Question

Let X be a continuos random variable with cumulative disribution function $F_X\left(x\right)$. Another random varible Y is defined as, $Y=F_X\left(x\right)$. Then mean square value of Y, i.e $E\left[Y^2\right]$ will be _____

• A. 0.33

Answer 1

The correct option is A 0.33
Any CDF will vary from 0 to 1 only. so, the random variable Y will also vary from 0 to 1.
$0\le y\le 1$
The probability density function of Y can be given as,

$f_y\left(y\right)=\frac{f_x\left(x\right)}{|\frac{dy}{dx}|}=\frac{f_x\left(x\right)}{|\frac{d}{dx}\left(F_x\right(x\left)\right)|}=\frac{f_x\left(x\right)}{f_x\left(x\right)}=1$ $0\le y\le 1$

So, $f_y\left(y\right)$ = {1 ; $0\le y\le 1$
{ 0 ; otherwise

$E\left[Y^2\right]={\int }_{-\infty }^{\infty }{y}^2{f}_y\left(y\right)dy={\int }_0^1\left(1\right)y^{2}dy=[\frac{y^3}{3}{]}_0^1=\frac{1}{3}$