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
"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"
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