Answer to Question #242561 in Statistics and Probability for PRS

Question

The probability density function of X is given by f(x)={ 6x(1-x) 0<x<1 0<y<1

{0 otherwise

find the pdf of Y=X³

Answer 1

$f(x)=6x(1-x),\space 0\lt x\lt1$

$0 , \space elsewhere$

And we have to find the $pdf$ $Y=X^3$

Let $G(y)$ be the value of the distribution function of $Y$ at point $y$, it is defined by,

$G(y)=p(Y\leqslant y)$

$=p(X^3\leqslant y)$

$=p(X\leqslant y^{1/3})$

$=\int^{y^{1/3}}_06x(1-x)dx$

$=\int^{y^{1/3}}_0(6x-6x^2)dx$

$=3x^2-2x^3|^{y^{1/3}}_0$

$=3y^{2/3}-2y$

In order to find the pdf of random variable $Y$ given by $g(y)$, then we differentiate $G(y)$ with respect to $y$ as below.

$g(y)=G'(y)=dG(y)/dy$

$=d/dy\space (3y^{2/3}-2y)$

$=2y^{-1/3}-2=2(y^{-1/3}-1)$

Therefore, the $pdf$ of $Y=X^3$ is given as,

$g(y)=2(y^{-1/3}-1),\space 0\lt y\lt 1$

$0,\space elsewhere$