\frac{f_X\left(y\right)}{1-F_X\left(0\right)}$
• B. $f_Y\left(y\right)=\frac{f_X\left(x\right)}{1-F_X\left(0\right)}$
• C. $f_Y\left(y\right)=\frac{f_X\left(x\right)}{1+F_Y\left(0\right)}$
• D. $f_Y\left(y\right)=\frac{f_Y\left(x\right)}{1-F_Y\left(0\right)}$

Answer 1

The correct option is A $f_Y\left(y\right)=\frac{f_X\left(y\right)}{1-F_X\left(0\right)}$

Given, $y=\frac{1}{2}(x+|x|)$
when $x>0$, $y=\frac{1}{2}(x+|x|)=\frac{1}{2}(x+x)=\frac{2x}{2}=x,y>0$
and $y=x,y>0$
$\therefore$ $F_y\left(y\right)=P[X\le y|X\ge 0]$
$=\frac{P[X\le y,X\ge 0]}{P(X\ge 0)}=\frac{P(X\le y,X\ge 0)}{1-P(X<0)}$
$=\frac{P(0\le X\le y)}{1-P(X\le 0)}=\frac{F_X\left(y\right)-F_X\left(0\right)}{1-F_X\left(0\right)}$
$\therefore f_Y\left(y\right)=\frac{d}{dy}{F}_Y\left(y\right)=\frac{f_X\left(y\right)}{1-F_X\left(0\right)}$